 This is an excerpt of the video of the landing of Apollo 11, the LM makes a brutal incoherent change of attitude as it is quite close to the moon.  In the video of the landing of Apollo 15, not only the LM has an incoherent attitude, but moreover its trajectory makes a brutal right angle as it is just about to land.  And when the LEM takes off (the ascending part), instead of starting vertically, it abnormally starts in oblique (yet the LM itself is horizontal).  In the video of the landing of Apollo 16, the LM seems to be filmed by a camera which is independent from it, and outside it; yet the camera is supposed to be mounted on the LM (unless these are aliens who are filming!). Someone has told me it was the shadow of the lem, but it doesn't look like the lem's shadow to me: This shadow looks too sharply cut, it would be more blurry if it was the lem's shadow; moreover, the footpads are too light, not dark enough; and we can also see the shadows of the legs; I have been told it was the shadows of the probes, but they seem too long to me to be the shadows of the probes. If we make a comparison between the Apollo 16 landing (left) and the Apollo 17 landing (right):  They look quite different. The lem's shadow is darker and more imprecise in the Apollo 17 landing.  And when the LM takes off in Apollo 16, can you explain me why there is all this haze in an environment which is supposed to be void???  This video, in which we see the lunar module suddenly turn from a horizontal attitude to a vertical one just before landing on the moon, makes absolutely no sense, for, when the lunar module is close to the lunar surface, it has a low horizontal speed, and has lost the centrifugal force which was allowing it to counter the lunar attraction, and it does not benefit of air force to sustain it, like for a plane or a helicopter on earth, which means that it can only counter it with its main engine, and it must absolutely have a vertical attitude for its main engine to counter this attraction; its lateral engines allow it to make maneuvers of rotation and translation, but they are not powerful enough to counter the lunar attraction (and moreover their thrust cannot be adjusted). If the lunar module was flying with a horizontal attitude, it would not be able to counter the lunar attraction, would be attracted by the moon and would crash on it. And, even if the lateral engines had been powerful enough to counter the lunar attraction (and their thrust could have been adjusted to exactly counter the lunar attraction), the lunar module would not have been able to remain perfectly stable like what we see on the video when it is turned from a horizontal attitude to a vertical one. So, this video does not make the least sense, it is physically impossible.  We can also see this loo late and brutal rotation of the LM in the video of the landing of Apollo 12. |
 This is the video which shows the landing of the LM in Apollo 16. The Apollo believers claim that the black shape we see in the video is the LM's shadow on the lunar ground, and I say it is not, that it is a model standing up instead. I have led an analysis which tends to prove what I assert.  This animation shows how the LM's shadow moves as the LM is landing; the LM's shadow is progressively coming closer to the landing point of the LM, and, if it keeps the same size, by the fact that it comes nearer to the camera, the camera sees it bigger and bigger as the LM comes closer to the lunar ground.  I consider the grey lines I have circled the shadows of the legs of what I consider to be a model of the LM; but the Apollo believers consider instead that these are the shadows of the probes of the LM. The probes hang vertically under the LM and are in charge of detecting the lunar ground a little before the LM touches itself the lunar ground; they warn the LM's system that the lunar ground is close, upon what lights go on on the astronaut's panels which warn him that he must cut off the engine before the effective landing.  But we can see that the direction of these "probes" is different from the direction of the vertical axis of the LM on the "shadow".  On this part of the video, the "shadow" practically does not move and neither grows, which shows that the LM is not moving vertically; yet we can clearly see artifacts of the moon move under the LM; so, logically, the LM must be moving horizontally. But this sequence is just before the landing, and the LM must absolutely be perfectly still horizontally when it lands on the moon.  If the LM was moving horizontallly when it lands, it would fall when touching the ground.  On this sequence, we can see a footpad of the shape move forward whereas the other footpad practically does not move; yet they should move the same...Unless this shape is making a rotation, but, in that case, if it is effectively the LM's shadow, it means that the LM is also making a rotation, and, in that case, the LM's shadow would not remain at the same place on the video, but would move to another place.  On these two images of two different moments of the video, we can see the same artifact of the ground; this artifact is not always visible, for, sometimes, there is much fog which hides the lunar ground. We can see that, on the second image, the size of the shape has increased; but the size of the artifact has not grown correspondingly, it has less grown than the shape between the two images.  This is now a comparison between an image at the beginning of the video and an image a little before the end of the video. We still can see the same lunar artifact on the two images. This time, the shape is quite more bigger, it has more than doubled between the two images. The lunar artifact has also grown but not as much as the shape; unlike the shape, its size has not doubled. Normally, if the shape was effectively the LM's shadow, that means it would be at the same level as the lunar artifact, and the two should grow the same; if the shape grows faster than the lunar artifact, it means that it is closer to the LM's camera than the lunar artifact, which confirms my assertion that it is not the LM's shadow, but a structure standing up instead, a model of the LM. |
 When the lunar module comes back to the command module, it follows a kind of parabola along which it slowly and regularly turns from vertical to horizontal. Initially it has a vertical attitude, for it must first gain a vertical velocity, then it slowly turns to horizontal for it must gain the horizontal velocity which allows it to orbit the moon; the command module is orbiting at an altitude of 110km.  It is the attitude of the lunar module which allows it to distribute the force created by its engine on the vertical and horizontal axes; the more it is vertical, and the more it gives a vertical force, and conversely, the more it is horizontal, and the more it gives a horizontal force.  The lunar module cannot afford to make unnecessary maneuvers, for it must be sure to save enough fuel to make the final maneuvers; if it was unnecessarily wasting fuel, it could result in the lunar module running out ot fuel before making the final maneuver, and then it would be unable to dock with the command module, and would orbit the moon forever.  Initially, at the start of the trajectory, the lunar module has a vertical attitude.  This video is supposed to show the lunar module (ascending part) lifting off from the moon, leaving the lander on the moon. Most people believe that this video shows a lunar module really leaving the moon.  This image is extracted from this video. What are these color spots we see on the video? Is this a lift-off from the moon or a firework? Fireworks work on earth because oxygen allows to maintain the incandescence of the projected bits; here, in the void of the moon, the projected bits should immediately go out.  But, there was in fact a complication. Indeed, for a smooth progression of the lunar module, the thrust would have needed to be perfectly aligned with the center of gravity of the lunar module.  In the descent module, if there was a disalignment along the descent as the tanks were emptying, this disalignment could be corrected by swivelling the main engine.  But such was not the case in the ascent module. In the ascent module, the tanks were disposed, taking into account their respective density, and also the rest of the structure, so that the center of mass was initially aligned with the thrust of the main engine.  But this balance could not be kept throughout the whole ascent; as the tanks were emptying, the structure had a stronger and stronger influence in the balance of the lunar module, and the center of mass was progressively shifted from the line of thrust.  The fact that the center of gravity was being shifted from the line of thrust could eventually have been corrected is the ascent engine could have been swivelled, like the descent engine, which might have allowed to realign the center of gravity of the lunar module with the line of thrust, and so create no torque making the lunar module turn. But curiously, whereas the descent module had the possibility of swivelling its engine, the ascent module didn't have this capability. There would even have been a simpler way to restore the disrupted balance which would have consisted in moving a weight along the lunar module. It means that the shift of the center of gravity with the line of thrust could not be mechanically corrected, and the result is that the thrust of the main engine was causing a torque which was making the lunar module turn clockwise. If this torque had not been corrected, it would have made the lunar module start a fatal pirouette which would have caused its crash.  