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#200
by
Hop_David
on 11 Oct, 2012 23:22
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And yours does not to me, see the Rocket equation.
The rocket equation is exponential. To illustrate exponential growth, the story of a chess wager is often used. Terms of wager: 1st square of chessboard 1 grain of rice, 2nd square 2 grains, third square 4 grains, doubling each subsequent square.
If using oxygen and hydrogen, each 3 km/s added to your delta V budget doubles initial mass.
As Martijn says, you get to start over at each propellant source. Here's a graphic of the chess wager vs a chess wager with a propellant source every 4th square:
With lots of aerobraking, round trip from LEO to Low Mars Orbit and back can be around 6 km/s.
Round trip from L2 to Low Mars Orbit and back can be around 3 km/s (using much less aerobraking than the LEO round trip).
You might say that from LEO it takes 3.5 km/s to reach L2 so total delta V is 6.5. And 6.5 is more than six.
But e
3/4.4 + e
3.5/4.4 < e
6/4.4If you're math-phobic here's an analogy that might help: You will save miles on a transcontinental trip if you stay on the interstate. But turning off the interstate occasionally to get gas is worthwhile even if it adds to the miles you have to drive.
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#201
by
QuantumG
on 12 Oct, 2012 00:28
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Using chemical rockets:
LEO to C3 = 3.22 km/s.
vs
LEO to EML2 = 3.47 km/s (9 day transit).
LEO to EML2 = 4.31 km/s (4 day transit).
EML2 to C3 = 0.14 km/s.
It may seem obvious that going directly to C3 is the better option than going via EML2, however SEP changes the equation. As can be seen above, if you have the option, it's better to use slow transits. That is especially the case with SEP, which have a very high specific impulse as well.
SEP allows you to build a large vehicle at EML2. You can use either SEP or chemical propulsion to do the small delta-v to C3 and onto Mars transit.
The crew will still go on fast transits, to avoid spending too much time in the Van Allen radiation belts. The 9 day vs 4 day options above are probably not all that important, but they give you the idea.
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#202
by
baldusi
on 12 Oct, 2012 01:41
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A big save is if you can reuse the habitat for transit, for example. Plus the reliability of a very thorough checkout.
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#203
by
mmeijeri
on 12 Oct, 2012 07:39
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SEP allows you to build a large vehicle at EML2.
True. So does chemical propulsion by the way.
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#204
by
QuantumG
on 12 Oct, 2012 07:53
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SEP allows you to build a large vehicle at EML2.
True. So does chemical propulsion by the way.
Yep. I think you get more bang for your buck if you can do a SEP tug.. but that said, we don't actually have one yet.
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#205
by
guckyfan
on 12 Oct, 2012 12:21
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SEP allows you to build a large vehicle at EML2.
True. So does chemical propulsion by the way.
Yep. I think you get more bang for your buck if you can do a SEP tug.. but that said, we don't actually have one yet.
Ok, thanks to all first.
The picture is not all clear yet to me. But at least it starts forming.
I am still not convinced that EML2 all chemical from earth is a better solution than directly from LEO. Especially for a one off flight. But as soon as other methods like SEP-tug come into the picture that would change.
About SEP-tugs, yes we don't have them. But the required technology is now there, if we have a real need for them. But to need them we would have to have some sustained beyond LEO program.
I thougt trajectories would be quite clearcut and dependant on the depth of the involved gravity wells and available delta-v only. Learning about the Oberth effect and understanding how it works made the first deep crack in that.
Those delta-v charts also make assumtions, sometimes unspoken assumptions.
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#206
by
strangequark
on 12 Oct, 2012 15:07
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SEP allows you to build a large vehicle at EML2.
True. So does chemical propulsion by the way.
Yep. I think you get more bang for your buck if you can do a SEP tug.. but that said, we don't actually have one yet.
Propellant transfer and storage are also easier because you're handled compressed gases. And we do have them. We're just in the bad habit of putting transponders on them and leaving them in GEO.
