Talk:Delta-v budget

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

Other locations[edit]

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)

Slingshot Maneuvers[edit]

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)

Delta-V diagram problems[edit]

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)

There is an apparent problem with the two delta-V's adjacent to "Mars transfer". Based on a Hohmann transfer orbit from (1) Earth's orbit around the sun to (2) Mars' orbit around the sun (this assumes the vehicle has already achieved Earth escape velocity), I calculate a necessary delta-V of approximately 2.94 km/sec on the way out, or 2.65 km/sec on the way back. This is in poor agreement with the figures shown on the graph.--stclairc 15 August 2008.

I haven't done the calculation, but it's probably something to do with the Oberth effect- if you do the Mars insertion burn close to the Earth then you get much better efficiency than if you burn for escape, and once escaped, you then do the Mars insertion burn. In this case the graph goes from the C3 orbit to Mars transfer, and the optimum place for this burn is close to the Earth, where you would be doing about 7.8km/s. Right?- (User) WolfKeeper (Talk) 20:14, 15 August 2008 (UTC)

The latest version of the Delta-V diagram is very thin. A wider version, like the previous ones, would be better. Andrew Swallow (talk) 05:15, 12 November 2010 (UTC)

To-LEO figures[edit]

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)

To-NEO figures[edit]

Closest NEO Asteroids stating 0.8 - 2.0

However the table referenced lists 3.813 km/s

Just wondering how the 0.8- 2.0 number was derived. —Preceding unsigned comment added by Simultaneous (talkcontribs) 15:20, 8 May 2008 (UTC)

C3 issues...[edit]

Ooops. I just noticed that some of the C3s are different. If you go from LEO to GTO to C3 you end up at C3 with a perigee at LEO. If you go from GEO to C3, you end up with C3 with Perigee at GEO. If you then go on to say Mars transfer, the delta-v to get from C3(GEO) to Mars transfer is hugely more than to go from C3(LEO) (Oberth effect means you do much better to do the burn close to the Earth). I'm probably going to have to fix the arrow diagram, it shows an arrow going from LLO to C3, but I'm sure that's not the same C3...- (User) WolfKeeper (Talk) 22:12, 15 August 2008 (UTC)

Shattertheearth's edit of November 2, 2012 shows that the Mars Transfer Orbit figures do not agree. He tried changing the 'Interplanetary' table to show LEO to TMI is not 4.3 km/s but 3.3 km/s.


The 'Earth–Moon space — high thrust' table shows LEO to C3=0 is 3.2 km/s. 'Interplanetary' table shows C3=0 to Mars Transfer Orbit is 0.6 km/s. 3.2 + 0.6 = 3.8 km/s also The 'Delta-vs between Earth, Moon and Mars' diagrams shows LEO to GTO is 2.5 km/s, GTO to Earth C3 is 0.7 km/s and C3 to Mars Transfer is 0.6 km/s. 2.5 + 0.7 + 0.6 = 3.8 km/s The numbers should agree or an explanation needs adding. Andrew Swallow (talk) 23:20, 2 November 2012 (UTC)

I don't know exactly, but the paper referenced that the 4.3 km/s comes from talks about a 1 km/s perigee burn; sounds like they're doing something suboptimal; the Earth perigee burn would normally be 3.8km/s. I think maybe they're starting from L2, falling down to perigee and then doing a 1km/s burn or something.GliderMaven (talk) 00:45, 3 November 2012 (UTC)
That's probably what they're proposing (and it is actually a very good idea), but I'm not sure they're looking at a Hohmann transfer. Andrew is right that the numbers should add up unless they somehow use a different scheme of burns. Martijn Meijering (talk) 09:27, 3 November 2012 (UTC)

Electric propulsion Delta-V[edit]

Since spacecraft using electric propulsion cannot (or have difficulty) using the Oberth Effect the Delta-V tables do not apply to them. I have added the value for Mars C3=0 to Earth C3=0. The values for Earth orbits can usefully be added if anyone can find a source. Andrew Swallow (talk) 04:51, 4 October 2009 (UTC)

I understand that electric propulsion doesn't use the Oberth effect much, but I see just internet discussion ("guesses") that show values like 5.5km/s for the delta-v from Earth-escape to Mars-escape. Since this is just a guess, I'm going to mark that figure as disputed until some real analysis or scholarly literature shows something different. Robotbeat (talk) 18:15, 12 May 2010 (UTC)

Just removed this text:

Electric propulsion vehicles going from Mars C3=0 to Earth C3=0 without using the Oberth Effect need to use a much larger deltaV of about 6km/s. The hohmann transfer from Earth C3=0 to Mars C3=0 without using the oberth effect is about 5.5km/s.<ref> {{cite web | url = | title = thread called "Is Phobos a realistic goal?" entries by Hop_David and Rklaehn}}</ref>{{Dubious|Electric propulsion Delta-V}}

Robotbeat (talk) 19:04, 12 May 2010 (UTC)

Earth orbit delta-v table[edit]

