ApolloHoax.net
Apollo Discussions => The Reality of Apollo => Topic started by: Obviousman on March 07, 2015, 06:27:32 AM
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A discussion came up elsewhere about the Apollo 10 practice landing; what would have happened if for some reason they were forced to actually land on the surface; could they rendezvous with the CSM again?
I know that the ascent stage did not have the fuel to make a nominal ascent but what if the CSM dropped into a low orbit? I know this was a contingency plan for later missions and practiced in the sim but would it have been feasible for Apollo 10?
Anyone have the full Apollo 10 flight plan?
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What would have forced them to land?
If they had, which they could have done, there would have been no way to return. There wasn't enough propellant on the ascent stage to make any kind of lunar orbit. The CSM can't rescue you unless you do that.
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I seem to remember reading somewhere that the LM didn't have the full suite of software loaded. The landing routines weren't included, so it physically couldn't land. An abort would be an abort-to-orbit or an abort to a position where the CSM could manoeuvre to recover..
I believe that this document gives the operational LM abort and rescue plan. it's not great quality though:
https://archive.org/details/nasa_techdoc_19740072573
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I seem to remember reading somewhere that the LM didn't have the full suite of software loaded. The landing routines weren't included, so it physically couldn't land. An abort would be an abort-to-orbit or an abort to a position where the CSM could manoeuvre to recover..
That's what I have read too, along with the post that ka9q made. The ascent stage simply did not have the fuel to establish lunar orbit, even if the CSM could manoeuvre to its minimum orbital altitude (I think this is the correct term).
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IIRC, one of the reasons for the under fueling was to compensate for the LM dry mass being overweight. Reducing the fuel kept the wet mass on budget. The overweight also meant that if they'd had a full fuel load, making rendezvous could not have occurred according to the plan for A11 anyway.
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What would have forced them to land?
As earthlings we are used to aviation where aircraft has actively to do something to stay up in the air. For an an aircraft in emergency, safety is making a forced landing before the plane falls from the sky.
LM of Apollo 10 made engine burn to lower their orbit but they were still in stable orbit. Should they have had problems they had time to solve them because they weren't in danger of crashing into anything. If their problem had been unsolvable the CSM would have been able to rescue the astronauts from the LM.
Lurky
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IF the LM was in a stable position - as in not spinning.
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IF the LM was in a stable position - as in not spinning.
If the vehicles couldn't dock, I believe the contingency was to have the astronauts transfer to the CM via EVA.
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A discussion came up elsewhere about the Apollo 10 practice landing; what would have happened if for some reason they were forced to actually land on the surface; could they rendezvous with the CSM again?
The good news is, you're the first humans in all of history to land on another celestial body!
The bad news is ...
LM of Apollo 10 made engine burn to lower their orbit but they were still in stable orbit. Should they have had problems they had time to solve them because they weren't in danger of crashing into anything. If their problem had been unsolvable the CSM would have been able to rescue the astronauts from the LM.
Just for me to confirm my understanding - so they were in a reasonably stable orbit (roughly circular?), attached to the CSM. Then they fire up the engine to make their orbit elliptical (or more elliptical), with the low point being much much closer to the surface. So for Apollo 10, I guess they just do nothing, riding down and then back up again, perhaps enjoying the view if they're not too busy. But in an actual landing scenario, they wait until they're at the low point of the orbit, then fire up the engines again to put themselves down on the surface.
So the main failure scenario for Apollo 10 would then be, something goes wrong with the engine, and they're unable to return to the orbit of the CSM, in which case the CSM would have to match their orbit to pick them up.
Is that at least approximately correct?
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Very accurate.
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I don't know if the descent stage on Apollo 10 was fuelled fully - and actually had the fuel needed to soft-land.
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I don't know if the descent stage on Apollo 10 was fuelled fully - and actually had the fuel needed to soft-land.
According to Apollo By The Numbers, the A 10 descent stage had 18,218lbm of fuel, which is pretty much the same as the other descent stages (http://history.nasa.gov/SP-4029/Apollo_18-28a_LM_Descent_Stage_Propellant_Status.htm).
The ascent stage was fuelled with 2,631lbm (http://history.nasa.gov/SP-4029/Apollo_18-28b_LM_Ascent_Stage_Propellant_Status.htm), which was 51% of the average of the other missions.
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Being a hypothetical, I'm starting from the point that the LM did manage to land on the surface (I didn't know the landing routines weren't loaded).
However by what people say, the 51% propellant load would have been insufficient to get it anywhere near an altitude where an emergency rendezvous with the CSM could take place?
Is this a guess or based on some known parameters?
Anyone got Orbiter or something that can run the numbers?
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However by what people say, the 51% propellant load would have been insufficient to get it anywhere near an altitude where an emergency rendezvous with the CSM could take place?
Is this a guess or based on some known parameters?
Anyone got Orbiter or something that can run the numbers?
The Apollo 10 ascent stage after staging had a mass of 8,273 lbm. With only 2,631 lbm of propellant, and assuming a specific impulse of 311 s, then it had a Δv of only 1,167 m/s. This is well below that needed to attain orbit. About 1,850 m/s Δv was needed to reach orbit.
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What I am seeing, that in the case of a hypothetical landing, say Stafford and Cernan had made a long shot bid at history there was one shot for a rescue. It would have been to shoot up in and acr that crossed the with the CSM's orbit. And hope the CSM could snag them during the brief time while the arc and orbit crossed. Probably by abandoning the LM and jumping into space with one chance to catch the CSM. If that didn't work the Young would have a few minutes to maneuver to them before the arc took them too far away.
That's the movie version anyway. I suspect the velocity differences would have made it impossible even if the craft crossed close enough. Bob would a Δv of 1,167 m/s been sufficient to make it to the minimum altitude for a CSM orbit?