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#207
by
Hop_David
on 12 Oct, 2012 15:59
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Using chemical rockets:
LEO to C3 = 3.22 km/s.
vs
LEO to EML2 = 3.47 km/s (9 day transit).
LEO to EML2 = 4.31 km/s (4 day transit).
EML2 to C3 = 0.14 km/s.
Someone might look at .14 km/s from EML2 to C3 and then .5 km/s from C3 to TMI and conclude a TMI burn from EML2 would be .64 km/s. Doesn't work that way.
Kinetic energy is 1/2 m * v
2. If v is large, a small boost to v can make a big energy boost.
So 1/2 m * (v + h)
2 can be a lot bigger than 1/2 m * v
2 even if h is small.
For example lets compare a boost of velocity boost 1 for 10 vs 100:
(10 + 1)
2 = 121, a boost of 21.
(100 + 1)
2 = 10201, a boost of 201.
Deep in a gravity well, orbital velocity is high so a small velocity change yields big increase in energy. This is the Oberth benefit.
Someone like Guckyfan will look at orbital speed at EML2 and note 1.19 km/s is pretty slow. He correctly notes there is very little Oberth benefit at this altitude. Indeed, TMI directly from EML2 is a little more than 2 km/s.
But when orbital speed is low, it takes only a small dv to brake the orbit and plummet to the earth. For example with Farquhar's 9 day route, it takes only .33 km/s to drop from EML2 to a very low perigee:
Once at the an altitude of 185.2 km, it's moving 3.14 km faster than orbital speed at that altitude (7.79 km/s). 3.14 + 7.79 = 10.93 km/s. 10.93 km/s is a big v! When you're moving almost 11 km/s, a mere .5 km/s burn suffices for TMI.
So this route takes 3 burns: .15 km/s to drop from EML2, a .18 km/s burn at perilune and then a .5 km/s burn at perigee for TMI. This totals between .8 and .9 km/s.
A fully fueled and stocked ship departing from EML2 has a huge advantage over a fully fueled and stocked ship departing from LEO.
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#208
by
baldusi
on 12 Oct, 2012 16:11
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There's some of your nomenclature that's confusing me. When we talk about C3, don't we talk about kmē/sē of inherent energy, while when we talk about delta-v we talk about the integration of an impulse maneuver?
For example, I could change the inclination of an orbit, which would consume delta-v and have zero C3 change. Or, if I have used a supersynchronous insertion and then circularize, I'd need a lot of delta-v (and two burns) for a very small C3 change. Am I right?
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#209
by
mmeijeri
on 12 Oct, 2012 16:17
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Propellant transfer and storage are also easier because you're handled compressed gases. And we do have them. We're just in the bad habit of putting transponders on them and leaving them in GEO.
It looks as if Alphabus could easily double its electrical power if you put extra solar panels on it instead of transponders. Can you confirm or deny this is the case? If so, it would mean 40kW tugs are more or less the state of the art.
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#210
by
strangequark
on 12 Oct, 2012 16:34
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It looks as if Alphabus could easily double its electrical power if you put extra solar panels on it instead of transponders. Can you confirm or deny this is the case? If so, it would mean 40kW tugs are more or less the state of the art.
Something around 50kW is state of the art with minimal changes to existing architectures. So, that sounds pretty reasonable. There are a good handful of ~20kW HETs that are qualified, or nearly so, and a 2 thruster setup is pretty common.
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#211
by
mmeijeri
on 12 Oct, 2012 16:42
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Something around 50kW is state of the art with minimal changes to existing architectures. So, that sounds pretty reasonable. There are a good handful of ~20kW HETs that are qualified, or nearly so, and a 2 thruster setup is pretty common.
This could also be useful with arcjets instead of ion propulsion. I remember reading about high power arcjets a while ago. I think it may have been ESEX. Is that a state of the art thruster, or has there been work on high power arcjets since then? You could power two on a 50kW tug.
