Apollo 13's Slickest Trick, "The Entrance"

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posted on Jul, 3 2012 @ 02:44 AM
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reply to post by JohnPhoenix
 


I agree JohnPhoenix. I believe at least you are siding with me on this most excellent point. The whole thing is beyond fascinating.

My physicist friend has called me 4 times tonight asking me questions about the Apollo ships. He knows nothing of them and is sort of struggling. He has lots of smarts, but little background as regards Apollo nuts and bolts, Apollo logistics.




posted on Jul, 3 2012 @ 03:13 AM
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reply to post by decisively
 


not by much at all.. to jettison the SM they had planned a half foot per second in the positive x-axis, jettison the SM than half a foot per second negative x-axis. that was planned to give good enough seperation before re-entry and time to take photos.

re-entry was about the 142hr 20min mark.. LM jettison was at around 141hrs 20min and SM jettison at around 138hrs 2mins..

half a foot per second over one hour will give seperation of about 1800 feet, SM was planned to be jettisoned atleast 1ft/second difference, atleast 3hrs earlier than the LM and may already be re-entering earth..
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posted on Jul, 3 2012 @ 03:32 AM
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Something qualitative, until we do the numbers for ourselves




The final safe entry conditions;

www.google.com... aHU2AWLkuC6CA&usg=AFQjCNE49Uz5_HfSCI6VaHgPlS3GvKSmxQ&sig2=MKlMqeQpK3xQt3kYNFnt2w

From the above;


The final safe entry conditions normally were achieved at the 400,000 feet altitude above Earth's sea level. At this point, the inertial velocity was just under 36,100 fps with a flight path angle of about 6.6 degrees below the local horizontal. A typical azimuth at this stage was 52.21 degrees. Gimbal angles were usually yaw of 0 degrees, roll of 0 degrees and a pitch of about 152 degrees.


We all can work these numbers out later, crunch them and see what it would take to bust this entry in various and sundry ways.

For openers, it would be worthwhile to see what it would take(not much obviously) to alter the angle of attack/atmospheric penetration by enough to botch the entry. It clearly would take very little to push that number (roughly 6 degrees) to less than 5, and so bounce, and it would take equally as little to alter an Apollo capsule's trajectory such that an unacceptably acute approach resulted with an angle of penetration greater than 7, this, resulting in a burn.

There is little leeway, and so if the LM separation is active, it does make it very hard to account for how Apollo 13 could have splashed so very accurately.

I would imagine several of us will have at least some of the relevant numbers run by sometime tomorrow.



posted on Jul, 3 2012 @ 03:40 AM
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reply to post by decisively
 


Last and most important point before tonight's closing for me, we should be looking at Duke's, Young's, Bostick's and the comments of others. Clearly they are concerned about several aspects of this "capsule approach" and entry. They have special concerns given their dealing with this "stack" of things, and special concerns given the need to separate them. This, per Duke. Quite frankly, I am not sure at all what it is they may be most concerned about. Best.............
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posted on Jul, 3 2012 @ 09:12 AM
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This is crazy. Their inertial nav system would have given them all the data they needed for the burn. Not to mention ground based tracking.



posted on Jul, 3 2012 @ 09:17 AM
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reply to post by decisively
 


the change in angle will not be dramatic, the jettison of the LM will have been taken into account in the trajectory calculations to ensure it stays within the tolerated approach angle.

collision with the SM is not an issue as that was jettisoned 3hrs prior to LM jettison.



posted on Jul, 3 2012 @ 10:14 AM
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Originally posted by decisively
If the objects separate, then by definition, the linear momentum has changed, and with it, so has the angle of attack.

No, it hasn't. Remember, acceleration due to gravity is the same for all object, regardless of mass. So the trajectory of the CM is going to be the same whether the LM and SM are attached or not, until some other force starts acting on the craft, like atmospheric resistance.



posted on Jul, 3 2012 @ 11:31 AM
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Why is it so far fetched that they JUST GOT LUCKY, why does there have to be proof. They lived, that's it. If you want to think this is a fraud no matter what is proposed it will not sway your opinion. I liken you to the 300lb girl at the beach wearing spandex, and she thinks she looks hot regardless of the mirror or the laughter.

You need to put your effort into a more worthy cause.



posted on Jul, 4 2012 @ 01:11 AM
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reply to post by mrwiffler
 


After the "PC + 2 burn " , onboard guidance is shut down



No big mrwiffler, but you need to check your facts. After the "PC + 2 burn", they shut everything off. All of the attitude alignments and so forth, from "PC + 2" on, are done "manually".
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posted on Jul, 4 2012 @ 01:29 AM
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Not Far Fetched

reply to post by DOLCOTT
 



This thread is most certainly NOT "far fetched" oriented DOLCOTT, though your point is in a general sense a good one, and if "far fetched " be the conclusion that is drawn, so be it. If it was luck, then it was luck. However, consider the following;

The capsule, the CM lands/splashes super close to the recovering carrier. Someone needs to double check me on this, but off the top of my head, Apollo 13 splashed closer to its "target" coordinates than any other capsule, Not 8, 10, 11, 12, 14,15,16,17 splashed closer. So that doesn't seem lucky. Even if it did not splash the closest, the point still stands, and stands well, as it splashed super close all considered.