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When I was a wee lad (well, a teenager) following the Apollo 10 mission on tv, I remember the newscasters repeatedly emphasizing that if the astronauts landed on the moon for any reason whatsoever, that they were dead ducks.
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With 1,167 m/s Δv, the LM could definitely get into a suborbital trajectory with a high enough apex, but that's not the problem. For the CSM to rendezvous with them, it would have to slow down and match the LM's velocity. This would put the CSM on the same suborbital trajectory. It seems highly doubtful to me that the CSM could rendezvous with the LM, transfer the astronauts, and then get back up to speed all within the brief window of time.
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With 1,167 m/s Δv, the LM could definitely get into a suborbital trajectory with a high enough apex, but that's not the problem.
Right. Orbits are all about velocity, not altitude.
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The problem is the CSM would have to match (at least approximately) the sub-orbital trajectory of the LM to do the transfer. So they'd have to slow down, dock with the LM, do the transfer, cut loose, and then get out of the death spiral they're in. Furthermore, all this would have to be done in a few minutes, because the sub-orbital trajectory is going to last less than 45 minutes (and half of that is on the way up) before the LM makes a new crater. And if they stuff it up, that's three dead astronauts instead of two.
I guess they could revert to a stable orbit with the LM still attached, assuming the CSM has enough fuel, but still - not a lot of time to work with, and a lot of potential for an even worse outcome.
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Speaking of stuffing it up, I didn't notice Bob B.'s post, which already said most of what I said :-[
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With 1,167 m/s Δv, the LM could definitely get into a suborbital trajectory with a high enough apex, but that's not the problem.
Right. Orbits are all about velocity, not altitude.
And velocity in the right direction. This needs to be explained to some people. I had to do the whole "common sense" (ETA*) explanation to a good friend and contemporary just the other day...well it was at the same time I got him to watch Saturn V footage (mostly Apollo 4 I'm pretty sure). The pitch over threw him...until I explained why 17,500mph-ish straight up wasn't a good idea.
(*ETA Falling "around" the Earth.)
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Thanks for the responses - the wealth of information makes this place so valuable.
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What if one of the astronauts had stayed on the surface, in a "I'm just going outside, I may be some time" manner? Would the ascent module have been able to make it then or would the reduced mass not make all that much difference?
A bit dark, I know, I'm just curious!
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What if one of the astronauts had stayed on the surface, in a "I'm just going outside, I may be some time" manner? Would the ascent module have been able to make it then or would the reduced mass not make all that much difference?
Too little to make much of a difference. Leaving behind an astronaut would gain less than 30 m/s, which is still far short of what would be needed.
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Too little to make much of a difference. Leaving behind an astronaut would gain less than 30 m/s, which is still far short of what would be needed.
That might only be a measly 30 m/s, but it brings it home how lots of little weight reductions here and there made a lot of difference. It amazes me how the engineers adapted to changes of design, and the impact this had on the whole space craft. System integration must have been an utter nightmare.
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What if one of the astronauts had stayed on the surface, in a "I'm just going outside, I may be some time" manner? Would the ascent module have been able to make it then or would the reduced mass not make all that much difference?
Too little to make much of a difference. Leaving behind an astronaut would gain less than 30 m/s, which is still far short of what would be needed.
Did you use the reduced mass of the half-fueled LM as basis for that? Seems like a very small velocity gain for the loss of 80 kg.
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Did you use the reduced mass of the half-fueled LM as basis for that? Seems like a very small velocity gain for the loss of 80 kg.
I got exactly the same figures as Bob, a little under 30 m/s is gained in dV for a loss of 80 kg, unless we're both doing something daft. Maybe someone can check both our figures and put up the equations.
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How much mass had to be lost before the LM could get to any orbit with half fuel? 1000 kg?
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Did you use the reduced mass of the half-fueled LM as basis for that? Seems like a very small velocity gain for the loss of 80 kg.
I used the figures in Reply #13, i.e. 8,273 lbm total mass (with 2 astronauts) and 2,631 lbm propellant. However, I just realized I made a mistake. I intended to use the mass of an astronaut with his suit and backpack, but I forgot that I was working in pounds-mass instead of kilograms. An astronauts in his EMU with PLSS backpack had a mass of about 350 lbm. Therefore the corrected numbers are,
With two astronauts,
Δv = 311 * 9.80665 * LN(8273 / (8273 - 2631)) = 1,167 m/s
With one astronaut,
Δv = 311 * 9.80665 * LN((8273 - 350) / (8273 - 350 - 2631)) = 1,231 m/s
So the difference is 64 m/s instead of 30 m/s.
Now that I think about it, were PLSS backpacks included on Apollo 10?
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How much mass had to be lost before the LM could get to any orbit with half fuel? 1000 kg?
About half its mass. Δv is a function of the ratio of the fully fuelled mass to the empty mass. If you cut the fuel mass in half, then you have to cut the empty mass in half in order the maintain the same ratio.
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Δv = 311 * 9.80665 * LN((8273 - 350) / (8273 - 350 - 2631)) = 1,231 m/s
Rechecking my first calculation, I did not get exactly the same figures as Bob (I can't do math when it's gone midnight, not any more).
I used the same calculation, but I converted Allan's 80 kg to lbm, so arrived with 31 m/s gain in Δv. Accounting for suit and PLSS, a gain of 64 m/s (as per Bob). Again, it goes to show the design margins. Incredible stuff really.
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I got exactly the same figures as Bob, a little under 30 m/s is gained in dV for a loss of 80 kg, unless we're both doing something daft. Maybe someone can check both our figures and put up the equations.
It's just a fluke that we got the same number - we were both off by about half but for different reasons. You were off by half because an astronaut's mass doubles when he's in his suit and backpack. I was off by half because I used kilograms when I should have used pounds.
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May I ask, why is the g-value being used related to the earth and not the moon?
I think it's because the weight of the fuel used in the calculations is based on earth weight but would like to be sure.