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#212
by
Hop_David
on 12 Oct, 2012 17:20
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There's some of your nomenclature that's confusing me. When we talk about C3, don't we talk about kmē/sē of inherent energy, while when we talk about delta-v we talk about the integration of an impulse maneuver?
(Googling...) Looks like C3 means
characteristic energy.
Up until now I thought it meant energy of a parabolic trajectory, in other words where C3 = 0.
A lot of people seem to use C3 as a synonym for escape or parabolic orbit. For example in a post above QuantumG says it's .14 km/s from EML2 to C3. From EML2, it takes about .14 km/s to achieve escape.
For example, I could change the inclination of an orbit, which would consume delta-v and have zero C3 change.
With my new understanding of C3, I see that you're right.
I'll reword some of my arguments hoping I'm using the term C3 correctly.
For a high altitude orbit with C3 close to zero, kinetic energy as well as potential energy are close to zero. With kinetic energy close to zero, velocity is close to zero and there is very little Oberth benefit. However with low velocity, it's less expensive to do plane changes. It seems to me a high altitude is the best place to change inclination.
Given a high altitude, low velocity orbit, it is also less expensive to change the flight path angle so this low energy orbit can have a low perigee.
Potential energy is mass * - μ/r. So shrinking r increases potential energy.
Since v
2/2 - μ/r is close to zero, increasing potential energy means increasing kinetic energy. Greater kinetic energy means greater velocity.
At a low perigee, a near parabolic orbit has nearly enough velocity for Trans Mars Injection, a .5 km/s perigee burn suffices for TMI.
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#213
by
guckyfan
on 12 Oct, 2012 18:22
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The Post by Hop_David made a lot of things much clearer to me. Thanks for that.
http://forum.nasaspaceflight.com/index.php?topic=12196.msg965593#msg965593In this trajectory the passes through the VanAllen Belt are quite fast so the radiation would not be too bad as well. Also I now understand how you can take advantage of the Oberth-Effect from EML2.
This would have been my next question. How exactly does such a trajectory look like. That is now answered.
BTW I am not allergic to math.
However over many decades my math exposure has been reduced a lot.
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#214
by
Robotbeat
on 12 Oct, 2012 21:47
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SEP allows you to build a large vehicle at EML2.
True. So does chemical propulsion by the way.
Yep. I think you get more bang for your buck if you can do a SEP tug.. but that said, we don't actually have one yet.
Propellant transfer and storage are also easier because you're handled compressed gases. And we do have them. We're just in the bad habit of putting transponders on them and leaving them in GEO.
+1
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#215
by
giggleherz
on 14 Oct, 2012 04:44
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Looking for an answer in all the wrong places.
I need an answer in regards to the dust problems that previous rovers have encountered. I know Curiosity is powered internally without the need for solar power, but with a slim chance that her lenses might get dumped on with dust dose she have any way to deal with it?
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#216
by
Jim
on 14 Oct, 2012 11:12
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Looking for an answer in all the wrong places.
I need an answer in regards to the dust problems that previous rovers have encountered. I know Curiosity is powered internally without the need for solar power, but with a slim chance that her lenses might get dumped on with dust dose she have any way to deal with it?
No
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#217
by
Robotbeat
on 14 Oct, 2012 16:21
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Opportunity has been on Mars fOr over eight years and still can see just fine.
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#218
by
strangequark
on 15 Oct, 2012 15:20
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This could also be useful with arcjets instead of ion propulsion. I remember reading about high power arcjets a while ago. I think it may have been ESEX. Is that a state of the art thruster, or has there been work on high power arcjets since then? You could power two on a 50kW tug.
ESEX is state of the art. There hasn't been much interest in high-power arcjets. Most of the focus for cislunar is on Hall Effect thrusters.
Fun fact on arcjets too. All modern thrusters use an exotic alloy for the thruster chamber/nozzle that is no longer produced. The entire world supply is currently sitting in a warehouse in Redmond, WA.
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#219
by
mmeijeri
on 15 Oct, 2012 15:36
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Ha, guarded by top men I bet! Top. Men.
