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#20
by
spacetraveler
on 07 Apr, 2011 06:41
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No, and if that is what you are reading into it I don't think you understand the system of low energy transfers.
I understand it enough to know that what I've read on it doesn't sound anything like what you have been saying.
The idea with a low energy transfer is that you get to an initial position and velocity that places you inside a stable manifold tube of your combined system, such that the combined gravitational effects of the Earth-Moon-Sun or Earth-Moon-Sun-Mars system eventually take you within the target destination's influence at which point you expend a small amount of delta-v and attain ballistic capture.
In getting to the desired initial point and initial velocity, there is no scenario I can think of where you would have to wait in a parking orbit for 3 weeks, do another small burn, then wait another 3 weeks. You can either, calculate the motion of the planetary bodies and time the launch date so that only a main initial burn places you on the proper trajectory, or if what you are saying really works where you can attain some kind of gravity assist by doing multiple small burns, then couldn't you time the launch so that you are where you want to be after the final burn in a sequence of consecutive ones?
IMO, that whole scenario you have described gets you a negligible benefit at best, certainly nothing on the order of magnitude you have claimed. The LET does let you do a transfer with less fuel, but I don't understand how it relates to everything that you just stated as far as the process.
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#21
by
Downix
on 07 Apr, 2011 07:01
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How about this. Let us do the total LEO first:
To a basic 185km orbit at 26 degree inclination, the Atlas 552 can bring 20,050 kg. Which means, at 13,600 kg payload, plus STAR-48's weight of 2134 kg, so a total of 15634 kg in total. Which means that your Centaur can retain 4416 kg of fuel. Now, doing a 20 second burn at every Perigee would expand the orbit, adding approximately 176 m/s DeltaV (154 m/s actual impulse, 22 m/s gravity assist). By limiting your burns to just this narrow window, you get full effect of the earths gravity to help you. This would also use 229.4 kg of fuel. The Centaur could as a result add a total of 3388 m/s Delta V from this once it's fuel is used up. EML1 is 3768 m/s, to remind you. Now you ignite the STAR-48, which with the gravity asist slingshotting through EML1, around the moon then earth adds another 984 m/s delta v, giving you a total of 4372 m/s Delta V. The total needed for TMI, mind you, is 4298 m/s. So there you are, 13,600 kg on TMI, and it only took a *lot* of burns, 19 of them in fact.
Ahh, I thought you might have been describing resonance effects with the moon.
"(154 m/s actual impulse, 22 m/s gravity assist)" this is just for the last burn which moves the orbit towards a apogee where lunar influence starts to dominate isn't it? Previous burns would give a smaller gravity assist I think, and a circular LEO would not be able to get any assist.
The stage starts at 185km orbit at 26 [28?] degree inclination, could you give a table of the burns and their assists?
Right. There is some assist on the ellipticals, the longer the elliptical, the better the boost. So, your initial burn has no assist, but as it goes on, you get slightly more, and more. The burns are done to maximize what little assistance you can get, which is why the exact timing of the burn is not always on the same orbit, which is why I say 2-3 weeks. You're using gravities effects, from the earth, moon, sun and several lagrange points to gently push your weight into alignment necessary for the STAR-48 to then do the kick burn.
I can turn the raw numbers into a table if you'd like.
I came up with the scenario about a month ago to teach myself about low energy transfers, before attempting to make such scenarios for AJAX test runs. It was thanks to studying them that I identified a mechanism to push an Orion + Bigelow Genesis module with a fully loaded DCSS onto a Venus flyby mission with free return trip, which did not require any newer technologies than to be able to loiter the DCSS for 24 hours. Very tricky alignment, however, as it only lines up right once every 8 years. (earth- moon- SEL2- earth- Venus- moon- earth)
I've grown to love the challenge of low energy transfer systems.
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#22
by
Downix
on 07 Apr, 2011 07:07
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No, and if that is what you are reading into it I don't think you understand the system of low energy transfers.
I understand it enough to know that what I've read on it doesn't sound anything like what you have been saying.