This does seem weird. How can one be minutes late and hit the bullseye ? If you are minutes late, then you have brought your capsule in too shallow. And if you come in too shallow, you'll consequently, come in long, overshoot your splash site.

As I emphasized early on however, the hope for this thread is that it becomes a place to play with physics, a thread in which one can have fun playing with the law of conservation of momentum.

I do mostly CT stuff. When it does comes to straight forum posting, I'll do math, philosophy, some biology, literature. I have unfortunately only been doing CT recently, Apollo stuff primarily, and don't get me wrong, I love it, but just wanted to post a thread here on ATS where I could do a bit of physics. I was going to crunch some numbers on this thing today, but just too tired , so maybe tomorrow or the next day.

What people do here is their prerogative based of course, but ideally, I was hoping the thread would be one where we all assume Apollo AUTHENTICITY, and work toward a solution, one that is NOT FAR FETCHED.

Thanks for your post DOCOTT, It is an important, a most excellent as I am fond of saying, point you make.
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posted on Jul, 4 2012 @ 03:28 AM
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By linear momentum, I of course mean the linear momentum of the CM component, and not the linear momentum of the once combined CM/LM system. The CM's velocity, AND SO ITS DIRECTION OF TRAVEL, will have changed. We all agree, to be sure, the CONSERVATION OF MOMENTUM LAW HASN'T ABDICATED ITS DUTIES, NOR HAS IT BEEN TEMPORARILY SUSPENDED BY FIAT.

reply to post by nataylor
 



If two objects separate by way of JETTISONING/CHARGES BLOWING THEM APART, AS INDEED IS THE CASE WITH THE APOLLO CM AND LM, the total momentum of the system remains unchanged, but at the same time, the CM component's momentum is not the same as that of the LM component. They sum to the original momentum of the CM___LM stack, but individually, they are now different, 2 different mass values, two different velocities. The latter , the velocity VECTOR, critically important here.

We have established that in the case of the CM______LM separation, charges are used. As such, ENERGY ( a scalar quantity) ENTERS THE SYSTEM. MOMENTUM ( a vector having magnitude and unlike energy also possessing a SPECIFIC DIRECTION/ a specific spacial orientation) IS UNCHANGED.

The point I made above, is not a point in dispute. It is simply a statement of the conservation of momentum law. If the CM and LM separate, the CM____LM stack has "lost" some of the original CM____LM paired momentum to the LM which now itself moves "independently". "Independently" is placed in quotation marks here to emphasize that the system as a totality of moving components can never become independent in such a way that the law of momentum conservation is violated. Once the reference frame is specified/established, then the total linear and total angular momenta of the system always sum to the same value, no matter how much energy is so imparted by virtue of the jettison charges being blown.

Indeed, this was the whole point of the thread, because the system's total momentum is a constant, the CM's velocity vector ASSUMES AN ENTIRELY DIFFERENT DIRECTION, ONE VERY MUCH OTHER THAN THAT OF THE ORIGINAL CM_____LM "STACK". This, because the CM/LM jettison charges will have blown the LM off on to a new course, one other than the original CM___LM stack. BEING BLOWN OFF, THE LM NOW HAS A VELOCITY/DIRECTION OTHER THAN THE PRESEPARATION CM______LM STACK DID. The CM and LM cannot separate if their velocities and so directions remain unchanged. In order to separate, each follows, must must must follow, as a matter of fact, a distinctly DIFFERENT DIRECTION.

Because post-separation (CM mass X Velocity of CM) + (LM mass X Velocity of LM) must always and rigorously be equal to the pre-seapration (CM_______LM stack mass) X (CM_____LM stack velocity), once those charges blow and separation occurs, the CM's linear momentum must change from that of the pre-separation CM______LM mass/velocity product.

The 2 so Separated objects ALWAYS ALWAYS ALWAYS carry the CM______LM stack's momentum, for ever and ever, and so once separated, The CM and LM have different momenta, different one from the other, and one from the once stacked CM____LM.
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posted on Jul, 4 2012 @ 02:27 PM
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reply to post by decisively
 


The mere act of the CM and LM separating does not alter the trajectory of the CM. The energy of the explosive bolts goes into breaking the bolts, not pushing the craft apart. The RCS on the LM moved the LM away from the CM. The fact that the CM weighs less than the CM and LM combined doesn't change the trajectory. The only force acting on the CM is gravity, and acceleration due to gravity is independent of mass. Look up the formulas for balistic trajectory, you don't need to know the mass of the object.



posted on Jul, 4 2012 @ 04:24 PM
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reply to post by decisively
 



The 2 so Separated objects ALWAYS ALWAYS ALWAYS carry the CM______LM stack's momentum, for ever and ever, and so once separated, The CM and LM have different momenta, different one from the other, and one from the once stacked CM____LM.


