 One of the arguments of the Apollo believers to prove that Apollo was real is that the lunar module was followed by tracking stations of the earth. |
 One of these tracking stations is Jodrell Bank in England. As this radiotelescope is located in England, and not in US, this radiotelescope is considered as an "independent source" proving that Apollo was real. However, the radiotelescope equipment of jodrell bank has been provided by NASA, and was controlled by NASA engineers. So, there is better for an "independent source". |
 The radiotelescope of Jodrell bank was using the Doppler effect to follow the lunar module. The doppler effect is what you can experiment when a horning car is passing along: As it moves away from you, the tone of the horn changes and becomes lower; this is due to the change of shape of the sound wave. This effect is called doppler effect, and is used by the police to measure the speed of cars. |
 The radiotelescope of Jodrell bank has recorded the move of the lunar module by using the doppler shift. Link to the radiotelescope tracking of Jodrell |
 In fact, at the location that the landing site of Apollo was, it is the altitude of the lunar module that Jodrell's radiotelescope was tracking; it could not follow the lateral moves of the lunar module. |
Now, the beginning of the graph is extremely shaky; if it was corresponding to the actual moves of the LM, it would mean that the LM was alternately going at speeds of several tens of thousand of kilometers per hour in both directions. |
 In fact, they explain that the first half of the graph corresponds to the adjustment of the radiotelescope receiver, and that only the second half of the graph is significative, and corresponds to actual moves of the lunar module. |
 But it makes no sense that making an adjustment of the radiotelescope receiver would change the frequency of the received signal that much; in fact, the radiotelescope adjustment cannot change the frequency of the received signal, but only its amplitude; and the amplitude of the received signal is not visible on the graph recorded by the radiotelescope. |
So, the fact that the frequency of the received signal would change that much during the radiotelescope adjustment is pure nonsense, and an obvious joke. |
 Now let's get interested in the part of the recorded graph which is really significative In fact this graph does not reflect the move of the lunar module relatively to the moon, but relatively to the radiotelescope. |
 On the significative part of the graph, we see some little perturbations, followed with a straight line; these perturbations correspond to reactions of Armstrong while he was flying over the lunar surface looking for a place to land on, taking altitude to avoid local obstacles. And the straight line which follows corresponds, according to the explanation which is given under the graph, to the "relative velocity between the telescope and the landing point of the eagle". Normally, the doppler shift represents a speed, and not a position, but the representation of the speed on the graph would not allow to get an easy idea of the way the lunar module was moving, that is why they were integrating the Doppler shift, in order to convert the recorded speed into a position. |
 The upper part of this graph represents a speed, and the lower part the integrated speed becoming a position; notice that it is important to have a horizontal line of the graph representing the null speed; a speed which is above this line means that the target is moving closer to the radiotelescope, and a speed which is under this line means that the target is moving away from the radiotelescope, and a speed on the line of null speed means that the target is stationary relatively to the radiotelescope. The lower part of the graph represents the integrated speed which becomes a position, knowing that, the higher the position is, and the closer the target is to the radiotelescope, and vice versa |
 This second example shows how it is important to know where the line of null speed is; on the upper part of the graph, the shape of the speed is the same as on the previous example, but the fact that the line of null speed is higher than on the previous example makes that the position, i.e. the integrated speed, has a shape which is completely different from the previous example. It means that, if the Jodrell's graph was directly representing the doppler, and not the integrated doppler, they would have represented the line of null speed, otherwise the graph is not exploitable, and does not talk. |
 Outside these considerations, there is another evidence that the graph can only represent the integrated doppler (i.e. the position), and not the doppler itself (speed). If I prolong the line of the move of the moon after the lunar module has landed (red line), the difference between this line and the graph represents the move of the lunar module relatively to the moon; the fact that the curve of this graph is almost always above the line of the move of the moon means that the moon moves faster away from the radiotelescope than the lunar module itself does; but, if the moon moves faster away from the radiotelescope than the lunar module, it means that the lunar module is moving up relatively to the moon, and, if the lunar module keeps moving up relatively to the moon, how can it land on the moon? In fact, if this graph was representing the doppler, then the curve of the graph would remain under the red line representing the move of the moon (like the blue curve I have drawn); if the lunar module is going down relatively to the moon, it would mean that it is moving away faster from the radiotelescope than the moon; before landing, it would reduce its vertical speed, which means that the graph would get closer to the red line representing the move of the moon. |
 This figure shows how the integrated doppler shift is represented on the graph according to the move of the lunar module relatively to the radiotelescope: - when the radiotelescope records a straight horizontal line for the integrated doppler, it means that the LM remains stationary relatively to the radiotelescope. - When the radiotelescope records a straight rising line, it means that the LM is getting closer to the radiotelescope, with a constant speed , and the more this slope is vertical, and the more the LM comes faster toward the radiotelescope. - When the radiotelescope records a straight falling line, il means that the LM is getting away for the radiotelescope, with a constant speed , and the more this slope is vertical, and the more the LM gets away faster from the radiotelescope. |
 If the recorded line is bent instead of being straight, it means that the speed of the LM is not constant, and that it is either accelerating or decelerating. |
Now, how can we interpret this graph? We know that the final straight line corresponds with the move of the moon, after the lunar module has landed, for it is clearly said in the comment of the NASA which is under the graph |
 We are going to get interested in the move of the moon relatively to the radiotelescope, which is represented on the graph after Eagle has landed, on the last part of the graph. On this part, we can see that there is a variation of 4 vertical ticks for 40 horizontal ticks; that makes 1 vertical tick for 10 horizontal ticks. However, the graph is not graduated, so we don't directly know at what speed this corresponds on the graph; but we can know it if we have a another way to know the speed of the moon relatively to the radiotelescope. From the moment that we know this speed, we'll be able to make measurements on the graph. |
 There are two facts which make the distance of Jodrell's radiotelescope move relatively to the moon. The first fact is that the orbit of the moon around the earth t is not exactly circular, but elliptical instead. The elliptical orbit has two extreme points, the perigee which is the closest point to the earth, and on which the moon appears in full, and the apogee which is the farthest point from the earth, and on which the moon is hidden from us. This figure exaggerates the difference between the perigee and the apogee, and also represents the elliptical orbit too flattened; in reality, the elliptical orbit is much less flattened, almost a circle, but the earth is excentered relatively to the center of the moon's orbit; it is however a focus of this orbit. |
 The perigee of the moon's orbit is averagely at 356000 km from the earth, and the apogee averagely at 407000km from the earth; the difference between the apogee and the perigee is enough for the moon to look 12% smaller viewed from the earth when it is at its apogee than when it is at its perigee. |
 In fact, this figure is more accurate; the moon's orbit is closer to a circle, and there is less difference between the perigee and the apogee than on the previous figure, but it is less obvious on this figure that the orbit of the moon is elliptical, and that there is a difference between the two extremities of the elliptical orbit. |
 On Internet, it is possible to find tables which give the phases of the moon. The interesting one is the one which gives the phases of the moon at time of the landing of Apllo 11. On this table we can see that the moon had roughly turned of 60° on its orbit after the previous new moon. |
