Bob B.
Bob the Excel Guru?
Posts: 3,072

Post by Bob B. on Sept 4, 2011 18:23:08 GMT 4
Thanks, Bob. Perhaps I knew a little more about it than I thought, because what he said didn't add up. I had always heard of "radial" as you describe it. In some applications, “radial velocity” means “lineofsight velocity” but that’s not the case here. Without considerable more research, I don’t know. I do know, however, that Patrick1000’s quote is wrong because he’s treating radial velocity as if it’s crossrange velocity by saying it would cause the LM to drift south. An error in radial velocity would cause the LM to land either long or short of the target. Unfortunately I haven’t the motivation or energy to deal with that guy anymore.



Post by Nowhere Man on Sept 4, 2011 22:55:06 GMT 4
Doesn't matter, he wouldn't listen. He's off on a "the LRRRs were placed so we could more accurately target our ICBMs at the Russkis" tangent. A complete waste of electrons.
Fred



Post by coelacanth on Sept 6, 2011 3:27:05 GMT 4
The following should give you what you want regarding weights of various components at various times: history.nasa.gov/SP4029/Apollo_1837_Selected_Mission_Weights.htmAs you probably know, the CM/LM was pressurized to just 5 psi. I don't know the exact size of the docking tunnel, but it was about 30 inches diameter. This means the force pushing on the LM was about 3500 lbf, which was enough to give it an initial acceleration of about 3.3 ft/s ^{2}. Of course, as soon as the vehicles undocked, that pressure was very quickly relieved; therefore, the duration the LM was subjected to any kind of a push was only a fraction of a second. I doubt the LM received even 1 ft/s deltav as a result of the tunnel pressure. At the time of CSMLM separation, the CSM's mass was only about a ton more than the LM, therefore it received just slightly less deltav in the opposite direction. Hi there, I am very much out of my element when talking about the dynamics of compressed gasses, so I hope these aren't foolish questions. But I wonder a) How can we determine the amount of energy required to compress a given mass of some particular gas from an extremely diffuse state to a particular density? Is this a simple thing to determine, or is it complicated? b) If the energy used to compress this gas were all imparted to the two undocking craft, then I could figure out how much velocity would be imparted to each because both energy and conservation of linear momentum must be conserved. (Working from the inertial frame the two docked craft were in before the undocking.) However, maybe not all the energy goes to the two craft  some could be imparted to the gas itself, and maybe the gas is hot, even after it diffuses. But would we be able to get a useful upper bound on the velocity imparted to the two craft this way?


Bob B.
Bob the Excel Guru?
Posts: 3,072

Post by Bob B. on Sept 6, 2011 10:29:18 GMT 4
a) How can we determine the amount of energy required to compress a given mass of some particular gas from an extremely diffuse state to a particular density? Is this a simple thing to determine, or is it complicated? It's easy if you have the right computer program. I took oxygen at 0.34 atmosphere (5 PSI) and a temperature of 22 ^{o} C and calculated its enthalpy, i.e. its internal energy, and got a value of 2,738 J/kg. If we now allow the gas to expand, I can calculated its new enthalpy. However, as it expands, it escapes the confinement of the tunnel and dissipates into space. The question is, how much can it expand and still do work? At what point do we say the gas is no longer transferring energy to the LM and CSM? Do we let it expand 2X, 3X, 10X? Just for an example, I said let's expand the gas to three times its original confined volume. I have no idea if that's right, but I've got to use something. After expanding its volume 3X, the enthalpy of the original gas is now 90,622 J/kg. That's a change of 87,884 J/kg. So, how much oxygen is trapped in the tunnel? At 0.34 atm and 22 ^{o} C, the density of oxygen is about 0.45 kg/m ^{3}. The volume of gas we're concerned with is just that between the two hatches. Off hand, I don't know what that is, but it can't be much more than a cubic meter. (Roughly, the tunnel's volume was about 0.5 m ^{3} per meter of length.) If we have 0.45 kg of gas and an energy release of 87,884 J/kg, then we have 39,548 J of energy. FYI, most of the energy is released in that first 3X expansion. For example, allowing the gas to expand all the way up to 10X releases only 4,450 J more. Interesting concept. I'll let you work out the next part of the problem. Perhaps you can also try to verify the distance between the hatches to come up with a better estimate of the oxygen mass.