Much as I hate to bump this thread, the momentum of the system will remain unchanged; that v*(mass CM) + v*(mass LM) = v*(mass CM + mass LM). In order for the vector component of the system to be changed laterally, an impulse must be applied at right angles to it. If you multiply out the mass of the system by its linear velocity, you will see that an extremely large impulse would be required to cause a noticeable deflection. No point in going into the details; it's just common sense, really. It would be like trying to swat a bullet.

Edit to add: Why Morphy? Capablanca was much the better player.
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posted on Jul, 4 2012 @ 11:42 PM
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Originally posted by JohnPhoenix

Originally posted by OccamsRazor04

Originally posted by decisively
reply to post by OccamsRazor04
 



The point is that Duke does not know what might or might not happen. This is an unknown. That is why Duke and presumably Young as well expresses worry. See the post just above with reference to their going to Kranz and saying, "enough cutting bait, let's fish".


Exactly, an unknown, as in it is unknown it will have an effect. You jump from unknown to it will have an effect, a known. Sorry unknowns stay unknowns until evidence shows they would have an actual, not feared risk. The monster in the closet is an unknown, prove its real.


He doesn't have to prove it's real - it IS an Unknown variable. Virtue of this itself is enough to make the astronauts stop dead in their tracks and get more data before making any kind of maneuver. Otherwise your saying NASA plays fast and loose with men's lives - after they had planned for months and had contingency after contingency in place for everything. I just can't see NASA doing this.. that's way out of the ballpark. It's very unscientific and dangerous.


There is a reason they used fighter pilots, fighter pilots go on their gut. I'm sorry he does have to prove it is real since he is saying this is proof it COULDN'T be real. If there is no effect then there is no effect and it is proof of nothing. He wants US to prove it is real for him, it does not work that way.



posted on Jul, 4 2012 @ 11:43 PM
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Originally posted by decisively

No proof of a change in linear momentum is needed

reply to post by OccamsRazor04
 


If the objects separate, then by definition, the linear momentum has changed, and with it, so has the angle of attack.



I am sorry, because I said so does not cut it. If two equal and opposite forces are both applied then by definition there is no change. I'm not doing your work for you.



posted on Jul, 5 2012 @ 10:34 PM
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reply to post by OccamsRazor04
 


I think we are misunderstanding one another. will try again a bit later. Best.



posted on Jul, 5 2012 @ 10:46 PM
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reply to post by OccamsRazor04
 

From The Peanut Gallery



Oh , I thought of something. Here is another way to say what I want.

When the Command Module/Lunar module combined stack separates, when the lunar module is jettisoned, the command module capsule then continues traveling toward the earth at a velocity of greater magnitude than that of the just separated Command module/Lunar module stack, and even more importantly for our considerations here, that velocity vector will be of a DIFFERENT DIRECTION THAN THAT OF THE JUST SEPARATED COMMAND MODULE/LUNAR MODULE STACK..

One of the things we are interested to know is first of all how much the direction changed. Did it changed a lot ?
If it did change a lot, what would that mean if anything ? From here, from the peanut gallery, it looks like a very interesting issue/problem.....



posted on Jul, 5 2012 @ 11:51 PM
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reply to post by decisively
 


Now this is where you should have started, you started with an outcome supported by zero facts. What you should have done is started with the facts, and then predicted the outcome. Then match that expected outcome to the "known" outcome and see if there are any discepancies. Then show your work and prove those discrepancies and that is where the debate lies. You can not say it would have gone differently than it did. The first step is work has to be done to PROVE what should have happened. Right now there is conjecture, nothing more. Those of us who know it is real, and do not care for the conjecture, aren't interested in the debate. Once you have actual proof, actual maths that can be scrutinized, then you will have a debate.



posted on Jul, 6 2012 @ 05:24 AM
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reply to post by OccamsRazor04
 


No incorrect, I started with NASA's story, that the two entities, the LM and the CM once together separated. After separation, the Apollo CM splashed very close to the recovery carrier. This is a given.This is NOT a CT thread. You seem to want it to be.
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posted on Jul, 6 2012 @ 05:50 AM
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reply to post by decisively
 



When the Command Module/Lunar module combined stack separates, when the lunar module is jettisoned, the command module capsule then continues traveling toward the earth at a velocity of greater magnitude than that of the just separated Command module/Lunar module stack, and even more importantly for our considerations here, that velocity vector will be of a DIFFERENT DIRECTION THAN THAT OF THE JUST SEPARATED COMMAND MODULE/LUNAR MODULE STACK..


In that case you are simply wrong. The velocity, a vector quantity, remains unchanged. Conservation of momentum, remember?





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