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It's just a fluke that we got the same number - we were both off by about half but for different reasons. You were off by half because an astronaut's mass doubles when he's in his suit and backpack. I was off by half because I used kilograms when I should have used pounds.
I see that now. I used 80 kg and multiplied by 2.2 to get lbm, which gave me 176 lbm. Your suit and PLSS in kg was close to my mass in lbm (being 204 kg). Complete fluke. :-[
It's too late for this now. :)
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The Earth G-value relates to the equation of rocket power. To get the exhaust velocity you take the specific impulse in seconds and multiply it with G in m/s^2 - and get the answer in m/s.
The number 311 in the above equation is the specific impulse of the LM's ascent engine.
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About half its mass. Δv is a function of the ratio of the fully fuelled mass to the empty mass. If you cut the fuel mass in half, then you have to cut the empty mass in half in order the maintain the same ratio.
Actually it's a little less than half because a fully fuelled ascent stage had more Δv than it needed just to get into some sort of minimally acceptable orbit. To get into a 9 nautical mile orbit, the LM needed a Δv of about 1,850 m/s. Therefore, we can use this Δv to compute the exact amount of mass the ascent stage needed to lose. We have,
1850 = 311 * 9.80665 * LN((8273 - X) / (8273 - X - 2631))
where X is the require mass reduction. Solving for X we get 2,488 lbm (1,128.5 kg).
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I think it's because the weight of the fuel used in the calculations is based on earth weight but would like to be sure.
Welcome to the board with your first post. It's a specific impulse and is a convenient geocentric quantity. The way I think about specific impulse is that it enables us to normalise the rocket thrust if we know the weight of propellants through the nozzle.
We have some aerospace engineers that write at the board, they will be able to explain it fully. This site helped me understand it a little:
http://exploration.grc.nasa.gov/education/rocket/specimp.html
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Now that I think about it, were PLSS backpacks included on Apollo 10?
Probably, to retain the contingency EVA capability in case they were unable to hard dock after rendezvous.
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May I ask, why is the g-value being used related to the earth and not the moon?
That's a question that I've seen frequently. Although others have already explained, I'll try my hand at it as well.
The equation that we're using to calculate Δv is called the Tsiolkovsky rocket equation, typically written as
Δv = Ve * LN( mo / mf )
where Ve is the velocity of the expelled exhaust gases, mo is the initial total mass, and mf is the final total mass. The difference between mo and mf is the mass of propellant burned.
In practice we don't actually use Ve, instead we use something called the effective exhaust gas velocity, denoted C. (Explaining the difference between Ve and C is a complication that I'd rather not get into right now.) Therefore, the rocket equation is more correctly written
Δv = C * LN( mo / mf )
The thrust of a rocket is given by
F = C * ṁ
where ṁ is the flow rate of the expelled mass. Rearranging we have
C = F / ṁ
The term F/ṁ is a very useful parameter in describing the performance of a particular propulsion system. In the SI system, F/ṁ has the units N-s/kg, while the equivalent in the imperial system is lb-s/slug. Further note that the value of F/ṁ is different in each system of units.
Fortunately there is a simplification. If we divide by go, standard gravity, then we obtain a parameter that has the units of seconds, as well as the same value in any system of units. This new parameter is called specific impulse, and is given by the equation
Isp = F / (ṁ * go)
where go equals 9.80665 m/s2 in SI units. We use standard gravity because 1 kilogram-force = 9.80665 Newtons by definition.
We can check units as follows, noting that 1 N = 1 kg-m/s2
(kg-m/s2) * (s/kg) * (s2/m) = s
Rearranging the Isp equation we get
Isp * go = F / ṁ
Therefore, by substitution
C = Isp * go
Tsiolkovsky's rocket equation therefore becomes
Δv = Isp * go * LN( mo / mf )
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Now that I think about it, were PLSS backpacks included on Apollo 10?
Probably, to retain the contingency EVA capability in case they were unable to hard dock after rendezvous.
That would be my guess as well.
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Now that I think about it, were PLSS backpacks included on Apollo 10?
Probably, to retain the contingency EVA capability in case they were unable to hard dock after rendezvous.
That would be my guess as well.
On later missions, they ditched the PLSS before lunar takeoff, and retained the OPS to allow for EVA transfer - and also for EVA to retrieve film from the SM during trans-Earth coast.
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May I ask, why is the g-value being used related to the earth and not the moon?
I think it's because the weight of the fuel used in the calculations is based on earth weight but would like to be sure.
We've seen some very mathematical answers to this, however, I thought it was simply that fact the the issue isn't the weight of the fuel but the mass. If a rocket plus its fuel load is 1000kg, it might be "weightless" in orbit, but its mass is unchanged; still 1000 kg.
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What if one of the astronauts had stayed on the surface, in a "I'm just going outside, I may be some time" manner? Would the ascent module have been able to make it then or would the reduced mass not make all that much difference?
Too little to make much of a difference. Leaving behind an astronaut would gain less than 30 m/s, which is still far short of what would be needed.
Not enough, even if he pushed some ...
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Just out of interest, once the LM had landed and all immediate post-landing activities had been performed, was there any way to reactivate the Descent Stage and use it as a sort of Lunar Ascent Rocket First Stage?
I have vague memories of this as the basis of a hypothetical Apollo alternate-history story I came up with many years ago, when I knew only the tiniest fraction about Apollo compared with now.
And while I think about it, I suspect I may have been inspired to think about the idea by a radio play I heard part of a long time ago (late 70s or early 80s). The scenario of the play was a CMP going rogue and threatening to ignite the SPS while the CDR and LMP were on the Moon. As I didn't hear the whole play (IIRC I had a battery powered transistor radio under the blanket late at night and the batteries died) I didn't find out what happened. If anyone knows anything about the play I'd be curious to learn more.
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The descent stage was safed and deactivated by venting the helium using to pressurize the tanks. Once that is done, the engine is dead, and can't be used for anything. And the amount of fuel in it was only a few hundred liters at maximum.
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The descent stage was safed and deactivated by venting the helium using to pressurize the tanks. Once that is done, the engine is dead, and can't be used for anything. And the amount of fuel in it was only a few hundred liters at maximum.
On average, there was 5% of the total average fuel loading left that was usuable.
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So around 400 kg? How much dV is that with the descent stage still attached? High enough to stage successfully?
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May I ask, why is the g-value being used related to the earth and not the moon?
It's an artifact of the utterly obsolete English system of units, and shows why using them ought to be made a capital crime.
Although rocket specific impulse is typically quoted in seconds, this involves implicitly canceling pounds-force (lbf) with pounds-mass (lbm).
An Isp of 300 seconds means that the engine can produce 1 pound-force of thrust from 1 pound-mass of propellant for 300 seconds. But a pound of force and a pound of mass are two completely different things. Only in standard earth gravity does a mass of 1 pound-mass exert a downward force of 1 pound-force, and these units remain the same regardless of the local gravitational acceleration and the actual weight in pounds-force of 1 pound-mass.
The correct measure of rocket engine performance is effective exhaust velocity, which is what you get when you don't incorrectly cancel units. In SI, the units of Isp are simply m/sec, which makes far more intuitive sense anyway.
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It's an artifact of the utterly obsolete English system of units, and shows why using them ought to be made a capital crime.
Yes, given we would hang people for accidentally standing on a ladybird with 4 spots, it seems surprising that we invented a system that when used was not punishable by death. I think it is possible that most people collapsed of an aneurysm when using the system, so no one was ever brought to trial for its use. ;)
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It's an artifact of the utterly obsolete English system of units, and shows why using them ought to be made a capital crime.
Although rocket specific impulse is typically quoted in seconds, this involves implicitly canceling pounds-force (lbf) with pounds-mass (lbm).
An Isp of 300 seconds means that the engine can produce 1 pound-force of thrust from 1 pound-mass of propellant for 300 seconds. But a pound of force and a pound of mass are two completely different things. Only in standard earth gravity does a mass of 1 pound-mass exert a downward force of 1 pound-force, and these units remain the same regardless of the local gravitational acceleration and the actual weight in pounds-force of 1 pound-mass.
The correct measure of rocket engine performance is effective exhaust velocity, which is what you get when you don't incorrectly cancel units. In SI, the units of Isp are simply m/sec, which makes far more intuitive sense anyway.
Technically (see http://xkcd.com/1475/ (http://xkcd.com/1475/)), the "Isp in seconds" issue can be created using both pseudo-SI and pseudo-FPS systems of measurement. You need the same "G" correction factor if kgf are used instead of N and the Isp is expressed in seconds by an SI user. This is just a less common error than using the non-consistent unit of lbm instead of the consistent unit of slug (the amount of mass accelerated at 1 ft/s2 by one lbf, which weighs ~32 lbf on earth).
Even less commonly, you can use lbm as your unit of mass and the poundal as your consistent unit of force. One poundal is equal to about 1/32 lbf.
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It's an artifact of the utterly obsolete English system of units, and shows why using them ought to be made a capital crime.
I don't have a problem with it. It's just another system. If you understand it and are practiced in using it, it's no more difficult than SI.
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The correct measure of rocket engine performance is effective exhaust velocity, which is what you get when you don't incorrectly cancel units.
It a way, I wouldn't say it's the correct way. There's nothing wrong with the English system way of calculating specific impulse, it's just different.
In SI, the units of Isp are simply m/sec, which makes far more intuitive sense anyway.
That's the same thing as effective exhaust velocity. I wouldn't call that specific impulse.
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It a way, I wouldn't say it's the correct way. There's nothing wrong with the English system way of calculating specific impulse, it's just different.
It's both different and wrong because force and mass simply don't cancel (equivalently, gravitational acceleration is not dimensionless). Specific impulse has units of velocity, not time, so to be correct the English way would give results in units of feet per second, not seconds.
Using the correct units can additionally provide insight into a fundamental tradeoff of rocket engine design by equivalently representing effective exhaust velocity as power per unit thrust with a dimensionless scale factor of 0.5. E.g., an effective exhaust velocity of 1 m/s implies an exhaust kinetic power of 1/2 W per N of thrust, and an effective exhaust velocity of 3 km/s corresponds to a power of 1500 W/N.
One might claim an intuitive meaning for Isp measured in seconds: the time that a rocket engine could hover in 1 g gravity. But even this doesn't work because the mass of the rocket doesn't remain constant as it depletes its propellant. Nor does it include the mass of the engine and payload.
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It's both different and wrong because force and mass simply don't cancel (equivalently, gravitational acceleration is not dimensionless).
You're thinking about it the wrong way; we're not canceling force and mass, we're canceling force and force. In the English system, specific impulse is defined as the impulse per unit weight ejected (standardized to Earth gravity). The weight of 1 pound-mass at standard gravity is 1 pound-force, thus we have force•time/force = time.
Specific impulse has units of velocity, not time, so to be correct the English way would give results in units of feet per second, not seconds.
The English system does use feet/second, it's called effective exhaust gas velocity.
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The descent stage was safed and deactivated by venting the helium using to pressurize the tanks. Once that is done, the engine is dead, and can't be used for anything. And the amount of fuel in it was only a few hundred liters at maximum.
So essentially during, the course of a descent to the surface of the Moon, there is not likely to be any abort situation where the preferred choice would be to land as opposed to pushing the throttle to max and starting an ascent back to orbit?
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It's an artifact of the utterly obsolete English system of units, and shows why using them ought to be made a capital crime.
I don't have a problem with it. It's just another system. If you understand it and are practiced in using it, it's no more difficult than SI.
SI is always easier, more logical, less error prone and for efficient.
