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Post by Ginnie on Aug 29, 2008 13:14:40 GMT -4
I see what you are saying, Ginnie. That would be a very different looking path. Is it necessary, though? Doesn't the orbit have to take into account one fixed position -- and in this case it is the Earth? I'm sure someone here knows. Oh, its absolutely unnecessary! Actually, it just confuses instead of clarifying. ;D I just thought it would be a fun thing to animate. But seeing how huge the orbital path is, I would need to make either a huge animated gif or use teenie weenie earth/moon/trajectory path. |
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Bob B.
Bob the Excel Guru?
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Post by Bob B. on Aug 29, 2008 13:58:54 GMT -4
Ginnie, I think what you’re showing is essentially correct. The direction the Earth is moving is incorrect, but the basic premise seems sound (in your illustration the lunar phase would be near full). Of course, no matter how you depict it, you have to pick a frame of reference. I did everything in X-Y-Z coordinates for which I needed to pick an origin for the axes. (Note that since I’ve made the orbits coplanar, the Z coordinate is always zero.) To me, the most logical origin was the center of the Earth, which is why my illustrations show a stationary Earth. It sounds like what you want to do is shift the origin to the center of the Sun. Personally, I wouldn’t do this. I’ve already made several assumptions that make this a “simplified” model, so why now add back in a needless layer of complexity?
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Post by Ginnie on Aug 29, 2008 14:08:46 GMT -4
Ginnie, I think what you’re showing is essentially correct. The direction the Earth is moving is incorrect, but the basic premise seems sound (in your illustration the lunar phase would be near full). Of course, no matter how you depict it, you have to pick a frame of reference. I did everything in X-Y-Z coordinates for which I needed to pick an origin for the axes. (Note that since I’ve made the orbits coplanar, the Z coordinate is always zero.) To me, the most logical origin was the center of the Earth, which is why my illustrations show a stationary Earth. It sounds like what you want to do is shift the origin to the center of the Sun. Personally, I wouldn’t do this. I’ve already made several assumptions that make this a “simplified” model, so why now add back in a needless layer of complexity? Oh, just for my own personal amusement BobB! It wouldn't help at all on this thread - only muddy it. ;D I think I may be getting the plane wrong in your illustrations. I'll have to re-read the OP. |
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Post by Ginnie on Aug 29, 2008 14:19:29 GMT -4
Now I was thinking:
The Earth is orbiting on its axis.
And orbiting the Sun at say, around 35,000 miles per hour.
And the Sun is orbiting the Galaxy at ...
And the Galaxy is moving from the centre of the Universe?
So how fast are we really travelling through space?
Thank goodness for gravity.
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Post by dmundt on Aug 29, 2008 14:26:48 GMT -4
The point where the trajectory is bent around the moon would remain essentially the same, would it not? It would be the Earth side that would be essentially open, with the return coordinates significantly different than the exit coordinates. I don't have all the language tools I need for this, obviously. But you'd end up with more widely spaced outbound and inbound legs of the Earthward side -- but the moonward side would look very much the same as it does in the Earth-centered version. I had to do this in paint -- so it's not very good:  |
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Bob B.
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Post by Bob B. on Aug 29, 2008 14:29:46 GMT -4
One of the things I find interesting when I model something like this is that there are often unexpected surprises. When I research things further I almost always find that my model is correct and it is my expectation that was flawed. I had one such surprise with is project. In the my first simulation ( seen here) I targeted a pericynthion altitude of sixty nautical miles. This resulted in the return leg of the trajectory missing Earth by several thousand kilometers. For my second simulation I wanted the spacecraft to intersect Earth at the correct angle for a survivable reentry, i.e. a free-return trajectory. To do this I had to adjust my aim point further in front of the Moon. The result of this was that the pericynthion altitude as the spacecraft looped around the back of the Moon was much higher than I expected – 1,446 km or about 780 nautical miles. I initially thought this was an inaccuracy in my simulation. I then came across this diagram in one of my books. For a hybrid trajectory, the spacecraft is initially placed on a free return but then switches to a non-free return after a midcourse transfer. The non-free return trajectory takes the spacecraft low over the backside of the Moon, but as you can see, the free return has a pericynthion altitude in the range of 100 to 1,500 nautical miles. My 780 mile figure is almost right down the middle of this range. What I thought was a mistake is near perfect! I like surprises like that. Also after modeling this I can now see perfectly how the moon-centered orbit is hyperbolic, which is something I early questioned. I guess it’s true that a picture is worth a thousand words. |
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Bob B.
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Post by Bob B. on Aug 29, 2008 14:33:20 GMT -4
Oh, just for my own personal amusement BobB! It wouldn't help at all on this thread - only muddy it. ;D Well then, by all means please enjoy. |
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Post by Ginnie on Aug 29, 2008 14:43:16 GMT -4
Oh, just for my own personal amusement BobB! It wouldn't help at all on this thread - only muddy it. ;D Well then, by all means please enjoy. What I liked about your illustrations is that since they are in eight hour increments, when animated it gives a sense of the speeding up and slowing down of the spacecraft. If they were four hour increments, an animated gif would display this facet even better. |
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Post by Ginnie on Aug 29, 2008 15:00:10 GMT -4
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Bob B.
Bob the Excel Guru?
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Post by Bob B. on Aug 29, 2008 15:09:13 GMT -4
What I liked about your illustrations is that since they are in eight hour increments, when animated it gives a sense of the speeding up and slowing down of the spacecraft. Yes, that was done on purpose. A few of the illustrations are outside this pattern because they are there to show special events, such as crossing the equigravisphere. If I can figure out how to make my own animated GIFs, I'll probably eliminate these from the sequence so we are always looking at equal increments of time. I'll probably also eliminate the arrows and writing near Earth on the T=0 image because it looks funny in the animation (it flashes on and off so quickly that I can't read it anyway). If they were four hour increments, an animated gif would display this facet even better. I was thinking about a 4- or maybe even a 3-hour increment. I think either would work nicely, though I'lll be able to re-use more of my existing illustrations if I go with 4 hours. |
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Bob B.
Bob the Excel Guru?
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Post by Bob B. on Aug 29, 2008 23:59:32 GMT -4
Success! I've figured out how to do animated GIFs in Photoshop. I've updated my web page to include a single animation in lieu of all the individual still illustrations. I used a time increment of 4 hours. Here's the link again: Circumlunar Free Return Trajectory |
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Post by Data Cable on Aug 30, 2008 1:45:22 GMT -4
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Post by ajv on Aug 30, 2008 14:18:21 GMT -4
Nice work again, Bob! I'm not sure about the equigravisphere definition. I would prefer to make a distinction between (literally) an "equal gravity" surface and the (arbitrary) transition between coordinate systems ("the Moon's sphere of infuence"). Combining the two senses has historically meant confusion by writers and conspiracy theorists alike. On the other hand, in the Apollo 12 transcript the PAO says: Apollo 12 entered the Moon's sphere of influence or equigravisphere at 68 hours, 30 minutes 22 seconds. which as we've seen is actually the 40000 (statute) mile mark. One thing that has been lost in your transition to the (great) animation is the nice snapshot of the instant where the spacecraft passed the 40000 mile point showing how far off the center-line the trajectory was at the point where they entered the Moon's sphere of influence. |
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Post by scooter on Aug 30, 2008 14:36:28 GMT -4
So, is it a safe assumption that the hybrid trajectory came from confidence in the hardware? ...aka the SPS? (or contingincy DPS capabilities)
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Post by JayUtah on Aug 30, 2008 16:28:44 GMT -4
Yes. The hybrid trajectory opens up more landing sites. But the SPS must work for LOI-1 and LOI-2 maneuvers, otherwise your crew is lost.
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