vq
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Post by vq on Aug 16, 2009 23:25:33 GMT -4
Yikes, this thread has gone macabre. So to sum up, the LM had to make some sort of orbital trajectory for the rendezvous to occur but the CSM had enough spare dV to reach the LM as long as they made it to orbit. I think the first part (LM has to make orbit) seems to be the consensus, although not based on any in-depth analysis. I'm not sure anyone has researched the second issue, about whether the CSM has enough fuel, has been established, although it does seem like the requirement to change lunar orbits is relatively small given its overall fuel demands. Does anyone have a good source for the quantity of fuel remaining for the SPS at the end of each mission?
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Post by homobibiens on Aug 16, 2009 23:42:27 GMT -4
Does anyone have a good source for the quantity of fuel remaining for the SPS at the end of each mission? I do not. I went Googling to see if I could find out, and one of the links brought me to the Pravda forum. Having survived this truly bizarre encounter, I will take a bit of time off from the search, to prevent accidental exposure to additional brain dead "information"
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Post by Kiwi on Aug 17, 2009 5:04:25 GMT -4
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Bob B.
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Post by Bob B. on Aug 17, 2009 9:42:43 GMT -4
Does anyone have a good source for the quantity of fuel remaining for the SPS at the end of each mission? I've got that information, though I don't currently have access to it. I'll make a post about it later when I get the time.
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vq
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Post by vq on Aug 17, 2009 21:09:34 GMT -4
Does anyone have a good source for the quantity of fuel remaining for the SPS at the end of each mission? Have you checked out Apollo By The Numbers with all it links -- the sticky thread at the top of this section... I had given it a brief look but didn't find anything. Upon closer inspection, most of the information we need is here: history.nasa.gov/SP-4029/Apollo_18-37_Selected_Mission_Weights.htmCalculated below (all masses in lbs): Mission | Initial mass | Final mass | dmass | %remain | A7 | 36419 | 23435 | 12984 | 68.0 | A8 | 63307 | 31768 | 31539 | 22.3 | A9 | 58925 | 24183 | 34742 | 14.4 | A10 | 63560 | 25095 | 38465 | 5.3 | A11 | 63473 | 26657 | 36816 | 9.3 | A12 | 63536 | 25444 | 38092 | 6.2 | A14 | 64388 | 24375 | 40013 | 1.4 | A15 | 66885 | 26323 | 40562 | 0.1 | A16 | 66923 | 27225 | 39698 | 2.2 | A17 | 66893 | 26659 | 40234 | 0.9 |
Initial CM mass is at EOI for A7, initiation of MCC1 for A8, and transposition for docking for A10-17. Final CSM mass is the mass just prior to CSM separation. This makes several simplifications, including ignoring expenditure of all consumables other than SPS fuel and the equipment/moon rock exchange that occurred during the landings. It also assumes that all SM's had 40,600 lbm of fuel (from Wikipedia but probably varied between missions; I know A7 did not launch with this much fuel). It will be interesting to see how this approximation stacks up against the actual numbers when we find them but I am guessing that the trend will be that this chart shows more fuel consumption than actual due to the mass of H2/O2/RCS consumed throughout the mission.
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Bob B.
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Post by Bob B. on Aug 18, 2009 17:06:07 GMT -4
I’ve been investigating one possible scenario to see if a rescue of the LM by the CSM was within the range of possibility. I’ve chosen Apollo 11 for the test. For my metric-minded friends, sorry about the use of nautical miles and feet, but these are the units appearing in the NASA documents. According to the Apollo 11 pre-launch press kit, the LM was to be launched into a 9 x 45 n.mi. (perilune x apolune) orbit with the insertion point being 18 degrees past perigee at an altitude of about 10 n.mi. At insertion, the LM would be 255 n.mi. behind the CSM, with the CSM in a 60 x 60 n.mi. orbit. I’m not sure along what line the 255 n.mi. distance is measured, but it seems to suggest the moon-centric angular separation between the LM and CSM was about 15 degrees. Following insertion, the LM would perform a series of maneuvers to catch up with and dock with the CSM.
The real-time numbers from the mission reports vary slightly from the pre-launch plan, but pre-launch figures are good enough for this “what if” scenario.
The scenario I’m assuming is that the LM was able to reach its initial planned orbit of 9 x 45 n.mi., but due to some undefined failure, was not able to perform the subsequent maneuvers necessary to rendezvous with the CSM. In this scenario, could the CSM become the active vehicle and rendezvous with the LM in a timely manner and still have enough propellant left over to return to Earth?
