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BTS Jungkook and Red Velvet Yeri Story ( BTSVELVET JUNGRI ) || Love Long Journey part 1

I’m really not sure how you will take in the stories

I’m telling you right now

I tried not to feel this way

But all I was a burden to you

It must have been heavy I guess the delusion thinking that we’ll understand each other was not enough No matter what we say, things are too complicated between us What can possibly describe what we were thinking Now that I think of you I honestly don’t know why I didn’t grab your hand in the warm breeze Now that I think of you I honestly don’t know why I left you alone in the winter that you love And I wanna go back to that time I don’t even know what I should tell you first This place I’m standing I know, in my heart that you’re the one I wanted to be bold and have more courage to hold you at least once I never gonna make it alone It is just too hard to erase you along with it Baby still I’m alone now What can I possibly describe why I cannot get over you Now that I think of you I honestly don’t know why I didn’t grab your hand in the warm breeze Now that I think of you

I honestly don’t know why I left you alone in the winter that you love And I wanna go back to that time

Parallel plane waveguide

Up till now we discuss the wave propagation in an unbound or semi bound media. We investigated specifically the refraction and reflection of electromagnetic wave across the dielectric boundaries. Lastly we also saw the reflection of an electromagnetic wave from a conducting boundary. Now we take a very important topic of electromagnetic waves and that is waveguides that the name suggests waveguide is a structure which can guide electromagnetic energy along it. We have variety of waveguides in practice For example the coaxial cable which you are seen as a transmission line also the wave guiding structure because it can guide electromagnetic energy from 1 point to another the parallel wire transmission line is also a wave guiding structure. However as we go to higher frequencies, these waveguides have excessive loss. And that is way you get the waveguides which are like hollow pipes which are more efficient for transmitting energy at higher frequencies So when you go to the frequencies like gigahertz or 10s of gigahertz, the time the coaxial cable are the parallel wire transmission line though they are waveguide in structures, become more lossy and then we get the waveguides like rectangular waveguides which is like hollow pipe, whose cross section is rectangular or a circular pipe who cross section is circular They can transmit energy from 1 point to another more efficiently. So in fact in communication the waveguide structure play the very important role. The communication engineer is always in search of would waveguide in structures So that he can transfer energy at higher frequencies very efficiently from 1 point to another So ideally it is waveguide structure should have as minimal loss is possible so that while transmitting the energy from 1 point to another, there is no loss of power Also these structures should have certain characteristic like they should be small in size; they should be compact; they should be moldable in the shapes and sizes if you want so you can carry the energy very easily from 1 point to another. Of course whenever you have a structure like this, we always start with the basics. That is, the Maxwell equations and solve the Maxwell equation for given boundary conditions and then we get the solution for the particular problem. However without going into that approach which is a routine kind of approach what we will try to do first we will try to understand, if you still confined our self the uniform plane wave propagation and try to capture this uniform plane wave between the boundaries, can we get a structure like a waveguide structure? At the sort precisely we are going to do here that with the understanding which you have develop from the reflection of a plane a wave from a conducting boundary. We will try to develop an understanding for a waveguide what is called the parallel plane wave guide. As the name suggests this waveguide essentially consist of 2 parallel planes which are infinite extend and the electromagnetic energy is trapped between these 2 planes. So the energy essentially propagates along this planes trapped between 2 planes. So we now in this lecture we essentially

discuss what are called the parallel plane waveguide As we have seen in the previous lecture we have taken a conducting boundary and on which the uniform plane wave are incident at some angle theta and there is a reflected wave And then by satisfying the boundary condition we saw the reflection coefficient for the electric field is minus 1. So this boundary was behaving more like a short circuit boundary and then the fields which you having in this media is superposition of the incident field and the reflected field So by superposition of these 2 fields essentially we obtain the electric field in the medium above the conducting surface and that essentially was given by this. So we saw that it consists of 2 types of fields 1 is the standing wave kind of field which is in x direction. And there is a traveling wave of field which is in z direction. So we had seen there is a pattern whose amplitude varies the electric field amplitude varies as a semisolid function in the x direction that is perpendicular to the boundary and along the boundary you are having a traveling wave which is having a phase constant which is beta times sin theta So this boundary as this demonstrates has a capability of guiding the wave along the z direction because we having a net traveling wave which is traveling along z direction Precisely that is the phenomena essentially we are going to make use of to get a more guiding structure. That means I will track the energy from both sides. Now at the moment the energy is not trap energy traveling into this semi infinite media we will try to trap the energy even from this side. So that energy propagates in the 2 planes something like this which we call as the parallel plane waveguide And then we said that the electric field which is tangential to the interface in this case has to be 0 at the interface that means Zx equal to 0. The total electric field should be 0 and that is what we got this condition from. So when beta x cos theta is 0 that means when x is equal to 0, the time the electric field goes to 0 since satisfy the boundary condition However what we also saw that only x equal to 0 but there are some other values of x also for which the tangential component of electric field will go to 0. So whenever we have a beta x cos theta is multiples of pi that m is an integer. So when beta is cos theta is equal to m pi or x equal to minus divide by beta cos theta for those values of x again the electric field would be 0 everywhere And since the electric field is oriented like this perpendicular to the plane of the paper which is the y direction, if you put a conductive sheet here which is tangential to this electric field which is 0 at this point it will not affect the boundary conditions. Because the electric field is 0 here and boundary condition demand that this tangential component should be 0 at the conducting boundary That means at this value of x if I insert a conducting sheet of infinite extend like this the fields will remain unaffected. What that the means is that not only the value x equal to 0 where the condition will be satisfy, but I will have some another value of x another value of x another value of x at which if I insert a conducting boundary like this These fields whatever we have got will remain unchanged I will get the same field distribution So I substituting the value for beta which is 2 pi by lambda here lambda is the wavelength in medium 1. I can get this x which is m lambda divided by 2 times cos theta. So if I am equal to 0 or 1 or 2 or 3 I get different values of x at which this boundary can be inserted without affecting the electric or magnetic field distribution So these boundary locations essentially as we saw the plot of the magnitude of electric field it would go 0 here, it would go 0 here, it would go 0 here, here, here. So I can insert a boundary at this location this location this location and the field distribution will be mentioned. However by inserting a boundary at this distance, essentially now the wave is trap between these 2 boundaries. So I have created a structure which is more close structure it is no more a semi-infinite medium. Now it is become a bound medium in x direction

So this was 1 in interface this is another interface and now the field disturb between these 2 boundaries. Similarly I can put the plane here so I can get a boundary like this or I can get here. And now the energy electromagnetic energy is trap between these 2 parallel planes Precisely that is the structure we are trying to investigate and the separation between this planes essentially is given by this condition So let me redraw of the structure this is the original boundary which we had below which we have a conducting medium this is the x direction this is the z direction. The y-direction is perpendicular to the plane of the paper and let us say, the wave is incident at an angle which is theta so the reflected wave also will be at angle theta. And we are considered a case here where the polarization is perpendicular that means the electric field was this. So this is Ei and then we have got here Er which was equal to minus Ei because the reflection coefficient was minus 1. So that these this reflected electric field has same magnitude as the incident field only thing is its negative that means its direction is reverse. So if this field was coming upwards these fields essentially should be going downwards and then the magnetic fields for this where this was Hi and this is Hr. And for these term we had write down the components of the electric field or magnetic field So we head 3 components for this case the electric field which was y oriented and the magnetic field which will be either z oriented or x oriented. So magnetic field now lies in the plane which is x-z plane it has 2 components and the electric field is y oriented which is perpendicular to the plane of incidents Thus the situation essentially we are considering And now we have got this condition here that x is equal to m lambda upon 2 cos theta at that location again the tangential field will go to 0. So we will have 1 location here where field will go to 0. I may have 1 location here or I may get next location where there which will be double of this and so on. So this distance is lambda upon 2 cos theta from here to here where this field will again go to 0 or if I go to double of that it will be lambda upon cos theta. So this distance again lambda upon 2 cos theta this distance So this electric field which is tangential to this interface is 0 here, it is 0 here, it is 0 here and so on. So in fact I have multiple locations of planes where this component tangential component of electric field will go to 0. So what we are saying now is that if the introduce a plane at this location then the field distribution will remain unaffected However only thing which is going to happen is that when this planes for not there the fields are distributed everywhere in the semi-infinite space. However when I introduce now a boundary at this location the fields are only confined to this region, we do not know what is going to be in this. So essentially we will have propagation of electromagnetic wave only in this region or in other wards 1 can save that the wave which was coming here at an angle theta, it goes and 1 it goes here it reaches this boundary again it will be making an angle which is theta with normal So we have gets reflected from here that is same angle. So it will appear as if the wave is moving coming here getting reflected, again getting reflected, when it reaches here again it is the boundary, then reflected. So the fields which are going to survive in this structure can be visualizes as multiple reflections of uniform plane wave. So wave simply bouncing

back and forth between these 2 planes which are conducting boundaries. So I conducting boundary here now between these 2 boundaries by multiple reflections the wave by this exact pair is going to travel along the structure for along this planes which are conducting planes. And the height over the separation of these planes is given by this. What now 1 get do is, 1 can invert the problem and 1 can say well here knowing the angle theta we could find out what should location of these planes. What should be separation between these planes, so that this field remains unaffected? If I invert this relation say well separation between the places is given and let us say that is equal to D. Then what are the angles at which if the wave is launch the propagation will be sustained That means it will satisfy the boundary condition and the wave will propagate by this multiple reflections around the boundary. See if I say that I have 2 conducting boundaries now and let us say separation between them is given by d. Then this d should be equal to m lambda divided by 2 times cos theta. Or if I invert this expression I will essentially get cos of theta that is equal to m lambda by 2 d. Now this is very interesting because m is a discrete number it is 1, 2, 3, 4, so it is the integers. So this quantity is discrete, what that means is that for given value of d that means for given separation between this conducting boundaries? This angle theta is no more continuous were launch at any arbitrary angle will not sustain in this structure Only if the, it is launch at particular angles which are discrete, because m is discrete, then only you can have a sustain propagation of electromagnetic wave in this structure Now this is a very significant departure from our original understanding of wave propagation Up till now when we talked about unbound media there was no restriction on the launching angles. We could launch ray in any direction When we had a semi infinite medium even then we could have any angle of incidence and we had some field distribution in the 1 medium or both media depending upon what is media wave. However now what we have saying is if we make the structure bound in this direction then I cannot launch an electromagnetic wave at an arbitrary angle of incidence. If I do that this structure is going to reject that way it is not going to except that way. If the wave is launch at these angles which satisfy this condition which are discrete now then and then only you can have a sustain propagation of electromagnetic wave. This behavior of the bounds structure is capture by what is called the model propagation. So what we are saying is that is, the wave which can be launch at discrete angles Superposition of this incident and reflected wave they are going to create certain electric and magnetic patterns in this. And you having a unique pattern created for electric and magnetic field for a given value of m. Say m equal to 0 and get 1 pattern is m is equal to 1 I will get another pattern if m equal to 2 I will get another pattern and so on So we have this discrete electric and magnetic field patterns which you can survive inside this bounds structure. That is what is called the model propagation of electromagnetic wave So we have migrated now from the continuum domain of theta to discrete domain and that is the significant departure which you have done in the wave of propagation. And then these discrete angles are going to create discrete electric and magnetic field patterns which we call as the modals patterns and this propagation in this bound media is called

the model propagation There are few things which we can which we observe at this point. One is since for physical angle theta the cos theta has to be between 0 and 1 for a given value of d I have only limited number of m’s. For example I can put m equal to 1 I can put m equal to 0 I can put m equal to 1, 2, 3 and whenever this quantity becomes greater than 1 that does not represent the physical angle theta. That means not only that the waves can be launch inside this structure, a discrete angle. But you also not that this discrete angles are finite number. That means for a given spacing between the boundaries only finite number of uniform plane wave can be launch at discrete angles which are given by this Or in other words this is going to create discrete finite number of electric and magnetic field patterns inside the bound structure which is what is called a parallel plane waveguide So this model propagation essentially is the core of all waveguide structure whether you take a parallel plane waveguide or you take a rectangular waveguide or you take optical fibers or dielectric waveguides. Whenever we have a bound medium like this the electromagnetic energy is going to travel in definite patterns And that is what is called a model pattern and that propagation is model propagation So model propagation is a very important aspect of electromagnetic wave of propagation in a bound structure which is called a waveguide Once we understand this then of course finding out the electric and magnetic fields that analysis becomes straight forward So let me right down the conclusion which we got from here that 1 wave survive at discrete angles in a parallel plane waveguide and there of finite number of angles at which the wave survives. Now in the original discussion we say this is the incident wave and this was the reflected wave However when we have a structure like this which is a parallel plane structure this way which is reflected from here goes and meets this boundary at this time. So as for as this boundary is concerned, this becomes say incident wave and this will become a reflected wave As for as this boundary is concerned, this is incident wave and this is reflected wave, so there is once we are having 2 boundary there is nothing like a incident wave and reflected wave depending upon which boundary you are considering The wave can be called the incident wave or a reflected wave. What that essentially means is there is nothing like what is you should call incident or reflected there are 2 steps of wave which are move. One set of wave which is moving like this which is having the wave vector which are parallel which are in this direction, other set of wave which are moving like that. And superposition of interference of these 2 waves plane waves 1 moving this wave and 1 moving this wave. Create the field distribution inside this bounds structure that is what is called the model pattern So what we have essentially is these are the 2 boundaries which are conducting boundaries and there is 2 sets of waves 1 moving this wave other set of wave which moves this wave And super positions of the fields are these 2 waves create the patterns which are the

model patterns. So essentially the propagation of electromagnetic wave inside this bound structure can be visualize as superposition of the 2 waves which are traveling like this Let us same angle with respect to this direction which is parallel to the boundary or perpendicular to the boundary. Once we get this then we can get now the field expressions very easily for this pattern. However before we get into that let us take 1 more observation about these fields And that is if I look at now this propagation where the multiple reflections are taking place and the net propagation of energy is in this direction which is along the z direction The electric field is now perpendicular to this plane of incidence what we call which is the plane of this paper. So when the reflection takes place, the electric field vector comes like this it get reflected go like that it get the like that and so on. So effectively we are having a wave of propagation now which is in z direction and the electric field is like that everywhere inside this medium where is the magnetic field is like this and this So it is a direction is like that when it comes here the direction get change it become like this and direction get change it will become like this. So the magnetic field if I look inside the space between this plane waveguides has 2 components. One is perpendicular to this that mean in the x direction 1 is in z direction and since the net propagation is taking place in z direction The magnetic field essentially has components in the direction of net propagation and in the direction perpendicular to it. Whereas, if I look at electric field the electric field is always perpendicular to direction of the propagation; no matter where I go I will always find this electric field is perpendicular to the net direction of propagation. That means electric field is having a special nature in this particular case. It always remains transverse to the direction of the net wave propagation. That is the reason we designate this propagation as the transverse electric propagation. So we call this mode whatever the pattern which are going to set up inside the structure. Since this patterns will have electric field which will be transverse to the direction of let wave propagation. We call this mode as the transverse electric mode So we have now propagation what is called the transverse electric mode. That means the electric field always remains transverse to the direction of net wave propagation inside the structure. And in short this is denoted TE. So that mode for the electric field is transverse to the direction of net wave propagation we call at that as a TE mode. Now if I look this structure again this thing here we can get either value of let us say m equal to 1 that gives me this spacing which will be lambda upon 2 cos theta. If I can take m equal to 2, I can get this spacing for a given launching angle that means either I can have a set of fields which will be in this or I can have a set of field which will be in this. And for this region m will be equal to 1 if I go here m will be equal to 2 and so on That means though the fields are having the same nature that means electric field is transverse I can have a different possibilities either I can have a region corresponding to m equal to 1. Or I can have a case which is m equal to 2 or m equal to 3 and so on whatever is maximum value per meter. That is the field distribution which you are going to get in this is characterize by this integer number m. So that is the reason when we talk about

the transverse electric mode, we put an index to this mode which is this value m as a suffix and we say that this transverse electric mode corresponds to this value of m which could be 1, 2, 3, 4 and so on. So we say that for a parallel plane waveguide this mode will be TEm mode where we will see little more the physical meaning what represents But at this point it appears that is m is going to tell at what angle this wave is launched for a particular value of m, so that we have a sustain propagation of the electromagnetic wave. Once we get this then we can go back to our fields which we have derived in the last lecture. Now the electric field is essentially given by this and cos theta now is equal to this quantity m lambda upon 2 d. So from here I can get first of all this quantity beta x cos of theta which is coming in the field expression for the electric and as well as in the magnetic field. So I can substitute for now cos theta which is m lambda upon 2 d. So I can write here beta x into m lambda upon 2 Now beta we know is the phase constant in medium 1 and now this no medium 1 as such it is the medium which is filling this waveguide that means the medium which is filling the phase between the 2 conducting planes. So that phase constant is 2 pi by lambda x to m lambda divide by 2 d. So lambda will cancel 2 will cancel so this quantity will essentially become m pi by d into x. So I have got this beta x cos theta which is m pi by d into x So this field expressions I can reply this quantity by m pi x by d. What about this quantity now since the cos theta is given by this I can get sin theta because I require sin theta here So from that beta sin theta that will be beta square root of 1 minus cos square theta. And cos theta is m lambda by 2 d. So I can write here this is 2 pi by lambda square root of 1 minus m lambda by 2 d square. So I have got now the phase constant which is in z direction which is this. That is what we get from this field expression. That is the phase constant beta times sin theta and beta x cos theta can be replace by this quantity. Once I get that then I can go to my electric and magnetic field which we derive last time and substitute these values for theta and we can get the fields for the transverse electric case And that will be now the electric field E that will be 2 j Ei sin of m pi x by d e to the power minus jz beta square root of 1 minus m lambda by 2 d square. Similarly I can write