So, the only way to counter this torque caused by the disalignment of the center of gravity was to create a counter-torque with the lateral engines. However, the thrust of the lateral engines could not be adjusted; they were not creating a torque equivalent to the torque which would have been necessary to counter the torque generated by the disalignment of the center of gravity. This means that they could not be permanently fired, and had to be discontinuously fired.  If they could have been controlled with a sufficient frequency, the torque might have been countered in a quite smooth way, despite the fact that their thrust could not be adjusted. But the guidance of the lunar module had a quite important period, two seconds, which was largely insufficient to insure a smooth control of the torque.  The result is that the lunar module could not avoid to have a noticeable oscillating move in its ascent after a while, when the balance was starting to be disrupted. The torque created by the disalignment of the center of gravity was making the lunar module turn clockwise and pushing it on the right; then when the lateral jets were fired, the counter torque was making the lunar module turn counterclockwise and pushing it on the left; then the lateral jets were shut off and the torque of the main engine was again making the lunar module turn clockwise and move on the right, and so on... The reaction time of the lateral jets was slow enough for the amplitude of this sinusoidal move to be relatively important. When the shift of the center of mass was slight, this amplitude was not too important yet, but, the more this shift was increasing along the ascent, and the more the amplitude of the sinusoidal move was becoming important.  Clavius on his site also talks about this disbalance which happens during the ascent of the lunar module, and says that it explains the sunusoidal move we see on the video of the ascent of the lunar module, since he has read it in the description of the ascent.  So, in fact, whereas the lunar module follows its parabolic ascending trajectory, it would make a stronger and stronger sinusoidal move around this trajectory, which would look like something like this (very approximative of course). The fact that the lunar module constantly has to counter the disalignment torque with its lateral engines means that it is going to consume an important quantity of fuel of the lateral jets.  Yet, it is not the interest of the lunar module to waste this fuel, for it will need it when it has to make its final flipover maneuver to dock to the command module. If the lunar module was running out of fuel for its lateral jets before it reaches the orbit of the command module, it means that it would be incapable to make its flipover maneuver to dock to the command module; it would be embarrassing! Now, what conclusions can we draw from what we see on the ascent of lthe lunar module in Apollo 17. 1) If you observe the way that the lunar module turns, it is very clear that the plane of its ascent is perpendicular to the plane of the camera (this is especially visible by the way its panel turns). So its trajectory is initially perpendicular to the camera's plane. 2) Then we see the lunar module descend; of course, it probably comes from the fact that the camera vertically turns, but why did the operator wait for so long before turning the camera? He waits for some ten seconds before turning the camera, when the lunar module is about to leave the image on the top, before turning the camera, whereas there is 2.5 seconds before a command he sends and the result he sees on the video. So, if he had normally manipulated the camera, the lunar module should never have been at more than a quarter of the image's height from the middle of the image. 3) Then we see the lunar module move horizontally, but it moves horizontally in a plane which is parallel to the camera's plane, whereas the trajectory initially was in a plane perpendicular to the camera's plane! This is absolutely impossible, the lunar module cannot brutally switch from a plane to a plane which is perpendicular to it; it is physically impossible to instantaneously convert a horizontal speed to a perpendicular one! 4) And the lunar module brutally starts to make an important swaying move, whereas it was not showing any before; this is not possible, the center of mass is not brutally shifted from the line of thrust, but very progressively; so the amplitude of the swaying move should progressively increase, and not look like what we see on the video. You could say that it was less visible when the lunar module was not moving in the same plane, but we would still have seen variations of altitude; it is not credible that the operator could have compensated these variations with moves of the camera given the delay between his action on the camera and the result he sees. So, we here have an incredible accumulation of absurdities, completely discrediting this ascent. The fact of not immediately moving up the camera is likely a trick to hide the absurdity of the video, to make it less visible.  And there is another still more interesting video of the lift-off, filmed from the lunar module this time. The camera which films this video has a fixed position relatively to the lunar module, for we see an edge of the module's window always at the same place on the video; that means that, when the view of the camera turns, the lunar module turns the same.  At the beginning of the video, we see the shadow of the lunar module move on the video.  In fact, the shadow of the lunar module is projected along the direction of the sunlight which does not change during the ascent of the lunar module. That means that, as long as the lunar module remains vertical, the camera will see this shadow under the same angle, and thus the LM's shadow will remain at the same place on the video, and will just shrink with the distance. The lander will also remain at the same place on the video if it is visible on it.  You could say: May be the LM has already started to turn, which would explain that the LM's shadow moves on the video. But, in fact, on some images, we can see both the lander and the LM's shadow, and we can see that the distance which separates them is constant on these images, does not increase. If the LM had already started to turn to horizontal, It would then have a horizontal velocity which moves it away from the lander; the angle under which it sees the lander would then change, and that would make the distance between the lander and the LM's shadow increase on the video; the fact that this distance does not change on the video proves that the LM is still vertically above the lander at this point of the video and does not move horizontally yet. So, instead of seeing this at the beginning of the video, the LM's shadow going through the field of view of the camera...  ...We should see this: The LM's shadow keeping the same position on the video, and just shrinking.  In fact, what we see on the video shows this: While the lander is still vertically above the lander, instead of remaining with a vertical attitude like it should, it rotates so that the LM's shadow first goes through the field of view of the camera, and then the lander appears and crosses the field of view of the camera at its turn.  Then, on the last part of the video, we see the lander behave in an erratic way. Instead of moving in a continuous regular way on the video like it should, it moves in a totally irregular way: It moves, then stops, then moves again, then stops...The ascent of the lunar module is such that it could in no way give this irregular move of the lander on the video.  Once again, we are swimming in full fantasy! |
 While the astronauts are looking for a place to land, when they are near the lunar ground, they have to constantly fire the main engine. Why? Because they constantly have to counter the lunar attraction to maintain the LEM above the lunar ground; indeed, they no more have the horizontal speed to create a centrifugal force to counter this lunar attraction. So, the longer they look for a spot to land on, and the more they will consume fuel. And they don't know in advance how long this phase will take. And on the moon, no rescuers, and no gas pumps! And they also have to save enough fuel for the return to the command module! In these conditions, it is absolutely obvious that they have to put all the chances on their side by not wasting fuel before getting close to the moon and saving as much of it as they can! Making the clowns before really starting for the moon could cost them their lives! This is especially true when we know that Neil Armstrong was almost out of fuel when he landed on the moon. This is what we can find in the relation of his flight: "Words of warning came from Earth: just 60 seconds of fuel left before he would have to abort the landing". So, he was almost out of fuel, fuel that he could not refill on the moon, and he would however have made the fantasy of flying over the command module?  In fact, this is the kind of maneuver which would have been made. The command module, with the lunar module attached, would first turn vertical. Then the command module would move down to allow the maneuver of the lunar module. Then the lunar module would turn clockwise to a horizontal attitude to prepare for the travel. After a pause which allows the command module to inspect the lunar module, and make sure that all the legs have properly deployed, the lunar module fires its engine to decrease its orbital velocity of the necessary quantity (~75 feet/s) to put itself on the transfer orbit. Someone told me that the fact that the command module was under the lunar module was allowing to save the lunar module's fuel.  But it is wrong. If the command module turns in the opposite direction, with the lunar module down instead of up, the command module moves up to allow the maneuver of the lunar module, the lunar module turns counter-clockwise to a horizontal attitude to prepare for the travel, and, after a pause which allows the command module to inspect the legs of the lunar module, the lunar module fires its engine to decrease its orbital speed. We have an equivalent maneuver which does not make the lunar module use one drop more of fuel than in the previous maneuver. So the argument of fuel saving does not stand at all to justify the maneuver we see in Apollo 11.  In reality, the separation was made much before the DOI, a half orbit before (and it takes almost an hour for the CSM to cover a half orbit). So my animation does not really show the correct separation maneuver; there is a too short time between the separation and the DOI on my animation.  But, if the CM remains under the LM during a half orbit, and that both spacecrafts orbit with their normal orbital speeds, the CSM is not going to remain on the vertical of the LM, but is going to shift horizontally instead, and take advance over the LM. Indeed, as the CSM is a little closer to the moon, its orbital speed is a little greater than the one of the CSM; furthermore, even it had the same orbital speed, it would still shift relatively to the LM, for it would have a little greater angular speed.  So, at the end of the half orbit, if both spacecrafts orbit with their normal orbital speeds, the CSM would have shifted relatively to the LM, and the astronaut's camera would no more see it under the LM. Eventually, the camera could catch it on the extreme left, but only is it was turned.  Now the CSM can also maneuver to always remain on the vertical of the CSM, but then its horizontal speed would be below the required orbital speed, which means that the gravity would pull it down. The CSM would start going down, and gradually faster.  This is besides confirmed by the photo AS11-37-5447, on which we can see the CSM quite distant under the LM. According to the NASA documentation, the CSM would descend till a distance of a half mile relatively to the LM. But the CSM also acquires a little vertical speed all along, that it will have to counter to stop it. Now, is there really an advantage for the CSM to maneuver this way... ..rather than the way that I promote, which is to turn the LM down, separate, turn the LM counter-clockwise in horizontal position, and after a pause of some seconds allowing the CSM to check the deployment of the legs, make the LM perfom the DOI? In my solution, the CSM consumes no fuel during the half orbit, it just follows its orbit with its natural orbital speed.  In my solution, the CSM can check the LM against the luminous background of the moon.  In my solution, when the LM performs the DOI, it is close to the CSM, and the CSM can easily track it while it does this maneuver. In the NASA solution, the CSM will have to burn fuel all along the half orbit to remain under the LM, for it would naturally shift forward if not controlled (because of the angular speed increasing which forces to decrease the horizontal speed to remain under the vertical of the LM). It's not a important quantity of fuel, but it is still fuel uselessly burnt.  In the NASA solution, the CSM will have to check the LM against the black of space, in less good conditions than in my solution, and it will have to make it shortly after the separation. When the LM makes its DOI, the CSM is too far from the LM to make a check of the LM.  In the NASA solution, the CSM is consistently farther from the LM than in my solution, which means that it will have more difficulty to track it while it makes its DOI.  In the NASA solution, the CSM has a vertical speed at the end of the half orbit which has to be countered, by firing either the main engine or the lateral thrusters (it will be longer with the lateral thrusters).  In the NASA solution, the CSM currently has a too weak orbital speed at the end of the half orbit, which means that it will have to increase it to go back to its original orbit (unless it decides to remain on the final orbit, but it will have to increase its too weak orbital speed anyway).  It means that there is not the least advantage to maneuver the CSM and manage the separation this way, only disadvantages: Waste of CSM's fuel, and less good checking and tracking of the LM. This maneuver is very obviously to be considered a joke from the NASA engineers. |
 I am also going to talk about sequences I have seen in the movie "In the shadow of the moon". This sequence is extracted from this movie. It shows the lunar module starting its descent toward the moon.  To initiate its descent toward the moon, according to NASA (and confirmed by my own calculations), the lunar module decreases its horizontal speed of around 75 feet/s (~ 80 km/h) in order to put itself on the transfer orbit which brings it closer to the moon on an orbit from which it will start the powered descent. This means that the lunar module is going to initially lag behind the command module. But, on this sequence, the lunar module does not appear to lag behind the command module; the way it is oriented, it should move up on the image; if it does not, as the camera does not move relatively to the window, there can only be one explanation: The command module is currently turning to follow the lunar module. Besides you'll observe that we see the lunar horizon and the end of the sequence, which confirms that the command module is turning. What's surprising is the remarkable stability of the LM on the image, that the CM can follow it so perfectly. In fact, the remarkable stability of the LM on the image is not the real point I want to talk about; my point is rather about the attitude of the LM as seen from the CM.  The fact that the CM turns to follow the LM lagging behind it means that its field of view rotates relatively to the LM. But, if the field of view of the CM rotates relatively to the LM, it also conversely means that the LM rotates relatively to the field of view of the CM.  Now, if I make another animation showing how the LM appears in the reference system of the CM (making that the angle of view of the CM has a fixed orientation in this reference system), we can see that the LM appears turning relatively to the field of view of the CM. This means that the LM should not have a fixed attitude on this sequence, and that we should see its attitude change all along the sequence. This is a very clear sign that this sequence is fabricated and not real.  This sequence is also extracted from the movie "In the shadow of the moon". It shows the lunar module returning to the command module.  Normally the lunar module should come back to the orbit of the command module as a parabolic arc, and this figure extracted from a NASA document confirms it. The lunar module has a more important angular speed than the command module as it comes back to the CM's orbit. Normally, if the attitude of the command module was steady, we should see the lunar module move on the image; instead of that it remains at the same place all along the sequence; the only possible explanation is that the command module turns to follow the lunar module. Again we may wonder that the command module can follow so perfectly the lunar module. But once again, my point is rather about the attitude of the lunar module. Like on the sequence of the departure of the LM, we should see a change of attitude of the LM. Yet, the attitude of the LM does not change at all, and appears to be vertical all along the sequence. |
 What I am going to talk about is described in the following document: Link to NASA document The first phase of the descent consist to go down to a lower orbit from which the different phases of the landing are successively going to be performed (starting with the braking phase). They say that, to go down to the lower orbit, the LM performs a DOI, that is a short retrograde maneuver which would allow it to lose 75 feet per second of horizontal velocity so to have the orbital velocity of the lower orbit. Let's see if it corresponds with reality. On the drawing of the NASA document, the trajectory of the LM looks quite elliptical. But they have cheated, for they have more than exaggerated the distance of the CM to the moon; they make it look like the CM was at a distance to the moon equal to the radius of the moon.  In fact, the CM was at a distance to the moon only representing a sixteenth of the moon's radius. I show here how its orbit around the moon was looking like in reality; I have of course exaggerated its size so it could be visible. That means that the trajectory of the LM was quasi circular, very slightly elliptical. In fact they have magnified the distances to the moon so that the problem becomes more clear. On their drawing the elliptical trajectory of the LM looks like the elliptical trajectories of satellites around the earth.  However, the elliptical trajectory of satellites around the earth follows a very precise rule: The earth's center is one of the foci of the satellite's ellipse. When the trajectory of the satellite is circular, the earth's center is also the center of the circular trajectory.  Now, the foci points of the ellipse they show can be calculated; if we call "a" the half long axis of the ellipse, "b", its half short axis, and "c" the distance of the focus points to the ellipse's center, we have the well known relationship: c²=a²-b². This allows us to place the two foci on the ellipse's long axis; I show them with red crosses. I show with a green cross the moon's center. And now we have a big problem: The green cross should coincide with one of the two red crosses, and very far from it, as you can see. The ellipse followed by the LM is not the normal orbiting ellipse of a satellite.  Now I move the moon so that its center coincides with a focus of the LM's trajectory, so that the LM's trajectory really becomes a normal satellite's trajectory. And you see the problem? That would make the LM bump into the moon!  If the moon's center is far for any focus of the ellipse on their drawing, it is because their ellipse is too flattened. The ellipse has to be closer to a circle so that the moon's center may become a focus of this ellipse, like on this drawing I corrected. In fact, because the LM is much closer to the moon than shown on the NASA's drawing, the trajectory of the LM is very close to a circle, very very slightly elliptical.  According to Kepler's law, the orbital speed along an elliptical orbit of the moon is equal to: V=SquareRoot(1.622*1731000²*(2/r-1/a)) where r represents the distance of the satellite to the moon's center and a the semi major axis of the elliptical orbit. The orbital speed of the CM at the point that the LM makes the DOI (62 nautical miles) is: V=SquareRoot(1.622*1731000²/(1731000+114824))=1622.66 m/s The elliptical orbit which must allow to go from the CM's orbit to the lower orbit of 15000m (50000 feet) must have a semi major axis such that: 2*a=2*1731000+114824+15000, so a=1795912m. The LM starts from a distance r to the moon's center such that: r=1731000+114824=1845824m. So, if we calculate the orbital speed of the elliptical orbit which must bring the LM from the CM's orbit to the lower orbit, Kepler's law says it is equal to: V=SquareRoot(1.622*1731000²*(2/1845824-1/1795912))=1599.95 m/s That makes a difference of 22.71 m/s with the CM's orbital speed; expressed in feet/s, it makes 75.7 feet/s. They say that the DOI makes the LM lose around 75 feet/s; it seems to fit with what I have calculated. And, if the reduction of speed becomes 85 feet/s, the LM will arrive at the level of the moon on the other end of the ellipse! It means that it is far from being a safe maneuver, for an error of the initial reduction of speed could have catastrophic consequences on the other end of the elliptical orbit. Furthermore, when the LM arrives on the lower orbit, if is does nothing, it will not remain on the orbit, but get away from the moon instead; in order to remain on the lower orbit, the LM will have to decrease its speed by approximately the same value it initially did.  Now, they say this in the documentation: "This reduction is accomplished by throttling the DPS to 10-percent thrust for 15 seconds (center-of-gravity trimming) and to 40-percent thrust for 13 seconds. An overburn of 12 fps (or 3 seconds) would cause the LM to be on an impacting trajectory prior to PDI." So, if we call Am the maximal acceleration of the thrust, we have successive decrements of speed coming from the following steps: 1) 10% of Am is applied for 15 seconds in counter direction of the orbit; that makes a decrement of speed equal to: Am*0.1*15 2) Then 40% of Am is applied for 13 seconds; that makes a decrement of speed equal to: Am*0.4*13. 3) Then a final reduction of 12 feet/s, during 3 seconds, is still applied. So finally, the total decrement of speed is equal to: Am*1.5+Am*5.2+12=Am*6.7+12  Now, do we have a way to know Am? Yes, we have thanks to the specifications of the LM: - According to NASA, the LM had a total mass equal to 15200kg, for the H-Series (the one of the first missions). - The LM's engine had a force of 43900 Newtons, according to NASA So, the LM's engine, with full tanks and maximal thrust, could give an acceleration: Am=43900/15200=2.888 m/s²=9.627 feet/s² So, now we can calculate the (minimal) initial decrement of speed caused by the thrust of the engine: Speed decrement=9.627*6.7+12=76.5 feet/s You'll say that it does not make much difference with the reduction of speed which must be applied, but a variation of one foot/s in the reduction of speed makes a variation of altitude of 1250 meters at the other end of the orbit, and this variation might make the powered descent fail. The final overburn makes the LM lose 12 feet/s in 3 seconds; with the maximulm thrust, the LM would lose 12 feet/s in: 12/9.627=1.246s. It means that to reduce the speed of 12 feet/s in 3 seconds, the LM applied a fraction of the thrust equal to 1.246/3=0.41, so 41%. Now, why does the way the speed of the LM is decreased make no sense? Because it starts with a low throttle (10% thrust) and ends with a more important throttle (40% thrust); it should have been the converse, the LM should have started to throttle importantly, and more moderately at the end of the speed decrease, so that it could adjust the speed decrease as precisely as possible. Indeed, a variation of a simple foot/s at the start of the transfer orbit causes a variation of altitude of 1250 m at the other end of the transfer orbit! So things have been made in reverse order!  Then they describe the operational phases of the landing, starting with a braking phase which allows the lunar module to lose most of its important horizontal velocity. The lunar module indeed has a very important initial horizontal velocity relatively to the moon of almost 6000km/h that it must absolutely lose before landing; Imagine what would happen if it was landing with a such velocity: It would be literally pulverized!  During the initial phase, the LM is oriented horizontally, because it must use the thrust of its engine to counter the horizontal velocity and make it decrease. Initially, the centrifugal force allows to counter the lunar attraction which means that it does not need to be countered. Progressively, as the LM loses its horizontal velocity, the centrifugal force decreases, which means that the lunar attraction tends to more and more attract the lunar module toward the moon and make the vertical speed increase. But, during the initial braking phase, the vertical velocity remains small relatively to the important horizontal velocity, and represents only a hundredth of the horizontal velocity, or around it. So, in this phase, the lunar module ignores the lunar attraction and consecrates the thrust of its engine essentially to decrease the important horizontal velocity, which means that it remains horizontal. After this phase, it starts turning to vertical, in order to start countering the lunar attraction before the vertical velocity becomes too important, moderately at first, then more and more as the horizontal velocity becomes smaller and smaller, to end up completely vertical at the end.  Here is the summary of the powered descent. A column (TFI) indicates the time, a column (inertial velocity) indicates the horizontal velocity relatively to the moon, a column indicates the altitude rate, that is the loss of altitude per second, a column indicates the current altitude, and a column indicates the DeltaV (the triangle is a greek symbol which means "Delta").. But how is the second column, the real velocity obtained?  With the radar, of course, by using the Doppler effect (this effect is used by the police radars to measure the speed of your car). Of course!  But Wait...  In the NASA document, they explicitly say that the velocity could be acquired by the radar only from the altitude of 22000 feet. That means that the radar was only allowing to acquire the velocities I have colored in green; the velocities I have colored in red could not be acquired by the radar, because corresponding to an altitude above the limit of acquisition of the radar (22000 feet).  And, concerning the altitude, they say that if could be acquired by the radar only from the altitude of 39000 feet. That means that the radar could only allow to acquire the altitudes I have colored in green, and not those I have colored in red.  