 I have made this figure on which everything is exaggerated, nothing is at scale: The earth and the moon are too big relatively to the moon's orbit, and the orbit is too flattened, but it is for an illustrative purpose. The distances of the apogee and perigee, and the fact that the earth is a focus of the elliptical orbit allow us to calculate the data of the ellipse: - The semi major axis was equal to 382263 km - The semi minor axis was equal to 381487 km - And the distance between the earth's center and the ellipse center was equal to 24338 km. You can see that there is not much difference between the two axes of the orbit, and that the orbit is effectively quite close to a circle. Between the apogee and the perigee, the moon got closer to the earth by a distance equal to: 406601-357925=48676km. There were 14.63 days between the perigee and apogee, which means that the moon got closer of 48676/14.63=3327km each day; and the moon got closer of 3327/24=138.6 km each hour. So the moon was getting closer to the earth at an average speed of 138 km/h. This speed was varying a little (tending to accelerate a little between the apogee and the perigee), but not far from the average value. At the moment of Apollo 11, when the moon was close to mid course, the speed of the moon relatively to the earth was quite close to the average speed. |
Now, if Jodrell was not moving relatively to the center of the earth, the line of the graph after the lunar module has landed should be moving up and not down |
 But, due to the fact that the earth spins, Jodrell was moving relatively to the earth's center. |
 At the time of the new moon, the moon is between the earth and the sun. So the point of the earth which faces the direction of the new moon is in the time zone of noon. The opposite point of the earth, which is in the obscurity, corresponds to the time zone of midnight. The point of the earth which is at mid distance between these two points corresponds to the time zone of 6 o'clock PM (18hrs). Now, when Apollo 11 landed, it was 20h17 UTC, which also happens to be the time of Jodrell. Between the time zone of 18hrs and the one of 20hrs, the earth turns of 30°. The earth is on the left of the minor semi axis (though I have exagerrated its distance to the ellipse's center, for I have oversized the earth), and the moon had not still reached its mid course, which means that it is on the right of this axis. It means that the direction of the moon from the earth's center makes with the sun's direction an angle which is around 60°. Conversely, at the time of Apollo landing, the direction of Jodrell's radiotelescope was making with the direction of the 18hrs time zone an angle a little over 30°, and with the direction of the sun an angle a little over 120°. In short, between the direction of jodrell from the earth's center, and the direction of the moon, there was an angle which was around 60°. The way that Jodrell was placed relatively to the moon, and according to the direction of the earth's rotation, it was making Jodrell move away from the moon, but at what speed? Was it a speed which was close to the speed that the moon was getting closer to the earth, which would make that the difference would not be important? In order to evaluate at what speed Jodrell was moving from the moon, I am going to take a reference system which is the orbital plane of the moon, and of which the x axis is in the direction of the sun; I am going to calculate the coordinates of the moon and Jodrell in this system, which will allow me to know the distance of Jodrell to the moon. If I call xj, yj, and zj the coordinates of jodrell in this system, and xl, yl and zl the coordinates of the moon in this same system, the distance of Jodrell to the moon will then be equal to: square root((xl-xj)²+(yl-yj)²+(zl-zj)²). And how can I deduce from it the speed of Jodrell relatively to the moon? Very simple: In one minute, the earth turns of 0.25°, so I add 0.25° to the angle of rotation of Jodrell and I do the calculation again; the difference between this distance and the previous distance is the variation of distance of Jodrell relatively to the moon in one minute; as there are 60 minutes in one hour, I multiply this difference by 60, and I'll then know the speed of Jodrell relatively to the moon in km/h. |
 Now, there is something which is going to complicate the calculation: the fact that the axis of self-rotation of the earth is tilted relatively to its plane of rotation relatively to the sun (which is also the one of the moon's orbit); it is tilted of 23.4°, and I have to take it into account if I want to make accurate calculations. When I made the calculations, I told myself that it would have been have been simpler if the earth had had no tilt, but, unfortunately it has one. The landing of Apollo 11 occured in summer, so the orientation fo the axis of self-rotation of the earth if oriented relatively to the sun is like the position I have circled. |