down expression for the magnetic field which we have got last time. So we get, this is the y oriented. So this field is essentially oriented in y direction and I can get the magnetic field which will be having component Hx which is minus 2 j Ei upon eta sin of theta which is which you get from here this quantity So sin of theta will be square root of 1 minus m lambda upon 2 d whole square I can write that will be just keep it for time being sin theta. Sin of this quantity which is same as this So sin of m pi x by d and this quantity same e to the power minus jz beta square root 1 minus m lambda by 2 d whole square and then I have the magnetic field z component which will be H of z will be minus 2 Ei upon eta cos theta which will be m lambda upon 2 d So this is m lambda upon 2 d cos of this quantity cos of m pi x by d in this term is same e to the power minus jz beta square root 1 minus m lambda by 2 d whole square. So in this configuration now we get the 3 field components which are given by that. So this is the y component and this is the magnetic field x 2 component 1 is x and other 1 is z. So the field which can survive in a parallel plane waveguide for an electric field vector which is perpendicular to the plane of incidence will have this 3 field components Now as we saw that this quantity here we have this is now showing you the amplitude variation in x direction whereas this is the 1 which is showing you the traveling nature of the wave which is in z direction. So we have a phase constant in the z direction in which the net wave is propagating that is nothing but this quantity which is beta into square root of 1 minus m lambda upon 2 d whole square So when the net wave propagation takes place now this quantity beta does not have a meaning because beta is the phase constant of a uniform plane wave which is bouncing back and fore between these 2 conducting planes. What phase constant essentially we are going to see for the net propagation that will be this quantity the total quantity So we say now that the phase constant for this mode which is superposition of 2 uniform plane wave as a phase constant which is this quantity. So we say that the mode is going to propagate with a phase constant which is this 1. So we say now for the phase constant for the mode and normally it is represented as beta z. So let is called beta z or let me just put as beta bar per beta bar gives me the propagation constant of the model fields along the boundaries and this quantity is nothing but beta into square root of this quantity 1 minus m lambda upon 2 d whole square And this quantity as we know is essentially sin of theta, so we have beta bar upon beta that is nothing but a sin of theta. That is what the expression would be Now first thing 1 should immediately note from this expression is that if I put m; equal

to 0 in this expression. Then the electric field will go to 0 because this quantity will make 0 for any value of x this quantity will go to 0 this quantity will go to 0. So the E y will go to 0 Hx will go to 0 Hz will not go to 0 because of this term but then when m equal to 0 this quantities 0. So these field will go to 0 so what that means is that for these fields to survive inside the structure I must have value of m which is integer but non 0. Because for, m equal to 0 all this fields will identically will go to 0 that means these fields cannot survive inside the structure. What does m equal to 0 mean if I go from here if I put m equal to 0 in these expressions that corresponds to this angle theta of pi by 2. That means if I put a wave which is parallel to the boundaries if the wave goes parallel to the boundaries. This wave cannot propagate inside this structure So if I have electric field likes this and if the wave goes like this See if I have a situation that is these are the 2 parallel plane boundaries and if I try to launch a wave inside this which is having an electric field which is perpendicular to this okay. So I am trying to put the wave inside the structure electric field is like this and m is equal to 0. So this is my Ei if m is equal to 0, this wave cannot survive inside the structure it cannot propagate this field cannot get excited inside this structure What that happens we can see little more physically by that happens and then happens because what this quantity m essentially telling. The m is telling you that if I take m equal to 1 then as x varies from 0 to d which is the spacing between the 2 planes. I get 1 half cycle variation of this field So what the m is telling you is inside a structure if I look at the field distribution amplitude variation of the field the m equal to 1 would corresponds to half cycle. So this is x equal to 0 this is x equal to d. So from here an x is equal to d by 2 this quantity will be pi by 2 when m equal to 1. So I will get the field maximum at this location which is d by 2. So I will get an amplitude variation for this which will be like that for, m equal to 1 if I take m equal to 2 then it will represent 1 full cycle variation as x varies from 0 to d. So if I take m equal to m equal to1, to b. So at x equal to d by 2 here when m equal to 2 this quantity will be pi so again sin pi will be 0 which will be this. So this will represent m equal to 2 Similarly as I go to higher values of m I can get more and more cycles in that. So that will be so this will represent m equal to 3 and so on. So what this m what we have now the index for this field tell you is how many half cycle variation of the electric field amplitude or for that matter magnitude field amplitudes would be the in the directions transverse to the direction of the net wave propagation So if m equal to 0 there is no variation That is what this would mean. So m equal to 0 means the constant field; m equal to 1 means half cycle m equal to 2 means 2 half cycles m equal to 3 means 3 half cycles and so on So m equal to 0 means constant. So that means

if I consider now a field for which m equal to 0, the field is constant in the transverse direction it does not vary And in this case the electric field is oriented like this. Now we know since these are the conducting boundaries the electric field must be 0 here it must be 0 here because the tangential component to the conducting boundary. So for, m equal to 0 we do not want any variation of the field and it should be 0 here and here that can only happen if the field is identically 0 everywhere. And that is what precisely these expressions are telling you that when you put m equal to 0 these field cannot survive So the lowest value of m which can survive the field which can survive for that lowest value of m, they will be corresponding to m equal to 1 So that means the modes which can survive for transverse electric mode the lowest mode which can propagate in this structure is the TE 1 mode this is the lowest mode. So for a given phasing of these conducting planes we must have this value m which is at least 1 then and then only the fields will exist inside the structure. If m equal to 0 then the field will not exist inside the structure and the wave will not. So, depending upon the size that the saw there be finite number of modes, so you can have first mode to propagate which is TE 1 with the size is sufficient you may get TE 2 TE 3 and so on. So, on a given structure you can have a multiple modes propagating depending upon how many angles can be satisfied for this m But for the lowest mode when m equal to 1, one can ask what is the wave which can survive in this. That means d this is the largest value we can have which is equal to 1 for theta equal to 90 degrees. So if I take this is 1 then lambda by 2 will be equal to d, so the separation between the planes is lambda by 2 for propagation of this TE 1 mode. It should be at least lambda by 2 if the separation is less than lambda by 2. I cannot have an angle d is less than lambda by 2 for; m equal to 1 this quantity will be greater than one I do not have a physical angle at which the wave can be longed So what that means for this lowest mode to propagate inside the structure my d should be greater than or equal to lambda by 2. And that is a very important conclusion what that means is once the separation between this 2 planes is given. That means the height of the parallel plane waveguide is given. I require a wave length which is small enough, so that this condition is satisfied or if I invert this essentially I get my lambda should be less than or equal to 2 d. So only those wave lengths since this has to be less than this That means the frequency has to this above certain value, so that this condition is satisfy then and then only the wave propagation will take place. If lambda is greater than 2 d or frequency is less than certain value then the wave propagation cannot take place inside this tangent. So from here we get a important concept of what is called cut off of a mode That is when this condition is satisfied d is equal to lambda by 2. That frequency or lambda is equal to 2 d for the equality condition That is the lowest frequency which can propagate on a waveguide whose separation ST So from here essentially we can get that lambda maximum is the largest lambda which can propagate on this that should be equal to 2 d or if I write in terms of the velocity and the frequency

is velocity of the wave in that medium. Let us say denote by c divided by the minimum frequency which can propagate on this that is equal to 2 d. So from here I get f minimum is equal to c divided by 2 d below this frequency the wave does not propagate and that is the reason this minimum frequency for this mode for m equal to 1 mode. We will call as the cut off frequency of that mode. So for a particular mode to propagate the frequency must be above that value and that is that value is what is called the cut of frequency of that mode? So we have concept what is called cut of frequency and that is the lowest frequency which will propagate inside the structure for a given mode number given m as we can see as the mode number increases when become T 2 then the frequency will be double and so on. So that is the reason we are saying this; the lowest mode because that is the lowest frequency which can propagate in this structure. And that will be in this in the mode which will be TE 1 mode. Now if I put this condition that this quantity your d equal to lambda by 2 if I put in our propagation constant expression. This is the phase constant which I hired for these fields and if I put m equal to 1 and lambda upon 2 this quantity here is equal to d then m lambda upon 2 d that will be equal to 1 So this quantity beta bar essentially will go to 0. So at this frequency when this condition is satisfy what we call as the cut off frequency At cut off frequency the phase constant of this wave goes to 0. And below this frequency this quantity is greater than 1 this quantity is imaginary. So beta becomes imaginary that means it is no more a phase constant. But it becomes attenuation constant and the wave losses the wave nature it represents only the field which exponentially dying down because behave only attenuation constant and there is no phase constant. So when you have a frequency below this cut off frequencies, it represents the fields which are exponentially decaying fields but there is no propagation of electromagnetic wave So what we see from this analysis is that when we have a bounds structure like a parallel plane waveguide firstly there are discrete angles at which the uniform plane waves can be launched inside this. Superposition of this uniform plane waves which are bouncing back and forth between 2 parallel planes gives you the field distribution which will be like this. And those field patterns we call as the model patterns and we also see that for a given mode number or to have a particular mode pattern excited inside the structure the frequency has to be about certain value what we call as a set of frequency So depending upon which order of the mode you want to excite you require a minimum frequency then and then only that particular mode will excited. So even if the waveguide is capable of supporting different values of m it will depend upon the frequency whether that value of m will be acceptable or not because whether this condition the frequency is greater than set of frequency is satisfied or not. So this is now a basic picture of model propagation in a bounds structure like a waveguide. So we will elaborate on this and then following this essentially we will try to see another mode when we meet in next lecture and then we will go to the more compact structures like the rectangular waveguides

Mod-06 Lec-36 Matrix Analysis of Plane and Space Frames

Good Morning This is lecture number 36, module 6, Matrix Analysis of Plane and Space Frames. If you recall, in the last session, we covered the reduced stiffness method; We will continue with that method, and also complete the flexibility method, as applied to plane frames This is covered in the chapter on plane and space frames, in the book on Advanced Structural Analysis So, this is the reduced element stiffness method You know that the size of the plane frame element stiffness matrix is 3 by 3. We have done a few problems This is how you do the transformation the T D matrix in the reduce element stiffness method Now, let us look at a problem with sloping legs, which is actually a complicated type of problem, when you apply the reduced stiffness method. So, you have to intelligently identify the sway degrees of freedom; we have discussed this earlier. And you need to express the chord rotations in terms of the identified sway degree of freedom So, let us look at this problem which we have solved earlier by the more regress method So, let us solve this by the reduced element stiffness method and let us a take advantage of the fact that we can ignore axial deformations We have solved this problem earlier by the slope deflection method. What is the degree of indeterminacy, kinematic indeterminacy? Shortest needed degree of kinematic indeterminacy refers to the active global coordinates. The absolute minimum when you have when you do not have axial deformation 8, 5, 8 Ok Five At the supports A and D, do you need to have anything? No, No No. We do not even bother about finding reactions We do the minimum work. with So, at B and C, how many do we have? At each of them You have a rotation at B; you have a rotation at C Vertical reaction And you have just one sway degree of freedom; that is what we did in slope deflection method; remember – theta B, theta C, and delta BC Correct, but we have done this problem by the conventional stiffness method. There the degree of kinematic indeterminacy was – at B and C, you had each 3 degrees of freedom; 3 plus 3; you also had a rotation at A and D; it was 8 So, we have, we are reducing 8 to 3, but we are ignoring axial deformation; that is a major reduction. The method of solving is, as we have discussed earlier, the steps are the same plus we take advantage of the fact that, we limit the considerations only to active degrees of freedom because we can always get the support reactions at the end from the free bodies So, let us do this problem. So, I have got 3 degrees of freedom; global coordinates 1, 2, 3; that is all I need. Local coordinates, again, I take advantage of the fact that the there is a hinge at A and D So, my 2 degrees of freedom B model will now become 1 degree of freedom model. What I have achieved is massive reduction in the quantum of work A, I am not going to use a plane frame element; I am using a beam element. A plane frame element has 3 degrees of freedom in the reduced formulation, but I am not worried about the axial degree of freedom. So, the 3 becomes 2 The plane frame element becomes a beam element and I am using that formulation for element number 2 because there is a fixed end force coming in there. There is an intermediate

load, and B and C are not hinged. But for elements 1 and 3, I take advantage of the fact that, I have a hinged support at the ends A and D where, the bending moments are 0. And so, you know, I can take advantage of a reduced stiffness. What is the element stiffness for that flexural stiffness? 3EI by L. So, I just need 1 degree of freedom which I marked as 1 star; is it clear? So, with this, I can proceed. I first need to write down my T D matrix T D A matrix Will you try that? To do that, you have to be very careful. When you are dealing with sloping legs, you have to correctly express the chord rotations in terms of the sway. So, the displacement D 1 corresponds to the delta in this figure right The figure at the bottom is a generic figure So, you have chord rotations of delta by the height for the elements 1 and 3. right And for the element 2, you have to work this out using trigonometry; it turns out to be as shown in that formula delta tan alpha plus tan beta by the span L. I am not going through this all over again because we did exactly this, when we did the slope deflection method So, if you plug in the values of tan alpha and tan beta because this the dimensions are shown there, you will get the chord rotation for the element 2 as plus 27 by 36 times the sway D 1 Just plug in the values of tan alpha tan beta; this, the span of that element BC is 3 meters So, you will get this. Shall we proceed? We have done this earlier. It is just a repetition We put a positive sign because you are getting an anticlockwise chord rotation for element 2, but please note – for elements 1 and 3, the chord rotation is clockwise; clockwise; chord rotation is clockwise. So, and it is given by delta by the respective heights Now, can we write down the T D matrix. We can So, the T D A matrix for the three elements are as shown there. If you apply D 1 equal to 1, then for the first element, the chord rotation is 1 by 4; it is a clockwise chord rotation. You get equivalent anticlockwise end rotations and it will be plus 1 by 4 because anticlockwise is assumed to be positive in the sign convention The same is true for the element 3. So, you get 1 by 3 because the height is 3 meters As far as the element 2 is concerned, it is 27 by 36. right It is We just worked it out know 5 2; it is 27 by 36, if you expand it, sorry 17 by 36, it turns out to be 0.4722 But it is a clockwise rotation. So, your equivalent end rotation will be, sorry, it is an anticlockwise chord rotation. You are you get clockwise equivalent end rotations. So, it turns out to be minus. This is the only difficult part; that first column in your T D matrix has to be very very carefully written because it is totally dependent on your correct assessment of the chord rotations; both magnitude and direction The others are simple. this If you put D 2 equal to 1, we are now talking about the second column in the T D matrix. Then, you will find that you get it will effect only elements 1 and 2. The right end of element 1 and the left end of element 2. So, you get… and element 3 is [un effective] So, you get 1 1 0 0, and similarly, for D 3; is it clear? This is easy to generate. Can I proceed? You got your T D matrix. Now, you have to carefully write down your fixed end forces We did this earlier, and remember, we just calculated the fixed end moments. But here, you have to be careful because you have only 3 degrees of freedom D 1, D 2, D 3, and you do not have those vertical forces which we put on the last occasion. right There are

no vertical degrees of freedom. So, you have to manage with rotational degrees of freedom and the sway degree of freedom. You have to handle the problem in that framework. So, how do you do that? Well, first of all, from this analysis, you can find out that there are going to be no fixed end forces for elements 1 and 3. Why? Because there is no intermediate load, and for element 2, the fixed end moments are 30 kilo newton meter minus 30 kilo newton meter and there is no axial force. So, this is what you get from this diagram, but how do you include the effect of axial force here? You have only 1, 2 and 3 in the global coordinates How would you transfer this understanding to equivalent joint loads, the global coordinates? What is In other words, what is F 1, F 2, and F 3? F 1 f, F 2 f, F 3 f. What is F 2 f? F 2 f is plus 30 because it comes from element 2. What is F 3 f? Minus 30. What is F 1 f? What is f one f? It is not 0. If it is 0, then you make a mistake What is F one f? 40 Yeah. You look at what is happening at B and C. You have restrained those degrees of freedom at B; you get a horizontal reaction to the left of 45 kilo newton at C; you get a horizontal reaction 40 kilo newton. What is happening to the whole beam? So, you will get a net force of 5 kilo newton acting to the Left or right Okay. To the left; you are right. But, when you find the net loads, you have to oppose everything; you put a minus sign. Have you got it? So, that is it Look at this. It is completely different from what we did earlier, and you are going to get the same results. So, there are so many different ways of solving these problems, but sloping legs problems are tricky. They are notorious; students generally make blunders; practicing engineers do it. But luckily, for you, today you have nice software which is the black box to most engineers. We just feed in that input and you get the output, and you hope all is well But we are learning and we are learning to do these problems by different methods including flexibility method. And we know one thing for sure – the answers are have to match the answers have to match. We will expect some difference between ignoring axial deformation case and including axial deformation case, and that will be a little more pronounced when you have sloping legs. We will see that; apart from that, you should get the same answers we got when we did the slope deflection method So, we proceed. We write down the element and structure stiffness matrices; very easy to write down; EI by L for all those values are known, and 3EI by L is what you will put for elements 1 and 3. And for elements 2, it is a standard 4EI by l, 2EI by l 2EI by L, 4EI by L; is it clear? We can generate this and then carry out these products, and do the final super position and assemble all the matrices. We will get the answers. This I am going fast because you know how to do it Now, you are ready to take any load. You give any D 1 D 2 D 3; you get any F; F 1 minus F 1 f; F 2 minus F 2 f; F 3 minus F 3 f; that is your net load vector. You will get the solution with the power of this matrix. So, we will do that. We will invert this matrix; plug in those values of 5 minus 30 and plus 30. You got those answers. And that is what your deflected shape looks like Now, you find that, that deflection, if you compare with the earlier result has the little discrepancy, but that is you have to live with that because that is the approximation you get from ignoring axial deformations, but this is exactly the answer you get when you do slope deflection methods. right It is also exactly the answer you get from flexibility method, ignoring axial deformation