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SI is always easier, more logical, less error prone and for efficient.
The only thing I find easier about SI is that standard measurements are in multiples of ten or a thousand. Otherwise I find it effectively the same.
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The only thing I find easier about SI is that standard measurements are in multiples of ten or a thousand. Otherwise I find it effectively the same.
The recurring confusion about which acceleration of gravity to use in specific impulse calculations is a pretty good example of a basic problem with English units. That's why I brought it up.
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The descent stage was safed and deactivated by venting the helium using to pressurize the tanks. Once that is done, the engine is dead, and can't be used for anything. And the amount of fuel in it was only a few hundred liters at maximum.
So essentially during, the course of a descent to the surface of the Moon, there is not likely to be any abort situation where the preferred choice would be to land as opposed to pushing the throttle to max and starting an ascent back to orbit?
We are of course discussing a hypothetical landing that would not be by the book. So there is no reason to assume that the engine would be safed after landing, by the book. Anything physically possible, even though well beyond normal procedures remains viable. That said, the method by which they could have landed without the guidance of the computer during the final stage is unclear to me. Particularity if Stafford and Cernan had not trained for it. Perhaps if the goal was just to land, rather than to land is a specific spot, it was possible.
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The only thing I find easier about SI is that standard measurements are in multiples of ten or a thousand. Otherwise I find it effectively the same.
The recurring confusion about which acceleration of gravity to use in specific impulse calculations is a pretty good example of a basic problem with English units. That's why I brought it up.
People who are properly trained and practiced in the use of English units aren't confused. Most confusion arises from the fact that there are two systems of units, not that there is something inherently wrong with English units. I don't doubt for an instant that there are non-scientists and non-engineers who have been exposed to metric units their whole life who would get confused by some uses of SI units. For instance, a layman probably doesn't know what a Newton is or how it's used. People who need to know do know because it is part of their training. Likewise, people who have learned and are practiced in the use of English units know what they need to know and aren't confused by it.
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We are of course discussing a hypothetical landing that would not be by the book. So there is no reason to assume that the engine would be safed after landing, by the book. Anything physically possible, even though well beyond normal procedures remains viable. That said, the method by which they could have landed without the guidance of the computer during the final stage is unclear to me. Particularity if Stafford and Cernan had not trained for it. Perhaps if the goal was just to land, rather than to land is a specific spot, it was possible.
Apollo 10's LM had a total mass of 30,735 lbm, with 18,218.7 lbm of decent stage propellant. The ascent stage at staging had a mass of 8,273 lbm, with 2,631 lbm propellant. Both the descent and ascent engines had a specific impulse of 311 s. Therefore the Δv of both stages was,
Δv (descent) = 311 * 32.174 * LN[ 30735 / (30735 - 18218.7) ] = 8,989 ft/s
Δv (ascent) = 311 * 32.174 * LN[ 8273 / (8273 - 2631) ] = 3,830 ft/s
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Δv (total) = 12,819 ft/s
For the six Apollo landings, the median Δv for powered descent and ascent are as follows (the variance from mission to mission was small):
Δv (descent) = 6,696 ft/s
Δv (ascent) = 6,063 ft/s
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Δv (total) = 12,759 ft/s
So, theoretically, it looks like Apollo 10 did have enough total Δv to just barely match that needed to land and takeoff again. Of course that doesn't mean it was possible; the LM wasn't designed to takeoff using the descent stage. I also can't conceive of any emergency that would compel them to land. They were already in orbit, so why land just to takeoff and get back to where they already were.
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Therefore the Δv of both stages was...
Δv (total) = 12,819 ft/s
For the six Apollo landings, the median Δv for powered descent and ascent are...
Δv (total) = 12,759 ft/s
So, theoretically, it looks like Apollo 10 did have enough total Δv to just barely match that needed to land and takeoff again. Of course that doesn't mean it was possible; the LM wasn't designed to takeoff using the descent stage. I also can't conceive of any emergency that would compel them to land. They were already in orbit, so why land just to takeoff and get back to where they already were.
I think there are the makings of a decent historical fiction in here:
During the A10 mission's LM test, an unforeseen and massive solar flare erupts. Stafford and Cernan, just having entered the moon's shadow, recognize that they will promptly receive a fatal radiation dose in the thinly-walled LM as they come across to sunrise so they perform an impromptu manual landing in the dark and out of radio contact with earth, expending too much fuel from the descent stage and dinging the DPS engine bell in a hard landing in the process. Mission control directs Young to return to earth alone, as he is getting dangerous doses of radiation on each orbit despite the shielding of the oriented CSM stack. Young declines to leave while there is a chance of rescue. Stafford and Cernan race against the clock of resource depletion and DPS helium rupture disc while attempting radical field modifications: reducing ascent stage mass by removing windows, hatches, backup equipment, and sections of the pressure hull and pounding the DPS engine bell back into a functional shape. Young, progressively sickened from radiation exposure, relays technical guidance to the astronauts on the moon with each pass and positions the CSM in an optimal orbit for rendezvous with the LM. Young eventually slips into a coma, and the CSM systems degrade as mission control struggles to keep the CSM (and Young) alive remotely and with intermittent radio contact. Stafford and Cernan, struggling with exhaustion after their two day (and counting) long spacewalk, complete repairs and attempt orbital rendezvous, staging the DPS to the APS in flight.
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I like that!
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I don't doubt for an instant that there are non-scientists and non-engineers who have been exposed to metric units their whole life who would get confused by some uses of SI units. For instance, a layman probably doesn't know what a Newton is or how it's used.
Because they don't understand that force and mass are two different things...
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During the A10 mission's LM test, an unforeseen and massive solar flare erupts.
For which the best response would be an immediate rendezvous and docking with the CSM, not a landing...
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During the A10 mission's LM test, an unforeseen and massive solar flare erupts.
For which the best response would be an immediate rendezvous and docking with the CSM, not a landing...