The idea with a low energy transfer is that you get to an initial position and velocity that places you inside a stable manifold tube of your combined system, such that the combined gravitational effects of the Earth-Moon-Sun or Earth-Moon-Sun-Mars system eventually take you within the target destination's influence at which point you expend a small amount of delta-v and attain ballistic capture.
In getting to the desired initial point and initial velocity, there is no scenario I can think of where you would have to wait in a parking orbit for 3 weeks, do another small burn, then wait another 3 weeks. You can either, calculate the motion of the planetary bodies and time the launch date so that only a main initial burn places you on the proper trajectory, or if what you are saying really works where you can attain some kind of gravity assist by doing multiple small burns, then couldn't you time the launch so that you are where you want to be after the final burn in a sequence of consecutive ones?
IMO, that whole scenario you have described gets you a negligible benefit at best, certainly nothing on the order of magnitude you have claimed. The LET does let you do a transfer with less fuel, but I don't understand how it relates to everything that you just stated as far as the process.
It's negligible to start, correct, but it slowly adds up. While you start off with less than 0.4% benefit, the next pass gives you 0.6%, then 1.1%, and so on. Each one accumulates over time. At no time did I claim it was a magnitude shift, if you check the Delta-V involved we're at less than 900 m/s difference between what Atlas V 552 w/ STAR-48 can normally place 13500kg to this. So, it's how to gain that extra 900 m/s in delta-v, that is the focus. While, yes, we're going it by a few meters/second here and there, it adds up. That is precisely why it takes so blasted long.
Something to point out, the GTO performance of Atlas V is it without the Star-48. The Star-48 adds an extra 575682 kgf-sec of impulse at the end. On a 13,500 kg load, that's a good kick in the pants, yes?
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#23
by
spacetraveler
on 07 Apr, 2011 07:08
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As an example, this was a mission that used a LET to the moon.
http://www.lpi.usra.edu/meetings/lpsc2007/pdf/1915.pdfIt took roughly one year and 2 months to do the transfer, and resulted in 288 thruster firings or around 1 every 1.5 days.
So clearly the moon's alignment over a 2-3 week period had nothing to do with it. The alignment mattered for timing the launch and capture.
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#24
by
Downix
on 07 Apr, 2011 07:09
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As an example, this was a mission that used a LET to the moon.
http://www.lpi.usra.edu/meetings/lpsc2007/pdf/1915.pdf
It took roughly one year and 2 months to do the transfer, and resulted in 288 thruster firings or around 1 every 1.5 days.
So clearly the moon's alignment over a 2-3 week period had nothing to do with it (other than timing the position for the launch and capture).
Right, but that was going to the moon. This is to Mars. Needs a different set of parameters.
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#25
by
spacetraveler
on 07 Apr, 2011 07:14
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As an example, this was a mission that used a LET to the moon.
http://www.lpi.usra.edu/meetings/lpsc2007/pdf/1915.pdf
It took roughly one year and 2 months to do the transfer, and resulted in 288 thruster firings or around 1 every 1.5 days.
So clearly the moon's alignment over a 2-3 week period had nothing to do with it (other than timing the position for the launch and capture).
Right, but that was going to the moon. This is to Mars. Needs a different set of parameters.
There's still nothing that would require you to wait 3 weeks between burns if all you are after at first is raising the orbit.
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#26
by
Downix
on 07 Apr, 2011 07:31
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As an example, this was a mission that used a LET to the moon.
http://www.lpi.usra.edu/meetings/lpsc2007/pdf/1915.pdf
It took roughly one year and 2 months to do the transfer, and resulted in 288 thruster firings or around 1 every 1.5 days.
So clearly the moon's alignment over a 2-3 week period had nothing to do with it (other than timing the position for the launch and capture).
Right, but that was going to the moon. This is to Mars. Needs a different set of parameters.
There's still nothing that would require you to wait 3 weeks between burns if all you are after at first is raising the orbit.
But that's just it, I'm not just raising the orbit. I'm also changing the plane and angel of the ellipse of orbit without using fuel, to get it into alignment for final TMI burn. Slight gravity tugs over weeks can do that.