If you do a strain energy formulation, you will get this. Then, you find the member forces when you draw your final bending moment diagram The bending moment diagrams do not change significantly; it is only if the deflection that ok You know how to handle this problem. You got a similar one in your assignment. You got a hinge, support, you have a support settlement also, but luckily, for you, you do not have sloping legs. They are vertical. So, please do those two assignment problems and you will get a good understanding of stiffness method We now look at the last remaining topic of flexibility method, not applied commonly in practice for frame. We do it for beams; we do it for trusses; frames it is not common, but the logic is the same, and at least conceptually, let us learn it. You will find that in your assignment. I have not given a problem on flexibility method; in your final examination, it is unlikely. I will ask you a question for plane and space frame, but, let us just go through the theory, and see how it works First, you need to identify the degrees of freedom. You know that, it is the same as you use for the reduced stiffness method; 3 degree of freedom system. You can generate the flexibility matrix from first principles You apply 1 unit load at a time. Here, you apply loads. You can do one thing. You can take your stiffness matrix and invert it; you will get the flexibility matrix, but that is no fun Flexibility is for people who want fun because you want to understand exactly what is happening So, here, when you pull it, you are not going to get any moments. It is just like an axial truss element. You know that the deflection, the lateral elongation you get is an extension in that element. It is the flexibility, axial flexibility l by E, Clear? And that is your f 1 star 1 star Next, you apply a rotation. Now, you are doing what you did for the beam. You get rotations of L by 3EI and L by 6EI. One will be anticlockwise; the other will be clockwise. You do not have any axial you do not have any axialchange in length. That is an assumption we are making because we are saying that the flexibilities do not interact. The axial degree of freedom and the flexural degrees of freedom are independent That is an assumption we are making here We will see through that assumption in the next module, and will bring in that interaction Now, we accept this and you can do it at the other end, and you can generate your flexibility matrix. Very easy to remember because you know what you did for a beam element. you are know You know it is L by 3EI minus L by 6EI L by 3EI and minus L by 6EI. You just have to add the axial stiffness as you did in the stiffness matrix L by EI Then, you do the transformations and standard procedure. let us This is for the beam element For the frame element, it will have 3 degrees of freedom; the rest of the procedure is the same You generate the structure flexibility matrix You write the compatibility equations. You solve and find the redundants; then, you can find the joint displacements if you want to In the stiffness method, you had no choice You have to find D A to proceed to get the member forces; here, you do not need to find By solving that second equation, you get the redundants. If you can draw the free bodies and get your member end forces, you finish everything So, this is the procedure. The displacements come at the end Now, let us do just one example with and without axial deformations. Now, please note, when we did the method of consistent deformation, when we did the theorem of least work, we assumed axial deformations were negligible So, we are now doing for the first time flexibility method with axial deformation So, let us see how to do that. What is the degree of kinematic indeterminacy? Here, we put in two complications, just to see how to apply this method; one is you have a support settlement; the other is you have an internal hinge. So, if you did not have that hinge, you know that the degree of static indeterminacy is 3. That hinge brings your moment release So, it drops to 2. Now, you can choose your

two redundants Which would you choose? You want the primary structure to be a cantilever Then, that third element will be dangling in the air. It will be unstable. Can you please think carefully? I mean you preferred keeping the fixed end, A as fixed end right; that is what you meant by cantilever. Mind you, you can still do that, but at D, you do not remove that support. What kind of roller? You want the roller to roll vertically or horizontally Horizontally, vertically You have a choice; not only that, you can choose any two. So, let us do something interesting in the next slide So, I am going to choose my redundants x 1 and x 2 as the vertical reaction, and the fix and the moment at the support D which was I have to in my primary structure. I have to I have to release those two corresponding restrains to displacements. So, what is the support I will put there? Put the roller in the vertical direction; put the roller in the vertical direction, and number your coordinates appropriately So, we know, you know that we have always put the redundants at the end of the list So, we have 1, 2, 3 degrees of freedom at B, 4 and 5 at C. There is no rotational degree of freedom. You cannot bring it there because you do not have a unique rotation there. Each of those two elements will rotate by different amount. So, you do not number it We did the same thing in the reduced element stiffness method, but here, we add 6 and 7 as redundant coordinate. So, this is how it will look, if you write the force vector We separate F A and F x. F A is the load applied It is 50 kilo newton because there is a horizontal load at B; rest of the quantity is the 0, and you have x 1 and x 2. x 1 and x 2 are unknown. Any question? ok You also have to write down the compatibility equations to solve for the unknown redundants x 1 x 2. What are What is D x? What is D 6 and what is D 7? Right D 6. that is why you bring in the support settlement D 6 is minus 0.01 meters and D 7 is 0; yeah. You have a question? I am not ignoring. Go back to the question We said we will first include; then, we will ignore, and then we will see the difference So, I had we will remove that in the next phase. Right now, we will include So, the procedure: First write down the force transformation matrix; find out the equivalent joint loads. You have to do the… This part is borrowed from stiffness method which is unfortunate because your flexibility method is not able to stand on its own. It has to borrow this concept because you just can handle intermediate loads in a matrix formulation Ah unassembled element flexibility stiffness matrix, structure flexibility matrix, T F transpose f star T F; write down the displacement equations, all the compatibility equations, get the redundants, get the member forces, and if you wish, get the joint displacements The procedure is very clear. The items 4 and 6 emerge from those set of equations. Then from the second equation, you get the redundant; from the first equation, you get the joint displacement. We have done this earlier; the procedure is the same So, this is interesting. Let us generate the T F matrix. T F matrix has two parts; T F A and T F x; how do we do that? You have to apply unit load at a time. So, it will look like, if you apply unit load F 1 equal to 1, take that huge structure and apply. Do you think anything will go to the support D? No. So, it will be, only element 1 will behave like a cantilever. right Am I right? If you apply F 1 equal to 1, and

no other loads, then, you need not draw the whole frame; just pull out that cantilever as it is; you got those results. right So, I am going to demonstrate here, how to generate the force transformation matrix F star is f star f plus T F A times F A net plus T F x times F x net. Right. We will work out that F A net and F x net later, after we handle the fixed end forces, but right now, our task is to write down T F A and T F x So, I have written down the first three columns, corresponding to F 1 equal to 1, and I will show you F 2 and F 3 shortly. But you look at that first column 0 4 0. Is it right? Because only the first element will have non zero items; second element will not be affected; third element will not be affected because this is a cantilever action for element one Agreed. So, that is why you got zeros here I am applying this load on this structure Here, only this element gets effected. Elements 2 and 3 remain unaffected. So, the first column corresponding to F 1 equal to 1 will have 0 0 0 0 0 0 for elements 2 and 3. For element 1, these are the reactions I get if you go back, and I have not shown the element coordinates Element coordinates would be… what would they be? They will be 3 by 3. right They will be 3 by 3 So, you will find that the first item here, what does this correspond to? What does f one star correspond to? Axial force. Will there be any axial force in this element? No. That is why I put zero there Second element corresponds to what? Moment at this. Well, it depends on how you put your element, but it is here 4 here, right and the third one, moment at B 0; it depends on start node end node. I put start node at A So, anticlockwise is clear; so, it is 4 Now, let us take the second case: F 2 equal to 1. Can you try drawing the response for F 2 equal to 1? I am going to show it to you; F 2 equal to 1 means you are going to… what is going to happen if you apply F 2 equal to 1? You will just get one axial force in First So, that is the simple thing. So, does it make sense? You get just one. You get just one here and the rest of it are all zeros Clear? Very easy to fill up. Next one, can you try? F 3 equal to 1; F 3 equal to 1; F 3 equal to 1, will it affect element one only or …? Only one It will affect only element one. Try drawing Will you get any reactions at D when you apply that unit format? No. No. I will tell you one way to understand. Remember, when we talked of the internal hinge, usually you have a parent child relationship. Now, clearly, element 1 element 3 is a child. It badly needs elements 1 and 2 for it to hang on to for its own their life. right So, the child cannot take loads when the parent is loaded. When the child is loaded, you will get something in the parent. Let that be very clear. right So, you can quietly remove the child from the picture and look at the parent Parent is a cantilever. Now, if you apply a moment at joint B in that L shaped cantilever, do you think anything will go on to element 2 and 3? No So, this is flexibility method; really have to enjoy the physics of the problem and get the answer. So, you just have to draw this In coordinate Yeah Um, one – axial degrees of freedom, and two – moments Right In the element one Right But element 2 has one axial different or one moment No. They are all frame elements; all are identical; all have one star as the axial, and because there is going to be an axial force in that element Sorry Element 2 there is no Ah We do not take advantage of it here. We do not take advantage of that hinge here because that we will do when we ignore axial degrees of freedom. You understand, this is here we

are not taking advantage of the release. There, you can do it. I am not taking advantage in this case, but I will take, I will take your suggestion in the next case, where we ignore axial deformation So, your point is valid. I can actually reduce the degrees of freedom for element 2 and 3 Right now, I have not done that. It is I will still get zero moments there. So, your point is good. We will do it later. Is it clear? Now, I will go faster Ah If you apply F 4 equal to 1, again the child is not affected. So, I have not shown the child. Here, child is element 3. So, can you generate the middle part as I have shown here? You can work it out. Next one, apply F 6 equal to 1 and F 7 equal to 1. You get the remaining parts. Here, the child is being pulled and the child is hanging on to the parent; parent also gets pulled. Clear? So, you have got all three elements activated when you apply . These are simple things These are statically determinate system. You can write down those forces, but it is fun writing down the T F matrix. Now, just look at this complete matrix. What we are saying is we have written a set of equilibrium equations; if someone gives us this vector which is a fixed end force vector, someone gives us this load vector including the redundants; we have drawn it for a primary structure; the redundant are treated as loads on the primary structure. F 6 and F 7 are loads on the primary structure. Is it clear? We get the bar force, get the member end forces, and we will do this in the last step. So, how do we proceed now? We have to write down the fixed end forces. Elements 1 and 3 have no fixed end forces. Now, these are the local coordinates by the way. So, this analysis, we have done earlier. We have done it for the reduced element stiffness method. Here, you do analysis where you do not get a moment at C which means you have to treat it like a prompt cantilever to get the equivalent joint forces. Is it clear? You get the vertical reactions and write down the F 2 star f vector. You have only one moment there because here you are accounting for the . At C, there is a hinge. So, you do not So, does not make sense. Now, I am taking take care of the hinge, but I am still keeping my full-fledged 3 degree of freedom matrix, but I will take the advantage in the next situation So, I have got my fixed end force vectors for the first and third. They are null vectors For the second one, I just have one moment there. 111.11 and that is anticlockwise, and I can write the F f vector. Clear? Next, I should get the net joint loads, resultant nodal loads. So, that is F f net is F A minus F f A and F x minus F F x. I just do this substitution. When I write the second one, I have to that is where I bring in those forces You see, this is given to me. This is written in the beginning. I have 50 kilo newton load and I have x 1 and x 2. I do not know. This is my F A F x vector This is my fixed end force vector which I compute by inspection. by inspection If I go back here, I see that, not only this will contribute to F 3, and these two reactions will contribute to F 2 and F 5. Is it clear? That is how I get these two quantities here Does it make sense? They are pointing upwards with a positive, but I put a minus sign, and now I get the final picture. This is the structure I am going to analyze. Is it clear? It is completely taken care of the F x of the intermediate loads How do I analyze this? I first generate my element flexibility matrices. I have taken them all. To have this format, it should be L by EA and all these values [are known/unknown] You can generate the flexibilities. What do you do next? Generate the structure flexibility matrix. You you have the unassembled flexibility matrix by putting them in a diagonal form You have the T F A matrix. You do that product You will get F x x. It is a 2 by 2 matrix which you can generate Please note, when you actually do it on matlab or on a computer, you have to be careful when you deal with flexibilities because stiffness

values are going to be large when you write in kilo newton meter units but the reciprocal of it, the flexibilities are going to be very small So, if you work with the absolute figure, you will have massive rounding of errors So, you have to work with significant figures So, you have to cleverly take out ten raise to minus 6 outside. So, you get large numbers inside and you do not truncate those digits This is what you have to be careful when you deal with flexibility. Is it clear? In stiffness, you do not really have that problem, but in flexibilities, you have. You understand? If you have this quantity and you write it in decimal format with ten raise to minus 6, you might write it as 0. So, 9800 becomes zero because you are used to writing by hand, point naught. Now, you get tired of writing so many zeros, you say, may be it is meaningless and you put zero. Is it clear? So, be careful because you you need to work with significant figure So, you do this, get the inverse of it, and you you are ready to do the final calculation Similarly, you can generate f x a and f x a T. All these can be done through matrix operations Now, you find the redundants by solving that second equation. You do not have any temperature effects or you know constructional errors So, you do not have any initial forces here You plug in those values and you finally get you get the F x net ok So, you get x 1 and x 2 in the last step x 1 and x 2 is 39 kilonewton and 74.18 kilonewton Find the member end forces. You remember those large equations we generated with the T F matrix. Now, you are in a position to put, and all those unknown values you get the final end forces exact When you draw your… if you compare with what we did earlier, earlier in the exact method, the conventional stiffness method or the reduced element stiffness method without axial deformation. You will get these answers absolutely; no error till the sixth or seventh decimal place; it is exact ok Then, you draw the free body diagrams draw the bending moment diagram and again remember we said that you can get the deflection straightaway you are really [interest in this way/interesting the sway] by recognizing that c d base like a cantilever we have done this earlier but, if you still want to work out displacements for the hard way you are welcome to do so do this calculation and interpret the results we will get exactly the same values we got earlier ok So with this we complete the flexibility method including axial deformation which is something new you discovered you have not done it earlier you are not done it in method of consistent deformation you are not done it in theorem of this work or column analogy method that is another method Now let us see how to take advantage of of axial deflection one way to get the solution if you writing a program is put e a tending to infinity there large value of e a you get the result of solution without axial deformation because if axial rigidity is infinitely large axial deformation is is small that is what you get But let us do it this way now what is interesting is your degrees of freedom now reduce you had seven earlier now you have four for reasons which we explained earlier so you have only one two and three four corresponds to your restrained degrees your redundant degrees Also for the beam element we will take the advantage that you had pointed out the two degree of freedom beam element for one anD 1 degree of beam elements for two and three can be used in place of frame elements So we we are going back to beams and we are taking advantage of the modified flexibility so we can do that for element one you still have two degrees of freedom but, because we want to take advantage of the internal hinge at c i put a roller support there so i have only one degree of freedom and my f a and f x vectors are easy to calculate it is smaller much easier to work out d x is the same only thing in earlier we wrote d six d seven now you write d three d four Force transformation matrix much smaller much easier f one equal to one f two equal to one

same results these two do not change f three equal to one f four equal to one same only we have renumbered them f six became f three f seven became f four your size of your t f a matrix is shrunk to a nice compact size right You can interpret the results very easily You got your T F matrix. Find the fixed end forces exactly as we did earlier. There is no change in this. Only your size of your vectors are become smaller. You got rid of things that were not required. You are dealing with the same system element flexibility matrices f 1 star is the same as we got earlier. But for elements 2 and 3, you take advantage of L by 3 EI. This is a clever trick you can do. 3EI by L for stiffness, L by 3EI for flexibility So, you get so I have for the first time, I have introduced this symbol f star tilde because it is like k star tilde. It is different You know you are working with a reduced flexibility right. Generate your flexibility matrix. This is the unassembled flexibility matrix Then, you do the operation and generate your You can do them separately; f x x is T F x transpose f star T F x, and f x A is T F x transpose f star T F A; f A A is T F transpose f star T F A. Is it clear? Or you can deal with it altogether and then partition it at the end, but this is much easier to handle Find the inverse of your f x x matrix. Find your redundants; plug in those values; you get your solutions as 39.048 and 74.286 now; what did you get earlier? 39.059 and 74.182; big deal So, that is why we do not waste time dealing with the axial deformation; the errors you make are negligible. And in the context of what we discussed yesterday, this is certainly accepted; this kind of… because they are going to be much larger errors in reality ok Find the member end forces; plug in those values as we did earlier. Find the displacements if you wish. Now, you get the same sway at both ends, but it is very close to what we got earlier. And the deflection at C earlier was 10.07 mm; now, it will go down only as much as the support D goes down; is it clear? So, we have tackled this problem in many different ways. We are very clear on what is going on Proof of the pudding is in the eating; which means, the answer should match. Completely different methods and same answers Finally, your free body diagram; bending moment diagram. Compare these two diagrams; exacts with axial deformations and without axial deformations It is all the same. May You are making errors in the decimal places; not worthwhile doing it. So, powerful methods when you do manually, do them the easy way out. When you are writing a program, do not do flexibility method; do the stiffness method. You have a choice – reduced elements stiffness method or conventional stiffness method With this, actually we have completed everything except space frame, which are really tough, but I am trying to make it simple. And again, you do not need to do anything in your assignment or your exams, but let us get a taste of it It is like moving from one dimension to two dimension to three dimension. The concept is the same, but you have to really think; especially, if you want to do the reduced method this which you should learn to do manually So, we will cover it. I am not expected do any example. We will do this in tomorrow’s class After that, the seventh module again is an exposure I want you to have. I want you to be present; not so much to learn, to solve it because it is really advanced; it is second order analysis, but you should have a flavor of it. And those of you, who are interested can actually sit and solve those problems I am planning to finish that in three sessions, and then, we will have one last concluding session. So, we are nearing the end of this course. Thank you

Mod-05 Lec-11 Intersection of Half Planes and Duality

Welcome to lecture eleven of computational geometry. So, we will start only new problem which is actually related to the problem of convex hull and I will establish that relation in the course of lecture, but let me first state the problem Ah. So, we are what we are given. So, we can talking about a this problem is intersection of half-planes. So, we are given a set of half-planes what is a half-plane look like is basically bounded by a line bounded by a line and you know 1 side to the line is the half-plane that we have we know will be pertaining too. So, you know I will just shared this indicating that you know this is the side of the half-plane that I am interested in That’s half-plane I am talking about the line divides the plane into 2 half-planes So, the shaded side is the one that I am pertaining to. So, this is 1 half-plane you know that is another half-plane lets called this H 1 H 2 you know and have more half-planes H 3 H 4 something H 5 and. So, on. So, forth So, given these half-planes by the way half-plane is a a very simple example of a convex set right half-plane is convex and you can prove it by definition convex the entire segment of the 2. lying inside the half-plane will also be inside the half-plane Right. So, half-planes is a convex set and what I am interested in is given a set of let us say n half- planes set S of n half–planes compute the intersection. So, H I 1 to n which basically means the region common to all half-planes now in the example that have given here I mean just the infection what does it look I mean should be common this this region should be common to all half-planes. So, for instance a a is this point common to all half-planes Now, because this point certainly does not satisfy h 5 right is this point common to all half-planes Now, because it certainly does not satisfy h 4 right h 4 is h 4 is this this thing and the point is to the other side of h 4 Does this satisfy all constrain all all half-planes there is region common to all half-planes Yes ok. So, if finds this 1 is common. So, what we are interested in is not just 1 or 2 points, but they entire region that is common to all the half-planes and because half-planes are convex by definition the intersection of half-planes will also necessarily be half-plane And what is the intersection of all half-planes and what kind of region is that it is a convex region and it may be a convex it basically a convex polygon right It is a region which is convex and is bounded by straight lines which are essentially the lines bounding the half-planes. So, intersection of half-planes is a convex polygon provided Well I am not even talk about n 1 n 1 is 1 situation yeah sure it is not a strictly speaking if the convex polygon yeah the polygon is essentially bounded this may not be bounded

that is one one problem What is the other problem Intersection can be null exactly the intersection may be empty right. So, so the common region is convex polygon in most cases, but caveats is may be unbounded and other problem is may be empty Fine how do we deal with a unbounded is a a simple way of dealing with unbounded. So, you do not have to construct a consider this as a special case when we are trying to design a algorithm Yeah we are saying that I put a kind of a window Right and the window should be a rectangular window let us say we should be large enough to contain Right see good. So, that is I was say So, since it is unbounded of course, the the rectangle region I am to be. So, the rectangular region suppose is some rectangular region what I what want is that want make make sure that even if it is unbounded Also this is not unbounded . So, this region looks like unbounded right. So, these 2 edges do not into converge what you know basically they they going to go. So, I can clip it using this rectangular window and I will be happy to set of and and it kind of it contains all the vertices of the of the of the intersection region right. So, this misses out of on the unbounded thing, but a as long it it has it has adequate description of this this intersection region because it it captures all the vertices and all the vertices corner points essentially going to be corner points of the they they will be described by 2 or more let us say lines bounding the half-spaces And therefore, you know you can actually in some sense compute somehow these smallest a if you compute the smallest and the largest ah intersection points you know both in x and y deduction that could be one that could find this rectangle because there we no vertices falling outside the rectangles So, that is one way to sort of deal with this this kind of situation where it is unbounded we now the question of poses how do you compute this smallest and largest you know let me not get into that part what just just. So, that you know we we can a we can we do not have to deal with unbounded regions is what I am saying. So, some way we can simplify the whole thing without having to talk about unbounded regions essentially all I am saying is that we have some extra kind of half-planes defined by these lines. So, along with the given half-planes I am also looking at essentially these 4 half-planes And then that will bound the entire intersection So, the entire intersection is become bound, but emptiness is something that we cannot push it away why. In fact, emptiness is a very important problem. So, so 1 motivating problem for this finding in intersection is a can you tell me give me 1 example of why are you interested in intersection of half-planes exactly. So, a basically the constraints a linear constraints are basically half-planes right. So, ah 1 of an and it is an known that

