Talk:Delta-v budget

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What's a "Taxing budget"? -- Anon.

1 Other locations
2 Slingshot Maneuvers
3 Delta-V diagram problems
4 To-LEO figures
5 To-NEO figures

[edit] Other locations

Could Delta-V requirements for more locations (lagrange points, other planets, etc.) be given in the Delta-V chart. Linguofreak 03:30, 19 March 2006 (UTC)

The movement inside Earth Moon system Delta-v figures only work if a heat shield is being used when returning to LEO, similar to the ones used when returning to the Earth's surface. Aerobraking or aerocapture can slow the spacecraft down by 3.2 km/s. The heat shield will add about 15% to the vehicles mass, for a 7 ton space craft that is 1.05 tons. Some heat shield designs are reusable. Aerobraking can take months so may not be suitable for manned flights. If thrusters are being used instead of a heat shield than the Delta-v is the same as the LEO to x figure. Andrew Swallow 04:49, 6 October 2007 (UTC)

[edit] Slingshot Maneuvers

It would be nice to add an example of the benefit of a slingshot maneuver (applying delta-v at a point of closest approach). E.g. add the delta-v needed to escape earth's gravity starting from apogee of a geostationary transfer orbit. Jrvz 14:52, 12 September 2006 (UTC)

[edit] Delta-V diagram problems

What is "Earth C3"? Should tha be "Earth L3"? Second, I see that the delta-v directly from LEO to lunar orbit is 4.1 km/s, but only 3.9 km/s going on the path to GTO, "Earth C3", lunar orbit. Something doesn't add up. The second path should be larger than or equal to the first path. -- KarlHallowell 10:55, 2 October 2007 (UTC)

You are correct, however the numbers are obviously rounded to one decimal place; this explains the discrepancy.WolfKeeper 13:34, 2 October 2007 (UTC)

Comparing with [1] ~~one of the numbers .7 should be 1.7, and C3 is L5 there.~~--Patrick 11:44, 2 October 2007 (UTC)

Those numbers seem to be correct on the graph in fact- the Strout graph has been flipped and rotated on the diagram so as to fit the Mars-Earth path on the diagonal axis.WolfKeeper 13:34, 2 October 2007 (UTC)

Unfortunately the source for the Mars numbers seems to have gone away.WolfKeeper 13:34, 2 October 2007 (UTC)

Oh, I see that the label, "Earth C3" is correct. This is a parabolic trajectory corresponding to the minimum velocity needed to escape from the Earth-Moon system. I suppose the other discrepacy could be also to the variation in LEO delta-v. I can't tell how standard it is. But round off error sounds likely. -- KarlHallowell 06:08, 5 October 2007 (UTC)

2.5 + 0.7 + 0.7 against 4.1 is not quite explained by rounding errors.--Patrick 09:36, 5 October 2007 (UTC)

The numbers are taken with good faith from the 2 sources. You're engaging in OR by adding them up! (I could probably list about 5-10 reasons why you're lucky that's it's as close as it is, including orbital planes, LEO altitudes, transfer windows, Oberth effect, rounding etc. etc. etc.) There are *reasons* why it says 'all numbers are approximate' at the bottom of the image!WolfKeeper 10:41, 8 October 2007 (UTC)

I understand, but inconsistency is somewhat confusing.--Patrick 11:48, 8 October 2007 (UTC)

[edit] To-LEO figures

It looks like these to-LEO figures are actually to-Earth figures, or equivalently, figures for reaching LEO altitude. Going from a higher orbit to an actual LEO (which is, by the way, probably a rare manoeuvre) there is no atmosphere, so delta-v should be equal to the figures for the other way around.--Patrick 09:00, 6 October 2007 (UTC)

From LEO to the Earth's surface there is a Delta-V of 9.3 km/s, this is normally absorbed by a heat shield. Since the heat shield is not a motor the table ignores the Delta-V.

A spacecraft slowing by aerobraking (or skip reentry) drops down into the atmosphere and uses friction to slow down a bit, then momentum returns the spacecraft to a lower LEO height. This is repeated until the desired velocity and height are reached.

A large reusable spacecraft returning from the Moon, Mars or high orbit could use this manoeuvre to avoid landing on the Earth. Since a small shuttle craft uses a lot less fuel flying back up from the Earth to the large spacecraft in LEO this can save money and fuel. Andrew Swallow 23:36, 7 October 2007 (UTC)

I see, thanks. So these to-LEO figures from higher orbits refer to possible manoeuvres that have never been carried out yet?--Patrick 06:41, 8 October 2007 (UTC)

Testing to see if the aerocapture theory works is likely to be worthwhile. A delta-v of 3.00 can require many tons of fuel, launching that can soon add up to millions of dollars. If tested using small satellites the payback period could be a little as 4 or 5 trips to EML1 and ELM2. LEO -> EML1 is 3.77; EML1 -> LEO is 0.77; 3.77 - 0.77 = 3.00 km/s Andrew Swallow (talk) 01:03, 20 January 2008 (UTC)