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Post by pleasedebunkme on Jul 18, 2011 22:04:28 GMT -4
Could someone please tell me why these two theories are wrong:
Antipode Re-Entry Corridor: This theory states that the Apollo re-entry into the Earth’s atmosphere on the far side of the Earth was not possible because the command module was traveling at parabolic escape velocity, and at that high rate of speed the command module would have swung out into parabolic orbit and not have landed at the antipode. This theory relies upon the fact that when the command module came back to Earth it was traveling faster than when it left. Thus, it was allegedly not possible to complete a half orbit around the Earth, because while traveling at escape velocity the command module would have escaped.
Pressure Differential: This theory states that the Apollo command module did not have the structural integrity to handle the extreme pressure differences when it entered the Earth’s atmosphere. According to this theory, the “lifting characteristics” of the command module as it plowed through the upper atmosphere would have blown it apart.
Thank you.
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Post by JayUtah on Jul 18, 2011 22:45:31 GMT -4
...and at that high rate of speed the command module would have swung out into parabolic orbit and not have landed at the antipode....if it had not encountered the atmosphere. That's what slows it down. The velocity is converted to heat. If there had been no atmosphere at perigee, the CM would have not been captured. Obviously perigee is where the entry interface occurs, because that's where the air is. Coincidentally that's the fastest part of the orbit. This theory states that the Apollo command module did not have the structural integrity...For that you need (a) a derivation of the CM's relevant strength, and (b) a derivation of the strength required, i.e., the load. Until your argument is quantitative and quantified, it's not complete.
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Bob B.
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Post by Bob B. on Jul 18, 2011 22:52:55 GMT -4
First off, a parabolic orbit is a very special case in which the velocity is exactly the escape velocity. Any slower than this and the orbit is elliptical, and any faster and it is hyperbolic. The Apollo spacecraft returned at speeds slower than escape velocity, though not by very much, therefore the trajectories were elliptical. The perigee of the elliptical orbit was on the side of Earth approximately opposite of were the moon was at trans-earth injection. Reentry took place near perigee, though this is somewhat subjective depending on at what elevation you define the atmosphere to begin. NASA defined entry interface as an altitude of 400,000 feet. Since perigee was lower than this, the spacecraft reach entry interface before reaching perigee.
The CM did have the structural integrity, so end of story.
(EDIT)
OK, I admit it, that second part was a little dismissive. But Jay is correct, without a quantitative argument there isn’t much we can comment on.
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Post by scooter on Jul 18, 2011 23:29:42 GMT -4
If I understand the CM design, I believe it's shape made it something of a "lifting body". Add to this was an off center CG, and apparently by slight rotation in the roll axis, they could tweak their descent rate, allowing them to do their "double dip" type reentry.
Did I get this at all right?
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Bob B.
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Post by Bob B. on Jul 18, 2011 23:39:32 GMT -4
Just to prove my point that the Apollo transearth trajectories were elliptical orbits… V esc = SQRT[2*GM/R] At entry interface, R = 6378140+(400000*0.3048) = 6,500,060 m Therefore, V esc = SQRT[2*3.986005E+14/6500060] = 11,074.5 m/s = 36,334 ft/s If you look at the following table, history.nasa.gov/SP-4029/Apollo_18-40_Entry_Splashdown_and_Recovery.htmYou’ll see that all the returning Apollo spacecraft had velocities below 36,334 ft/s at Earth entry. |
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Bob B.
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Post by Bob B. on Jul 18, 2011 23:53:26 GMT -4
If I understand the CM design, I believe it's shape made it something of a "lifting body". Add to this was an off center CG, and apparently by slight rotation in the roll axis, they could tweak their descent rate, allowing them to do their "double dip" type reentry. Did I get this at all right? That sounds about right. The Apollo CM had a lift-to-drag ratio of about 0.3, so the capsule could generate lift to shallow the descent and decrease the g's and heating. Trivia... In the movie Apollo 13 we see ballast being transferred from the LM to the CM to "get the weight right" because the "trajectory is still shallowing." The real reason to transfer ballast was to get the center of gravity right so the CM could generate enough lift during reentry. |
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Post by JayUtah on Jul 19, 2011 2:27:48 GMT -4
If the entry angle is on the steep end of the window (a wedge, really), the CM does a double-dip flight path in which it briefly gains altitude to allow the heat shield time to cool. On the shallow end of the window the altitude monotonically decreases although the CM is developing lift. By rolling the CM left and right, course-change moments could be generated.