I set my starting point (T=0) as the moment of LM ascent engine cutoff, i.e. orbit insertion. The starting conditions at T=0 are as follows:
Lunar Module Orbit: 9 x 45 n.mi. True anomaly: 18 degrees Flight path angle: 0.32435 deg. Velocity: 5,530 ft/s Altitude: 9.849 n.mi.
Command/Service Module Orbit: 60 x 60 n.mi. True anomaly: N/A Flight path angle: 0 deg. Velocity: 5,342 ft/s Altitude: 60 n.mi.
The LM trails the CSM by 15 degrees. Separation distance: 258.92 n.mi.
Based on these conditions, I simulated the orbits of the two vehicles. If no further maneuvers are performed, the LM catches up with the CSM in 103 minutes, but passes about 50 n.mi. beneath it. Clearly the CSM needs to lower its perilune so it can drop down and intercept the LM. But not only must it drop to the elevation of the LM, the maneuver must be timed so that the two vehicles reach the same space at the same time.
Through trial and error I found that a 66 ft/s SPS burn at approximately T=70 min. gets the job done. This places the CSM in a 12.3 x 60 n.mi. orbit. Although the CSM initially slows down, as it drops toward perilune, it speeds up. This means it takes longer for the LM to catch up with it. The paths of the two vehicles cross at approximately T=121 min. and at an altitude of 13.8 n.mi. The rendezvous takes place 25.5 degrees west of the LM’s initial insertion point.
Although both vehicles now exist in the same space and time, their velocities do not match. For the CSM to dock with the LM, it must match exactly the LM’s velocity. Not only are the vehicles traveling at different speeds, they are also traveling in different directions – their flight paths cross at about 1.2 degrees. The CSM must perform a 119 ft/s SPS burn to alter its velocity vector. This burn occurs with the engine nozzle pointed downward toward the moon.
After docking and crew transfer, the LM is jettisoned. The CSM is now in a 9 x 45 n.mi. orbit. It may be possible to perform transearth injection from this orbit, but let’s assume we have to return to the initial 60 x 60 n.mi. orbit before TEI can be attempted. This requires two burns, which can be performed in a couple different sequences, with a total delta-v of 122 ft/s.
I don’t know if this is really doable or not. I doubt there has ever been a situation in which the rendezvousing vehicles have approached each other at such a large difference in velocity and requiring such a dramatic maneuver on the part of the active spacecraft. Normally rendezvous are set up many steps in advanced with the vehicles approaching very slowly with only small braking maneuvers required. I’m sure it would be possible to plan a proper rendezvous between CSM and LM, but time is of the essence. The LM has limited consumables, so my scenario is intending to get to the LM as quickly as possible.
Even though other better methods are surely possible, I suspect the delta-v required to maneuver to a properly phased rendezvous is not much different than I calculate for my scenario. And this is really the main issue, i.e. how much delta-v, and hence propellant, is needed to perform a rescue of the LM. Based on my scenario, the total delta-v required to rendezvous and then return to the original orbit is 307 ft/s.
Do we have enough propellant to do this? The mass of the Apollo 11 CSM at the time of docking with the LM was 36,847 lbm. If we take this as our initial mass prior to the start of the rescue maneuvers, we can use Tsiolkovsky’s rocket equation to calculate the propellant required, where the SPS specific impulse is 314 seconds:
Mass of propellant = 36,847 – 36,847 * EXP[ –307 / (314*32.174) ] = 1,103 lbm
According to the mission reports, the mass of SPS propellant remaining at mission end was 4,467 lbm; therefore we used only about 25% of the SPS propellant margin.
Although I have not simulated any other scenario, I doubt the delta-v requirements would be significantly different had the LM reached a lesser orbit, for instance, a 9 x 20 n.mi. orbit instead of the planned 9 x 45 n.mi. As long as the LM reaches some sort of reasonable orbit, it appears the propellant margin may be enough for the CSM to go after and rescue the LM crew. I think the more critical issue is the time required to properly plan and execute the maneuvers. Can the LM be reached before the consumables run out?
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vq
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Post by vq on Aug 18, 2009 21:24:00 GMT -4
Thanks for the analysis! What was the source for the final fuel quantity? I feel like I should have been able to find it, but I haven't seen the actual value. As for the LM duration, it seems like Apollo 13 taught us that resources can be stretched given the right circumstances. I would imagine the ascent stage's endurance could have been extended significantly beyond the design requirement if absolutely necessary (assuming that its endurance was not compromised by whatever failure prevented it from completing the rendezvous as planned).