Post by coelacanth on Sept 6, 2011 20:21:23 GMT 4
It's easy if you have the right computer program. I see! Well I guess that is the case for a lot of things If we now allow the gas to expand, I can calculated its new enthalpy. However, as it expands, it escapes the confinement of the tunnel and dissipates into space. The question is, how much can it expand and still do work? At what point do we say the gas is no longer transferring energy to the LM and CSM? Do we let it expand 2X, 3X, 10X? I agree, that is the big question mark I see. That energy which does not go to moving the LM and CSM, where does it go? It could go into the kinetic energy of the gas itself (maybe the gas comes hissing out and starts going sideways at quite a high velocity?), or the gas could retain some of the heat from its compression. (Does it? I don't know.) And maybe there are some other destinations for this energy that are escaping me at this moment. Interesting concept. I'll let you work out the next part of the problem. Fair enough I'll get to it at some point this week, I think. Perhaps you can also try to verify the distance between the hatches to come up with a better estimate of the oxygen mass. I don't know where to look for that information, but I'll start with Google



Post by ka9q on Sept 6, 2011 21:14:57 GMT 4
Seems to me that relatively little of the compressed gas energy would push the two spacecraft apart, as they only have to move a little before the gas is directly vented to space. It's not like there was a big cylinder and piston that allowed the gas to expand to many times its original volume while doing work.


Bob B.
Bob the Excel Guru?
Posts: 3,072

Post by Bob B. on Sept 6, 2011 23:13:21 GMT 4
That energy which does not go to moving the LM and CSM, where does it go? It could go into the kinetic energy of the gas itself (maybe the gas comes hissing out and starts going sideways at quite a high velocity?), or the gas could retain some of the heat from its compression. (Does it? I don't know.) And maybe there are some other destinations for this energy that are escaping me at this moment. The gas will retain some heat, but most of the energy will transform into kinetic energy, thereby increasing the velocity of the gas itself. I've looked at a few illustrations and it appears the distance between the two closed hatches can't be more than one meter.


Bob B.
Bob the Excel Guru?
Posts: 3,072

Post by Bob B. on Sept 6, 2011 23:20:09 GMT 4
Seems to me that relatively little of the compressed gas energy would push the two spacecraft apart, as they only have to move a little before the gas is directly vented to space. It's not like there was a big cylinder and piston that allowed the gas to expand to many times its original volume while doing work. I tend to agree. But as coelacanth said, this process might be adequate to give us an upper bound on the deltav.



Post by coelacanth on Sept 7, 2011 5:24:10 GMT 4
Seems to me that relatively little of the compressed gas energy would push the two spacecraft apart, as they only have to move a little before the gas is directly vented to space. It's not like there was a big cylinder and piston that allowed the gas to expand to many times its original volume while doing work. I tend to agree. But as coelacanth said, this process might be adequate to give us an upper bound on the deltav. I'm somewhat reluctant to mention this in public, but if the reason for this is to debunk somebody's rubbish argument, and the upper bound turns out to be really big, then that might not be so helpful . . .



Post by chew on Sept 7, 2011 8:37:04 GMT 4
I tend to agree. But as coelacanth said, this process might be adequate to give us an upper bound on the deltav. I'm somewhat reluctant to mention this in public, but if the reason for this is to debunk somebody's rubbish argument, and the upper bound turns out to be really big, then that might not be so helpful . . . It's not to debunk somebody; it's just a physics exercise. Besides, we go were the science leads us even if it supports the Moon hoaxers.