the problem of finding weather there is a feasible solution and the problem of actually solving the linear program they not be very different in in complexity So the problem of finding intersection is almost equivalent to finding actually the optimum point So, So, just to find out if a given set of half-planes has a non non-prevail intersection that is the it is not empty that itself is is is a is a basic fundamental problem and you cannot push it away So, given a set of half-planes just the problem tell us whether I mean to to determine whether or not it defines a feasible there is non-feasible region is a basic problem. So, I cannot push that away right So, whenever. So, in to solve this problem find in the intersection half-spaces I have to deal with this 1 possibility that the intersection can be empty and we would not know in advance intersection is empty that will discover over the course of the algorithm algorithm should be able to tell us whether or not it is a intersection. So, you can see actually this problem you know simply stated in 2 dimensions you know as you move to higher dimensions and it could become quite complicated. In fact, the linear programming is nothing, but the the constraints in a linear constraints in arbitrary direction arbitrary dimensions So, you have these half-planes which are constraints in d dimensions and then there is some kind update functions and that is you know. So, fiinding the finding 1 point if the feasible region that we finding the optimal point in the feasible region and I am saying I am claiming right now is the not very different in complexities if you can find 1 you can find the another, but you know this problem that we discuss in today I will be discussing independent of polymer linear programming, but they this certainly you know some relevance to linear programming except that in linear programming we do not have to find the entire region So, actually linear program into some ex10d you can you you may think about perhaps it is simpler than this problem because I will be happy to find only 1 feasible solution which is optimal solution I do not necessarily have to find the description of the entire feasible region which is which could be a convex polygon in 2 dimensions or a convex poly tope in high dimensions although the we have not discuss this issue before this, but let me just a mention that the convex all in higher dimension which is that intersection of n half-spaces in in dimension d what is the kind of descriptive complexity it may have I mean that that that structure you know how complicate that we in 2 dimension it is a convex polygon. So, that description that structure is defined by let us say n corner points or the n adjacent So, forth in 3 dimensions may have already brought it of once right. So, that that the description is very similar to a In 3 dimensions if I take the intersect of n half-planes in 3 dimensions and suppose that intersection is non empty that region is a convex 3 dimensional convex poly tope and what is the how did you describe in 3 dimension convex polygon How many phases how many corners and how many edges yeah, but what is the total you know description and what what is the complexity of that that structure no no what is 4, but vertices for 4 triangles no no no no you are talking about tetra hydrons am not talking about tetra hydrons intersection of 4 half half half spaces in 3 dimensions I am I am I am mentioning a given n half-spaces in that is in 3 dimensions the intersection is suppose is non-empty it is a convex 3 dimensional convex poly tope what is the how do describe the what is the complexity of the structure in how many phases how many edges how many corners n c 2 n c 2 will be already n square kind of complexity right exactly. So, essentially the 3 dimensional convex poly tope you know is is nothing, but the plane a graph which for. So, that is you know you can euler’s formula follows that you know that you know a a a 3 dimensional structure Ah a plane. So, because this all convex need not only the whole thing is convex the whole intersection is convex if I if I limit if you limit yourself a plane look at 1 of the planes and look at the projection just look at the intersection of the remaining half-planes with this planes

Suppose I take the intersection I restrict myself to 2 dimensional plane 1 of the phases right. So, this is the 3 dimensional structure Let let us say I take 1 phase and I only look at the the the structure of 1 phase that is that is a 2 dimensional structure. So, that itself again will be a will be convex right so. So, now, from here it kind of follows that you are not going to have you know a the plane can only describe at most 1 phase you cannot have the same plane describe to 2 disconnected phases because that would not be convex So, therefore, it follows that the number of phases of this intersection is bounded by the number of number of planes yeah n right So, so there no more than n phases in the of the 3 dimensional structure and in the n phases by euler’s formula and a plane a planarity for the what were the plane and graph formula it follows the number of phases and number of vertices and a number of edges in a graph there are kind of linearly related ok So, the entire structure this 3 dimensional convex poly tope described by the intersection of n half-planes still has a linear structure it is it is it is not n choose to vertices I can say that right So, So so in 3 dimensions as a 2 dimensions you still have a linear structure in 2 dimensions very simple it is exactly an in 3 dimensions it is a planer graph kind of structures. So, it is a order of an, but there after you know things become moiety So, you a when you wants to go to 4 dimensions you cannot visualize the first you have to resolve other ways of you know figuring out what the what the complexity is and without getting into too much details let me say that the complexity grows roughly has n to the power of d over 2 So, it is like in. So, higher dimensional convex poly tope defined by intersection of n half-spaces that grows as roughly n to the power d over 2. In fact, there is a kind of flow on that right And this is known to be theta. So, there are actually convex poly topes which can have that many number of its it is not the phases its its everything its dimension 0 is vertices is dimension 1 that edges. So, all the facets of dimension 0 to dimension d minus 1 So, that can grow exponentially essentially So right So, if d is 2 or 3 you can see this is essentially order n then moment d becomes 4 it becomes n square and there after it really it really becomes the you know quit heavy And therefore, we actually solve linear programming you are not going to compute this entire feasible region because of feasible can be can have an exponential size and no way you can actually if you if you want to do it in polynomial time you cannot the 4 to compute this So, I am just trying to draw a line between that the fact that I am we are not really trying to attack the linear programming by constructing the intersection although intersection contains a all the information about linear programming, but you know we cannot construct it But of course, in lower dimension like we are dealing within 2 dimensions 3 dimensions we can we are just looking up on as it as a problem which is interesting you know other applications also And this has and the and the size is at most order n intersection. So, one thing let me do again ah like we did for this a convex hulls. So, this seems to be at least intuitively some sort of connection between convex hulls and an computing the intersection of half-spaces in the sense that both of them eventually give us some kind of convex polygons right But what is real relation of why should you know something that is discribed by points related to something will displayed in a half-way in when we we cannot we cannot appreciate that right now, but we did one thing when we try to compute the convex hull you know at least in in terms of conceptual simplicity we were able to dealing the upper hull over n 10 we we growly describe the algorithm in

terms of upper hull or lower hull right So, here I will take a similar root where I do the poly I will distinguish between half-planes that are downward pointing I will just say what I mean by downward pointing and upward pointing of course, I miss not something something which can either downward or not nor upward any way first let us say what were downward point is if this half-plane we have a half-plane say essentially now it is this what I am saying downward pointing and this is a half-plane that I am saying is a upward pointing One quick test for this is the half-plane is going to be describe some kind of inequality a linear inequality a quick test could be does this contain a the point you know y is equal to minus infinity a something This is essentially contains y equal to plus infinity. So, just see that whether or not this satisfies the inequality. So, I am I am just dividing the set of given half-planes into these 2 of course, there this kind of thing which is neither upwards or downwards since I cannot handle this you know we will say that these do not happen And why can why cannot this happen will just do a know random notation good. So, you are; obviously, the same language you know. So, these kind of bad things do not happened to us right right. So, now, now I am just coming to that I am just coming to that. So, now, once I have I have I have separated them out these 2 kinds of things Then we talk about essentially a the intersection of downward half-planes and intersection of upward half-planes right. So, we compute them separately which means that the downward the first case you know you can imagine that all contain minus infinity y equal to minus infinity So, it will looks like something like this and clearly if there is there are any of these downward half-planes the intersection is non empty portion Because all them will contain that minus infinity point. So, in somewhere we have also done away with this emptyness problem by separating half-planes and the upward ones we look like you know something like you know again this will be non empty and after we have computed them what do you do because we have to find out the intersection a yeah you know you know it is not pasting it is not pasting let me point out this is not going to be pasting into simple pasting It it could be more complicated than that, but eventually for the final answer will have to compute you know this intersection point and whatever is the intersection point and by the way this could actually be more complicated like this this could go on like this this and. So, on so, but finally, you know this is the region that we are interested in after you compute the intersection the downward half-planes after we compute the half-planes will again have to deal with finding the intersection of these 2 objects, but then you can see after all these objects has nice structure you know these are both kind of chains these are you know one one as you know this this kind of a structure is usually called convex chain right The red one is a convex chain red block one is a kind of convex chain there also monotone ok So, we are too nice you know monotone chains and therefore, claim that intersecting these will be fairly easy and this is also a problem for in one of the these one of the assignment problem where a in generally your asked to find out the intersection of 2 convex polygons So, this can be dealt with in that frame work

also this is this kind of viewed as the problem of finding intersection of 2 convex polygons and whatever time it takes the suppose you know it takes order n time its order n times it is a log n time it is log n time So, finally, let us let we let we just say that it can certainly in order n time that is for you figure out how to do it exactly and. So, will not bother about this final step of you know how do you find intersection of the convex and the concave chains or the regions regions actually Ah chains sorry upward region and the downward region. So, will now focus on just one of them let us say intersection of downward half-planes now I could try to not describe or define or develop an algorithm from scratch which at this point you know I would not attempt to do not now I will to just find out or explore does this have any kind of relation with what we have done just before this that is that the convex hull computationok Now; obviously, these are too looks like very 2 different 2 very different problems you know in 1 case dealing with points when we are talking about convex hulls So, convex hulls has point input and intersection problem has half-plane input which can also we thought of thought with the sort with the half-planes are actually described by lines let us say It is 1 side of the boundary line right in 1 case it is a point input that other case kind of a line input. So, somehow we need to be able to do some kind of correspondence between lines and points to be able to to to able to find out if there is any relation between these 2 structures at all All right. So, for that we will will resort to what is called some kind of duality relation some duality function let us say a duality map into more precise So this duality mapping let me define as d of a. So, what this duality mapping will do for us it will map points 2 lines and vice versa what is a point a point is a usually defined by a coordinates right. So, point So, we are this we are dealing with 2 spaces right the space of points as space of lines So, to be able to define a kind of mapping So, what do you mean by point to point is a order pair of coordinates a b whats a line a line has basically a parameter equation right So, a x plus b y plus c is an one possibility, but then a there is because I want to let it to the points. So, actually the line is a 2 dimensional structures again you should have only 2 parameters. So, if you think about the y is equal to m x plus c this is basically parameterized by the slope on the intersection my space of points of course, is a obvious choice of pair order pair of coordinates and for the lines we again choose the. So, now, we are choose the m comma c representation So, that we are dealing with 2 dimensional phases. So, in both phases. So, it is a mapping from 2 dimensional phases to 2 dimensional phases right So, this is what this duality mapping will do for us now in literature there are lots of ah you know different kinds of duality mapping and they have different kinds of properties So, I will first try to describe what are the preferred properties for the kind for the duality function that we used and then later instantiate one’s specific function that achieves all these properties So, let me first try to elaborate an what kind of properties you know this mapping should

satisfy what that if if you satisfies then we are in good share So, one thing that will do first is is say that itself inverse. So, desirable properties I will say. So, d of d’s of x equal to x x is either a point or line they want it to be self inverse ah So, if I apply the dual transform if x is a point apply the dual transform get a line and if I apply the dual transform for this line we get the point vice versa If I x is a line apply a dual transform line I get a point and again apply the dual transform I get back the same line So, this is 1 property would like to have another property is that you know is a is let us say 1 is to 1 most good mappings would have this properties I mean useful mappings have this properties. So, I want to trying to list the desirable properties I have not even said what function actually will actually satisfies the properties ok Now, here is a very important thing till now you know these all fine incidence. So, incidence properties is a following consider a point p and a line l incidence property says that if p is incident on l then d is of l which is a point is incident on d’s of p is a line. So, its incidence preserving the mapping that we are in interested in we will be incidence preserving if a point happens to be an a line and I take the dual transform of the point then it is a line a dual transform is a point and again that point should be incident of a line. So, whatever function you interested in should satisfy this property This kind of implies a following if l 1. So, l’s will be like line l 1 and l 2 intersect in p. So, p is a points l’s of lines in p is a points ok If 2 lines l 1 and l 2 intersect in p right l 1 l 2 p then d’s of p which is the line can you completed yeah should pass through d’s of l 1 and d’s of l 2. So, does is follow from 3 follows some 3 because the incident property if it incidence a is preserved then this point should lie on both lines and therefore, this line should pass through both these points So, it is a consequent of 3 it is not a separate property, but just to highlight something that you know will be able to use make use of this property When we talk about this connection between l’s n intersection of half-spaces right So, so one more thing. So, we are talking about incidence of what is the point does not lie on the line So, it is either below or above. So, we need to say something about again the orientation So, above sorry above or below property So, say that if p lies above l now the above I thing the same way that we talking previously

that you know plus. So So, here is a line So, it is a line l and p basically is above l just what do you mean by p above right So, p lies above l then d’s of l which is a point l ok is exactly on what kind of function finally, choose so, but I will just prefer this 1. So, above below. So, if a point is a above l d l d’s d’s of l will be below d’s of p You just interchanging the orientation, but its consis10t suppose all points and lines this will represent. So, again let me not even get into what function we satisfy all these properties I just use the properties to establish suppose such a function exist such a duality mapping exists then will try to use this to show the connection between hulls and intersections So, again we have limiting our hulls to when we talk about intersection we only talking about this you know downward planes or upward planes. So, here is basically what we are try to establish So, given set of downward half-planes let I of h denote intersection now consider the duals of the lines describing the half-planes h and denote it by S right. So, we have given a set of h downward half –planes the half-planes are described essentially by s Bounding line and which side of the line it is we know that it is a downward side that will obtain now we consider the duals of the bounding lines that describe the half-planes that duality is defined right where this a line to point duality that those points I am calling it S let c h of S is denote the convex hull of S So, just to draw the distinction we have this is my downward intersection of downward half-planes right now I am talking about convex hulls of points which are the duals of those lines described in the half–planes and here is my convex hull or the duals of the of those

lines the points of the duals of the lines So, what relation thus these blue structure have this structure will does not look like mind that blue structure looks about half that of the rate structure In fact, is half that of the rate structure So, what will establish is that you look at the chain which is that the the lower hull consider only the lower hull. So, lower hull is this structure if this again some kind of convex chain upward convex chain So, blue one is downward convex chain the black one is the upward convex chain which is the lower hull of this set of points right and now the claim is that at this is 1 to 1 correspondence between this lower hull and that blue structure Right namely that the half-planes that describe the boundary the half-plans that describe the boundary or in 1 to 1 correspondence with the points that describe the lower hull. So, this is the The yeah Exactly. So, the downward convex chain describing the intersection of half-planes is in 1 to 1 correspondence with the points which are duals of the lines describing h this is the lower hull Namely suppose I am just saying suppose this is half-plane half-plane 3 and half-plane 2 and half-plane 10 and half-plane 20 and half-plane you know 50 and somewhere that these points would correspond to the dual So, this point will be dual of the half-plane the line describing the plane 3 So, like dual of 3 this is dual of 2 and dual of 10 and. So, on. So, forth dual of 50 suppose this is true before even a we have approved it suppose this is true then how will you construct the intersection of half-planes you just take the then you will go the dual space basically you know even half-planes Look at the lines described in the half-planes look at the duals of those lines which are points use your favorite convex hull algorithm to construct the convex hull of these in the dual space which is which is the set of points and then the lower hull of that is in 1 to 1 correspondence with the intersection. So, once I know this is you know these are the these are the 3 d’s of 3 d’s of 2 are the points then I immediately get the I can from there I can simply get this structure that this convex chain is the intersection has first it first start with 3 then then then the boundary is defined whether the half-plane 2 with the boundary is defined with the half-plane 10 and. So,on. So, forth So, we recover the structure right away from there if this this claim is true why is this came true let me switch to the we will; obviously, try to use the the properties that we claim you know some there is some mapping that will have this properties and and that is the kind

of dual duality transform that we have using So, these are all downward this is my intersections this particular half-planes maybe I should actually draw them in red. So, this half-plane is even not even part of that boundary this So, the the intersection the the description of this intersection you know is essentially the the chain the chain of the half-planes that describe the boundary of the intersection if that is what we that is what we are interested in. So, the red some like this these half-planes do not even figure in the intersection. So, the description of the intersection does not even contain these half-planes So, those are basically I am saying like you know some points are going to be on the hulls some points are not on the hull this is equivalent of the corner vertices We now we have a what is the property of this boundary this boundary. So, why do thing that you know this half-plane in is part of the boundary where as you know the red is not you know some something something that is here Is not no no lets talk to the dual we are just trying to characterize this only in terms right in other wards you can think about it like this you know if you looking at you know suppose I am trying to draw plot a function which is let us say at any point x what is the minimum what is the minimum y that we are looking at. So, suppose these are functions these these these these half-planes like are functions you know these are linear functions any point of time I am looking at the min of the the the lines I mean which is ar the lowest. So, if you draw a lines these are the intersections. So, this is the lowest So, that is why it is part of the intersection a a half-plane. So, of course, this this this you know you move this x to here and it still you know this function is the lowest, but some point a next intersection points it changes next this function becomes a lowest. So, it is it something what is called a lower profile It may not does not even to be linear functions you can describe you you can define in terms of any kind of function I can have any any arbitrary function ok and I can look cause the and define the lower profile in the same way that any point point wise minimum among all the functions is the overall lower profile is what we calling So, this is one way to characterize that you know something will become a part of the intersection something some half-plane will not be a part of intersection so now we are going to now now let us go to the dual space I mean are are you happy with this characterization then only basically So, some half–plane will be a part of the output only when it becomes a minimum for some x So, now, when you go to the dual space where you know this line becomes a point all all all lines bascally some points When is a point a. So, so given a convex a set of points. So, a point is a part of the a you know a you know a final description of the convex hull or a corner point if we can draw a half-plane or a line through this point a tangent namely say this is size is point you can draw tangent through the point such that all other points are above this line right Can you do now can you see the correspondence between the a dual transfer preserve well

preserved means is that just switches the upper and the lower thing ok So, I am able to draw a line through the point now use the the incidence property the dual transform of these point this point Is one of these lines that that is how I got these points from right this line is some arbitrary line that passes through the point So, a dual of this line dual of this line will be some point will be some point on this line right. So, this one will become some point in the line may be it is a d of t. So, there is a point basically where all other points lie above this above this tangent and at the same d’s d’s of t point they satisfies it satisfies all the half-planes it is below of the half-planes that is all that is the proof no I its above below right. So, these points are above and this point is below all the all the lines its we are switch the upper and a above the line and below the this things if the if the point is above the line the dual of the line will be below the points So, just switched that is why we are looking the lower hull and the and the downward chain So, I will stop here today You know is need meditate a little about it this, but tomorrow hopefully and you know in the next class will be continue about this