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Bob B.
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Post by Bob B. on Aug 18, 2009 22:30:47 GMT -4
What was the source for the final fuel quantity? I feel like I should have been able to find it, but I haven't seen the actual value. It comes from NASA's Post Launch Mission Operation Report No. M-932-69-11. Apogee Books has assembled many of the pre-flight reports, post-flight reports, crew debriefings, etc. and published them in their "The NASA Mission Reports" series of books. Each book includes an end of mission consumables summary. Below is the SPS propellant data for each mission: Mission | Launch Load | Prelaunch Planned Remaining | Actual Remaining | Apollo 11 | 40,600 lbm | 4,304 lbm | 4,467 lbm | Apollo 12 | 40,614 lbm | 3,339 lbm | 3,473 lbm | Apollo 14 | 40,796 lbm | 2,202 lbm | 1,977.1 lbm | Apollo 15 | 40,497 lbm | 1,379 lbm | 1,509 lbm | Apollo 16 | 40,544.2 lbm | 1,681.6 lbm | 2,399 lbm | Apollo 17 | 40,539.8 lbm | 1,731.8 lbm | 1,630 lbm |
No data for Apollo 13.
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vq
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Post by vq on Aug 18, 2009 22:50:27 GMT -4
Thanks for the data! I might have to look into that book.
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Post by slang on Aug 19, 2009 7:08:10 GMT -4
Just musing, slightly tangential: would it have been possible to restart the descent engine, after landing? If possible, would the slight push with remaining fuel be significant wrt to highest possible orbit?
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Bob B.
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Post by Bob B. on Aug 19, 2009 14:37:51 GMT -4
I’ve found a far better solution to this rendezvous problem. The new method takes a little longer but can still be completed in 4 hours and requires less delta-V. I’m again using T=0 as LM engine cutoff/orbit insertion. By T=48.6 min. the CSM will be directly over the apolune of the LM. At this point the CSM makes a 58 ft/s burn to lower is perilune to 18 n.mi. One-half orbit and 57.6 minutes later, the CSM makes another burn of 8 ft/s to lower its apolune to 54 n.mi. These maneuvers place the CSM in an 18 x 54 n.mi. orbit with its perilune and apolune phased to those of the LM’s orbit. In other words, the CSM’s orbit is the same shape as the LM’s orbit but exactly 9 n.mi. higher. At the time of the second phasing burn (T=106 min.), the LM trails the CSM by 4.9 degrees and a separation of 152 km. We now have to wait for the LM, which is moving faster in its lower orbit, to catch up with the CSM. By T+185 min. the LM has closed to 42 n.mi. At this time the CSM performs a 13 ft/s burn to begin a 50 minute descent to meet up with the LM (and changing the CSM orbit to 11 x 51.6 n.mi.). The CSM intercepts the LM at T+235 min. and at an altitude of 14.4 n.mi. The interception point is 25.5 degrees west of the LM’s orbit insertion point. The CSM must again make a burn to match the LM’s velocity; however, the velocities of the two vehicles are much closer than the previous example. A modest 15 ft/s burn is required to place the CSM in an orbit matching the LM. Add it up and we have a delta-V of only 94 ft/s. Of course, if the CSM needs to return to its original 60 x 60 n.mi. orbit, we must add the same 122 ft/s as before, bringing our total delta-V to 216 ft/s. There is one more issue to consider. This example assumes the initial orbit of the CSM is 60 x 60 n.mi. circular. In reality, the orbit was not circular – it was 56 x 64 n.mi. This means that a change in velocity direction as well as magnitude is likely needed to correctly phase the CSM orbit to the LM orbit. Unfortunately I don’t know where the CSM perigee was located, so I can’t determine the angle change. I can, however, determine the worse case scenario, which occurs when the initial CSM perigee and LM perigee are 90 degrees apart. Thankfully this adds only 4 ft/s to the delta-V of the phasing burns. Our maximum delta-V, therefore, is 220 ft/s. Previously I calculated the amount of propellant required based on the mass of the CSM at the time of the rendezvous maneuvers. Although correct for that instant, this doesn’t accurately reflect how much of the propellant margin is used. Because we’re burning off some propellant to perform the rendezvous, the CSM is less massive for the subsequent maneuvers of TEI and MCC. This means the latter maneuvers will require less propellant than they would have had we not performed the extra rendezvous. A more accurate determination of the propellant requirement is to use the mass of the CSM after all the other maneuvers are completed. In essence, we are simply tacking on another 220 ft/s to the total delta-V budget of the SPS. The final mass of the CSM just prior to pre-entry separation of the SM was 26,657 lbm, so let’s use this as the initial mass in our propellant calculation: Mass of propellant = 26,657 – 26,657 * EXP[ –220 / (314*32.174) ] = 574 lbm It appears, therefore, that the CSM shouldn’t have too much problem rendezvousing with the LM in the case of an emergency. It also appears the rendezvous can be performed in a relatively short and timely manner. In fact, the four hours it takes in this example is only 1/2 hour longer than the duration of the nominal rendezvous and docking. The problem comes if rendezvous cannot be achieved within this 4-hour window. If the LM passes by the CSM within rendezvousing, it will take days before the LM catches back up with the CSM for a second attempt. This is likely a one shot deal. (edit) Here is what the rendezvous looks like graphically, plotting vehicle altitude vs. time. The yellow diamonds are where the propulsive maneuvers occur.