Bob, I don't think you were too abrupt. But at the same time I don't mean any rudeness to the original poster. Arguments of the "not enough" variety are inherently quantitative. Without the associated quantities, there is simply no argument.
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Post by chrlz on Jul 19, 2011 7:20:20 GMT -4
Could someone please tell me why these two theories are wrong: Why do you think we should? Are these *your* claims? If so, why have you not fleshed them out properly? The way they are worded suggests barely a passing knowledge of the topics. If they are the claims of others, why do you not cite where you encountered them? In science and engineering (which is sorta what this forum is about), a 'theory' is a possible, plausible explanation of some observation/phenomena, that has been developed using a proper methodology, eg the proverbial scientific method.There's also this thing called burden of proof, and the burden, I'm afraid, is yours. What you have posted there, doesn't exactly cut it... |
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Bob B.
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Post by Bob B. on Jul 19, 2011 10:21:47 GMT -4
The perigee of the elliptical orbit was on the side of Earth approximately opposite of were the moon was at trans-earth injection. I want to take a moment to explain the use of the word “approximately” in the above quote. The antipode of the perigee point is the apogee point, but the Moon was not located at the apogee – the apogee was located many thousands of kilometers beyond the orbit of the Moon. The elliptical transearth orbit and the Moon’s orbit crossed at just one point, which was a point located past apogee on the Earth-inbound half of the ellipse. We can describe where in an orbit a spacecraft is by its true anomaly. True anomaly is the earth-centric angular distance of a point in an orbit past the point of perigee. The true anomaly of the perigee is 0 degrees (or 360 degrees) and the true anomaly of the apogee is 180 degrees. All the Apollo orbits were similar, but I have studied Apollo 11 specifically in great detail. The true anomaly of Apollo 11 at TEI was about 191 degrees, and the true anomaly at Earth entry (entry interface) was about 347 degrees. Therefore, the spacecraft traveled through approximately 156 degrees of true anomaly from TEI to Earth entry. When people say the spacecraft reentered at the antipode from where it started, they are making a simplification to convey to laymen in easy to visualize terms generally what happened. When working with real orbital mechanics, however, simplifications aren’t good enough. |
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Post by pleasedebunkme on Jul 19, 2011 11:43:28 GMT -4
The burden of proof is always on the proponent of the initial theory (that humans landed on the moon and made it back safely).
The velocity of 36,194 fps is close enough to parabolic escape velocity to produce a large orbit. And, such velocity is greater than the velocity Apollo 11 used to break out of its orbit for the translunar injection. Whether that orbit might be an escape orbit or a return orbit does not change my question.
With a target angle of re-entry at minus 6.50 degrees at 400,000 feet, the perigee had to be below the 400,000 feet height cited by NASA as the point of re-entry. And, with such a flat re-entry angle, it would seem that the orbital component of the return trip had to be more of a sweeping arc, rather than a straight return. NASA says that re-entry was at the antipode. So, in order to get to the antipode, with a re-entry angle of minus 6.50 degrees, there had to be a large orbital component to the return trip.
Your responses above present two more obvious questions: First, with the orbital component in the return trip, how was it possible for the command module to accelerate its velocity up to 36,194 fps? Second, with this orbital component, and with the added distance, how was the command module able to return back to Earth as quickly as it did?
So, the way I see it, if the command module had a relatively straight return to Earth, at that velocity, and then swung around to the far side, it would have been flung out into some type of large orbit. But, if the command module had a large sweeping orbital arc to its return trip, then it never would have accelerated its velocity up to being (almost) parabolic escape velocity.
About the antipode, when NASA specifically says “antipode” then we must assume that NASA meant the antipode. You do not get to rewrite the official story forty-two years later and make the claim that the antipode really was not the antipode. And, that is my point: You cannot get to the antipode, like NASA said it did, using the return trip parameters that NASA has claimed. The response above seems to agree with this point.
With respect to the structural integrity of the command module, I have been unable to discover any data on its torsional strength. The only thing I can find would be a statement from Michael Collins that the command module could not withstand a pressure difference greater than 8 psi. It would seem to me that the leading edge of the command module would have had tremendous pressure that vastly exceeded 8 psi.
I am asking for your help, not your snide comments. It was my understanding that: “The goal of the website is to use factual information to counter the claims that the Apollo moon landings were faked.” I am asking for that factual information. I even said “Please.”
Thank you for your consideration.
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Bob B.
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Post by Bob B. on Jul 19, 2011 11:59:34 GMT -4
About the antipode, when NASA specifically says “antipode” then we must assume that NASA meant the antipode. Why must we assume that? I don't recall ever reading anything from NASA saying the reentry point was at the anitpode unless it was just some very basic and simplistic description. NASA has provided the orbital data and it clearly tells us where the reentry points were and they were not at the anitpode. As I said previously, the entry point for Apollo 11 was 13 degrees before perigee, per NASA's own data (the other missions were similar). You're splitting hairs over the use of a single term rather than looking at the real data. The data has to govern. You do not get to rewrite the official story forty-two years later and make the claim that the antipode really was not the antipode. I’m not rewriting history, the data I’m using is that released at the time of Apollo. |
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Post by gillianren on Jul 19, 2011 12:03:08 GMT -4
The burden of proof is always on the proponent of the initial theory (that humans landed on the moon and made it back safely). Actually, that's not quite true. In this case, the burden of proof has been met and it is those who claim that the landing was impossible who have the burden of proof in this case. Besides, saying something is impossible is an original claim with its own burden. |
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Bob B.
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Post by Bob B. on Jul 19, 2011 12:06:36 GMT -4
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Post by pleasedebunkme on Jul 19, 2011 12:16:28 GMT -4
From the Apollo 11 press kit, page 64:
The latitude of splashdown depends upon the time of the trans- earth injection burn and thedeclination of the Moon at the time of the burn. A spacecraft returning from a lunar mission w i l l enter the Earth's atmosphere and splash down at a point on the Earth's farside directly opposite the Moon. This point, called the antipode, is a projection of a line from the center of the Moon through the center of the Earth to the surface opposite the Moon. The mid-Pacific recovery line rotates through the antipode once each 24 hours, and the transearth injection burn w i l l be targeted for splashdown along the primary recovery line.
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Post by JayUtah on Jul 19, 2011 12:36:53 GMT -4
The burden of proof is always on the proponent of the initial theory (that humans landed on the moon and made it back safely).Not necessarily. If someone else proposes a different hypothesis to account for the same observations as the initial proposition, then that new proponent has the burden of proof. That's what's known as an affirmative rebuttal. It affirms something different, rather than simply attacking or disputing the original claim. You can't affirm something without incurring a burden of proof. In this case, the burden to prove Apollo authentic has been carried to the satisfaction of all but a few noisy and ill-informed critics. The believers in Apollo include all the relevant scientists, engineers, and credible historians. That doesn't mean it's above disputation, but it does mean that the disputation has to consider all the relevant facts, including those that may require considerable expertise. "I don't understand how it could have worked" is not a valid disputation from a layman. NASA says that re-entry was at the antipode.Where exactly does NASA say this? How is "re-entry" defined in that context? I know at least three commonly used definitions of an entry interface. One is where the spacecraft reaches a certain point along its planned orbit. Another is where the spacecraft descends to a particular geodetic altitude. And a third is where the aerodynamic deceleration reaches a certain value. Entry is parameterized differently for different purposes. With respect to the structural integrity of the command module, I have been unable to discover any data on its torsional strength.Do you know what torsion is? How does torsion (as opposed to tension, compression, shear, and bending) apply in the structural design of the command module? Why is it relevant to this question? The only thing I can find would be a statement from Michael Collins that the command module could not withstand a pressure difference greater than 8 psi.Do you understand the difference between static contained pressure and aerodynamic pressure? Do you understand the difference between the pressure vessel and the aerodynamic outer shell? It's pretty clear from your additional information that the claim in question really doesn't know what it's talking about. It confuses two completely different regimes of gas behavior, and two completely different structural components of the spacecraft. Unless you're prepared to delve deeply into some pretty complicated numbers, it's safe to say that question simply doesn't make sense. I am asking for your help, not your snide comments.Nearly all the people who come here asking questions eventually end up trying to propose a hoax theory. You need to be aware that the "I'm just asking questions" approach is the most common way to start a hoax debate. We're asking you to cut to the chase, if there's going to be one. I am asking for that factual information.You're doing more than that. You're asking to have two specific claims addressed. And we're asking whether you're making a hoax claim. That's the other part of the site's charter.
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