Blackbird: The Fastest Spy Plane (Extended Cut) – SR-71

>> [Background Music] Learning occurs in many ways, reading, writing, observing and doing Montgomery College’s Access to History is for the doers Our guests have participated in actions or events that we now recognize as part of history In the studio, they sit down, mic up and have conversations with faculty and students from Montgomery College There is no host, no flashy set, and no commercials, just people and their stories Sometimes the stories date back many years and sometimes the story is current It’s all historical and it’s part of Montgomery College’s Access to History [ Music ] >> [Background Music] In this rare episode of Access to History, we leave the studio for the Steven F. Udvar-Hazy Center where student veterans who are part of Montgomery College’s Combat to College program got up close to aircraft that made history They also spent time with retired Air Force Colonel Joe Kinego who recorded over 900 hours piloting the famed Lockheed SR-71 “Blackbird” in military reconnaissance missions all over the world during the Cold War Traveling up to 17 miles above the earth at over three times the speed of sound, foreign powers tried to shoot down the Blackbird but none were successful Colonel Kinego’s presentation to the students during this visit contained information that at one time was top secret [ Music ] [ Silence ] [Background Music] US Military veterans now enrolled at Montgomery College were given a unique opportunity as part of the Combat to College program They were taken to the Steven F. Udvar-Hazy Center near Dulles Airport in Chantilly, Virginia The center is a companion facility to the Smithsonian’s National Air and Space Museum located along the National Mall in downtown Washington Home to about 200 aircraft and 150 large space artifacts, the building’s hangar-style floor area is as big as a Nimitz-class aircraft carrier Some of the most famous aircraft who have ever flown are here At the center of the entire facility is a plane unlike any other It’s painted black with a cartoon skunk on its tail Although it’s a military plane, it carries no missiles or bombs Its weapons are primarily cameras and incredible speed It’s the Lockheed SR-71 “Blackbird.” Retired Air Force Colonel Joe Kinego was one of its pilots On this day, he took the students in for a presentation that told its story [ Music ] >> [Background Music] First of all, welcome to Udvar-Hazy Center I want to congratulate you and thank you all for your service I know you’re retired– turning veterans, I understand, and all enrolling in college and that’s just absolutely fantastic I have always been a big believer that the military is a great place to get a start And then after that, when you kind of have a little bit of a feeling as to what you want to do with your life to go ahead and start your college education So, congratulations to all of you for that A lot of people think that the SR-71 was developed as a result of Gary Powers being shot down in his U-2 on May 1st, 1960 And as you’ll see as we go through the dates, that’s not actually true It’s not untrue but it’s not a true fact There’s more to it than just that July 5th, 1955 was– ’56 rather, was the first Soviet Union overflight by a U-2 aircraft done by the Central Intelligence Agency Two important things during the Cold War and you’re all are familiar with the Cold War I’m sure to some extent, the Soviet Union was a closed society We needed to know what was going on in there with their weapons development, with their missile development, and what were they doing We felt the only way to really do that was overfly the Soviet Union >> The U-2 he refers to is the Lockheed U-2, a high-altitude reconnaissance plane developed by aeronautical engineer, Kelly Johnson Earlier during World War II, Johnson led the design of the P-38 Lightning P-38s were used during Operation Vengeance where the US located and shot down Japanese Admiral Yamamoto over the Pacific Yamamoto was then the leader of the Japanese Imperial Navy

and one of the key architects of the Japanese attack on Pearl Harbor An original P-38 is currently on display at the museum The U-2 was designed to fly above 70,000 feet, safe from the range of the Soviet MiG-17– their best interceptors at the time, and safe from Soviet radar and missile defense systems Or so it was believed >> First flight was in July 5th, 1956 Two very important things happened at that point The first one was they found out our intelligence experts and our scientists found out that the information that the U-2 brought back was second to none It was fantastic It was– it was exactly what they were looking for Great, pictures, they could see the missiles, they could see what was going on in the Soviet Union, they could see the military bases very well That was the good news The bad news was that the Soviets knew the U-2 was there They could see it on the radar They could track the U-2 They knew it was there So our scientists at that point felt that probably four to five years before Soviet missile technology caught up to the U-2 And boy, they hit that one pretty much right on the head because it wasn’t– but you know, four years later, that Powers was shot down over the Soviet Union, it is U-2, May 1st, 1960 Following that, in 1962 when President Eisenhower worked with Khrushchev to get Powers back, there were no more manned overflights of the Soviet Union So, the U-2 didn’t overfly Soviet Union anymore The SR-71 never overflew the Soviet Union You will hear a lot of people say, “Oh, yeah, the SR-71 was overflying the Soviet Union.” Never happened because Khrushchev said in 1962 that he’d release Powers as long as we did stop all the manned overflights of the Soviet Union We agreed to that so we didn’t do that anymore, but that’s the time that the satellites were back online and we had unmanned vehicles that were being able to do the same sort of thing >> So between 1957 and ’58, well before CIA pilot Gary Powers was shot down over the Soviet Union, Project GUSTO was launched to develop a successor to the U-2 reconnaissance aircraft >> This is a handwritten note that Kelly Johnson did, you know, probably at the Lockheed cafeteria in Burbank And you see up here it says basically redesign the U-3 At that point he was calling it not a U-2 but a U-3 He didn’t know what to call it He had all those parameters and requirements, high altitude Down here, he did things that engineers do which I don’t even understand, lot of formulas and all down there that he was putting together But this was all handwritten This is Kelly Johnson’s handwriting from 1958 And then up here, he started a report and he did what’s unthinkable now of a person this important It was– he started writing the introduction to the proposal that he was going to give to the government This was his handwritten– his handwritten proposal there But I wanted you to see this because it’s just– you don’t get to see this very often But that’s how the whole program started right there That led to the A-12 OXCART This is the CIA version of the SR-71 It’s very much like an SR-71 You put it next to each other, you wouldn’t be able to tell the difference It’s about seven feet shorter than an SR-71 But this is the first airplane that the contract was built for and this was in 1959 It was May of 1960 when Powers was shot down So the actual contract to build this aircraft was prior to the shoot down of Powers in the U-2, so that’s the point I was making there, that it was after the U-2’s first flight but prior to the shoot down that this whole development process started >> The A-12 OXCART had a small radar cross-section, about one square meter And in the early days of radar, it would appear on the enemy’s scopes only once or twice before its incredible speed got it out of range Because this was before the advent of computers, the design of the aircraft was done entirely using a slide rule The A-12 was America’s first stealth aircraft During flight, extreme heat was caused by air friction and the aircraft’s engines Due to this, 93 percent of the aircraft is constructed with titanium alloy Titanium is lightweight It can withstand extreme temperatures both hot and cold And it gets stronger during these conditions >> I will tell you, sitting right there where I sat in a pressure suit, and so I’ve got the pressure suit on with the gloves and I’ve got about an inch and a half

to two inches of glass and I could put my hand against the cockpit window for about eight seconds before I had to pull it away because it got that hot But now I was comfortable in the cockpit because we could pressurize the cockpit and I could cool the cockpit and I was in a pressure suit so I could cool myself up also But if you touch the surface like I did, about eight seconds before you have to pull your hand away >> But having titanium as its primary construction material presented one key obstacle >> The United States had no source of titanium back in the late ’50s and the early ’60s The only country that did, and this was during the Cold War, of course, was the Soviet Union So the Central Intelligence Agency opened basically a storefront in Maryland, an international business that then opened a storefront in Europe as an international business, and then went into the Soviet Union and bought titanium and that it shipped back to Europe, and at which time they then shipped it to Southern California where they use it to build the SR-71 that then flew in the Cold War against the Soviet Union It was just a great story It’s just like your– yeah, just like your Saturday afternoon spy stories And then the heat required special fuels We use the fuel to cool the aircraft actually, special hydraulic fluids Hydraulic fluid also has a consistency of molasses You have to heat it before you can actually run it through the engines and all So, there’s a lot that goes into getting the airplane ready to fly but that’s all as a result of the heat >> Heat created by the aircraft cruising at around 90,000 feet above the Earth at Mach 3.2 which is about 2,100 or 2,200 miles per hour, or 36 or 37 miles a minute If it was possible to fire a high-powered rifle like the .30-06 from coast to coast and maintain its muzzle velocity, the Blackbird would arrive about five minutes before the bullet >> We always had a– we always had a joke that you didn’t know– it was nothing like being lost at the three times the speed of sound because– because you really get lost quickly >> The obvious main source for all this speed, the two massive engines >> Pratt & Whitney J58, its continuous afterburning engine which was unheard of at the time and still is one of the state-of-the-art things The air goes in there, this is where the air goes in, and as the air starts hitting the compressor stages, most of it gets bypassed around on this bleed bypass valves, if you will, they go around And then it reenters the engine there in the afterburner section where it gets reburnt and that equates for the ramjet cycle which means that at Mach 3 and above, 80 percent of my power was basically a ramjet, which meant that that engine and the inlet system associated with it was actually pulling us through the air as opposed to the engine pushing us through the air So you save a lot of fuel by doing that I mean we burnt an awful lot of fuel in the SR-71, don’t get me wrong But the airplane became more fuel-efficient the faster you flew This is an engine on a test stand and an engine run And I showed you this because the maintenance guys, the maintenance guys would run engines a lot out of Beale Air Force Base in the SR-71 And about twice a year when the weather was nice, they would set up bleachers out at the engine test stand and invite people to come out and watch an engine run, give them all head protectors and they’d sell hot dogs and sodas and popcorn to make money for their ready runs and all And this is what it looked like It got– the ground would shake You’d feel yourself vibrating but the engine will get so hot– so there’s the front part of the engine There is the afterburner part of the engine It became almost translucent You can almost see through it through the engine actually And the way you could tell that it was operating on all cylinders and everything was burning perfectly was 13 of these diamonds If there were 13 diamonds and they would count ’em, the engine would be operating actually at its max capacity And that’s how they’ve done They’d shut it down and we’d all go away [ Music ] >> [Background Music] Designing the aerodynamics for any aircraft is critical But for one capable of traveling over three times the speed of sound required unique characteristics >> The sonic boom on all airplanes comes off of the nose And then when the sonic booms comes off of the nose and the air is disrupted, you want the air to hit all the other surfaces basically straight

on like a regular airplane flying at 25 or 35,000 feet So everything has to be built so that that air hits the wings, the engines and all straight on So, if you look at it, look at the engine These engine spikes that are out in the front, they are cantered down and end That’s so that the air coming off here hits them directly The wings, you see the bending of the wings, that’s so the air would hit those wings directly in that two-degree angle of attack You could see the vertical stabilizers here that cantered in a little bit Everything was built to make the airplane precise and efficient at that altitude But in doing that, the airplane has no square corners, so without square corners, very little radar reflectivity going back So, unbeknownst to them or maybe beknownst, ’cause they were all professional engineers, they built an airplane that had very little radar cross section just because of the way they wanted to make it fly Then they put the black paint on it which absorbs some of that radar energy and they mixed some carbon in the bottom of the airplane here to misdirect some of the other radar So as a result, very little radar cross-section on the SR-71 Now, in the cockpit, I could tell if I was being tracked by radar I could tell if they locked on and I could tell if they launched a missile You can tell all that in the cockpit just through lights People ask me if I ever seen a missile I never saw a missile in flight I had several missions where I actually saw launch lights in the cockpit I looked out but I didn’t see– I did not, myself, see a missile but we’ve had guys see a missile SA-2 missiles, which back when I was flying, could fly 75 over 80,000 feet, no problem Now, they kind of go ballistic up there but they’re up there with you They don’t actually chase you They just come up and hit you, but the problem is you’re going so fast they would actually have to shoot that missile off a launcher long before you got into the range of the missile or the range of the radar that actually guides that particular missile So, you know, they have to know exactly where you’re going to be No SR-71 was ever hit with a missile There is talk that one of the CIA A-12 airplanes over North Vietnam came back with a little shrapnel in the tail from an SA-2 missile But the belief is that they were probably barrage launching on the B-52 bombers and it just went up to altitude and blew up and the A-12 aircraft happened to be in the area I don’t know that part for a fact but that seems to be [ Music ] >> [Background Music] The term “Blackbird” refer to the family of aircraft designed and built by Lockheed’s Advanced Development Programs division more commonly known as Skunk Works The Skunk Works logo can be seen on the vertical stabilizers of the SR-71 located in the museum There were a total of four different aircraft in the Blackbird family Fifty Blackbirds in total were built Of those 50, 13 were A-12s and 32 were SR-71s Aeronautical engineer, Kelly Johnson, was the primary designer of this aircraft >> Back during the Cold War, unlike budget scenarios now, back during the Cold War, if you had an idea and you were a smart engineer that the Department of Defense and the government respected like Kelly Johnson, money was no problem So, every time Kelly had an idea for something else to develop here in this Blackbird family, he had all kinds of money But I will tell you, when he got to the SR-71, he delivered the SR-71– how many times you hear this– he delivered the SR-71 several months early and many million dollars below budget And he actually gave that money back to the government for the SR-71 OXCART very quickly, the first flight of the CIA aircraft was in 1962 Area 51, you’ve always heard those mythical stories, yes, there is really an Area 51 I’ve been there several times What they would do is they’d build these airplanes in California They drive them over the mountains and then they would take them up into the desert at the Area 51 and they’d put them together and fly ’em After the Skunk Works, Lockheed moved out of Burbank into Palmdale, California in high desert They would build them there and then just drive them not many miles up into the high desert for Area 51 and those areas Even back in those days, in Area 51 and probably still today, they know exactly when the bad guy satellites are breaking the horizon And when they’re breaking the horizon, they stop everything If you go to Area 51 for whatever reason and I didn’t, I never had to do this because when I went out there ’cause of the position I was in and what I was doing But they will foggle you It’s called foggling When you get off the airplane, you put on glasses and the glasses are set up so that you can see down So when you’re walking along, you can see about three feet

in front of you but you can’t see anything else So when you’re going from building to building and you’re walking outside, you put on these foggles so that you can’t see what’s going on The important date is May 29th, 1967, operations began at Kadena Air Base, Japan This is in the A-12 now flown by the CIA, the first SR-71, to put in the time perspective The first SR-71 had already been delivered to the Air Force So there’s a lot going on at this point The A-12s were delivered to the CIA, they were going through checkouts The Air Force wants to get onboard so they put out a contract to build SR-71 So SR-71s are being built actually and have been delivered to the Air Force before the A-12 by the CIA flies its first operational mission over Vietnam by the CIA So in the fall of 1967, the A-12 and the SR-71 have a fly off Back in those days, you could say it right now, you cannot pick a name like this, that was called “Pretty Girl” was the name of the fly off that they gave it to And they basically flew the A-12s and the SR-71s against each other, taking pictures and comparing all the different sensor takes and sensor actions And between the two of those, the Air Force’s SR-71 model did get the go ahead In 1966, the SR was first delivered to the Air Force In March of 1968, the first SR-71 operational mission was flown also out of Kadena Air Base, Japan At that point, they had CIA A-12s there and Air Force SR-71s there, and then when the SR-71 was successful, they started flying the A-12s home Everything was very classified in those days, so people watching didn’t know that there were two different airplanes operating out of there They didn’t know one was an A-12, one was an SR-71 And in 1968, the A-12 program was terminated [ Music ] >> [Background Music] The SR-71 was the final design of the Blackbird family of aircraft It could travel at speeds of Mach 3.2, although Mach 3.3 could be reached if necessary Its maximum altitude was 85,000 feet >> At those altitudes, you can see the curvature of the Earth, about 350 miles At those speeds, you fly faster than the Earth rotates, so it was not uncommon for me at night to fly in a westerly direction and actually see a sunrise come up in the west During the day, the sky is a deep, deep blue, not quite black You can see the different shades of blue as you’re looking down into the black You can see some of the brighter stars during the day and my guess is what you’re seeing are probably the planets, Venus and Mars and all At night, you cannot see constellations like we’re seeing here, you know, Orion and Pegasus and those things, you can’t see those, the Big Dipper, because at night, the entire sky at that altitude is just one big Milky Way I mean it’s nothing but stars It’s just the most beautiful sight you’ve ever seen in your life It’s absolutely incredible I was flying a mission over the North Pole once and because of the way the Earth is shaped and longitude and latitudes up there, got over the North Pole I had about an 18-minute transit to go all the way across the pole And in 18 minutes, I had a full moon rise over my right shoulder in the cockpit and set to the left three complete times The moon just went around the cockpit like that three complete times It was absolutely the most amazing thing I had ever seen >> At over 80,000 feet in speeds exceeding Mach 3, SR-71 pilots have ejected In fact, every Air Force crew member who ejected from one survived [ Music ] >> [Background Music] Some of the records that the airplane has set, the important thing about these records set a lot of reference These all still stand today but the important thing about these records I like to bring up is we did this everyday We didn’t– we set these records ’cause the country was set, you know, and we had the records and the crews were there to make sure that everything was done correctly But that’s how we flew everyday in the airplane We had no subsonic missions We took off subsonic, we refueled subsonic behind the tanker and then we went supersonic And we just stayed supersonic the whole time and then we came down I want to just take a minute here and just tell you some of the missions we flew in the SR-71 It was the Cold War, some of the fun missions we flew and as veterans, my very first mission out of Okinawa, Japan was over North Vietnam And we had cameras onboard and all