 OK, now we have our earth tilted down of 23.4°. Jodrell was at a latitude of 53.3°. The direction of the moon makes with the x axis (sun's direction) an angle which is a little less than 90°, and which will be a parameter in the calculation. The current distance of the moon is not far from its average distance, and the moon is close to its mid-course; I'll take 380000 km for it, but it will also be a parameter of the calculation. The angle of Jodrell's direction with the x axis is a little more than 120°, it will also be a parameter of the calculation. |
 The center of the section of the earth which contains Jodrell is at a distance which is equal to: R*sin(l), in which R is the radius of the earth, and l the latitude of Jodrell; if I call t the tilt of the earth (23.4°), the coordinates of the center of the earth's section which contains Jodrell are: xa=R*sin(l)*sin(t) za=R*sin(l)*cos(t) Now, we are going to calculate the coodinates of Jodrell relative to this point. I consider the horizontal plane which passes by the center of the earth's section which contains Jodrell. The direction of Jodrell from the center of this section makes which this plane an angle which is equal to: t'=t*(90-a)/90, where t is the earth's titlt and a the rotation angle of Jodrell relatively to the x axis (around 120°). I project Jodrell on the horizontal plane; the distance between the center of the earth's section which contains Jodrell and the projection of jodrell on the horizontal plane is equal to: d=r*cos(t'). r is the radius of the earth's section which contains Jodrell, and is calculated from the earth's radius as: r=R*cos(l) (l=Jodrell's latitude). Then the coordinates of jodrell relatively to the center of the earth's section which contains it are: xj'=d*cos(a)=r*cos(t')*cos(a) yj'=d*sin(a)=r*cos(t')*sin(a) zj'=t*sin(t') and finally the coordinates of jodrell in our system become: xj=xa+xj' yj=yj' zj=za+zj' And, if we call D the distance from the earth to the moon, and al the angle of the moon's direction to the x axis (a little less than 90°), the coordinates of the moon are: xl=D*cos(al) yl=D*sin(al) zl=0 And finally, the distance of Jodrell to the moon is equal to square root((xl-xj)²+(yl-yj)²+(zl-zj)²). As I said, I redo the calculation after having added 0.25 to the angle of Jodrell to the x axis, and I make the difference with the previous distance, and then I multiply this difference by 60 in order to obtain the speed that Jodrell goes away from the moon (the moon being considered immobile in this calculation). |
 Now, there is a little problem; I said that the orientation of the tilt axis is the one of June 21 because Apollo 11 landed in July; but, between the landing of Apollo 11 (July 20th) and June 21st, there is a month, or almost, and during this time, the earth has turned of 30° around the sun. |
 The consequence is that the direction of the earth's tilt makes with the sun an angle of 30°; but this can easily be turned around by adding 30° to both Jodrell's rotation angle and the moon's direction before making the calculation (and it does not change the result in an important proportion). |
 Now, I have all I need to make the calculations; however, I did not trust myself to make all these calculations on a calculator; it would have been too fastidious, and the chance of error too important. That's why I made a little program to make the calculations in which the parameters are: - The angle that Jodrell's direction makes with the x axis (direction of the sun). - The angle that the direction of the moon (from the earth's center) makes with the x axis (direction of the sun). - The distance between the earth's center and the moon's center (around 380000km). - the variation of the tilt angle since June's 21st (30°); this variation is added to both the Jodrell's direction and the moon's direction before making the calculations). After having imputted all these parameters, the "Compute distance" button must be clicked to make the calculation. The calculation then displays Jodrell's distance to the moon, and the speed that Jodrell moves away from the moon, which is in fact what we are really interested in. The initial calculation is made with an angle of 60° for the moon's direction, which is the maximum value for this angle, for it is in fact a little less, and a distance to the moon equal to 380000km, the rotation angle of Jodrell is set to 120°, which corresponds to the time zone of 20hrs. It gives a speed of 878 km/h for the speed that Jodrell moves away from the moon (with a stationary moon). Link to my (compressed) program to compute the distance of Jodrell to the moon, and the speed of variation of this distance |