Where are the figures for going from Earth directly to the other orbits? Spacecraft don't normally make a stop in LEO before traveling to other orbits as this maneuver of circularizing to LEO then decircularizing to reach a higher orbit requires a higher delta-v than a direct trajectory. The difference between stopping over in "intermediate" orbits and direct trajectories is of crucial importance to delta-v budget planning that should be in the article and in the table. For example, spacecraft traveling to the most popular orbit, GEO, launch directly from earth to the Hohman transfer orbit, GTO (Geostationary transfer orbit, stage there, and then finally circularize to GEO. By the way, GTO as the most popular staging orbit should be added to this chart. Flegelpuss (talk) 02:58, 3 March 2010 (UTC)

Good comments! Obviously this chart has a human spaceflight bias: as you point out "stopping over" in LEO is not common for robotic missions, but all human spaceflight missions beyond LEO to date have stopped over in LEO. Your comment about GEO is good as well, but note only equatorial launches go direct to GTO; others must perform inclination change burns. (Proton/Briz can do as many as five of them IIRC!)
IMHO the article really needs to enable readers to perform their own delta-v calculations, rather than rely on tables, especially for "simple" cases like Earth-orbit maneuvers. (sdsds - talk) 04:52, 3 March 2010 (UTC)

IANARS, so I'm having trouble reading the delta-v charts. Do you add each each stage to the previous ones to get the total cost? If so, in many other mathematical contexts (statistics) it is more common to show the combined sum of each step. This makes finding the "step" as cheap as a single subtraction, and finding the total trivial. The current system makes finding the step trivial, but finding the total quite expensive. Or I may be completely misreading the diagram.

Either way, we should explain how to interpret the table. — Preceding unsigned comment added by Riventree (talkcontribs) 21:43, 7 July 2011 (UTC)

Oh, and the charts themselves are unitless. They're probably km/s, and that should be put in the label somewhere. Riventree (talk) 21:46, 7 July 2011 (UTC)

Yet another nit: The transfer orbits should probably specify what they center on. As a non-rocket-scientist, I wouldn't know if a mars transfer orbit were around earth or the sun. I'd guess the sun, but we might as well specify.

Riventree (talk) 21:58, 7 July 2011 (UTC)

"Delta-v needed to move inside Earth–Moon system are given in km/s units" Right from the page. The table is clear enough: read it like a mileage table on a map. From GEO to L1, start at GEO on your left, read across til you find L1. I presume the Mars orbits, frex, are Earth-centric, as launched from here. (No launchers elsewhere yet...) I don't feature swinging by Sol, if that's what's implied by a Sun-centric orbit (also NARS ;p Just longtime space buff). On the inter-Martian orbit maneuvers, the dV will be additive. "The current system makes finding the step trivial, but finding the total quite expensive." True. However, it also illustrates the cost in-system, which may be of interest. The number of permutations in the inter-Martian system alone could create an enormous table. TREKphiler any time you're ready, Uhura 23:54, 23:57, & 00:01, 8 July 2011 (UTC)

So a Mars Transfer Orbit is earth-centered? I'd never have guessed that. — Preceding unsigned comment added by Riventree (talkcontribs) 05:33, 8 July 2011 (UTC)

I added units to the table. Readers shouldn't have to hunt for this.--agr (talk) 02:21, 8 July 2011 (UTC)
Just reading through this, surely any planetary transfer orbit is sun centric by definition - the planets do orbit the sun after all. Lucien86 (talk) 11:24, 22 January 2012 (UTC)

Bad link[edit]

This is coming back bad for me... TREKphiler any time you're ready, Uhura 06:24, 22 August 2010 (UTC)

EML-1 to HALO orbit around EML-1[edit]

A surprisingly high delta-v is needed to go from the point EML-1 to a HALO orbit around it. A spacestation will probably be in the orbit since staying at a single point that moves is difficult. This delta-v may need adding to the tables and a double check making to ensure that it is not already in the figures. Andrew Swallow (talk) 21:08, 6 December 2011 (UTC)

Very interesting. How high is the added delta-V? Does it depend on the amplitude of the halo orbit? What about the delta-V requirements for transfers between an EML-1 Halo and other orbits, e.g. LEO, GEO, LLO and escape--are they higher or lower than for L1 proper? And of course, can you point to a source?--agr (talk) 12:10, 7 December 2011 (UTC)

Too technical[edit]

This article seems to be written from the point of view of the physicist, rather than the engineer. That's unfortunate, since the concept is widely used in human spaceflight. And no real-world examples are given; it would be nice to have the delta-v of the Gemini and Apollo spacecraft, Space Shuttle, etc. JustinTime55 (talk) 20:03, 8 February 2012 (UTC)

Original research[edit]

This paragraph is not well written:

"Current electric ion thrusters produce a very low thrust, so the Oberth effect cannot normally be used. This results in the journey having a higher delta-V and frequently a large increase in time. The high specific impulse of electrical thrusters may significantly reduce the cost of the flight but due to the extra time are not viable for humans (interplanetary flights are an exception). The delta-v is in km/s, normally accurate to 2 significant figures and will be the same in both directions unless, say a heat shield is used."