but the weather didn’t matter, or normally the weather did matter and you wanted a clear day and you delay sometimes and all But the mission was over North Vietnam and it was primarily to go over the Hanoi Hilton The SR-71, ’cause I told you, the sonic boom coming over the nose of the airplane here creates that sonic boom and then the spikes right here, they create a second sonic boom So when the SR-71 flies overhead, you get a very distinct “boom, boom.” [ Explosion ] Don’t ask me why it’s that distinct and why it’s that separated I really don’t know but it’s not like a [inaudible], it’s a “boom, boom” very distinct, double sonic boom And when you go through the sound barrier and when I’m taking tours around here, a lot of people don’t realize that when you break the sound barrier in an airplane and you go supersonic, that supersonic sonic boom follows you everywhere you fly So it’s not just a one-time occurrence when you go through the speed of sound So if you broke the sonic boom over Los Angeles, the people in LA would hear a sonic boom but all the way across country, every place you flew over, Kansas, Colorado, or wherever, people in the ground would hear that sonic boom And the same with the SR-71, I was going to hear a double sonic boom everywhere you go So it’s not just a one-time event, everybody hears a sonic boom >> Do you hear it? >> No, no You don’t hear anything in the cockpit Nor is it like the movies you see where you start to shake and your teeth go back and everything, you kind of get– you kind of get down there and you feel what– it’s kind of like a pause but it’s not in the aircraft so much It’s in the instruments ’cause the instruments pause and then they go supersonic and then they all pick up where they left off So you don’t get any shaking and rattling and anything like that But anyway, so the purpose of this mission– and we flew several of these in years, was to fly over the Hanoi Hilton where our POWs were being kept so that they would hear that very distinctive double sonic boom [explosion] to let them know that people were thinking about them Years afterwards when I go to happy hours and all and had the different– the clubs and all in the Air Force, you’d run into some of the POWs and they said they did hear that and that was kind of, you know, it’s uplifting as it can get when you’re in that kind of a situation So that was one mission we– one type of mission we flew Another type of mission we flew which was very much fun was over Cuba, and with the explicit approval of the Department of State And that was primarily when they knew that Fidel was going to be out there having a big parade or it was a Cuban holiday and they were going to be marching all the troops by and waving and doing all those things that they do We’d fly right down the middle of Cuba right when the parade was– was parade hitting, you know, stage center and give them that big double sonic boom >> Yeah >> So that they’d hear it on the ground And then the other types of missions we flew were more of the standard missions where we flew reconnaissance missions against the Soviet Union, against China, North Vietnam, North Korea, you know, all those types of things where we’re collecting photo imagery and electrical intelligence >> The SR-71 carried numerous sensors Among them were high-resolution cameras, side-looking radar, ELINT or electrical intelligence recorders, and COMINT or communication intelligence recorders >> Now, we had a good COMINT recorder onboard but it’s funny because we flew so fast You couldn’t catch a lot of a conversation that people were having But that proved to be very good for the electrical intelligence recorders because we could fly by a, say a Soviet new radar that was out there that our intelligence people thought was onboard And if they turn that radar on even for a couple of seconds and turned it off, we had traveled such a distance that not only could we collect the radar waves from it but we could also– we’d travel far enough to give a little bit of a triangulation so that they could actually see not only what it was but where it was We had an astro-inertial navigation system onboard It tracked three stars Not unlike Star Trek, we had a stardate We put a stardate in the airplane and the airplane system would know what stars to track and once it saw the sky, it would lock on to three stars We promised the president, because of the sensitivity to our missions, that we would never be more than 300 feet now, 300 feet off the centerline of the mission track we were supposed to fly, and that astro-inertial tracker would keep us that The reason we did that is like our flights along the Soviet and Chinese border Back in those days, I don’t know what they do now but in those days, the Soviets declared supremacy out to 15 miles that they owned out there We said we gave them out the three miles, international law So our missions were at the three-mile point along the borders That’s why we get a lot of reactions from the Soviets,

airplanes coming out trying to come up and intercept, doing those types of things Here’s another mission that was really neat This was during the Arab-Israeli wars in the end of ’73, early ’74 You all are too young but that was the timeframe where we’re having another gas crisis back in those days Because of the political ramifications, our European allies weren’t allowing us to overfly or fly out of their countries because they didn’t want to upset the Arabs, of course, for the oil and the gas So these are missions that had to be flown by the SR-71 from the west coast to the east coast of the United States We moved them from Beale Air Force Base to North Carolina, to the east coast It’s about 11,000 miles roundtrip, ten plus hours Probably, as I say, ten plus hours, probably more like 12 hours But you can see it goes across the entire Atlantic Ocean through the Strait of Gibraltar, down the Mediterranean, over the Sinai for about five minutes, and then comes all the way back and lands We flew nine of those missions and every one of them flew just as programmed No problems at all The rules of engagement that we had was that if you got a launch light in the cockpit, you push the throttles forward to accelerate to maximum Mach which is 3.2 So in this case, you’d accelerate from Mach 3.0 to Mach 3.2, not a big deal but a little bit of acceleration And then this– and then the second technical thing we would do is look out the window to see if you see a missile That’s what the checklist said And then if you saw a missile, if you saw a missile, what you didn’t do was try to fly the airplane to avoid the missile because they felt you are more likely to lose the airplane trying to duke around up there than the missile was to hit you >> In the unlikely event that the crew was shot down, limited resources were available to them >> In the survival kit of the SR-71, we had a small fold-up .22 [ Laughter ] >> That’s a big– >> Yeah, yeah I’m not sure exactly what that was for but we had a small, we had a small .22 in there that we carry with us And we couldn’t carry a wallet or anything We wore a name tag and the U-2 still were– now with your name on it and your rank It would say like “Captain Kinego” or “Major Kinego.” And that was so you weren’t considered a spy >> You have to wear your dog tags? >> Did not wear dog tags for these missions >> For SR-71 pilots who would fly long missions that span the globe and lasted multiple hours, a common question was asked >> I need to go to the bathroom [ Laughter & Inaudible Remarks ] >> I’ve been asked that a lot but I’m not sure I’ve been asked on camera before [ Laughter ] >> Number two, you were just on your own I mean, you just– you just– and we had people do it People just went– [ Inaudible Remarks ] I mean you just went, and we had people that just got sick or something and just had to go But that, you just go Number one, it was easy You put on a thing called a urinary control device It’s like a condom basically with a tube coming out the front, you put it on It comes outside your suit and it goes out a tube and it was in your pocket, and in your– goes into a plastic bag in your pocket and it’s got a sponge in there [ Inaudible Remarks ] No, no, no Well, no, no, yeah, you left the whole– the UCD on Yeah, you put that on before you– you had to put it on first before you put your pressure suit on You got in the airplane and you got– you’re all hooked up ready to go If you had to go to the bathroom, then you would put a little air on it and then you kind of open up the valve and you go to the bathroom and then you could turn it off, shut it down [ Inaudible Remark ] I’ve been asked that question on 6th grade tours out here [ Laughter ] Everybody wants to know that >> Yeah >> Yeah, you got to be careful how you explain it to those 6th graders but– But no, you need to– yeah, I mean that’s what we did >> After the presentation, the student veterans left the conference room and Colonel Kinego took them on a brief tour at the museum >> This airplane that you see right here, it’s a Nieuport 28 In 1914, World War I started, right? In 1918, we sent our first soldiers and airmen over to Europe for World War I When our airmen got over there, we had no airplanes We had no airplanes because we didn’t develop the aircraft industry at all We had to buy airplanes from the French and so the first airplanes we flew was this Nieuport 28 Nieuport 28 is a complete fabric airplane This is all fabric over wood It had a pesky problem that the French didn’t like, in that the fabric on the top of the wing would come off

and when the fabric came off, the airplane would crash and the French pilots didn’t like this So as a result– As a result, the French sold it to the Americans This is the first airplane in the United States that first went into combat That Uncle Sam hat you see there with the gold ring around it, if you recall in your history the, you know, we threw our hat in a ring to go over to war in Europe Well, that was the Hat-in-the-Ring Squadron It’s the 94th Squadron, threw their hat in the ring to go in flight combat Do you ever stop and wonder why when you talk about airplanes, you talk about fuselage, empennage, pitot-static system, those are all French names, yeah That’s because we invented the airplane but we didn’t actually develop the airplane The United States didn’t have an aircraft industry French and the Europeans developed the airplane because the winds of war were blown over there during that timeframe coming up to World War I [ Music ] [Background Music] OK, this is Enola Gay This is the airplane that dropped the first nuclear weapon on Hiroshima And a lot of people ask me, “Is this a mockup? Is this the real airplane?” This is the real airplane That wing tip to the far wing tip is about 142 feet That’s over 20 feet longer than the first flight that the Wright brothers made when they developed the airplane Anybody know why it was named Enola Gay? Colonel Paul Tibbets was the pilot Most of the people in the squadron had no idea what they were training for Colonel Tibbets did The night before the mission, he went out to the airplane, he wanted something on the airplane, so he had them put on “Enola Gay” which is his mother’s maiden name Right in the very front, it had this very new development called the Norden bombsight The Norden bombsight in World War II was completely state-of-the-art It allowed the bombardier to put cross hairs on a target that they want to drop on and then lock on, and the airplane will actually fly, the airplane would actually drop the bombs That’s why, you know, you hear about Colonel Tibbets but you never hear about who is the bombardier that pushed the button and dropped the nuclear bomb because the bombsight actually did that President Truman who had taken over from President Roosevelt and the allies did send a memorandum to the emperor of Japan saying, we need your immediate surrender unconditional or something very dramatic will happen Didn’t say what it was going to be We never heard anything back from the emperor So on the 6th of August 1945, seven B-29 Silverplate taxied out and took off from Tinian Island When they got to Hiroshima, dropped the bomb, it was an air burst, went off at about 1,500 feet, went back to Tinian, waited President Truman heard nothing from the emperor, so on the 9th of August now, 1945, six B-29s took off out of Tinian and what do you think the primary target was then? [ Inaudible Remarks ] Nagasaki is too easy, too easy, yeah The primary target was Kokura, Japan Enola Gay was one of the weather ships on that mission, got to Kokura and it was overcast so they left Kokura and went to Nagasaki and dropped the second weapon, same as the first That was on the 9th of August Finally, on the 15th of August, they heard from the emperor that Japan had surrendered [Background Music] This is the restoration hangar This right here is Sikorsky and it’s a Sikorsky JRS-1 This particular airplane that’s being restored right now is basically a flying boat You can tell from the front here, and the way the gear go up and it’s got platoons you can put on on the side here It’s a flying boat and this one survived Pearl Harbor This actually survived the bombing and took off, and its job was to go out and look for the Japanese fleet during the days after Pearl Harbor ’cause you recall, they were still pretty much convinced that the Japanese were going to invade Hawaii [ Music ] [Background Music] The space shuttle we’re going to see here is Discovery Discovery is the workhorse of the fleet There were 135 shuttle launches and Discovery flew 39 of them One of the better views you’d get of it is from right here That’s where the crew sits The flight deck is up there The crew lives down here That’s the cargo bay Cargo bay is 60 feet long You can put a school bus in there, a Winnebago, and it’s 15 feet in diameter Saturn V rocket are the rockets we used to send people to the moon This is the ring right here that carry the avionics and navigation It has less power– less power than your iPhone right now, and less computing capability than your iPhone, and that’s what– how it’s been

That engine, if you recall, got its fuel only from main fuel tanks so when the main fuel tank was gone, that engine never burned again So when the shuttle reentered, came back to land, it has no power It was a glider coming back in >> Main gear touch down >> And the tail, there’s a split tail The pilot can open and close it as an air brake This arm that you’re looking at right here, that’s called the Canadarm, probably one of the greatest robotic achievements exported out of Canada When we launch things into space, a satellite then would be actually attached to the end of that arm so they would execute that arm out and then we simply lay the satellite on its orbit [ Music ] [Background Music] Here is the SR-71 They built the airplane with like expansion joints like you would build a bridge So on a hot summer day, you know, they put them together and it doesn’t buckle As the airplane gets very hot, the combination of all those expansion joints coming together is about six to eight inches Most airplanes have fuel tanks and then they have bladders that sit in the fuel tanks that the fuel goes in Back in the day when they were building this airplane, there was no technology like that that would keep those bladders from melting, so there’s no bladder So the fuel goes in the fuel tank but just sits right against the skin of the airplane And you have those expansion joints, so because of that, the fuel didn’t leak out Now, you’ll hear various stories about pours out It didn’t pour out but it leaked out pretty significantly Your first five missions in the SR-71 are in the B model This is an A model you’re looking at here This is an operational model The B model doesn’t have a navigator It has an instructor pilot It sits back there but the rear cockpit is elevated so that the instructor pilot can see out the land and all You only get five missions in the B model to train and the fifth mission is your checkride So, you really only get four missions to train and you got to learn how to go Mach 3, climb, refuel, do all those things So, you had to come there and be able to fly fast airplanes You had to come here knowing how to air refuel You know, people ask me what did it feel like to fly at that speed now Well, the answer is it didn’t feel like anything I’m sitting so far forward, I can’t turn my head far enough to actually see the airplane, so I don’t see any airplane So I’m in the pressure suit breathing hundred percent oxygen, so I don’t smell the airplane And I’m in that pressure suit with the air blowing in my head, I don’t hear an airplane So it’s like going 36 miles a minute on a telephone pole and it’s very smooth [ Music ] >> [Background Music] For student participants in Montgomery College’s Combat to College program, it was an experience unlike any provided in a classroom Spending time with a retired Air Force colonel who piloted one of the most famous planes in history and walking through a museum that gives them access to history >> Being in the military, you get to work alongside of modern technology But here, you actually get to see the history of all that technology, and what you worked with and how it got there So it’s really cool It’s interesting >> My first semester here at Montgomery College, I did a report on Hiroshima and then to see the Enola Gay is kind of– brought back the feelings of me first coming back to school as an adult and it’s just spectacular >> I don’t think without Combat to College, you know, I would not be doing as well as I am in school >> The Combat to College program is fantastic for veterans especially when they’re coming from a very structured lifestyle and then just kind of transitioning, integrating back into “normal” society It’s one of the only schools that I’ve seen or spoken to that actually help veterans with that issue, which is just kind of a way for that veteran to kind of feel appreciated for the service when the college wants to do this type of activities [ Music ]

5.4, 6.1 Area of Regions in the Plane

– Depending on the context that we’re working with, the area underneath the curve in a graph can represent the amount of money gained or lost or the amount of time lost or the efficiency of a product And so we’re often very interested in finding the geometric equivalent of the area under a curve in a plane So that’s going to be our question today is how do we find area under a curve? And the answer to that is that the definite integral is a representation of area Here’s what I mean by that If we’ve got some curve and we’re interested in the area between a and b that’s underneath this curve, that turns out to be the integral from a to b of our function, saying that function is f of x dx That is the way we find area is we take that definite integral between a and b So for example, if I want the area under x squared minus 10 between x equals 4 and x equals 6, what that means I’m going to find is the integral from 4 to 6 of that x squared minus 10 dx So we just need to evaluate that, and we know how to do that The antiderivative of x squared is x cubed times a 1/3 minus the antiderivative of 10 is 10x And we’re going to integrate that from 4 to 6 And so that just means we’re going to plug 6 in for the x So we have 6 cubed minus 4 cubed– since it’s a polynomial, we can just do the subtraction together– minus 10 times the 6 minus the 4 And if we plug that all in the calculator, we’ll end up with 92/3 for the total area between 4 and 6 underneath the x squared minus 10 So really what we’re doing is we’re just giving some context to what we learned how to do in the previous video Let’s say, for example, we want the area under f of x equals e to the negative x plus 2 that falls between x equals 1 and x equals 5 Well, area under a curve is an integral, and we’re integrating from 1 to 5 because that’s where I want the area to be of my function e to the negative x plus 2 dx So let’s evaluate this The antiderivative of e to the stuff is e to the stuff, and then we just have to divide by the derivative of negative x, which is negative 1 And we’re integrating this from 1 to 5 which means we plug those values in So we get negative e to the negative 5 plus 2, minus a negative, which makes it a plus, e to the negative 1 plus 2 Simplifying that becomes negative e to the negative 3 plus e to the first power And that gives us the area that’s under our curve between 1 and 5 Let’s try one more example Let’s find the area under the curve f of x equals e to the x times e to the x plus 2 cubed between x equals 0 and x equals 2 Well again, we know we have to integrate from 0 to 2 of our function e to the x times e to the x plus 2 cubed dx This one we can’t find the integral directly,

but we can use substitution if we make that e to the x plus 2 equal to the u That way it becomes u cubed And we can do that because its derivative is multiplied in there of e to the x dx So now we have the integral from– make sure we plug these limits into our u equation So we’ve got e squared plus 2 It’s a funny looking limit, but it’s a limit If we plug 0 in, we get e to the 0, which is 1, plus 2 is 3 So we’re integrating from 3 to the e squared plus 2 The e to the x dx all becomes du The rest becomes u cubed And we know we can increase the exponent by 1 and divide by the new exponent– 1/4 u to the fourth, integrated from 3 to the e squared plus 2 limit Plugging those limits in, we’ll end up with 1/4 times e squared plus 2 raised to the fourth power minus 1/4 times 3 to the fourth power And simplifying that, we end up with 1/4 e squared plus 2 to the fourth power minus 3 to the fourth is 81/4 So we end up with our solution– the area under the curve– which sounds pretty straight forward But there is one little problem that comes up that we need to address A problem and a solution And it’s just something we need to be careful of when we’re talking about area under a curve Let’s take a look at the area under f of x equals x squared minus 4 between x equals 2 and x equals negative 2 Well, we know to do this we just have to integrate from our limits The lower limit, the smallest number’s negative 2 up to 2 of our function, which is x squared minus 4 dx And we can take that antiderivative quick enough at this point That’s x cubed times 1/3 minus 4x And we know we want to integrate from negative 2 to 2 So we plug those limits in 1/3 times 2 cubed minus 4 times 2 And then we subtract, which will change the signs, plugging the negative 2 in 1/3 times negative 2 cubed Changing the sign, now we have plus 4 times negative 2 Plug all that in, and we’ll end up with an area of negative 32/3 How is area negative? Area can’t be negative Something’s wrong In fact, the problem can get even worse Let’s take a look at a problem like this where I want to find the area under f of x equals x cubed between x equals negative 2 and x equals 2 Well, using our old methods, we have the integral from negative 2 to 2– negative 2’s the small, 2’s the big– of our function x cubed dx which is 1/4 x to the fourth integrated from negative 2 to 2 So we plug the top value in– 2 to the fourth times 1/4 minus– plugging the bottom value in– 1/4 times negative 2 to the fourth And if we do that math, we end up with 0 How is there zero area? There is no area between negative 2 and 2?