 To show that the distance to the moon is not very important and does not have much effect on the result, I input a distance of 370000 for the moon's distance, and the resulting speed is the same as the result obtained for 380000km. |
 And with a distance of 390000km for the moon's distance, we obtain the same speed as for a distance of 380000km. So, you see, it is not important al all that I did not use the exact distance for the moon's distance. |
 If I use 65° instead, the speed that Jodrell moves away from the moon becomes 835km/h. |
And if I increase a little Jodrell's rotation angle (for the landing happed at 20h17), the speed that Jodrell moves away from the moon still increases. |
Now, we have not taken into account the fact that the moon moves closer to the earth. If we subtract the speed that the moon moves closer to the earth from the speed that Jodrell moves away from the moon, we obtain a speed around 700 km/h. |
Now, since we could obtain the speed that the moon moves away from Jodrell, and that we know that this speed corresponds to a variation of one vertical tick for 10 horizontal ticks (4 vertical ticks for 40 horizontal ticks), we can take measures on the graph. For instance, when we have a variation of n vertical ticks for m horizontal ticks, it will correspond to a speed of around: (n*10*700)/m km/h. |
 Now we are going to get interested in the "wiggles" of the lunar module which would come from Armstrong piloting the lunar module. I identifies three of them: - One with a varation of 2 vertical ticks for 3 horizontal ticks, which would correspond to an average speed of (2*10*700)/3=4666km/h. - One with a variation of 2 vertical ticks for 4 horizontal ticks which would correspond to an average speed of (2*10*700)/4=3500km/h. - And one with a varation of 2 vertical ticks for 3 horizontal ticks, which would correspond to an average speed of (2*10*700)/3=4666km/h. And, as the slope of these wiggles is opposed to the slope of the moon's move, it means that the speed of the moon's move should still be added to these speeds to obtain the actual speeds of the LM relatively to the moon. Now you'll say that it is not very precise, and extremely approximative; we can hardly read a fraction of a tick on this graph, but it is still obvious that Armstrong maneuvered the LM at a speed which was largely more than 1000 km/h. |
 Now, is it realistic that Armstrong, while he is flying on the lunar surface at moderate speed, looking for a spot to land on, would go at speeds over a thousand km/h each time he has to avoid an obstacle? |
 Seriously! |
Furthermore, if we could accept that the two last bumps would come from maneuvers of Armstrong to avoid craters, as the moon fans assert it, it is obvious that the first bump cannot be explained that way, because the lunar module was still too high at this bump to be bothered by a crater of the moon. |
 It is not all: On the part of the graph representing the flight of Armstrong, we have a global variation of 28 vertical ticks for 34 horizontal ticks, which would correspond to an average speed of (28*10*700)/34=5765 km/h. It makes more than 5600 km/h after we subtract the moon's speed. |
 Yet, in the table of the powered descent, the maximum vertical speed is the one of the "High gate" event, and it is equal to 145 feet/s, which corresponds to 156 km/h, around 36 times less than the average speed indicated by the graph of Jodrell! |
 In fact, the vertical speed is null at the beginning ot the powered descent, then starts to increase as the lunar module is oriented to decrease the great horizontal speed, then starts to decrease again as the lunar starts to turn vertical to also counter the vertical speed before it becomes too important; the horizontal and vertical speeds must both be null at landing. It means that the trajectory as seen by the radiotelescope of Jodrell should look like what I show in red on this schema; of course, it is indicative, and does not claim to be exact, just to give an idea. |
 The NASA engineers must have had much fun putting all these incoherences in the radiotelescope trackings, and I can easily imagine the fantasy explanations they gave to the astrophysicists of Jodrell Bank to explain these trackings. They must have had a great difficulty to refrain from laughing while they were giving these explanations (it is a real exploit, for, personally, I could not have helped it, I would have rolled on the floor laughing before the awed astrophysicists). But, if they could not refrain from laughing, they must certainly have found an explanation to justify it: The joy that Apollo 11 landed on the moon. |
 A new NASA lie is falling into the water! |