So, what was allowing to acquire the velocities and altitudes I have colored in red in the table, if the radar could not acquire them? Yes, what??? In fact it was possible to acquire the position and speed of the LM by using the inertial platform, i.e. the gyroscopes and accelerometers, but the accelerometer has a little error which is integrated to obtain the speed; and, to obtain the position, the speed is integrated, and so is the error on the speed. It means that the indications of speed and position given by the guidance platform are much less precise than the the ones given by the radars. And, besides, the radar does not give its indication from a given altitude; it gives it as soon as there is an echo coming back.  You'll notice that the altitude rate keeps increasing till the high gate, for the LM makes no effort to counter the vertical velocity till that moment, or not sufficient.  Between the Events I have colored in blue (Rotate to Windows-up position) and green (LR altitude update), there is a difference of time equal to 4:18-2:56=82 seconds; at the first of these two events (Rotate to Windows-up position), the altitude rate is indicated as -50; at the second of these two events ((LR altitude update), the altitude rate is indicated as -89; that means that the altitude rate is comprised between -50 and -89 between these two events; so the minimal loss of altitude is equal to 82*50=4100 feet, and the maximal loss of altitude is equal to 82*89=7298 feet; now the difference of altitude between these two events is equal to 44934-39201=5733 feet; 5733 is comprised between the minimum (4100) and the maximum (7298) we have computed from the rate, so we have a coherence here.  Now more embarrassing. Between the two events I have colored in blue (LR altitude update) and green (Throttle recovery), there is a difference of time equal to 6:24-4:18=126 seconds; at the first of these two events (LR altitude update), the altitude rate is indicated as -89; at the second of these two events, the altitude rate is indicated as -106; that means that the altitude rate is comprised between -89 and -106 between these two events; so the minimal loss of altitude is equal to 89*126=11214 feet, and the maximal loss of altitude is equal to 106*126=13356 feet; now the difference of altitude between these two events is equal to 39201-24639=14562 feet, so greater than the maximum loss of altitude (13356 feet) given by the altitude rate. And, in fact, the average loss of altitude given by the altitude rate is close to the mid-sum of the minimum and the maximal loss of altitude given by the rate of altitude, which is 12285 feet, still farther from the effective difference of altitude (14562 feet). Here we have a clear incoherence!  The table indicates that the initial horizontal velocity of the lunar module is equal to 5560 feet/s. As the intial vertical velocity indicated by the table is initially null, this represents an orbital velocity. I have computed the orbital speed at this altitude in meters per second (~1668m/s). I was wondering what conversion factor the NASA engineers were using for a foot. A foot is usually considered 30 cm long; however, an article in google says that is is very precisely equal to 30.48 centimeters. However, if I apply a conversion factor of 0.3048, I find an orbital velocity a little smaller than the one indicated in the table, 5473 feet/s, whereas, if I use a conversion factor of 0.3, I precisely find the orbital speed of 5560 feet/s indicated in the table; from this I deduce that the NASA engineers considered that a foot is exactly 30 cm long, and not 30.48 cm. We also know from this that the velocities indicated in this column are absolute and not relative to the lunar surface, for, if the orbital velocity was relative to the lunar surface, 15 feet/s second should be added to this velocity, and it would be of 5575 feet/s instead.  Now, the last velocity of the table, as the LM is at 12 feet from the lunar surface (less than 4 meters) is indicated as -15 feet/s. If this velocity was absolute, as the prior velocities of the column are, it means that it would represent a velocity of 15 m/s in counter-direction of the CM's orbit; as the CM orbits in counter-direction of the moon's rotation, it means that the LM would be turning around the moon in the same direction as the moon's rotation, and also at the same speed as the moon turns on itself at a speed of 15 feet/s.  So, it means that, if the last velocity of the table is absolute, the LM turns the same as the moon's rotation and is thus stationary relatively to the lunar surface, which means that the LM could safely land on the moon.  But, and it is the trick, the author of the document has added an indication on this speed which means that it is not absolute, but relative to the lunar surface (the absolute velocity is between brackets).  It means that the lunar surface is moving at a speed of 15 feet/s relatively to the lunar module because of its self-rotation of 15 feet/s, which means that the LM has a null absolute speed; if the moon was not rotating on itself, the lunar module would be stationary relatively to the lunar surface, but because the moon does rotate on itself, the lunar surface is moving relatively to the lunar module, which amounts to the same as saying that the lunar module is moving relatively to the lunar surface (like a plane on earth which goes at the same speed as the earth's rotation, and in counter-direction of the earth's rotation; relatively to the sun, this plane has an absolute speed which is null, with the result that the time does not change in this plane while it flies; but, relatively to us, on earth, we see this plane fly fast). So, because they specify that the last velocity (along with the velocity which precedes) is relative to the lunar surface, and not absolute, it means that the lunar module is really moving at a speed of 15 feet/s (~16 km/h) relatively to the lunar surface.  But the LM cannot land with a lateral velocity, it must be perfectly stationary relatively to the moon when it lands. Landing with a lateral velocity of 15 feet per second would be a perfect passport for tipping over when touching the ground! Now you might think that it still has the time to null this lateral speed before landing? When the lunar module has this lateral speed, it is at distance of 12 feet from the ground and descending at a speed of 3 feet per second; that means that it only has 4 seconds left before touching the ground; as it is laterally moving at a speed of 4.6m/s (correspondance of 15feet/s), it would mean that it has to produce a deceleration of 1.15m/s². The powerful central engine is itself able to produce a maximal deceleration of 3m/s² at full thrust. The lateral engines are considerably less powerful than the central engine, and are not able to produce this deceleration of 1.15m/s², which means that they won't be able to null the lateral speed of the lunar module before it touches the ground. What does exactly represent the DeltaV, which is displayed on the last column on the table? The DeltaV is a combination of a Horizontal DeltaV, I'll call DeltaVH, and a Vertical DeltaV I'll call DeltaVV. The horizontal DeltaV is the difference between the horizontal speed the LM would have if it had not been countered by the engine and the current horizontal speed. Likewise, the vertical DeltaV is the difference between the vertical speed the LM would have if it had not been countered by the engine and the current vertical speed. Once that the horizontal and vertical DeltaV are known, the DeltaV is simply calculated by summing the squares of these two DeltaV, and then taking the square root of this sum: DeltaV=SquareRoot(DeltaVH²+DeltaVV²). Since the horizontal velocity that the LM would have, if it had not been countered, is the initial orbital velocity, the horizontal DeltaV is calculated as: DeltaVH=Initial orbital velocity- current horizontal velocity. The speed acquired by the lunar attraction can be calculated by the formula: V=1.622*t, t being the time elapsed since the start of the landing procedure ; this gives the vertical speed coming from the lunar attraction expressed in m/s, so it must be divided by 0.3 to obtain it in feet/s. So, we now have: DeltaVV=1.622*t/0.3-Current vertical speed. And, with the knowledge of the two DeltaV along the two axes, we can calculate the DeltaV by the above formula.  So, let's take the event "High Gate": - The current horizontal speed is 506 feet/s, so the horizontal DeltaV is equal to: DeltaVH=5560-506=5054 feet/s. - The current indicated time is 8:26=506 s, and the current vertical speed (Altitute rate) is 145 feet/s; so the vertical DeltaV is equal to: DeltaVV=1.622*506/0.3-145=2590 feet/s. Now we can calculate the DeltaV: DeltaV=SquareRoot(DeltaVH²+DeltaVV)=SquareRoot(5054²+2590²)=5679feet/s. But the current DeltaV is indicated as 5375 feet/s only; that makes a difference of more than 300 feet/s.  Now, let's take the event "Low Gate": - The current horizontal speed is 68 feet/s (it is the number between brackets, the other one is said being relative to the lunar surface), so the horizontal DeltaV is equal to: DeltaVH=5560-68=5492 feet/s. - The current indicated time is 10:26=626 s, and the current vertical speed (Altitute rate) is 16 feet/s; so the vertical DeltaV is equal to: DeltaVV=1.622*626/0.3-16=3368 feet/s. Now we can calculate the DeltaV: DeltaV=SquareRoot(DeltaVH²+DeltaVV²)=SquareRoot(5492²+3368²)=6442 feet/s. But the current DeltaV is indicated as 6176 feet/s only; that makes a difference of more than 266 feet/s; it is less than the previous difference, but still relatively important, more than 200 feet/s.  