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#27
by
spacetraveler
on 07 Apr, 2011 07:52
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But that's just it, I'm not just raising the orbit. I'm also changing the plane and angel of the ellipse of orbit without using fuel, to get it into alignment for final TMI burn. Slight gravity tugs over weeks can do that.
The moon's gravity over 3 weeks is not going to change your orbital plane.
Also, this is an example LET to mars proof of concept that I found, and from what I can tell it basically just does an initial burn to escape from a circular LEO.
http://www.technion.ac.il/~dssl/papers/49810_proof2.pdf
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#28
by
Downix
on 07 Apr, 2011 08:00
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But that's just it, I'm not just raising the orbit. I'm also changing the plane and angel of the ellipse of orbit without using fuel, to get it into alignment for final TMI burn. Slight gravity tugs over weeks can do that.
The moon's gravity over 3 weeks is not going to change your orbital plane.
Also, this is an example LET to mars proof of concept that I found, and from what I can tell it basically just does an initial burn to escape from a circular LEO.
http://www.technion.ac.il/~dssl/papers/49810_proof2.pdf
'I studied that one awhile back. And you're right, over 3 weeks, the change would be slight. But as with thrust, over time, it accumulates. Remember, we're talking 18 burns, over almost 40 weeks. In 40 weeks you'll have changed your plane by 3 degrees. Without using any fuel. That alignment if paired with a precise final burn will save you almost 70 m/s of Delta V. Normally, nobody would care, but this is not a normal scenario.
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#29
by
Downix
on 07 Apr, 2011 08:04
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You know, this should show me not to get too smart for my own good. Just because something *can* be done does not make it something that *should* be done. Would do a lot better job if we just launched an empty Atlas 401 to use it's half-full Centaur as a second booster.
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#30
by
MikeAtkinson
on 07 Apr, 2011 08:08
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How about this. Let us do the total LEO first:
To a basic 185km orbit at 26 degree inclination, the Atlas 552 can bring 20,050 kg. Which means, at 13,600 kg payload, plus STAR-48's weight of 2134 kg, so a total of 15634 kg in total. Which means that your Centaur can retain 4416 kg of fuel. Now, doing a 20 second burn at every Perigee would expand the orbit, adding approximately 176 m/s DeltaV (154 m/s actual impulse, 22 m/s gravity assist). By limiting your burns to just this narrow window, you get full effect of the earths gravity to help you. This would also use 229.4 kg of fuel. The Centaur could as a result add a total of 3388 m/s Delta V from this once it's fuel is used up. EML1 is 3768 m/s, to remind you. Now you ignite the STAR-48, which with the gravity asist slingshotting through EML1, around the moon then earth adds another 984 m/s delta v, giving you a total of 4372 m/s Delta V. The total needed for TMI, mind you, is 4298 m/s. So there you are, 13,600 kg on TMI, and it only took a *lot* of burns, 19 of them in fact.
Ahh, I thought you might have been describing resonance effects with the moon.
"(154 m/s actual impulse, 22 m/s gravity assist)" this is just for the last burn which moves the orbit towards a apogee where lunar influence starts to dominate isn't it? Previous burns would give a smaller gravity assist I think, and a circular LEO would not be able to get any assist.
The stage starts at 185km orbit at 26 [28?] degree inclination, could you give a table of the burns and their assists?
Right. There is some assist on the ellipticals, the longer the elliptical, the better the boost. So, your initial burn has no assist, but as it goes on, you get slightly more, and more. The burns are done to maximize what little assistance you can get, which is why the exact timing of the burn is not always on the same orbit, which is why I say 2-3 weeks. You're using gravities effects, from the earth, moon, sun and several lagrange points to gently push your weight into alignment necessary for the STAR-48 to then do the kick burn.
I can turn the raw numbers into a table if you'd like.