That doesn’t make sense Let’s see if we can get a visual idea of the reason behind what’s happening with these two graphs, and maybe that reason will help us find a solution Our first function was f of x equals x squared minus 4 And we know x squared minus 4 is a parabola that’s shifted down 4, and actually has x-intercepts of negative 2 and positive 2 So when we found the area between negative 2 and positive 2, that entire area covered all those negative y values It’s below the y-axis, making it a negative value And similarly, when we did f of x equals x cubed– we should be familiar with the graph of x cubed as one that comes up, levels off, and takes off, and it’s perfectly symmetrical So when we go from negative 2 to positive 2, we end up with a positive area and a negative area connected to the x-axis And that negative area is exactly the same size as the positive area, which gives us a combined total of 0 But that doesn’t mean those areas are 0 and negative We need a better way to attack these And so our solution is going to be to subtract the negative parts In other words, when we did the problem– the first example, where we found the area under x squared minus 4 between x equals negative 2 and 2, looking above, we see that graph is entirely negative And so what we’ll do is we’ll add an extra negative, subtracting that integral from the negative 2 to 2 of our function x squared minus 4 dx That negative there is going to change the entire thing to a positive Basically, it gives us the other part of the graph shifting the other direction, which now becomes positive And that negative sign is going to make all the difference in the world because now when we take the antiderivative, we get the opposite of x cubed times 1/3 minus 4x integrated from negative 2 to 2 And now when we plug in the lower limit, we get the negative 1/3 times 2 cubed minus 4x– not 4x 4 times 2– plugging in the upper limit– minus the negative, which is now positive 1/3 times negative 2 cubed minus 4 times negative 2 And now when I put that value in the calculator, we’ll end up with a positive 32/3, and that’s a much more reasonable value being positive for the area The second example, though, we need to be a little bit more clever with because what we’re looking for here is the area– we’ve got that negative area– looking at the graph above– to the left, and the positive area to the right So we’re going to treat that like two separate integrals Notice, it switches at 0, 0 So we’re going to do the left side from negative 2 to 0, and we’re going to subtract that integral So again, we’re finding the area under x cubed between x equals negative 2 and x equals 2, and because the first part is negative, we’ll put a negative sign again So a negative negative makes a positive But we’re only going to integrate from negative 2 up to where the negative part stops, which is 0, of our function x cubed dx

The next part is already positive, so we’ll have a positive integral from that changing point up to our limit of 2 of our function x cubed dx And that will give us our total area, all positive, because we took the negative part and flipped it, taking the opposite sign So let’s find this solution We’ve got negative 1/4 x to the fourth, integrated from negative 2 to 0, plus a 1/4 x to the fourth integrated from 0 to 2 Plugging in those values, we have negative 1/4 times 0 to the fourth Minus a negative makes it a positive 1/4 times negative 2 to the fourth Plus, plugging the limits into the right half, plugging first 2 in– 1/4 times 2 to the fourth minus the lower limit– 1/4 times 0 to the fourth And now, if we work this out, the 0’s go away pretty quick That’s nice 2 the fourth is 16, divided by 4 is 4 So this becomes 4 plus 4, for a total area of 8 And now, instead of 0, we have a more reasonable area of 8 What we’ve been doing so far has been looking to find the area underneath a curve, like we did in our introduction where we said, hey, all the area goes down to the x-axis But sometimes we don’t want to go all the way down to the x-axis We only want to go down to another function Sometimes we want to find the area between two curves The idea here for the area between two curves is we have one curve up above and one curve down below, and we want the area between a and b that’s between those two values Well, if we were to integrate f of x from a to b, that would give us all the area underneath f all the way down to the x-axis But we don’t want all of it because we don’t want the stuff that’s below the second graph We’ll call that g of x We don’t want any of that lower area Well, we know we can calculate that lower area by integrating g of x dx over the area from a to b, but that’s the area we don’t want And so what we do for the part we don’t want is we’re going to subtract off that area When we subtract that area, we end up getting only the area between the two curves So again, the way we find the area between two curves is we integrate from a to b and subtract the bottom function from the top function Let’s see if we can take a look at an example where we do that, and we’re going to use our calculators a bit to help us identify which one exactly is the top function that needs to come first So we’re going to start by finding the area between f of x equals x plus 5, g of x equals x squared minus 2x plus 2, and x equals 0 and x equals 3 Well, first we need to know which graph is on top Going to my calculator, if I hit y equals, we can type in both these equations x plus 5 is the first equation, and x squared minus 2x plus 2 is the second equation Now, when I hit Graph, I’m going to watch the graph be drawn because the top equation is going to be drawn first There’s the top equation And there’s the bottom equation Notice, we’re only interested in going from 0 to 3

So between 0 and 3, what we notice is the line is on top and the parabola is on the bottom between 0 and 3 So we’re going to set up our integral just like that We’re going to integrate from 0 to 3 the top function, which we said was the line x plus 5 Then we will subtract the bottom function, which we said was the x squared minus 2x plus 2 dx This integral will then give us the area between the two curves Let’s simplify first by distributing that negative through the parentheses, giving us x plus 5 minus x squared plus 2x minus 2 dx, and combining like terms to get negative x squared plus 3x plus 3 dx Evaluating is pretty straightforward We’ve got x cubed times negative 1/3 plus x squared times 3/2 plus 3x integrated from 0 to 3 Plugging those values in, then, we’ve got negative 1/3 times 3 cubed plus 3/2 times 3 squared plus 3 times 3 And then we subtract or change the signs as we plug the lower limit in The negative becomes positive 1/3 times 0 cubed minus 3/2 times 0 squared minus 3 times 0 And I’d probably do that all on my calculator to end up with 27/2 is the area that’s between my two curves Let’s try one more example before we wrap up Let’s find the area between 2x minus 3 and negative x squared plus 4x We’ll go into the calculator Hitting y equals, we’ll clear out the two functions The first function is 2x minus 3 The second function– negative x squared plus 4x And when I hit Graph, there’s the line There’s the parabola And we’re looking for this area in between right here Well, one thing we need to know is where they intersect We’re going to have to solve this equation for where they intersect, and then the parabola, the second function’s on top So first, to figure out where they intersect– those are the limits– they intersect when the two equations cross each other– when 2x minus 3 is equal to the negative x squared plus 4x Well, if we add x squared to both sides and subtract 4x, keeping the minus 3, we can factor that to x minus 3x plus 1 So they intersect at 3 and negative 1 So that tells me that we’re going to integrate from negative 1 to 3, from low to high of the top function we said was the squared, so negative x squared plus 4x minus the bottom function And the bottom function was the 2x minus 3 dx And now we have an integral we can solve to find the area between the two curves Simplifying by distributing the negative through We have negative x squared plus 4x minus 2x plus 3 dx Combining like terms, we have negative x squared plus 2x plus 3 dx Taking the antiderivative, we end up

with negative x cubed times 1/3 plus x squared divided by 2 clears the 2 plus 3x integrated from negative 1 to 3 So we plug in our values Plugging the top one in, negative 1/3 times 3 cubed plus 3 squared plus 3 times 3 Subtracting, changing the signs, we now have a plus 1/3 times negative 1 cubed minus negative 1 squared and minus 3 times negative 1 And if we plug that all into the calculator, we’ll get the area between the two curves which comes out to be 32/3 So area really is just a major application of that definite integral, finding the amount of stuff between the values Be careful if you’ve got a negative area We’ll have to figure out how to put the negative sign in front to make it actually positive area so we don’t end up with 0 or negatives Also, when we want the area between two curves, we just need to subtract the two curves– one from the other So take a look at the homework assignment to practice some of these We will see you in class to discuss applications and practice these further

The Girl Who Talked To The Stars – The Incredible Journey

I’m here at Mataranka the capital of the Never Never, this is the original site of Elsey station- one of the biggest cattle properties in the Northern Territory. In 1902 Jeannie Gunn arrived here with a new husband to live and work on Elsey station, she had traveled all the way from Melbourne in Victoria This is the place that inspired the writing of two classic and much-loved Australian books The Little Black Princess and We Of The Never-Never Although the books were released as novels everyone knew that they were based on a cast of real-life characters, one by one these characters were identified by the public who fell in love with the stories of the Never-Never Only one character’s identity remained a mystery, the little black princess Bett-Bett Who was this little black princess of the Never Never? This was a question that readers of the books asked for decades, the true identity of Bett-Bett stayed unknown until she herself revealed it, more than 60 years after the books had first been published, together we’ll follow the footsteps of a little black princess and discover the true story of Bett-Bett’s life It’s a story we should all hear as it reminds us of some of the darker days of our history as a nation, but more so of the power of kindness and love to transform a person’s life, come with me on a journey into the Never Never In the era of the cattle kings this area was home to Aeneas Gunn and his young bride Jeannie who arrived here in 1902 to live and work on Elsey station. Aeneas, known as The Maluka, was a part owner and manager of the property, Jeannie was one of the first pioneer women of the outback and had traveled all the way from down south in Melbourne. Elsey station was one of the earliest and largest cattle stations established in the Northern Territory, it consisted of a homestead, kitchen, outbuildings, staff quarters, cattle yards and a small cemetery to the north. Over the years the old Elsey homestead has become a part of the legend of living in the harsh Australian outback many kilometres from civilisation This landscape and the people Mrs. Gunn encountered inspired the writing of the classic Australian books The Little Black Princess and We Of The Never- Never. What is the Never Never? Well it’s a term that was first used in the late 19th century and describes the vast remote outback regions of Australia. It was under trees like these near the billabong and just down the slope from the homestead that Mrs. Gunn first met Best Bett, one of the central characters of The Little Black Princess who also appears in We Of The Never-Never Although the books were released as novels everyone knew that they were based on a cast of real-life characters One by one these characters were identified by the public who fell in love with the stories of the Never Never, only one character remained a mystery, Bett Bett. Who was this little black princess of the Never-Never? For decades readers of the book continued to wonder after Jeannie Gunn left Elsey station princess Bett-Bett seemed to have just melted away into a lonely place in the bush. However the true identity of Bett- Bett wasn’t known until she herself revealed it more than 60 years after the books had first been published Elsey station was on the traditional lands of the Jawoyn people, a tribe of hunter-gatherers who lived and wandered over the vast catchments of the Katherine and Roper rivers. In 1894 a baby girl was born to a Jawoyn woman who immediately rejected the baby because her light-colored skin showed that her father was a white man, the baby girl was rescued by her aunt and named Dolly,

little did anyone realize that this unwanted little half-caste Aboriginal girl was to become known and loved throughout the world as the little black princess. Dolly spent her childhood wandering the lands of her people and living in a simple portable bark shelter Here at Bitter Springs she helped her aunt search for yams, water lilies, sugar bag and wild honey, goannas, snakes and crocodile eggs, she loved to play in the water and learn the dances and stories told by the tribal elders, but her nomadic childhood ended in 1902 when her aunt told Dolly that she was to live at Elsey station, Dolly was eight years old and didn’t fully grasp the significance of this, she assumed that she would simply rejoin her aunt in the great outback at some point in the future. Much of the Northern Territory has never been occupied by white people as the nature of the soil and the climate rendered most areas unsuitable for farming, many commercial ventures were tried but ultimately failed. What was moderately successful was the pastoral industry, there were enormous cattle stations, bigger than many small European nations, Aboriginal men and women were vital in maintaining these stations which would have failed without them Elsey station was one of the biggest cattle stations and Dolly was to join a large Aboriginal group who lived and worked at the homestead Dolly soon fitted into a new life at the homestead, sweeping floors, watering the garden, helping the Chinese cook and doing other household chores, she delighted Mrs. Gunn with her stories and adventures Mrs. Gunn tried to get Better-Bett to sleep on a bed, but to no avail, she preferred the stars and a dog on the verandah. Best-Bett found comfort under the stars and she would spend hours watching them, they were her friends and she would whisper her secrets to them and share the events of the day. Mrs. Gunn was impressed with Bett-Bett’s energy and drive, her curiosity and wild ways, the two developed a close relationship and a bond was formed which lasted their lifetimes. Bett-Bett soon became a part of the Gunn household Mrs. Gunn was to have a profound effect on the girl’s life, she started to teach her how to read and write and sew clothes Both Jeannie and her husband Aeneas were children of church ministers, they loved the Bible and its message and Jeannie told Bett-Bett stories from the Bible, and so from a young age she developed a love for the Bible Tragically after only 18 months at Elsey station Jeannie Gunn’s husband died of Blackwater fever, the severe form of malaria Now Mrs. Gunn was alone in a tough, vast land, the outback. There was no reason or way for her to remain on Elsey station and she was forced to return to Melbourne alone, but the outback remained with her, she wrote over time here and of the characters that have become famous through her books. Jeannie wanted to take Dolly, her Best-Bett, with her but Dolly wasn’t allowed to leave the Northern Territory, instead she was introduced to her father, an Englishman who worked on the Overland telegraph line, Dolly traveled with him for a while before he decided that Dolly should live with a white family in Darwin, this marked the beginning of a succession of foster families for Dolly, the first foster family treated Dolly harshly, she was forced to work hard, sleep on the floor in the veranda and she was not given any clothes or toys, she was also beaten regularly. During these difficult times Dolly found solace at night in her friends, the stars, and shared her burdens with them, it was comforting to know that they were the same stars she’d looked at when she’d known perfect happiness at Elsey station with Mrs. Gunn. When the unkind foster family moved inter state Dolly was placed with the family of a Mr. Ward who worked at the post office in Darwin. The Ward family treated dolly as a daughter, she was given a bed to sleep in and clothes to wear and to

Dolly’s delight was taken to this church Today it’s the oldest church building in the Northern Territory, it’s been part of Darwin’s history since 1897, this church building has survived the bombing of Darwin during World War two, cyclones, termites and the hot, humid tropical climate and it was in this church the Dolly learned more about Jesus and the Bible stories that Mrs. Gunn had shared with her, it was then that Dolly decided she would like to know more about Jesus and the Bible She had an inquiring mind and so had many questions about her favorite book, the Bible; Who wrote it? Where was God? Who made the world? and why did Jesus have to die if he was such a good person? Mrs. Ward would open her Bible and patiently try to answer Dolly’s questions as best she could, Dolly was delighted when, one evening, Mrs. Ward pointed to the brilliant stars, the same stars that dolly looked at every night and explained that the God of the Bible had made them all, now she loved the stars more than ever Mrs. Ward became another mother to dolly and the Wards her new family, but although Mrs. Ward was a school teacher she firmly believed, as many people did in those days, that education was wasted on black children and so Dolly never learned to read properly. In her nightly talks to the stars she told them that if ever she had children of her own she would make sure that they would be well educated. While living with the Wards Dolly had the opportunity to travel with them to Sydney, Adelaide and Melbourne. In Melbourne Dolly was reunited with Jeannie Gunn who was again trying to get government permission to bring Dolly to Melbourne to live with her, once more permission was denied, but Dolly was happy to continue living with the Ward family, she’d grown into their hearts and they loved her. To Dolly it was a time of settling into a new routine, a time to learn about Darwin and its people, little did she know that her life was about to take you to another remarkable twist When Dolly was 15 the good years ended, Mr. Ward was transferred to Adelaide, the same law that had frustrated Jeanie Gunn now stopped the ward family from taking dolly with them, their request was denied by the government and Dolly was moved to yet another foster family, as a teenager Dolly was now viewed by this family as a source of labour and income, she was treated like a slave and beaten regularly, these were the dark years of Dolly’s life. When she was 17 Dolly was forced to work in the bar of a local pub, the foster mother worked there also and decided to rid herself of responsibility for dolly by selling her to an old man as a teenage bride. Summoning all her courage Dolly decided to walk across Darwin to the office of the government appointed protector of Aborigines, she was terrified, but after describing her situation and fear of being forced to get married, the government officials supported Dolly’s position, which resulted in greater anger and abuse from her foster family So Dolly decided to break the law and run away, during the two years since the ward family had returned to Adelaide Dolly had continued to attend a local church, she was about to run away when she was offered lodging and work by a kind woman named Mrs. Tyndall, now this was still considered breaking the law as Dolly was supposedly still in the care of her guardian foster family. The punishment for running away was to be sent to one of the mission homes or state institutions. Instead of forcing Dolly to return to her foster family the protector of Aborigines looked at her case and determined that now Dolly was old enough to care for herself. She was finally free of her abusive foster family No one had the faintest idea the dolly was the Bett-Bett of The Little Black Princess fame. Mrs Gunn’s books had now been published and were high on the best sellers list, everyone in Australia it seemed was talking about Bett-Bett and

wondering what had happened to her Without being aware of it Dolly was probably the best-known Australian Aborigine in the world, it was the secret she guarded well, Dolly spent the next three years with Mrs. Tyndall, working with her and attending the church in Darwin. It was at this church that Dolly met a handsome young Englishman, Joe Bunsen. Joe was a quiet and gracious man but it seems that when he met Dolly it was love at first sight, they were married in 1918 and had five children Dolly and Joe quietly raised their five children in the suburbs of Darwin Dolly’s promise to the stars so many years ago was kept to the letter. All of her children learned to read and write as soon as they were old enough and received a good education. It was by sheer chance the Dolly’s daughter, Florence, found out that her mother had a secret [Florence] Well, one day I happened to be going through an old tin trunk and lo and behold there was a book there, plus some letters. After going through the book now I came to the final chapter, there was a little chapter written and I saw my mum’s name come up and I thought: this is who this book is written about, it’s about my mum and that was the first time I realised that mum had a history that I didn’t even know about [Gary]Flo asked her mother what they were all about, a reluctant Dolly explained that she was Best-Bett, the little black princess of two world famous books, she begged Flo not to tell anyone about the discovery because only Joe, her husband, and a very few other people knew, she explained to a daughter that Joe and Dolly had agreed before they were married that the secret be kept as long as possible so that they could live as normal people Over the years the Bonsons became known and respected throughout the community as a loving and sharing family The street Bonson Terrace is named after their family. Dolly and Joe had a very happy marriage and they remained inseparable for forty years until Joe’s death. During her eventful life Dolly experienced great oppression and abuse, but she also experienced great kindness and love. Prejudice and poorly formed government policy were barriers that repeatedly caused Dolly pain, but key individuals in Dolly’s life showed her unconditional love and tried to fight these barriers. Jeannie Gunn repeatedly tried to get permission for Dolly to live with her in Melbourne, the Ward family raised dolly as their own daughter, but couldn’t get permission to take Dolly with thim when they moved from Darwin to Adelaide and Mrs. Tyndall opened her home to Dolly as an escape from horrible violence. Before she turned 18 Dolly had experienced the best and the worst of human nature, her challenge was to decide what kind of woman she wanted to become, what example would she follow? Would the hardship, violence and loss she’d experienced result in a life of anger and bitterness for princess Bett-Bett? Or would she find something that gave her strength in the dark years from the faith of those who loved her? The people who showed Dolly love all had something in common, they were all active Christians. Jeannie Gunn introduced Dolly to the Bible right here at Elsey station, whilst the Ward family took Dolly to church with them and so did Mrs. Tyndall After gaining her freedom Dolly continued to attend church, she considered it a safe haven, a place that had given her a small glimmer of hope during the difficult times, but she still had questions questions that were at the very heart of her search for happiness, could God accept a person who couldn’t read the Bible? Could God love a half-caste someone who was neither black nor white? Sometime after Joe’s death Dolly had a dream [Florence] Well my mum had a dream of the second coming of Jesus and I didn’t know about the dream until one day we were able to see it up on a screen and she revealed to me that she had already had this

dream where she saw the second coming of Jesus and she was so thrilled about it [Gary] Determined to learn more about this picture Dolly asked to talk with a preacher, when they met she flooded him with the questions that had bothered her for so many years, she wanted to know about the second coming of Jesus and if he would accept her, sensing her fear and urgency the pastor reassured Dolly that everyone is equal in the eyes of God, regardless of whether they can read or not, regardless of the colour of their skin, regardless of their culture, the only thing that really matters is their relationship with Jesus Christ and Jesus himself said in John chapter 14: “Let not your heart be troubled: you believe in God, believe also in me. In my father’s house are many mansions: and if I go and prepare a place for you, I will come again, and receive you to myself; that where I am, there you may be also.” Finally she discovered the peace and assurance she’d been looking for, the promise of Christ’s return gave her hope, the second coming of Jesus became her passion and focus [Florence] Her greatest desire was to share what she had found, and when she found Jesus she want to share it but a biggest problem and caring was for her own culture and it seemed an impossible task then, but God had a plan where people of her culture are learning to know about Jesus and to love him and to realize that he was there for them too – like he was for her As Dolly grew older she moved to Humpty Doo where she lived with her children, she loved to sit on the veranda in the evening and watch the stars, they were still the same old friends that had witnessed her long and eventful life. Now they could share her inner peace in knowing that she’d always been accepted by Jesus and was part of the family of God and that Jesus is coming back for her Dolly’s story illustrates why we need to talk about and be aware of the poor treatment of Aboriginal people, the mistakes of prejudice and bad government policy and why many thought it was important that our government apologise to Aboriginal people in 2008. But your past circumstances don’t have to determine your present or your future, the story of Mrs. Gunn and Bett-Bett is a powerful illustration of this Yes, Jeannie Gunn could be seen as just a white colonial woman who tragically lost her husband after one year of marriage, Dolly Bonson could just be seen as a child who grew up without her biological family and childhood security, but their legacies say otherwise Jeannie Gunn’s journey led her to publish two best-selling books and she worked tirelessly for War veterans in Victoria during and after the two world wars. Dolly Bonson’s journey led her to 40 years of a happy marriage and five much-loved children and in the last decades of her life Dolly also found a spiritual peace in a faith that she’d been searching for since Mrs. Gunn had first told her about the God beyond the stars when she was a child here on Elsey station. Here at Elsey Cemetery just a short distance from the original homestead is the grave of Aeneas Gunn and a memorial to Jeannie Gunn, who died in 1961 and was buried in her hometown, Melbourne, there’s also a memorial here to Dolly Bonson, in the heart of the Never Never, in the landscape she loved, the words are as follows: “In memory of the last survivor of We Of The Never-Never, Beat-Bett, the little black princess, Dolly Bonson, who died in Darwin on the 3rd of March 1988, she sleeps awaiting her Saviour’s return and the gift of eternal life.” When Dolly died at the grand old age of 95 she was secure in her faith that Jesus is coming for her and that her journey would continue at the resurrection when He returns and that she would reach her final home, with Jesus, among her friends the stars. If you would like to experience the faith and assurance that brought peace to Dolly’s life, why not ask for it right now as we pray?