Now, let's take the last event "Touchdown": - The current horizontal speed is 0 feet/s (the number between brackets), so the horizontal DeltaV is equal to: DeltaVH=5560 feet/s. - The current indicated time is 11:54=714 s, and the current vertical speed (Altitute rate) is 3 feet/s; so the vertical DeltaV is equal to: DeltaVV=1.622*714/0.3-3=3857 feet/s. Now we can calculate the DeltaV: DeltaV=SquareRoot(DeltaVH²+DeltaVV²)=SquareRoot(5560²+3857²)=6766 feet/s. But this time, the indicated DeltaV is very close to the theoretical DeltaV, and even slightly greater, instead on being consistently smaller like in the previous events (more than 200 feet/s difference), for it is equal to 6775 feet/s. So, what happened, how is it that the final DeltaV could be so close to the final theoretical DeltaV when the previous values of the DeltaV were consistently smaller than the theoretical DeltaV??? In fact the answer lies in the document; they say this: "In addition, 85 fps is added to assure 2 minutes of flying time in the landing phase" "Also, a 60-fps AV is added for LPD operation in the approach phase to avoid large craters (1000 to 2000 feet in diameter) in the landing area." It means that they artificially add 145 feet/s to the calculated DeltaV for fancy reasons. If they had not artificially increased the DeltaV, the final DeltaV calculated by the guidance would be consistently smaller than the theoretical DeltaV, like for the previous events.  Now, it they add an extra DeltaV each time they fall below an given altitude, this DeltaV is going to make them go up again, and, when the LM falls again under this altitude, a new extra DeltaV will be added again, which will make the LM go up again, and so on... This way, I'm afraid the LM will never land on the moon!  Now, the normal way for the AGC to proceed is to compute, step by step, the current thrust from the indications given by the radars (altitude and moon relief) and the accelerometers; this is a continuous process. The guidance must certainly not arbitrarily add a fictional DeltaV in expectation of craters that it is not even sure they will really be here! Now, I think the NASA engineers have neglected the deceleration coming from the centrifugal force, and I don't think they intended it as an intentional error, I think that they just didn't take it into account; I think this because, if it is not taken into account, the final indicated DeltaV fits with the theoretical DeltaV, but it is only because they artificially increased it for fancy reasons, and it is the joke they meant. But, how to take into account the deceleration coming from the centrifugal force? Isn't it a bit complicated to integrate it in the computation of the vertical speed. In fact, not so much if the vertical speed coming from the lunar attraction is calculated differently; If this speed is calculated, not from the current time, but from the current loss of altitude, then the result which is obtained is smaller than the result obtained by using the current time, but this result takes into account the deceleration coming from the centrifugal force, so it is more accurate. An object falling in free-fall on the moon from a height H will reach the lunar ground in a time T which is such that: 1.622*T²/2=H. (H expressed in meters). So the time it will take for the object to reach the ground is: T=SquareRoot(2*H/1.622). During this time, this object will have acquired a speed equal to: V=1.622*T=1.622*SquareRoot(2*H/1.622) The result is in m/s (H being expressed in meters). If the unit is the foot instead of the meter, the speed obtained by an object in free fall on the moon after falling from a height H is: V=1.622*SquareRoot(2*H*0.3/1.622)/0.3 The obtained result is smaller than the result obtained from the current time of the landing procedure, but, unlike the previous calculation, it does take into account the deceleration coming from the centrifugal force, so it is accurate for calculating the DeltaV. Now, if we recalculate the final DeltaV, we have: DeltaVV=1.622*SquareRoot(2*(48814-12)*0.3/1.622)/0.3-3=723 feet/s. So, the final DeltaV becomes: DeltaV=SquareRoot(5560²+723²)=5606 feet/s. That means that the theoretical DeltaV which takes into account the deceleration coming from the centrifugal force is consistently smaller than the applied DeltaV (6775 feet/s), more than a thousand feet/s smaller!  So, it seems that the AGC was doing a real lousy job at evaluating the DeltaV! But it seems that the AGC was more busy showing alarms than correctly doing its job. So, how could the LM safely land with a so deficient guidance?  I think there is the hand of god in this! |
 In order to land on the moon, the LM decreases a little its orbital speed (of around 75 feet/s) to take a transfer orbit which allows to bring the LM on a lower orbit. According to the mission report of Apollo 17, the LM of Apollo 17 would reach the lower orbit, on the pericythion of the transfer orbit, at an altitude of 6.2 miles. I initially assumed that the LM initiated its powered descent as soon as it reached the lower orbit, which means that it would then start the powered descent at an altitude of 6.2 nautical miles; 6.2 miles is equivalent to 38274 feet.  But according to the powered descent of Apollo 11, this altitude would not even represent the braking phase of the powered descent of Apollo 11. The braking phase is the phase during which the LM reduces its very important orbital speed, without caring about its vertical speed which remains small during this phase. The altitude lost in the braking phase is the difference between the altitude of the beginning of the powered descent and the altitude of the High Gate Event, which ends the braking phase, and this difference is equal to 41299 feet. So, if the LM of Apollo 17 was starting its powered descent at an altitude of 38274 feet, it could not even end the braking phase.  It would crash on the lunar ground a little before the end of the braking phase, with both quite important horizontal and vertical speeds. And there is no possibility for the LM of Apollo 17 to make this braking phase shorter, for it is a little heavier than the LM of Apollo 11, but has the same engine, which means that it has more difficulty to reduce its horizontal speed than the LM of Apollo 11.  The conclusion is that, if the LM was starting its powered descent as soon as it reaches the lower orbit, at an altitude of 6.2 miles, it would be sure to crash on the lunar ground.  But the table of the trajectory parameters (3-III) indicates that the LM started in fact its powered descent at an altitude of 8.7 miles and not 6.2 miles. 8.7 miles is equivalent to to 53708 feet, so is higher than the 48814 feet of the powered descent of Apollo 11. It is apparently sufficient to make the powered descent (unlike the 6.2 miles).  The table of trajectory parameters informs us on the moon coordinates of the point the DOI was made. There are coordinates for the start and the end of the DOI.  The end of the DOI corresponds with the moment that the LM is on the apocynthion of the transfer orbit, so the coordinates of the apocynthion of the transfer are those of the DOI's end, so 19.2° Latitude South, and 166.77° Longitude West.  The pericynthion of the transfer orbit, which is the point that the LM reaches the lower orbit, is diametrically opposed to the apocynthion, and its moon coordinates are symmetrical to those of the apocynthion; the latitude of the pericynthion is the same as for the apocynthion, but relative to the opposite pole (i.e. North become South, and vice versa); in what concerns the longitude, the longitude of the pericynthion is the complement to 180° of the longitude of the apocynthion, and relative to the other direction (i.e. West becomes East, and vice versa). For instance, if the moon coordinates of the apocynthion are n° Latitude South and m° Longitude West, the moon coordinates of the pericynthion are n° Latitude North and 180-m° Longitude East.  So, the latitude 19.12° South of the apocynthion of the transfer orbit becomes 19.12° North at the pericynthion of this orbit.  And the longitude 166.77° West of the Apocynthion becomes 180-166.77=13.23° East at the pericynthion of this orbit.  So, by the knowledge of the coordinates of the apocynthion (19.2° South, 166.77° West), we know the coordinates of the pericyntion of the transfer orbit (19.2° North, 13.23° East). The Trajectory parameters also inform us on the moon coordinates of the point that the LM started its powered descent; they are 19.13° Latitude North and 48.75° Longitude East. So, the point that the LM reached the lower orbit, and the point it started its powered descent, have very different longitudes (13.23° East and 48.75° East), but the same latitude or almost (only 0.01 ° difference). The LM started its PDI with a difference of longitude of 35° before it reached the pericynthion.  I first thought that it was impossible that the PDI could have the same latitude. However, I reasoned like only the axis apocynthion-pericynthion was making an angle with the horizontal plane of the moon, and the perpendicular axis was in the moon's horizontal plane.  