I came up with the scenario about a month ago to teach myself about low energy transfers, before attempting to make such scenarios for AJAX test runs. It was thanks to studying them that I identified a mechanism to push an Orion + Bigelow Genesis module with a fully loaded DCSS onto a Venus flyby mission with free return trip, which did not require any newer technologies than to be able to loiter the DCSS for 24 hours. Very tricky alignment, however, as it only lines up right once every 8 years. (earth- moon- SEL2- earth- Venus- moon- earth)
I've grown to love the challenge of low energy transfer systems.
I'm also interested in how you are doing the math.
Its a 3 body problem (actually 4 as there is the Sun involved as well), so there are no exact solutions. I would have thought you would need to optimise each burn duration by tracking the entire trajectory until the next burn and hill climbing. But that would only optimise individual burns, I would have thought there were overall better sequences of burns where each burn need not give the maximum delta-v boost.
I am sceptical about whether it will actually lead to higher TMI payloads in practice, clearly it would require a completely new stage, perhaps ACES would be capable of it. Even with zero boil off, things like engine losses due to start/stop transients and residual drag at perigee could reduce the advantage to zero.
On optimising single launch performance it is probably best not to start in a circular orbit, but that depends on many factors.
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#31
by
Joris
on 07 Apr, 2011 08:09
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How about this. Let us do the total LEO first:
To a basic 185km orbit at 26 degree inclination, the Atlas 552 can bring 20,050 kg. Which means, at 13,600 kg payload, plus STAR-48's weight of 2134 kg, so a total of 15634 kg in total. Which means that your Centaur can retain 4416 kg of fuel. Now, doing a 20 second burn at every Perigee would expand the orbit, adding approximately 176 m/s DeltaV (154 m/s actual impulse, 22 m/s gravity assist). By limiting your burns to just this narrow window, you get full effect of the earths gravity to help you. This would also use 229.4 kg of fuel. The Centaur could as a result add a total of 3388 m/s Delta V from this once it's fuel is used up. EML1 is 3768 m/s, to remind you. Now you ignite the STAR-48, which with the gravity asist slingshotting through EML1, around the moon then earth adds another 984 m/s delta v, giving you a total of 4372 m/s Delta V. The total needed for TMI, mind you, is 4298 m/s. So there you are, 13,600 kg on TMI, and it only took a *lot* of burns, 19 of them in fact.
Filling in an the rocket equation:
DV=9.81*451*ln(20050/15634)
DV=1100 m/s
This is nowhere near the 3388 m/s you stated.
DV=9.81*451*ln(20050/19820.6)
DV=50.9
This is nowhere near the 176 m/s you stated.
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#32
by
Downix
on 07 Apr, 2011 15:49
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How about this. Let us do the total LEO first:
To a basic 185km orbit at 26 degree inclination, the Atlas 552 can bring 20,050 kg. Which means, at 13,600 kg payload, plus STAR-48's weight of 2134 kg, so a total of 15634 kg in total. Which means that your Centaur can retain 4416 kg of fuel. Now, doing a 20 second burn at every Perigee would expand the orbit, adding approximately 176 m/s DeltaV (154 m/s actual impulse, 22 m/s gravity assist). By limiting your burns to just this narrow window, you get full effect of the earths gravity to help you. This would also use 229.4 kg of fuel. The Centaur could as a result add a total of 3388 m/s Delta V from this once it's fuel is used up. EML1 is 3768 m/s, to remind you. Now you ignite the STAR-48, which with the gravity asist slingshotting through EML1, around the moon then earth adds another 984 m/s delta v, giving you a total of 4372 m/s Delta V. The total needed for TMI, mind you, is 4298 m/s. So there you are, 13,600 kg on TMI, and it only took a *lot* of burns, 19 of them in fact.
Filling in an the rocket equation:
DV=9.81*451*ln(20050/15634)
DV=1100 m/s
This is nowhere near the 3388 m/s you stated.
DV=9.81*451*ln(20050/19820.6)
DV=50.9
This is nowhere near the 176 m/s you stated.
Let us compare to the Falcon 9 Heavy. It looks to use half of it's upper stage propellant in getting to LEO, so let us work from that:
DV = 9.81*342*ln(25000/13500)
DV = 2067
Nowhere near TMI
So, there is something amiss in the calculation, or both of us (myself and SpaceX) are wrong.