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Post by Czero 101 on Aug 20, 2009 4:38:11 GMT -4
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Bob B.
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Post by Bob B. on Aug 20, 2009 13:33:01 GMT -4
If you look at Figure 3(b) on page 40 (PDF page 44), you’ll see exactly the method described in my Post #55. The CSM does a Hohmann transfer into a coelliptical orbit and then performs a terminal phase maneuver. The obvious difference between my scenario and the illustration is the timing and location of the rendezvous. Referring to Table II on page 28 (32), we see that the NASA plan takes place over about 5 hours and 2.5 orbits, while my scenario takes around 4 hours and 2 orbits. The difference is apparently due to the initial assumptions. For instance, I placed the LM in a 9 x 45 n.mi. orbit with the CSM initially 15 degrees ahead, while the data in Table II tells us the LM is in a 9.1 x 28.5 n.mi. orbit with the CSM initially 23 degrees ahead. Making up the larger gap will obviously take longer. It would be interesting to see how my calculations match up if I change the initial assumptions.
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Post by gonetoplaid on Aug 20, 2009 20:57:41 GMT -4
Hi Bob B,
I would like to see your results after you change the initial assumptions. This is really neat stuff.
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Bob B.
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Post by Bob B. on Aug 20, 2009 23:02:07 GMT -4
I would like to see your results after you change the initial assumptions. This is really neat stuff. My numbers are close, but my method apparently varies on the terminal phase part of the rendezvous. With a bit more work I could probably match the NASA numbers more closely. Using the data in Table II (page 28) I've constructed the following table. The times have been adjusted to set LM insertion at T=0. Where a delta-V is given, the indicated orbit is that resulting from the propulsive maneuver. Maneuver | Vehicle | Time (h:m:s) | Delta-V (ft/s) | Orbit (n.mi.) | Insertion | LM | 0:00:00 | ----- | 9.1 x 28.5 | Insertion | CSM | 0:00:00 | ----- | 58.1 x 58.3 | CSI | CSM | 0:51:00 | 54.1 | 18.8 x 58.3 | CDH | CSM | 1:48:38 | 27.6 | 18.8 x 38.2 | TPI | CSM | 4:10:22 | 16.8 | 13.3 x 33.0 | TPF | CSM | 4:50:41 | 22.6 | 9.1 x 28.5 |
CSI = coelliptic sequence initiation CDH = coelliptic differential height TPI = terminal phase initiation TPF = terminal phase finalization Below is the data from my simulation: Maneuver | Vehicle | Time (h:m:s) | Delta-V (ft/s) | Orbit (n.mi.) | Insertion | LM | 0:00:00 | ----- | 9.1 x 28.5 | Insertion | CSM | 0:00:00 | ----- | 58.1 x 58.3 | CSI | CSM | 0:51:39 | 54.1 | 18.8 x 58.3 | CDH | CSM | 1:49:12 | 27.5 | 18.8 x 38.2 | TPI | CSM | 3:59:00 | 13.6 | 15.9 x 31.4 | TPF | CSM | 4:54:25 | 13.9 | 9.1 x 28.5 |
The NASA method waits longer for TPI and then uses a relatively large vertical delta-v component (I used almost no vertical component). I think the NASA method is as it is because of the way the active spacecraft must sight on the target. I don't fully understand it, but surely they do what they do for a reason (those guys are pretty smart).
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