Bob B.
Bob the Excel Guru?
Posts: 3,072

Post by Bob B. on Sept 7, 2011 8:41:29 GMT 4
I'm somewhat reluctant to mention this in public, but if the reason for this is to debunk somebody's rubbish argument, and the upper bound turns out to be really big, then that might not be so helpful . . . I don't think it's going to be really big. I previously estimated about 40,000 J of energy, but that was based on one cubic meter of gas. After looking at some drawings, I think it's more likely half that amount. So we're now at 20,000 J, and that is roughly split 50/50 between the LM and CSM. So we have about 10,000 J added to a 15,000 kg LM. That's not going to impart a lot of velocity.



Post by chew on Sept 7, 2011 12:10:11 GMT 4
So this problem is like the opposite of the Mythbusters testing the myth that 2 cars colliding at 50 mph is the same as 1 car colliding at 100 mph into an unmovable object?
Why the concern the undocking impulse would affect the accuracy of the landing point? Aren't the state vectors updated with the accelerations measured with the guidance system?


Bob B.
Bob the Excel Guru?
Posts: 3,072

Post by Bob B. on Sept 7, 2011 15:18:52 GMT 4
Why the concern the undocking impulse would affect the accuracy of the landing point? I’m not concerned about it, but since it was brought up, I’m looking at it as an interesting problem to solve. In fact, I just came up with an answer by treating the problem in the same way as my many simulations. I figured the tunnel between the closed hatches to be 762 mm (30 inches) diameter and one meter in length. I pressurized this space to 34,474 Pa (5 psi) with oxygen at 295 K, which produced a force of 15,721 N pushing against each vehicle. I used Apollo 11 as the example, so the LM mass was 15,279 kg (33,683.5 lbm) and the CSM mass 16,818 kg (37,076.8 lbm). I then released the latches and allowed the vehicles to push apart, analyzing the sequence in one millisecond intervals. As soon as the vehicles start to move apart, a small gap opens up that allows the oxygen to vent to space. Unfortunately I’m not an expert on gas dynamics, but I do know that when a gas flows through an orifice, the flow will become “choked” when the gas velocity reaches the speed of sound. That is, the gas can flow through the orifice no faster than the local acoustic velocity. (This is what happens at the throat of a rocket engine.) I presumed this would also be the case at the narrow annular space between the two vehicles, therefore I figured the gas was venting at the speed of sound. Calculating the acoustic velocity, the gas density, and the area of the annular space, I could determine the flow rate of the gas through the gap. I could then determine the amount of gas remaining in the tunnel, revise the gas pressure, and recalculate the magnitude of the pushing force. I also took into account the temperature drop of the gas as the pressure was released. This process was done over and over in very small time increments. The pushing force drops to below 1 N just 319 milliseconds after the vehicles start to separate; however, the vast majority of the work is completed in half that time. 99% of the oxygen (by mass) is evacuated in 178 milliseconds. When all is said and done, the deltav of the LM is 0.1064 m/s (0.349 ft/s) and that of the CSM is 0.0966 m/s (0.317 ft/s). After 1/2 second the vehicles are 89 mm apart (3.5 inches) and separating at a relative velocity of 0.2030 m/s (0.666 ft/s). If anybody sees any fault in this analysis, please let me know.


Bob B.
Bob the Excel Guru?
Posts: 3,072

Post by Bob B. on Sept 7, 2011 22:41:45 GMT 4
I just discovered a mistake in my analysis. To calculate the gas flow rate through the small gap between the vehicles, I used the pressure and temperature of the gas inside the tunnel. However, as the gas rushes out the opening, its pressure and temperature drops. I should have used the P&T at the exit, not that inside the tunnel. This results in a lower mass flow rate, thus more of the gas is retained inside the tunnel for a longer period of time. This means the pushing force drops off more slowing, resulting in a greater velocity.
Below is a reproduction of the last part of my previous post with the corrected numbers. I feel reasonably confident in this new analysis, but I won't swear by it. Thermodynamics is definitely not my strong suit.
(EDIT)
It looks like I've got to edit the results once again. I haven't found anything else wrong with my method per se, however the computer program I was using to derive the gas temperature was yielding unreliable results for temperatures below 200 K (I was working outside the temperature range the program was intended to be used for). As a result, I had to revise my temperature formula (I'm treating it as an isentropic process). Below are the new results; I hope I have it right this time:

