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#20 by alexw on 25 May, 2011 10:52
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I've always just thought of them as a resampled Fourier series (sampling density proportional to sqrt(1 - x^2)). [ ...] Because of the increased sampling density at the periphery, a fit using Chebyshev polynomials is resistant to Runge's phenomenon.
Yeah, that's a sweet property and makes them worth using instead of the "I know, computers will let me brute-force it the simple way using a 20th-order Taylor series!" response. But they still experience Gibb's phenomena ... just taking a guess, that seems reasonable because Gibbs is obviously sorta fundamental in a Fourier-information sense. So Runge's is an inexact analogy?
-Alex -
#21 by alexw on 25 May, 2011 10:55
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This seems like an appropriate thread.. anyone know the minimum required thrust/weight ratio to enter into lunar orbit?
Why would there be one?
For low-lunar orbits, the orbital stability time due to mascons may put a lower bound on the practical T/W ratio, but that would be highly trajectory dependent. But for high orbits, or (IIRC) polar orbits?
-Alex -
#22 by QuantumG on 25 May, 2011 11:15
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Well clearly you have to deliver the required amount of delta-v on the right side of the Moon.. so there has to be a minimum burn duration. That's typically derived from thrust/weight of the engine.
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#23 by JohnFornaro on 26 Oct, 2011 19:40
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Well, now I have a few new questions. I'm trying to understand the way shadows are cast at the lunar poles.
Here's the scenario. There's a fairly detailed PDF of Whipple crater at the north pole, floating on these forums and the internets. I have parsed out the contour lines of that drawing and have created an AutoCAD R14 drawing, attached, showing the contours of that area in 3-D. At the top of the crater is a 400m mini-plateau, where I'd place a solar power generator. If you strike a line from the north pole at 5 degrees +/-, it shoots right down the wall of the crater to the "deep end", where supposedly ice is to be found. This is a tentative road location to the crater floor.
However, more or less due north, over the pole and beyond, is another plateu, which may shade my solar power generator every two weeks or so, depending on all the ecliptical conditions, which are fairly convoluted.
So, the Earth is tilted 23.44 degrees +/- from the ecliptic. The Moon is tilted 5.14 degrees from the ecliptic. Then, the Moon's axis is tilted 6.68 degrees from its orbital plane, and libration adds or subtracts about 1.5 degrees from that! Sheesh.
Knowing from the PDF where the peaks are, I'm trying to create two or four conditions to understand the two or four extreme conditions of solar incidence and thus figure out the shadows. This will also indicate the extent of the crater which is permanently in shadow.
I'm thinking it's a relatively simple exercise in geometry.
1. Place the Moon, Earth, and Sun at one axial extreme.
2. Investigate the shadow creation between the two peaks.
3. Place the Moon, Earth, and Sun at the other axial extreme.
4. Investigate that shadow creation between the two peaks.
Hopefully, this is the "easy" approach? It's not meant to predict dates, just to understand the limits of the shadows at the poles.
It's not orbital mechanix, I guess, but what is this investigation called?
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#24 by Proponent on 27 Oct, 2011 05:31
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I think for this purpose, you can ignore the fact that the moon revolves around the earth (provided you're willing to ignore lunar eclipses). Just think of the moon as a body which rotates every 29.5 days with respect to the sun about an axis that is inclined 1.5424 degrees to the ecliptic (according to Wikipedia). If you're standing right at a lunar pole, then over the course of a 29.5-day lunar month, the sun will appear to execute a circle around you, its center rising as much as 1.5 degrees above the horizon and dropping as low as 1.5 degree below it. The directions in which you see the sun above the horizon will depend on the time of year.
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#25 by Rhyshaelkan on 10 Mar, 2012 02:51
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So speaking of Cis-Lunar Orbital Mechanics. I was troubling over a lunar mass-driver launching cargo to EML1. However it donned on me that launching a cargo from the lunar surface towards EML1 would put it the vector completely wrong to be captured into a halo orbit.
Could position of the mass-driver on the lunar surface, and aim point at launching(slightly leading the target) allow for miniscule vector change? Or will decent amount of burn be necessary to achieve a halo orbit around EML1? -
#26 by David GREENFIELD on 07 Apr, 2012 23:29
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Not sure if this should be asked here or in historical spaceflight.
Does anyone know the actual velocities of the Apollo spacecraft (Moon and Earth relative) on crossing the boundary of the "Moon's sphere of influence" during both trans-Lunar and trans-Earth coast? If memory serves me right I think that they were of the order of a few hundred miles per hour, but I'm curious to know more accurate values. -
#27 by sdsds on 09 Apr, 2012 03:06
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Ooh, that's a great question! Sorry I have no immediate answer.