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#33
by
Joris
on 07 Apr, 2011 15:52
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Let us compare to the Falcon 9 Heavy. It looks to use half of it's upper stage propellant in getting to LEO, so let us work from that:
That is not known, they haven't said anything about the upper stage.
We don't even know its mass, wet or dry...
DV = 9.81*342*ln(25000/13500)
DV = 2067
Nowhere near TMI
So, there is something amiss in the calculation, or both of us (myself and SpaceX) are wrong.
Rocket equation is correct, so you must be wrong.
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#34
by
Downix
on 07 Apr, 2011 16:11
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Let us compare to the Falcon 9 Heavy. It looks to use half of it's upper stage propellant in getting to LEO, so let us work from that:
That is not known, they haven't said anything about the upper stage.
We don't even know its mass, wet or dry...
DV = 9.81*342*ln(25000/13500)
DV = 2067
Nowhere near TMI
So, there is something amiss in the calculation, or both of us (myself and SpaceX) are wrong.
Rocket equation is correct, so you must be wrong.
If the equation is correct, then how did we get Hiten to the moon from it's original 3192 delta-v:
DV = 9.81 * 248 * ln(197/174)
DV = 302 m/s
if the equation is correct, Hiten never made it to the moon. But it did.
As for upper stage, the F9H appears to use the same US as the F9, if you check the rocket's new height according to the SpaceX website. No stretching of the US, and the US is the same diameter. The F9's upper stage is a known quantity, with 43mT of fuel and an unloaded mass of 4.1 mT.
So, now the question is, how much Delta-V is the F-9H imparting to the upper stage? That is the unknown here I would think, yes?
Not arguing, I am genuinely curious.
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#35
by
Downix
on 07 Apr, 2011 16:16
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I'm going to make a bit of a leap here. Slight redirection of topic:
What methods could be used to push more mass to TMI from an Atlas, would you say? I've been thinking on it some, if we needed to get 10mT to mars, and had a single Altas 552 launch, it means we could get 20mT into LEO. What could we do to give enough throw to push something in the direction of Mars, hmm?
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#36
by
Joris
on 07 Apr, 2011 16:23
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If the equation is correct, then how did we get Hiten to the moon from it's original 3192 delta-v:
DV = 9.81 * 248 * ln(197/174)
DV = 302 m/s
if the equation is correct, Hiten never made it to the moon. But it did.
http://nssdc.gsfc.nasa.gov/nmc/spacecraftDisplay.do?id=1990-007AHiten was launched into highly elliptical Earth orbit on a Mu-3SII-5 rocket from Kagoshima Space Center in Japan at 11:46:00 UT (20:46:00 JST) on 24 January 1990. Injection velocity was 50 m/s less than the nominal value, resulting in an apogee of only 290,000 km compared to the expected 476,000 km
Seems like only a small push is need to send it to TLI from an orbit with a apogee of. 290.000km. And not 3192m/s
The MU-3SII can launch 800kg to LEO, and HITEN weighed 200kg.
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#37
by
Downix
on 07 Apr, 2011 16:25
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If the equation is correct, then how did we get Hiten to the moon from it's original 3192 delta-v:
DV = 9.81 * 248 * ln(197/174)
DV = 302 m/s
if the equation is correct, Hiten never made it to the moon. But it did.
http://nssdc.gsfc.nasa.gov/nmc/spacecraftDisplay.do?id=1990-007A
Hiten was launched into highly elliptical Earth orbit on a Mu-3SII-5 rocket from Kagoshima Space Center in Japan at 11:46:00 UT (20:46:00 JST) on 24 January 1990. Injection velocity was 50 m/s less than the nominal value, resulting in an apogee of only 290,000 km compared to the expected 476,000 km
Seems like only a small push is need to send it to TLI from an orbit with a apogee of. 290.000km. And not 3192m/s
The MU-3SII can launch 800kg to LEO, and HITEN weighed 200kg.
Then, what orbit should Atlas be putting it's payload in, hmm?