Dear Lord, the story of Dolly is both heartbreaking and inspiring, but we can also take encouragement from Dolly’s story knowing that you love us and care about us and that our past circumstances don’t have to determine our present or our future. You have a plan for our lives and you have the power to change them for the better, help us to trust you and give us the peace and assurance that comes from knowing Jesus. In Jesus’ name we pray, amen The story of Dolly Bonson is a fascinating and inspiring piece of Australian history, if you’ve enjoyed exploring the places and details highlighted in Dolly’s story and would like to experience Dolly’s story more deeply, you’ll want to make sure you receive the special gift we have for all our viewers today, it’s a book called The Girl Who Talked to the Stars, just before she died Dolly contributed to a biography about her life and shared more about the faith that helped her through the dark years In its pages you’ll learn more about the joy and sorrow that Dolly experienced and be inspired by her faith, you’ll also discover the source of her peace and happiness and learn how it changed her life forever and it could do the same for you Remember, there is no cost or obligation whatsoever, The Girl Who Talked to The Stars is absolutely free, here’s the information you need: Phone us now on 0481 315 101 or text us on 0491 222 999 or visit our website: to request today’s free offer, so don’t delay contact us right now If you’ve enjoyed today’s journey be sure to join us again next week when we will share another of life’s journeys together and experience another new and thought-provoking perspective on the peace, insight, understanding and hope that only the Bible can give us. The Incredible Journey truly is television that changes lives Until next week remember the ultimate destination of life’s journey, “Now I saw a new heaven and a new earth… and God will wipe away every tear from their eyes, there shall be no more death, nor sorrow, nor crying, there shall be no more pain, for the former things have passed away.”

Shipwreck Coast – The Incredible Journey

This site is where one of the greatest tragedies of the Shipwreck Coast happened, it was here that the Loch Ard sank and only 2 of the 51 people onboard survived Here is where a young sea captain’s last words were a message to his wife, and where a young girl lost her parents, three sisters and two brothers. Today we remember their stories The drama, beauty and wilderness of this part of Victoria’s coastline is breathtaking, here the land doesn’t gently slope down to meet the sea, rather the sea repeatedly attacked the cliffs, carving chunks of rock away until the land is left a solitary pillar of rock, eventually the pillar gives in to the repeated pummelling of the waves and crumbles to join the reefs below, reefs that seethe with foam and salt spray in storms and give this area its other name ‘The Shipwreck Coast’ There are approximately six hundred and thirty eight known shipwrecks along Victoria’s coast and only around 240 of them have been discovered, the small coastal traders and large ships carrying vital cargo and immigrants between Europe, America and the new colonies of Australia often battled severe storms, not only in the treacherous waters of Bass Straight but also at anchor in the precarious safety of Portland Bay This anchor was retrieved by divers from the wreck of the Falls of Halladale. In 1908 she joined the many ships that have come to grief on the reefs of the Shipwreck Coast. On the night of the 14th of November 1908 a navigational error caused the Falls of Halladale to sail through dense fog directly onto the rocks. The crew of 29 safely abandoned ship and all made it safely ashore by boat, leaving the ship foundering with the sails unfurled. For weeks after the wreck large crowds gathered to view the ship as she gradually broke up and then sank in the shallow water, today the Falls of Halladale is a popular destination for recreational divers, some of the original cargo of 56,000 roof slates remained at the site of the wreck along with corroded masses of what used to be coils of barbed wire, 22,000 slates were salvaged in the 1980s and used to provide roofing here at the Fagstaff Hill Maritime Village Not far down the coast from where the Falls of Halladale sank is the site of one of the greatest tragedies of the Shipwreck Coast, it was here that the Loch Ard sank and only 2 of the 51 people on board survived The Loch Ard left England on March the 1st, 1878, the ship was under the command of 29 year old Captain Gibbs and was full to its capacity of 17 passengers, 37 crew and cargo. On the 1st of June there was much excitement aboard the Loch Ard as after months at sea the captain and passengers were expecting to see land, the coast of Victoria, but when the fog lifted at 4 a.m

Captain Gibbs discovered that the ship was much closer to the cliffs of Victoria’s Shipwreck Coast than anticipated, ordering as much sail to be set as possible he desperately tried to turn the ship out to sea, but the ship soon stopped, the anchors were dropped, but failed to hold and the Loch Ard was tossed and pulled by the waves. Despite the frantic efforts of the captain and the crew the Loch Ard struck a reef connected to Muttonbird Island, waves broke over the ship and the top deck was loosened from the hull, water flooded the cabins, the passengers screamed in terror as the ship began to disintegrate. The mast and rigging came crashing down knocking passengers and crew overboard, there was pandemonium as the crew struggled to launch the lifeboats, when one was finally launched it crashed into the side of the Loch Ard and capsized Tom Pierce, the young ship’s apprentice who launched the lifeboat managed to cling to its overturned hull and sheltered beneath it for hours, he drifted out to sea and then when the tide turned at dawn he was swept into what is now known as Loch Ard Gorge He left the boat and swam to shore, bruised and dazed he found a cave in which to shelter, he was all alone Eva Carmichael was immigrating to Australia with her mother, father, three sisters and two brothers, while waiting to board a lifeboat she spoke with Captain Gibbs he said “If you were saved Eva, let my dear wife know that I died like a sailor,” Eva was then washed off the ship and into the sea, floating in the waves, terrifyed, calling out for help. After five hours in the water Eva was near unconscious and was carried by the waves into the gorge Tom Pierce saw her and swam out to bring her in, there was a case of brandy washed onto the beach which Tom used to revive her. Tom then climbed out of the gorge and found help from nearby Glenample Station. There, with much care and attention, the two shipwreck survivors gradually recovered and were nursed back to health Tom and Eva were the only two survivors of the 54 people on board the Loch Ard, all the other passengers and crew perished. Eva lost her parents, three sisters and two brothers Despite heroic efforts, only five bodies were ever recovered from the wreck of the Loch Ard and four of them are buried here, in this cliff top cemetery above Loch Ard Gorge, the fifth was buried on the beach where it was discovered. Eva was devastated by the loss of her entire family on that fatal shore, she was now alone in a foreign land and longed for her extended family back in Ireland, however she was devoted to Tom and forever grateful to him for rescuing her, Tom Pierce became a national hero and was awarded the gold medal of the Humane Society in front of 5,000 people on June the 20th 1878 at the Melbourne town hall The romantic sentiment of the time was that Eva and Tom should marry, but this was not to be, within three months Eva had returned to Ireland and they never saw each other again Two days after the shipwreck a wooden packing crate washed onto the beach of the gorge, it contained a life-size sculpture of a peacock, here at the Flagstaff Hill Marine Village in Warrnambool we can see the magnificence of the Loch Ard peacock, it’s the centrepiece of the museum’s collection of cargo that was salvaged from the wreck of the Loch Ard. The Minton peacock was the largest and grandest of the items in the Loch Ard’s cargo, which was destined for display at the Melbourne International Exhibition of 1880. The cargo carried by the Loch Ard revealed much about the affluence of Melbourne in

the era of the gold rush; items such as perfumes, pianos, clocks, linen, candles, confectionery, umbrellas and straw hats were on board together with heavier more industrial items such as railway irons, lead, cement and copper. At 144 centimeters tall the peacock is quite big and very fragile, that something so fragile could have survived the violence of a shipwreck is quite remarkable, I like to think it represents the hopes and aspirations of so many of those migrants who came out in similar circumstances aboard a ship and those of course who didn’t make it About 15 metres off Warrnambool breakwater there is the La Bella reef the reef got its name when the La Bella sank here on November 10th 1905, the La Bella approached Warrnambool at the end of a 37 day voyage, she was carrying a cargo of timber from New Zealand, the seas were heavy and mist hung low over the bay as the captain steered La Bella into the channel, here the ship was tossed onto its side by heavy breakers and ran aground on the reef, the sea was so rough that it wrenched a 1.5 ton anchor from the vessel. Several attempts were made by volunteers in lifeboats to rescue the stricken sailors but the rough conditions were too difficult and the boats returned to shore. The La Bella’s crew became exhausted and sailors were being washed overboard, one by one, by sunrise only 5 of the 12 crew still clung to the wreck 25 year old William Farrier was a local fisherman who wanted to help. In the morning he rode his small dinghy through the heavy seas and managed to rescue the captain, a volunteer lifeboat rescued a further three sailors, there was one terrified sailor left on the wreck, William made a final attempt and was able to reach the sailor just before the ship broke up and sank. [AVIS]The weather was stormy, no moon, all they knew that the waves were very high breaking over the reef, no lights of course there wasn’t electricity to have search lights out playing on the water or anything, so at 3 o’clock in the morning it was very dark. I think William Farrier was the one who had the least nerves, never took orders from anybody, did his own thing completely and he just left the lifeboat when the first two attempts turned them back and that’s when he got his own little skiff, he went out to the wreck twice after he’d got the first fellow off he brought him back around the breakwater into Lady Bay and handed him over and just promptly turned around the road back again and he knew he was capable of doing that, he didn’t want interference he was his own man, he could do it so he just went and did it [GARY] William Farrier became a national hero he was awarded the silver medal for bravery by the Royal Humane Society and was honored by the Prime Minister and the governor. Farrier’s rescue efforts are one of the most heroic in Victoria shipwreck history, the wreck now lies in 13 metres of water and is home to an abundance of marine life, William Farrier eventually left Warrnambool and became a lighthouse keeper. Light houses were the saviour for thousands who journeyed along the shipwreck coast Cape Otway light station is the oldest lighthouse on the Australian mainland, it’s operated continuously since 1848 Before Bass Straight was discovered by Matthew Finders around 1799, ships had to sail around Tasmania, Van Diemen’s land back then, taking an extra week to ten days, but the path between King and Flinders Islands in the mainland is still treacherous, crew on sailing ships call it ‘threading the needle’

In the Bass Straight the mighty Southern Ocean is forced through a passage merely 90 kilometres wide and up onto the continental shelf where the sea bottom becomes relatively shallow, in these parts the wind blows and swells of 10 to 20 meters aren’t rare. On a typical day the swell is about 6 meters During the 1840s increased immigration and direct mail services from England meant that shipping through Bass Strait was on the increase, the number of wrecks along the Bass Straight coast clearly indicated the urgent need for a lighthouse but action to build one wasn’t taken until after the 1845 Cataraqui wreck, this immigrant ship ran onto the west coast of King Island and all 399 passengers and crew died, it remains Australia’s worst marine disaster, and then the Cape Otway light station was built between 1846 and 1848 the lamp was finally lit on the 29th of August 1848, it was manufactured in London and was brought ashore at Cape Otway through crashing surf in small boats. The light mechanism consisted of 21 polished reflectors and lamps mounted on a frame, originally it was fuelled by whale oil, then kerosene and later electricity, the light shone nearly 50 kilometres out to sea and gave a really bright light its brightness is equivalent to 1 million candles. About 30 ships were wrecked off the coast just out from Cape Otway, from the lighthouse Two of the most significant of these ships were Jenny and Eric the Red Jenny was sunk in 1854, now just a few years after this lighthouse was completed in 1848 gold was discovered in Ballarat and workers throughout Australia went AWOL as they searched for their fortune at Ballarat, they left their jobs and even some of the assistant lighthouse keepers here at Cape Otway left their position, left their post, and made their way to Ballarat. Now it was during this time that the head lighthouse keeper had taken full responsibility for keeping the lamp burning and during this time, one morning while walking on the nearby beach, he found a large section of fresh mast and he knew that a ship had sunk, he then went searching and sure enough, at a beach not far from here, he discovered the survivors of the Jenny, he brought them back here to the lighthouse station at Cape Otway and cared for them using his own supplies until help was provided In 1851 Victoria had a population of 77,000 people, by 1861, just ten years later, the population of Victoria was 540,000 people, which was half the total population of Australia. People were rushing to Victoria in the hunt for gold, most of them arrived by sea with ships carrying people and ships carrying the supplies they ordered in from overseas and ships carrying the gold they dug up and sent back to England Each of these ships face the same treacherous conditions along the Shipwreck Coast, each of the passengers and crew knew all about the tragic shipwrecks but they had no other option for travel, so continued their journey, placing their faith in the captain and hoping that the weather would be mild and they would make it through the dangerous waters safely, when the weather was rough every person on board would strain their eyes to see the light shining from the Cape Otway lighthouse, the Lady Bay lighthouse in Warrnambool and the other lighthouses dotted along the shipwreck coast We’re all on a journey of some kind, hoping that we’ll have good conditions

as we make choices about work or study or family, as we travel through life Sometimes our journey is through thick fog, the waves are rough, there’s a storm, and you feel in your bones that danger must be close by, but you just can’t see it, then the bright beacon of a lighthouse cutting through the mist and showing the way is the saviour John chapter 8 and verse 12 says “I am the light of the world, whoever follows me will never walk in darkness but have the light of Life.” Like the passengers and crew on the ships that sailed along the shipwreck coast, to thread the needle to reach Melbourne, you may feel you’re traveling in a dark storm or thick fog, whatever storm you may be facing right now remember that psalm 119 verse 105 says “Your word is a lamp to my feet and a light for my path.” Today Jesus is still seeking and rescuing the lost and calling men and women, boys and girls everywhere and offering to take them to a place of peace and safety. If you feel you are drowning under the burdens of life, if you are being tossed about in a stormy sea of despair and heartache, if you are being blown around by the winds of strife and pain, then remember Jesus offers security, happiness and fulfilment and what a great difference that makes to a person’s life. If you would like to experience that difference in your life, if you’d like to be part of the greatest rescue story and have Jesus rescue you, why not ask him right now as we pray? Dear Heavenly Father, thank you for your love and goodness to us, the Bible is one big story of your rescue plan and it takes the whole Bible to tell this story, the story of our rescue and in this story Jesus is always at the centre because he is our rescuer. Father we’re often buffeted by the winds and storms of life, thank you for loving us so much and for sending Jesus to rescue us, we want to be part of your rescue plan and have you save us and take us to a place of safety and security in Jesus We ask this in Jesus’ name. Amen We all love rescue stories and we all love heroes, they stir our emotions. Some of the most dramatic, amazing and exciting rescue stories ever are found in the Bible. The stories of Daniel, Noah, Jonah, Joseph, Rahab and others have been shared, told and loved for generations, they never lose their appeal and are as popular today as ever, but the most amazing and incredible story is when

Jesus Christ rescued you and me. We are part of the greatest rescue story and he is our rescuer, Jesus offers security, happiness and fulfilment and what a great difference that makes to our lives If you’d like to experience that difference in your life, I’d like to recommend the free gift we have for all our viewers today, it’s a booklet called: Finding Courage to Meet Life’s Challenges There’s no cost or obligation Finding Courage to Meet Life’s Challenges is absolutely free, there is no cost or obligation whatsoever, so please don’t miss this wonderful opportunity to receive the free gift we have for you today. Here’s the information you need: phone or text us at 0436 333 555 or visit our website to request today’s free offer and we’ll send it to you totally free of charge and with no obligation. So don’t delay, call or text 0436 333 555 in Australia or 020 422 2042 New Zealand or visit our website to request today’s offer Write to us at PO Box 5101 Dora Creek, New South Wales 2264 Australia or PO Box 76673 Manukau, Auckland 2241 New Zealand Don’t delay, phone or text 0436 333 555 in Australia or 020 422 2042 in New Zealand to request today’s free offer, call or text us now If you’ve enjoyed today’s journey be sure to join us again next week when we will share another of life’s journeys together and experience another new and thought-provoking perspective on the peace, insight, understanding and hope that only the Bible can give us. The Incredible Journey truly is television that changes lives Until next week remember the ultimate destination of life’s journey, “Now I saw a new heaven and a new earth and God will wipe away every tear from their eyes, there shall be no more death, nor sorrow, nor crying, there shall be no more pain, for the former things have passed away.”

Traveling Pakistan by Train Faisalabad to Lahore Railroad Journey

Reisen mit dem Zug von Faisalabad nach Lahore über Chak Jhumra, Sangla Hill, Farooq Abad und Shekhupora Der Bahnhof von Faisalabad ist eingeschaltet Khanewal Wazirabad Branch Line Passagiere aus Karchi, Lahore Rawalpindi, Islamabad, Quetta Peschawar und viele weitere Städte und Städte von Pakistan Zugang in alle Teile Pakistans von Bahnhof Faisalabad in den Zügen der Pakistan Railways Shalimar Express, Pakistan, Nachtkutsche, Karakoram, Akbir, Badir und Ghouri Express