Consider if you will the rotating frame of reference in which the Earth and Moon are at fixed locations. Would you be satisfied with finding the minimum velocity along the trajectory when viewed in that frame? I ask because the answer to that seems easy enough to estimate. We know the times Apollo took to traverse those trajectories, so we could recreate the trajectories in a CR3BP approximation using numerical analysis, (e.g. Runge–Kutta methods) and find the velocity minima. This approach wouldn't require explicitly calculating a sphere of influence.... -
#28 by sdsds on 09 Apr, 2012 03:12
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So speaking of Cis-Lunar Orbital Mechanics. I was troubling over a lunar mass-driver launching cargo to EML1. However it donned on me that launching a cargo from the lunar surface towards EML1 would put it the vector completely wrong to be captured into a halo orbit.
Could position of the mass-driver on the lunar surface, and aim point at launching(slightly leading the target) allow for miniscule vector change? Or will decent amount of burn be necessary to achieve a halo orbit around EML1?
Why not have the lunar surface mass driver propel the cargo onto a ballistic trajectory that leads onto a halo orbit, with no additional propulsive maneuver required? Parker showed trajectories like these can be found with transit times on the order of 100 days. The velocity required upon leaving the lunar surface might be higher, but wouldn't the payload simplicity make that worthwhile? -
#29 by David GREENFIELD on 09 Apr, 2012 09:49
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Ooh, that's a great question! Sorry I have no immediate answer.
Consider if you will the rotating frame of reference in which the Earth and Moon are at fixed locations. Would you be satisfied with finding the minimum velocity along the trajectory when viewed in that frame? I ask because the answer to that seems easy enough to estimate. We know the times Apollo took to traverse those trajectories, so we could recreate the trajectories in a CR3BP approximation using numerical analysis, (e.g. Runge–Kutta methods) and find the velocity minima. This approach wouldn't require explicitly calculating a sphere of influence....
Thanks for the reply, that would be a great start. What frame of reference would you use? One of the things I was particularly intrigued to see was the difference in the values wrt Earth and the Moon. -
#30 by sdsds on 09 Apr, 2012 20:26
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It's easy to choose the frame if we make the simplifying assumption that the Earth and Moon are each in circular orbits with the same period, centered at their combined gravitational barycenter. Since the orbits are circular it is possible to use the barycenter as the origin and rotate the reference frame such that the Earth and Moon do not move.
Kirk Sorensen provided for awhile a java based web app that used this approach, and it was highly informative. (It was at astrojava.com, I think, but it's gone now.) He seems to have mostly moved on from spaceflight to solving the global energy crisis, but his java source code if it were available would be a fine place to start. -
#31 by David GREENFIELD on 12 Apr, 2012 16:53
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Pitty its no longer available, I wonder if another more basic approach would work?
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#32 by Robotbeat on 12 Apr, 2012 18:09
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Pitty its no longer available, I wonder if another more basic approach would work?
Patched conic sections. -
#33 by David GREENFIELD on 16 Apr, 2012 23:02
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Thanks.
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#34 by KelvinZero on 15 Oct, 2012 13:19
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Hi, I was interested in understanding (in laymans terms) what was involved, from an orbital mechanics perspective, of landing something at the lunar poles.
* especially with regard to the relative difficulty of accessing poles and equator.
* and where the answer differs for manned and unmanned.
(I couldnt find this in the Q&A but perhaps I just did not know the terms to search for, in which case a link would be appreciated, and make sense to include here anyway) -
#35 by Proponent on 15 Oct, 2012 15:02
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Regarding getting to the poles versus getting to the equator, here's how I like to think about it. Forget that figure-of-eight diagram we've all seen and try thinking of it this way....
The usual way to get a spacecraft from the earth to the moon on a fast (few days) trajectory is to first put the spacecraft into a highly-elliptical earth orbit, with an apogee somewhat beyond the moon. Departure from earth (or from low earth orbit) is timed so that the spacecraft will pass not far in front of the moon as the spacecraft nears apogee.
At apogee, the spacecraft is moving slowly with respect to earth -- maybe a couple of hundred meters per second. The moon, however, is orbiting at about 1 km/s. With respect to the earth, then the spacecraft is just kind of hanging there and the moon comes by and sweeps it up. From the moon's point of view, the spacecraft is zooming roughly toward it at about 1 km/s. That's at a distance where escape velocity is below 1 km/s, so the spacecraft must slow down with respect to the moon if it is to go into lunar orbit.