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#38
by
Joris
on 07 Apr, 2011 20:59
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If the equation is correct, then how did we get Hiten to the moon from it's original 3192 delta-v:
DV = 9.81 * 248 * ln(197/174)
DV = 302 m/s
if the equation is correct, Hiten never made it to the moon. But it did.
http://nssdc.gsfc.nasa.gov/nmc/spacecraftDisplay.do?id=1990-007A
Hiten was launched into highly elliptical Earth orbit on a Mu-3SII-5 rocket from Kagoshima Space Center in Japan at 11:46:00 UT (20:46:00 JST) on 24 January 1990. Injection velocity was 50 m/s less than the nominal value, resulting in an apogee of only 290,000 km compared to the expected 476,000 km
Seems like only a small push is need to send it to TLI from an orbit with a apogee of. 290.000km. And not 3192m/s
The MU-3SII can launch 800kg to LEO, and HITEN weighed 200kg.
Then, what orbit should Atlas be putting it's payload in, hmm?
Hiten was first launched to HEO, and than HITEN performed a very small TLI. Which was probably about be 300m/s.
And I don't know what orbit Atlas should launch into, but that is not the point.
The point is that your calculations were not correct and you still have no evidence for your statement about Atlas payload.
How about this. Let us do the total LEO first:
To a basic 185km orbit at 26 degree inclination, the Atlas 552 can bring 20,050 kg. Which means, at 13,600 kg payload, plus STAR-48's weight of 2134 kg, so a total of 15634 kg in total. Which means that your Centaur can retain 4416 kg of fuel. Now, doing a 20 second burn at every Perigee would expand the orbit, adding approximately 176 m/s DeltaV (154 m/s actual impulse, 22 m/s gravity assist). By limiting your burns to just this narrow window, you get full effect of the earths gravity to help you. This would also use 229.4 kg of fuel. The Centaur could as a result add a total of 3388 m/s Delta V from this once it's fuel is used up. EML1 is 3768 m/s, to remind you. Now you ignite the STAR-48, which with the gravity asist slingshotting through EML1, around the moon then earth adds another 984 m/s delta v, giving you a total of 4372 m/s Delta V. The total needed for TMI, mind you, is 4298 m/s. So there you are, 13,600 kg on TMI, and it only took a *lot* of burns, 19 of them in fact.
Filling in an the rocket equation:
DV=9.81*451*ln(20050/15634)
DV=1100 m/s
This is nowhere near the 3388 m/s you stated.
DV=9.81*451*ln(20050/19820.6)
DV=50.9
This is nowhere near the 176 m/s you stated.
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#39
by
Downix
on 07 Apr, 2011 21:15
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If the equation is correct, then how did we get Hiten to the moon from it's original 3192 delta-v:
DV = 9.81 * 248 * ln(197/174)
DV = 302 m/s
if the equation is correct, Hiten never made it to the moon. But it did.
http://nssdc.gsfc.nasa.gov/nmc/spacecraftDisplay.do?id=1990-007A
Hiten was launched into highly elliptical Earth orbit on a Mu-3SII-5 rocket from Kagoshima Space Center in Japan at 11:46:00 UT (20:46:00 JST) on 24 January 1990. Injection velocity was 50 m/s less than the nominal value, resulting in an apogee of only 290,000 km compared to the expected 476,000 km
Seems like only a small push is need to send it to TLI from an orbit with a apogee of. 290.000km. And not 3192m/s
The MU-3SII can launch 800kg to LEO, and HITEN weighed 200kg.
Then, what orbit should Atlas be putting it's payload in, hmm?
Hiten was first launched to HEO, and than HITEN performed a very small TLI. Which was probably about be 300m/s.
And I don't know what orbit Atlas should launch into, but that is not the point.
The point is that your calculations were not correct and you still have no evidence for your statement about Atlas payload.
I don't see that. I've been crunching my numbers some more, so I am not against discussion, but here I see only "you're wrong" without anything else.
So, I ask again, what orbit should Atlas launch into? If you don't have an answer, you've brought what you had to the table and should leave it at that and let us move the discussion onward shall we?