The statement about "not viable for humans", and then the exception for interplanetary flights, is unsupported and probably opinion, in the absence of a citation. JustinTime55 (talk) 22:22, 8 February 2012 (UTC)

Heat shield?[edit]

Also, the last sentence is unclear; how does a heat shield (what kind?) affect the directionality of the delta v? "Both directions" is unclear and also basically trivial (of course forces are symmetrical in free space). Also, the statement about kilometer-per-second units is unwarranted: magnitude is indeterminant; no law says one has to use that unit, and the accuracy statement is unfounded. JustinTime55 (talk) 20:03, 8 February 2012 (UTC)

A heat shield is required for aerobraking, this makes entering a low orbit cheaper in delta-v, although the heat shield may be impractically large for manned applications. Martijn Meijering (talk) 20:06, 8 February 2012 (UTC)
The effects of heat shields are described in the 'High Thrust' section, which is just above the 'Low Thrust' paragraph.
An alternative to the kilometer-per-second is miles-per-hour, in which case all the numbers would have been 2,237 times as big. NASA has gone metric so km/s are used on this page and in its source documents. Andrew Swallow (talk) 21:47, 8 February 2012 (UTC)

I've tried to address the issues JustinTime55 has helpfully raised. I think some of it was unclear writing, e.g the units and accuracy sentence seems to have been intended to refer to the following table, not the subject in general. I've expanded the intro to be more introductory and I'll try to find real world examples; others are welcome to do so also, of course. --agr (talk) 23:05, 26 February 2012 (UTC)

Thanks. That's much better. JustinTime55 (talk) 21:39, 27 February 2012 (UTC)

delta-v as a velocity change[edit]

The problem is, in essentially no practical case is the delta-v actually the velocity change over the burn. For example if you launch from the Earth's surface into space, the delta-v is 10 km/s but the change in velocity is about 7.4km/s. Delta-v and velocity change is only actually the same if you have infinite acceleration.GliderMaven (talk) 14:19, 29 February 2012 (UTC)

This issues isn't so much infinite acceleration as it is infinitesimal duration, and many planned rocket burns on real spacecraft are short enough that the difference is quite small. "Change in velocity" is the ordinary engineering meaning of delta V and It may be better to explain the subtleties in the text rather than the fight over lede wording.--agr (talk) 16:35, 29 February 2012 (UTC)
The point is, if you don't know what delta-v is (and what the heck are you doing at this article anyway?), you can simply click on the link which takes you to where it's covered. What you don't want to do in this or any article is imply something that is not true; if we give the impression that it's change in velocity people will tend to assume that it's change in velocity over the whole mission, and get totally non sensical results or misunderstandings. If anything we should be pointing out that it's NOT change in velocity, rather that stating that it is. I mean why on Earth would you want to do that?GliderMaven (talk) 17:14, 29 February 2012 (UTC)
Do not assume. Check!
As for the whole mission - when navigating a car if information is given between two specified points it means between those two points. The same applies in space. Andrew Swallow (talk) 17:31, 29 February 2012 (UTC)
Um what? Delta-v is 100% definitely not the actual change in velocity.GliderMaven (talk) 20:35, 29 February 2012 (UTC)
Look delta-v is what you get out of the rocket equation. If a rocket stands on its tail for 5 minutes above the Earth, it's related to the propellant it's burnt, not the actual change in actual speed (which in that case would be ... zero!) In another case, the same rocket, with the same amount of propellant might have taken off from the Moon and gone into orbit. Delta-v : the same, same amount of propellant, same exhaust velocity, same rocket equation. But change in velocity, a couple of kilometers per second not at all the same!GliderMaven (talk) 20:35, 29 February 2012 (UTC)
That's how it's defined, with the rocket equation or the integral of thrust over mass.GliderMaven (talk) 20:35, 29 February 2012 (UTC)
When you hover you are not in the same location relative to the stars at the end of the hover that you were at the beginning. If you moved you had a velocity. Andrew Swallow (talk) 21:11, 29 February 2012 (UTC)
Do it either way, hover relative to the fixed stars, or the ground, delta-v is the same. It depends on the exhaust velocity, mass ratio and that's all. What it ISN'T is the change in velocity relative to anything. The change in velocity isn't particularly useful in astrogation, for example when you're in an elliptical orbit, the speed varies continuously, but you're not burning any propellant, and not expending any delta-v.GliderMaven (talk) 21:30, 29 February 2012 (UTC)
For example, if it worked the way you said, when you change from one elliptical orbit to another, the change in actual velocity would depend on the arbitrary point in the orbit where you said you started and the point you said you ended. It would be no use.
But with the way delta-v is defined, for any particular, actual sequence of burns, the delta-v is the same wherever you say you started or stopped, it depends only on the total of how long and how hard your burns were.GliderMaven (talk) 21:30, 29 February 2012 (UTC)