Ya, but this is the Hollywood version. I've had similar Hollywood speculations, too and wondered about how to present a plausible rescue. The real problem in gaining even Hollywood correspondence to the real world is the fact that the CSM is traveling in excess of 3000 miles an hour. Hollywood plausibility says that you could do a sub-orbital LM launch that would, in a one in a million chance, cross paths with the CSM but overcoming the speed differential for astronaut transfer strains my imagination to produce a plausible scenario. It's not like Young could have thrown out a line for them to jump into space and grab on to.
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I don't doubt for an instant that there are non-scientists and non-engineers who have been exposed to metric units their whole life who would get confused by some uses of SI units. For instance, a layman probably doesn't know what a Newton is or how it's used.
Because they don't understand that force and mass are two different things...
I learned about Newtons in high school physics. It's a safe bet that anyone who needs to understand the difference between force and mass will know whata Newton is.
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During the A10 mission's LM test, an unforeseen and massive solar flare erupts.
For which the best response would be an immediate rendezvous and docking with the CSM, not a landing...
Ya, but this is the Hollywood version. I've had similar Hollywood speculations, too and wondered about how to present a plausible rescue. The real problem in gaining even Hollywood correspondence to the real world is the fact that the CSM is traveling in excess of 3000 miles an hour. Hollywood plausibility says that you could do a sub-orbital LM launch that would, in a one in a million chance, cross paths with the CSM but overcoming the speed differential for astronaut transfer strains my imagination to produce a plausible scenario. It's not like Young could have thrown out a line for them to jump into space and grab on to.
Ironically, this is an issue discussed by the characters in "The Martian".
ETA: Would some variant of a Skyhook work? I'm not sure what equipment in a LM could be fashioned into a skyhook, but hey, Hollywood!
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Ya, but this is the Hollywood version. I've had similar Hollywood speculations, too and wondered about how to present a plausible rescue. The real problem in gaining even Hollywood correspondence to the real world is the fact that the CSM is traveling in excess of 3000 miles an hour. Hollywood plausibility says that you could do a sub-orbital LM launch that would, in a one in a million chance, cross paths with the CSM but overcoming the speed differential for astronaut transfer strains my imagination to produce a plausible scenario. It's not like Young could have thrown out a line for them to jump into space and grab on to.
[cartoon]If you get the position and orientation just right, you could have the CSM come up behind them with the hatch open, and scoop them up.[/cartoon]
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....with a dV of 300 m/s which would make quite a mess.
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....with a dV of 300 m/s which would make quite a mess.
Ever seen those jet engine tests where they fire a chicken from an air cannon into a running engine to check for bird-strike damage?? :o
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Yes. But two astronauts in EVA-gear is much bigger than a chicken. And the CSM is not designed for that kind of impact. It would not survive as a viable spacecraft.
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ETA: Would some variant of a Skyhook work? I'm not sure what equipment in a LM could be fashioned into a skyhook, but hey, Hollywood!
The scenario goes pretty quickly to a "Space Shuttle to the moon" type of event. You have to make so many exceptions and modifications that the resulting spacecraft ceases to have any plausible fidelity to where you started. Perhaps one could make a story out of a rescue from an errant landing of a damaged, hypothetically better equipped LM, but Snoppy would not have had the resources.
Maybe Apollo nostalgia will come aground to the point someone is willing to make a rescue movie about Apollo 21 or 25 or whatever would have been the start of building a lunar base. A bad landing, or even better, perfidious sabotage, causes damage to the helium pressurization system that prevents rendezvous and is still leaking pressure. A rush against time to lift off while also needing to assemble a method of transfer through open space....
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I don't doubt for an instant that there are non-scientists and non-engineers who have been exposed to metric units their whole life who would get confused by some uses of SI units. For instance, a layman probably doesn't know what a Newton is or how it's used.
Because they don't understand that force and mass are two different things...
I learned about Newtons in high school physics. It's a safe bet that anyone who needs to understand the difference between force and mass will know whata Newton is.
You guys are confirming the point I'm trying to make. There is nothing instinctive about either system of units, they must be learned. Most layman can use a tape measure, know how far it is to the next town, can measure out a portion of food, and know when it is time to loose weight because of what their bathroom scale tells them, but other than that they're largely ignorant of units of measure. To know more than the basic everyday life stuff, it must be studied. In one system a person learns that there are basic units of length (meter) and mass (kilogram), and there's a derived unit of force (Newton). In the other system a person learns that there are basic units of length (foot) and force (pound), and there's a derived unit of mass (slug). Once those basics are learned, what's the problem? It is no more difficult to solve a problem using English units than it is to solve a problem in SI units. I just don't accept the argument that one system is intuitive while the other is some screwed up mess. People who need to know do know, and those who don't know are going to be confused by some aspects of either system.
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It is no more difficult to solve a problem using English units than it is to solve a problem in SI units. I just don't accept the argument that one system is intuitive while the other is some screwed up mess. People who need to know do know, and those who don't know are going to be confused by some aspects of either system.
This reminds me of the untis in electostatics, I studied with SI units. My lecturer kept waffling on about other units but said we did not need to know them. As a physics student we were studying the subject from a purely esoteric perspective and learning other unit systems was never really critical, but the engineers learned CGS.
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I don't doubt for an instant that there are non-scientists and non-engineers who have been exposed to metric units their whole life who would get confused by some uses of SI units. For instance, a layman probably doesn't know what a Newton is or how it's used.
Because they don't understand that force and mass are two different things...
I learned about Newtons in high school physics. It's a safe bet that anyone who needs to understand the difference between force and mass will know whata Newton is.