The problem is that the perpendicular axis could also make an angle with the moon's horizontal plane, nothing guarantees that it is in the horizontal plane of the moon. I chekched in the mission report of Apollo 17 if it could be confirmed that the minor axis of the transfer orbit was in the horizontal plane of the moon, but I must confess that I found nothing. I was hoping to find a statement like: "After having a quarter of a turn around the moon, we could see the moon's equator", but, unfortunately, I found nothing such, nothing which could allow me to prove that the fact that the PDI and the pericynthion had identical latitudes was abnormal. I could always say that it would be some coincidence if the PDI and the pericynthion had the same latitude, but the Apollo believers will tell me, and with reason, that I must prove that this coincidence cannot happen, and I must admit that I can't. So, I remained puzzled. Obviously the fakers meant something with identical latitudes for the PDI and the pericynthion, but in the same time they gave me no way to formally prove an anomaly. What does that mean?  And finally the light came. And if, instead of being intended as an anomaly, the identity of the latitudes of the PDI and the pericynthion was intended as a clue allowing to prove another anomaly?  Indeed, what does the fact that the PDI and the pericynthion have the same latitude mean? It means that the difference of longitude between the two also represents the angle between the PDI and the pericynthion (the angle represented in green on the schema) relatively to the moon's center in the reference system of the transfer orbit, which would not be the case if they had different latitudes! If I call "Angle" this difference of longitude (equal to 35.52°), and xc and yc the coordinates of the PDI in the reference system of the transfer orbit relatively to the moon's center, we have the relationship: yc/xc=tan(Angle). If I call x and y the coordinates of the PDI relatively the ellipse's center, a the the semi-major axis of the ellipse, b its semi-minor axis, and c the distance between the ellipse's center and one of its foci, thus the distance between the ellipse's center of the moon's center, since the moon's center is one of its foci, we have the well known relationship: x²/a²+y²/b²=1. But we also have: x=xc+c, and y=yc. So, if we use the coordinates of the PDI relatively to the moon's center, we obtain: (xc+c)²/a²+yc²/b²=1. But we also have: xc=yc/tan(angle), which allows to replace xc in the formula which becomes: (yc/tan(angle)+c)²/a²+yc²/b²=1. And what do be finally obtain? An equation of the second degree in yc, and I suppose that most of you know how to solve a such equation. Of course, we need to know the parameters a, b and c of the ellipse, but they are easy to obtain. First we have: 2*a=2*moon radius+altitude of Apocynthion+altitude of pericynthion. According to the apollo 17 mission report, the altitude of the apocynthion is 59.6 nautical miles, which gives 110379 meters, and the altitude of the pericynthion is 6.2 miles, which gives 11482 meters, and the radius of the moon is 1731km. It finally gives: a=1791930 meters. c can be obtained with: c=a-moon radius-altitude of pericynthion, which gives c=49448 meters. And, as there is the relationship: a²-b²=c², we can calculate b with the formula: b=square root(a²-c²), which gives b=1791247 meters. We now have all we need to calculate the coefficients of the equation of the second degree in yc, and solving this equation gives: yc=1017443 meters. We can obtain xc with: xc=yc/tan(angle), which gives xc=1425351 meters with the angle of 35.52°. And, knowing xc andf yc, we can calculate the distance of the PDI to the center of the moon with the formula: Distance=square root(xc²+yc²); after having subtracted the moon radius from this distance, we finally obtain the altitude of the PDI: Altitude of PDI=20233 meters; converted in nautical miles, it gives 10.92 nautical miles. But, in the trajectory parameters of the mission, they tell us that the altitude of the PDI was 8.7 miles 8.7 miles instead of 10.92 miles; that makes more than two miles difference (2.22), more than 4 km!  Now we have our anomaly, and it could be proven thanks to the fact that the PDI and the pericynthion of the transfer orbit had the same latitude! So, if the fact that the PDI and the pericyntion have the same latitude is not a proven anomaly, it allows to prove a real anomaly on the other hand!  Now, the Apollo believers still have a hope; it is a known fact that the center of the mass of the moon is a little different from its geometrical center; there would be a distance of several kilometers between them; this distance is generally considered equal to 2 kilometers, but it may be a little more. The center of mass is closer to the earth than the geometrical center, so closer to the pericynthion of the transfer orbit. It means that the distance between the center of the ellipse and one of its foci is increased by the distance between the two centers, since the center of mass is the real focus of the ellipse; the consequence is that the semi-minor axis is a little smaller (since we have the relationship: b²=a²-c²), and the transfer orbit a little flatter; this of course is going to reduce a little the altitude of the PDI.  What I have said opens of course a window of hope for the Apollo believers. Is the altitude of the PDI going to be reduced enough to match with the altitude indicated on the trajectory parameters? I am going to be generous; the generally accepted distance between the two centers is 2 kilometers, but, if I take that distance, the Apollo believers will tell me: You are not sure that this distance is the exact distance, it might be a little more. That's why I am going to take consistently more; I am going to take 10 kilometers for this distance, five times more than the accepted distance. This way, I am sure that the distance I take will not be smaller than the real distance between the two centers. And now let's do all the calculations again, this time taking into account the (exaggerated) distance between the two centers of the moon. The distance between the ellipse's center and its focus becomes: distance between ellipse's center and geometrical center + (exaggerated) distance between centers=59448 meters. If I call c' the distance between the ellipse's center and the center of mass, we have the relationship: b²=a²-c'², which gives b=1790944 meters. After having recalculated the coefficients of the equation of the second degree, and having solved it, we find: yc=1017386 meters. This allows us to calculate the new value of xc: xc=1425271 meters. After having calculated the distance to the (geometrical) center of the moon, and subtracted the moon's radius, we finally obtain the new altitude of the PDI: 20135 meters; in miles, it gives 10.87 nautical miles. So, by taking into account the distance between the two centers of the moon, and largely exaggerating it, we have only gained 0.05 miles, not even a tenth of a mile, when we needed to gain 2.22 miles to erase the anomaly.  We can call that death of a deceived hope! |
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On the way back to the command module, the Lem gets rid of its lander it doesn't need any more; this allows it to be lighter and easier to maneuver, and also to consume less fuel.  The LEM also follows a parabolic trajectory as on the way to the moon, but inverted: Indeed, instead of losing horizontal speed, it must at the contrary gain horizontal speed so that it can arrive on the orbit of the command module at the same horizontal speed as the command module. As the LEM has a horizontal speed which is smaller than the one of the command module, except when it arrives on its orbit, the LEM must take off before the command module is over it, so that, when it arrives on the orbit of the latter, it is just a little before. Note that this time the LEM will arrive with an orientation which is the opposite of the one it had when it left the command module for the moon. The problem is that the LEM can only communicate with the command module through its top; on its bottom, there is the engine, and no possible communication; this means that this time, when the LEM is arrived on the orbit of the command module, it has to rotate a half turn, in order to present its top to the latter, before joining the command module.  If the LM arrives after the CM instead of arriving before, then it will be unable to dock to the CM. Thence the direct rendezvous requires an excellent control. The engineers thought that the technology they used was not able to guarantee this level of control. This is how they explain it: "The most obvious solution, to ascend directly from the surface with one burn, and meet the passive vehicle shortly after launch, can be quickly rejected for several reasons. Most importantly, such an ascent requires that the launch be at precisely the specified time, and the engine, radar and guidance systems must perform absolutely perfectly (mere fantasies in the eyes of any engineer!). Anything other than such exactness and the ascending vehicle will miss its target. Even under ideal conditions, the all-important terminal phase (closing, braking and docking) would happen very quickly, and no control would be possible over the lighting conditions which could very well be total darkness. A final difficulty was the lack of easy solutions for recovering from any trajectory errors or systems failures during ascent." So the direct rendezvous I showed has been rejected not because it is theoretically impossible but because of the high level of control it requires.  