If the spacecraft is in the plane of the moon's orbit when it reaches the lunar vicinity, it will go roughly into an equatorial orbit about the moon. If left earth on an ellipse sightly inclined to the moon's orbit, then on reaching the moon it may be a few lunar radii above or below the plane of the moon's orbit. In this case, when it's swept up by the moon, it may pass over one of the lunar poles and go into a polar orbit. If it goes into a polar orbit, it won't pass behind the moon as seen from earth for the first few orbits. As the moon revolves about earth while the orientation of the spacecraft's orbit remains pretty much fixed, the spacecraft will eventually pass behind the moon as seen from earth. A quarter of a lunar month after arriving at the moon, it will be spending about half of each orbit behind the moon, if it's in a low orbit. This means, of course, that a polar orbit in general lines up for easy return to earth only twice a month, whereas in an equatorial orbit there's a departure opportunity on every orbit.
The other key point is that reaching the lunar poles does not require departing from a polar orbit around earth.
Robotic spacecraft might take chaotic trajectories involving multiple passes by the moon. This can allow them to use less propellant to get into lunar orbit, but the trip times are measured in months (Ed Belbruno, inventor of the technique, gave a FISO lecture about it). LCROSS used a trick like this to impact the lunar pole on a nearly-vertical trajectory.
Another difference between robotic and human missions could be a desire for a free-return trajectory for the human mission. I've read that for polar mission, free return requires quite a distant orbit about the moon, with a pericynthion of well over 10,000 km. -
#36 by JohnFornaro on 15 Oct, 2012 23:11
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Proponent, thanks for a very easily visualized explanation of the orbit, using the English language. Prompts me to ask a few questions myself:
I've attached P.324-325 of BMW.
r bar. First, is that how you say it?
2. r bar is a vector from the center of the Earth to the center of the Moon. It varies according to the eccentricity of the Moon's orbit. If you subtract the Moon's radius, and the Earth's radius, that's the distance from surface to surface at that time?
3. What's the value of r bar? What does knowing it enable you to calculate?
4. The Moon rotates from West to East around the Earth, from the Earth's POV. But the line of nodes rotates East to West, once every 18.6 years? Does this affect launch windows much? Particularly to the poles?
5. X sub-e is the Earth's vernal equinox. Is that the direction of the Earth's rotation around the Sun, or does it point to a star? It's supposed to be a fixed inertial frame, right? What does Z sub-e point to?
Now the hard one.
6. In launching from the Cape at 28.5 degrees to a parking orbit, where and when do you make the plane change so that you end up being grabbed by the Moon in a near polar orbit? Are any further inclination adjustments necessary once you get in LLO?
If you have the time and inclination, so to speak.
Thanks
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#37 by Proponent on 16 Oct, 2012 20:49
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I've attached P.324-325 of BMW.
r bar. First, is that how you say it?
Well, that's how I say it, but I'm no authority.Quote2. r bar is a vector from the center of the Earth to the center of the Moon. It varies according to the eccentricity of the Moon's orbit. If you subtract the Moon's radius, and the Earth's radius, that's the distance from surface to surface at that time?
Sounds right to me, with the very minor proviso that neither the earth nor the moon is perfectly spherical, so the effective "radius" will vary a bit with orientation.Quote3. What's the value of r bar? What does knowing it enable you to calculate?
Well, for example, if you've got a spacecraft in cis-lunar space, you're going to need to know where the earth and the moon are in order to determine their gravitational pulls on the spacecraft.Quote4. The Moon rotates from West to East around the Earth, from the Earth's POV. But the line of nodes rotates East to West, once every 18.6 years? Does this affect launch windows much? Particularly to the poles?
As the line of nodes regresses, the moon maintains a fixed inclination of about 5o with respect to the ecliptic (the plane of the earth's orbit about the sun). That means that the inclination of the of the moon's orbit with respect to the earth's equator varies from about 18o to about 28o as the nodes regress. That obviously affects the trajectory one needs to reach the moon, but I'm not sure I'd say it affects launch windows as such. It seems to me it should be a little easier to get to the moon when its inclination matches the latitude of your launch site, but I don't think it's going to be a big deal. More about this later.Quote5. X sub-e is the Earth's vernal equinox. Is that the direction of the Earth's rotation around the Sun, or does it point to a star? It's supposed to be a fixed inertial frame, right? What does Z sub-e point to?