You guys are confirming the point I'm trying to make. There is nothing instinctive about either system of units, they must be learned. Most layman can use a tape measure, know how far it is to the next town, can measure out a portion of food, and know when it is time to loose weight because of what their bathroom scale tells them, but other than that they're largely ignorant of units of measure. To know more than the basic everyday life stuff, it must be studied. In one system a person learns that there are basic units of length (meter) and mass (kilogram), and there's a derived unit of force (Newton). In the other system a person learns that there are basic units of length (foot) and force (pound), and there's a derived unit of mass (slug). Once those basics are learned, what's the problem? It is no more difficult to solve a problem using English units than it is to solve a problem in SI units. I just don't accept the argument that one system is intuitive while the other is some screwed up mess. People who need to know do know, and those who don't know are going to be confused by some aspects of either system.
I learned both systems and problem solving in SI I find much, much easier. So to the vast majority of people.
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I learned both systems and problem solving in SI I find much, much easier.
I've learned and used both systems extensively. SI is easier to understand in the way the base units are defined, and in the use of decimal multiples. SI just makes more sense. However, once learned, I find almost no difference in using one system versus another in problem solving. What's easier about plugging meters into an equation instead of feet?
So to the vast majority of people.
I can't testify to what other people think.
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As a physics student we were studying the subject from a purely esoteric perspective and learning other unit systems was never really critical, but the engineers learned CGS.
I'm an electrical engineer but I don't think I was ever formally taught cgs. MKS (meter-kilogram-second) was already the basis of SI when I was in high school in the early 1970s, so those were the units we learned.
If EEs ever used cgs-based units, it was before my time. Volts, amperes, watts, ohms, farads and henries are all MKS/SI. I can recognize cgs units on the rare occasion I see them in old physics papers, but I usually have to look up their meanings.
When I got to Cornell I was astounded to see some of the mechanical engineering professors use English units. I made so many mistakes with them that I routinely converted to SI, performed the calculations there, and then converted the results back to English units if that's what they wanted.
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My .02: when I fly, I use knots and nautical miles for speed and distance, record times in tenths of an hour, use feet for altitude, inches of mercury for barometric settings, and Celsius for temperature. It's all arbitrary.
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For which the best response would be an immediate rendezvous and docking with the CSM, not a landing...
This is a fictional scenario, but I handwaved that with the imaginary solar flare so severe that staying in the moon's shadow was the only means of survival available to the LM crew. This assumes that a rendezvous with the CSM would have required about a half orbit and exposure to the sun.
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What's easier about plugging meters into an equation instead of feet?
Well, the fact that so many important derived units are based on meters, not feet, with unity conversion factors.
It's far easier to remember that a newton is the force that accelerates a kilogram at 1 meter per second per second than to remember that 0.03108095 pounds-force accelerates 1 pound-mass at 1 foot/sec/sec, or that 1 pound-force accelerates 32.174049 pounds-mass (1 "slug") at 1 foot/sec/sec.
And then you have the ambiguity of the common word "pound", as opposed to lbm or lbf.
But my biggest daily gripe about the English system is its widespread use of fractions. True, nothing inherently requires this, but that's what everybody does. So, quick, which is bigger: a 17/64" socket or a 1/4" socket? How about a 6 mm socket vs a 7 mm socket?
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My .02: when I fly, I use knots and nautical miles for speed and distance, record times in tenths of an hour, use feet for altitude, inches of mercury for barometric settings, and Celsius for temperature. It's all arbitrary.
This is similar to my experience as a mechanical engineer in the United States. I encounter and use a large number of units of measurement, some of them less organized than the foot-pound system. For example, I routinely use or encounter all these units of energy or heat transfer rate: ton (of refrigeration), Btu/hr, kBtu/hr (frequently abbreviated MBH), W, kW, HP, therm/y, kWh/y. I do not think this is ideal and certainly is more prone to errors than SI, but to do my job I need to be comfortable using units from whatever system of measurement I encounter.
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Yeah, heat transfer is probably the very worst for unit proliferation. Energy supply is sometimes almost as bad. My pet peeve are those who give the output of some new power plant in "kilowatts per year". This phrase would be valid only if you were talking about a plant that produces solar panels or wind turbines.
Another peeve is the (more correct) citation of plant output in gigawatt-hours/year. Why not just give the average power?
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Yeah, heat transfer is probably the very worst for unit proliferation. Energy supply is sometimes almost as bad. My pet peeve are those who give the output of some new power plant in "kilowatts per year". This phrase would be valid only if you were talking about a plant that produces solar panels or wind turbines.
So the derivative of power output with respect to time? Maybe they're phasing the plant in slowly?
Another peeve is the (more correct) citation of plant output in gigawatt-hours/year. Why not just give the average power?
Works a lot better in SI, because there are ten hours per day, ten days per month, and ten months per year :P
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But my biggest daily gripe about the English system is its widespread use of fractions. True, nothing inherently requires this, but that's what everybody does. So, quick, which is bigger: a 17/64" socket or a 1/4" socket? How about a 6 mm socket vs a 7 mm socket?
I could answer them both pretty quickly, and I think my answers were correct.
I don't mind this particular aspect of the English system, but I did rather resent having to have two complete sets of tools. And often having to use both of them. On the same device ::)
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Another peeve is the (more correct) citation of plant output in gigawatt-hours/year. Why not just give the average power?
I suppose it depends on the audience and what what "gigawatt-hours/year" actually means. As a financial analyst looking at a bond issue of a merchant power plant, I would be very interested in the proposed generation output per year. More so than in the potential capacity of say, running full time and wide open. Because I'd want to examine the regional spark spread to determine the potential annual cash flow generation. "Gigawatt-hours/year" could also mean the maximum designed capacity assuming normal downtime, which is also of interest to an investor, who could look for an independent judgement on the need for electricity on the regional grid and match that to the type of plant in making a credit assessment.
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Well, the fact that so many important derived units are based on meters, not feet, with unity conversion factors.