That's why to the direct rendezvous they preferred a more gradual approach involving several orbits around the moon and involving the Hohmann transfer.  The hohmann transfer consists, after a first phase of powered ascent in which the LM acquires a horizontal speed which is a little greater than the speed which is required to orbit the moon, to let the horizontal speed slowly pull the LM away from the moon and closer to the CM's orbit with the engine shut (and thence consuming no fuel); at a given moment, the LM's horizontal speed becomes unsufficient for the LM to go on drifting away from the moon, so the LM momentarily fires its engine to acquire a new horizontal speed which will allow it do drift again away from the moon; once this horizontal speed has been acquired, the LM shuts off again its engine and goes on with its engine shut (and consuming no fuel). So, the Hohmann transfer allows the LM so progressively come closer to the CM in a spiral around the moon with a minimal consumption of fuel.  The LM may also have to momentarily fire its RCS when its orbital plane crosses the CM's one, and its orbital plane is different from the CM's one, in order to join the CM's orbital plane.  This animation shows how the lem behaves when it arrives on the command module's orbit. The LEM makes a half turn with a push on the lateral vertical engines in opposite directions; during this half turn, the command module remains behind the LEM. When this half turn is over, the LEM now gives a little push of the two vertical engines, in the same direction this time, to slightly decrease the horizontal speed and get closer to the command module; a push on the bottom horizontal lateral engine allows to control the altitude of the LEM so that it remains on the orbit of the command module, and does not fall under this orbit. This half turn does not need to be fast, it can be slow, and it has better be slow for the LEM has a slow computer which could not control a fast rotation.  Now, you are going to say: "But if the lem rotates, the sequence of photos in Apollo 14 in which we see the command module rotate might be normal, for, if the LEM rotates, from its point of view, it will see the command module as rotating!" In fact, no, this sequence is still incorrect, and we are going to see why.  First, at the start of the half turn, the LEM cannot see the command module, it is not in the field of vision of the cameras.  In fact there was a window in the lunar module which was allowing to see the command module when the lunar module was facing the command module frontally, that I have circled in red on this schema, but this window was only for allowing the astronauts to make the alignment with the command module, and was not allowing to take full field photos of the command module with the hasselblad cameras.  And it is also case at the end of the half turn.  That means that the photo above (AS14-66-9344) is impossible; there is no way that the astronauts could see the command module from the LEM under this angle. Now, when the LEM is in the middle of its half turn, there is a little range around which the astronauts will be able to view the command module.    The three stereoscopic views above show how the LEM views the command module respectively a little before a quarter of turn, at a quarter of turn, and a little after a quarter of turn. Note that, if the LEM rotates in the other direction, the astronauts will view the command module the same, for they will see it on their left instead of seeing it on their right, and what they will take in photo will in fact appear the same. This is not at all what we see on the photos of the command module in Apollo 14; the command module has wrong positions on them; it always keeps the same position on the photo. Finally, this couple of photos (AS14-66-9352 & AS14-66-9356) is wrong:  Why? Because we can see the command module get closer to the LEM before the LEM has completed its half turn; when the LEM sees the command module this way, it means that it has currently made a quarter of turn only, and not the complete half turn. The command module should wait for the LEM to complete its half turn, before getting closer to it.  The way the command module gets closer to the LEM on the Apollo photos, it means that it would try to join it the wrong way, on a side of the LEM which cannot communicate with the command module. So the sequence of photos of the command module in Apollo 14 is definitively wrong. In Apollo 15, we have another sequence of photos of the command module getting close to the LEM on the return back from the moon of the latter:  On this sequence, we see the command module getting progressively closer to the LEM...and suddenly, instead of docking to the command module, the LEM goes over the command module and starts flying over it, thus wasting the fuel it might be short of when attempting to dock to the command module!  When the LM is close to the CM, it should be horizontal; the only moment it makes a rotation is when it makes its flipover maneuver (to present its nose to the CM), but this rotation is made around the horizontal axis. On these photos of Apollo 16, the LM (taken from the CM) is in vertical position and making a rotation around the vertical axis! ONCE AGAIN, DOES IT MAKE SENSE, SERIOUSLY, DOES IT MAKE SENSE???  To conclude, these sequences of photos are a real laugh. When I see these photos, I have a great difficulty to take these missions seriously. |
 The deorbit maneuver, which was making the descent stage crash on the moon, was made by first orienting the lunar module in the direction of the descent trajectory, by using the RCS (lateral thrusters), and then pushing with its main engine along this direction to decrease the lunar module's speed so it would start to take this descent trajectory; after the lunar module has cut off its engine, the lunar module must continue to follow the trajectory it has followed during the deorbit maneuver. It means that the direction of the trajectory between the cutoff of the engine and the impact of the descent stage should be the same as the one between the ignition of the engine and its cutoff. According to the mission report of Apollo 17, the latitude of the engine's ignition is 0.24° South, and the latitude of the engine's cutoff is 2.02° North; the difference is 2.26° (the latitudes must be added, and not subtracted, for one is North, and the other one is South). The longitude of the engine's ignition is 86.97° East and the longitude of the engine's cutoff is 81.68° East, and the difference of longitude is 5.29° (this time, they must effectively be subtracted, for they are both East). The ratio between the difference of latitude and the difference of Longitude is 2.26/5.29=0.427; it corresponds to an angle of 23.13° relatively to the orbital plane of the moon (obtained by the arc tangent of this ratio). We should normally find the same ratio between the difference of latitude between the engine's cutoff and the impact point and the difference of longitude between the engine's cutoff and the impact point, if the lunar module has continued to follow the descent trajectory.  The mission report of Apollo 17 also gives the coordinates of the impact point of the descent stage. Its coordinates are 19°, 57 minutes, and 58 secondes latitude north, and 30°, 29 minutes, and 23 seconds longitude East. This corresponds to a latitude of 19.96611° North and 30.48972° East (the fractional part is obtained by multiplying the minutes by 60, adding the seconds, and dividing the result by 3600). The latitude difference between the impact point and the engine's cutoff of the deorbit maneuver is 19.96611-2.02=17.94611°, and the longitude difference between the impact point and the engine's cutoff of the deorbit maneuver is 81.68-30.48972=51.19028°. The ratio between this difference of latitude and this difference of longitude is 17.94611/51.19028=0.35058; so a ratio quite different from the previous one, and which corresponds to an angle of 19.32° relatively to the equatorial plane of the moon.  So, it means that, after the engine's cutoff of the deorbit maneuver, instead of continuing to follow the descent trajectory which has been determined by the deorbit maneuver (green line), the lunar module would have made a change direction of around 4° (red line), without a new maneuver having been made with the engines, which is physically impossible! If the lunar module had continued to follow the normal descent trajectory, like it should have, it would have crashed farther from the landing site, more in the north. And there is another surprise. The mission report of Apollo 17 also states that the impact point of the descent stage was at around 9.9 kilometers from the landing site; if they give a decimal digit, it means that the distance is precise at 100 meters at least.  But we also know the coordinates of the landing site of Apollo 17.  And, if we input the coordinates of the impact point and the ones of the landing site of Apollo 17 in the lunar distance calculator, we find a distance of 10.54 kilometers instead; 640 meters difference with the distance indicated in the mission report, while this difference should not logically have exceeded 100 meters! |