Xε is the direction from Earth to the sun at the moment spring begins in the northern hemisphere. Yε lies 90o from Xε in the plane of Earth's orbit. It points approximately opposite the Earth's velocity (if the Earth's orbit were perfectly circular, then it would be exactly opposite). Zε is perpendicular to both Xε and Yε.QuoteNow the hard one.
6. In launching from the Cape at 28.5 degrees to a parking orbit, where and when do you make the plane change so that you end up being grabbed by the Moon in a near polar orbit? Are any further inclination adjustments necessary once you get in LLO?
Bear in mind that the inclinations of the trans-lunar trajectories for low lunar polar orbit and for low equatorial lunar orbit differ by an amount on the order of the lunar radius divided by the distance from the earth to the moon, i.e., by about (1700 km)/(380,000 km) = 0.045 radian = 0.26o. Anyway, it seems to me it would be optimal to take off from the Cape into the plane you want. So, you make the plane change during ascent to the parking orbit. But the plane of the trans-lunar orbit isn't all that important anyway. The important thing is to arrive in the vicinity of the moon at the right time and place. To be captured, it certainly helps to have your velocity on arrival lined up with the moon's velocity so that you velocity with respect to the moon is minimized. But since, in the earth-centric frame, you're moving slowly on arrival in the moon's sphere of influence, it's not that crucial -- most of the moon-centric velocity arises from the moon's motion around the earth. -
#38 by KelvinZero on 16 Oct, 2012 22:10
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Thanks Proponent.
Those look like some pretty significant difficulties for manned polar missions. If I understand it:
(*) Difficulty of free return. (though I still dont really have any quantitative idea of how difficult, anyone care to speculate?)
(*) Lack of any-time return. You only have an option twice a month (unless your ascent vehicle can also take you all the way home without having to meet anything in lunar orbit)
On the other hand, robotic does not sound much harder. Anyone prepared to estimate the difference in difficulty of robotic equatorial vs polar missions as a factor of cost?
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#39 by JohnFornaro on 17 Oct, 2012 02:33
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Many thanks. BMW leaves many significant questions to the teacher to answer. I'm old fashioned in the sense that I ask the book to tell me everything; when it doesn't, I struggle.Quote from: JF
5. X sub-e is the Earth's vernal equinox. ...
Xε is the direction from Earth to the sun at the moment spring begins in the northern hemisphere. Yε lies 90o from Xε in the plane of Earth's orbit. It points approximately opposite the Earth's velocity (if the Earth's orbit were perfectly circular, then it would be exactly opposite). Zε is perpendicular to both Xε and Yε.
This is not mentioned in BMW at all. I guess the authors figure everyone knows that already.
My understanding has always been that the Earth rotates counterclockwise around the Sun, seen from the Earth's north pole. Therefore -Yε is in the direction of the Earth's motion! This is starting to make sense! [Edit: -Yε !! Not enough attention to detail.]4. The Moon rotates from West to East around the Earth, from the Earth's POV. But the line of nodes rotates East to West, once every 18.6 years? Does this affect launch windows much? Particularly to the poles?
As the line of nodes regresses, the moon maintains a fixed inclination of about 5o with respect to the ecliptic (the plane of the earth's orbit about the sun). That means that the inclination of the of the moon's orbit with respect to the earth's equator varies from about 18o to about 28o as the nodes regress. That obviously affects the trajectory one needs to reach the moon, but I'm not sure I'd say it affects launch windows as such. It seems to me it should be a little easier to get to the moon when its inclination matches the latitude of your launch site, but I don't think it's going to be a big deal. More about this later.
A key concept with lunar launch windows is the antipode, the point on the Earth's surface opposite the Earth-Moon vector. TLI must occur at or near the point where the orbit crosses the antipode. The antipode moves east-to-west along the Earth's surface, largely driven by the Earth's rotation but slightly counteracted by the moon's motion around the Earth. If the moon's declination is less than the launch site latitude, a due east launch (90 deg azimuth) will intersect the antipode twice per day. For Apollo these were called the Atlantic and Pacific injection opportunities. Apollo used variable-azimuth launch targeting (72-108 degrees, generally) to widen the launch window to around 2.5 hours.
EOR changes things a bit. Because the orbit plane is fixed, there's no way to widen the launch window - you get two injection opportunities per month, when the orbit plane crosses the antipode. And that only if the orbit inclination is greater than the moon's declination. So you don't want to do EOR for a lunar mission in an orbit of less than 28 degrees inclination. Otherwise there will be portions of the moon's nodal cycle where the antipode *never* crosses the orbit plane.
Thanks for this too, but it is harder to understand. For one thing BMW's discussion doesn't discuss the antipode at all.