That get's back to the point I made early that the vast majority of error and confusion comes from the fact that there are two systems of units. No matter what system you're most accustomed to, there are going to be times when you encounter sources that give other units. I think incorrect conversions is where most errors are made. The problem is having to keep straight two systems of units in your head. It's not because there is something inherently evil about English units. If the world worked exclusively in one system or the other, I doubt there'd be many problems either way.
It's far easier to remember that a newton is the force that accelerates a kilogram at 1 meter per second per second than to remember that 0.03108095 pounds-force accelerates 1 pound-mass at 1 foot/sec/sec, or that 1 pound-force accelerates 32.174049 pounds-mass (1 "slug") at 1 foot/sec/sec.
When I worked extensively in English units, the only number I had to keep in my head was 32.174 ft/s2. That was no harder than remembering 9.80665 m/s2. The numbers that one needs to remember becomes so engrained through constant use that I don't see how anyone can consider it difficult.
And then you have the ambiguity of the common word "pound", as opposed to lbm or lbf.
I went through all my years of schooling, got my engineering degree, and I don't remember ever encountering the term "pound-mass". It wasn't until 20 years ago when I started reading old NASA and rocketry literature that I came across extensive use of the term. When I was in school and solving problems in English units, the correct unit of mass was the slug, no exception. We would typically be given a problem in which the weight of an object was given in pounds. We would instinctively divide by 32.174 to convert to slugs and away we went. Pound-mass is just another unit is mass (i.e. the mass that has a weight of 1 pound in standard gravity) that has to be converted into slugs. It's no different than having to convert tonnes into kilograms, except in that case you can make the conversion in your head because you're simply dividing by 1000 instead of 32.174.
The only grip I have with the old NASA and rocketry literature is that they would often put the lbm to slug conversion factor right in the equation. This allows lbm to be entered directly into the equation without having to first convert into slugs. This went against my prior training and I had to adjust. Whenever I used any of these equations in my web page, I stripped the conversion factor out of them. This makes the equations equally useable in either English or SI units; however, when using English units, mass must be expressed as slugs, as it should be.
A good example of the above is the specific impulse equation, which is often expressed in old literature as Isp = F/ṁ. This is a case in which the lbm to slug conversion has already been included. However, if you use the proper unit of slugs, the equation is Isp = F/(ṁ*go). In this case, if we are given mass in units of lbm, we covert to slugs by diving by go and we see that go cancels out.
Isp = F/((ṁ/go)*go) = F/ṁ, where ṁ is in lbm.
I admit this can cause a brief moment of confusion, but once it's explained, everything is good. Once one becomes familiar with this way of doing things, there shouldn't be any difficulty going forward. Of course somebody accustomed to SI units might get confused, but that comes back to the point about most problems arising from having two system of units.
But my biggest daily gripe about the English system is its widespread use of fractions. True, nothing inherently requires this, but that's what everybody does. So, quick, which is bigger: a 17/64" socket or a 1/4" socket? How about a 6 mm socket vs a 7 mm socket?
That can sometimes be a problem, though I'm pretty use to it. Through frequent use I have most decimal equivalents memorized. For instance, with barely having to think about it I know that 2 5/16" is 2.3125". And I know that 1' 5" is 1.41667'. The difficulty comes in having the convert something like 1'-2 5/16" to feet. Of course it is no more difficult than having to work with time or degrees. I find that having to convert something like 26o 23' 51" into decimal degrees to be a far more annoying problem than working with feet and inches.
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This reminds me of the untis in electostatics, I studied with SI units. My lecturer kept waffling on about other units but said we did not need to know them. As a physics student we were studying the subject from a purely esoteric perspective and learning other unit systems was never really critical, but the engineers learned CGS.
My college years were 1976-81. I don't remember if I did anything in CGS, if I did it was likely chemistry. I only had one class in electrical networks I don't remember what we used. Thermodynamics, as I recall, was all MKS. My mechanics and engineering courses were mostly FPS, though with a good bit of MKS thrown it. I remember when I took mechanics it was very common to switch back and forth between systems of units. For instance, problem #1 on a test might be in FPS and problem #2 was in MKS. We were forced to become very proficient at both. When switching back and forth I found no difference in one versus the other in terms of difficulty.
SI units make more sense than English units, and I don't doubt that SI is probably easier to learn for a person is starting from scratch. However, if one it truly proficient at both, I just don't understand how one can describe SI as "much, much" or "far" easier. In my experience that is not the case. Conversions are easier within SI, but other than that, it's just different names for the same thing.
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As I may have mentioned before, I spent most of my working life using both SI and Imperial, basically because we had two major projects on the go, one dating to before and the other after the UK switch to SI. Once I'd switched from aerospace to F1 I could forget the Imperial units. As Bob says, you can work happily in either system.
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How much mass had to be lost before the LM could get to any orbit with half fuel? 1000 kg?
About half its mass. Δv is a function of the ratio of the fully fuelled mass to the empty mass. If you cut the fuel mass in half, then you have to cut the empty mass in half in order the maintain the same ratio.
I just realized something important when I re-read this thread. The RATIO of descent fuel to the mass of the complete LM and the ratio of ascent fuel to the mass of the ascent stage is . . . . .
the same (close, anyway).
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How much mass had to be lost before the LM could get to any orbit with half fuel? 1000 kg?
About half its mass. Δv is a function of the ratio of the fully fuelled mass to the empty mass. If you cut the fuel mass in half, then you have to cut the empty mass in half in order the maintain the same ratio.
I just realized something important when I re-read this thread. The RATIO of descent fuel to the mass of the complete LM and the ratio of ascent fuel to the mass of the ascent stage is . . . . .
the same (close, anyway).
Yep, that's exactly right. Good observation. Since both descent and ascent systems had to provide about the same delta-v*, and since both system has about the same specific impulse, both systems required about the same mass ratio.
* The descent stage actually needed a little more because it spent time essentially hovering as it came to a soft landing. The ascent stage could just takeoff and go and not fight gravity as much.