Calculation of �v (delta V) from launchpad to LKO (80km)

[modwizcode]

I thought I would ask a question about how the launch ðÂݴv figures are calculated for Kerbin since I haven't been able to come to the answer myself.

I have done the calculation for the transfer ðÂݴv added to the ðÂݴv required to combat gravity due to potential energy. This produces a higher number than what's possible because the gravity ðÂݴv will decrease as the craft rises.

However since air resistance isn't factored into the equation I was using it should probably balance out.

However for a 80km circular orbit I get about ðÂݴv = 5932.1614 as the total ðÂݴv.

I get transfer ðÂݴv of 2278.9316 and a gravity ðÂݴv of 3653.2298.

transfer ðÂݴv = 2278.9316 km/s

2

gravity ðÂݴv = 3653.2298 km/s

2

total ðÂݴv = 5932.1614 km/s

2

[Kryxal]

Partly it's just experience ... a good launch takes 4500-odd delta-v.

Anyway, your gravity figure has you fighting the full force of gravity for better than 6 minutes, that doesn't happen in a launch ... by maybe 2 minutes in you have something that can maybe be called an orbit, by 3 minutes it's a sure thing (though you'll intersect the ground if you don't fix it). That knocks 1800-2400 m/s off that number, but that will be without atmosphere which might add maybe 600 back.

If you want to see what things would REALLY be like without atmosphere, start with going into an orbit at ground level, then shift to a transfer orbit using the orbital mechanics equations, then circularize at 80 km again ... the numbers are going to be a LOT different (and seriously low, by nearly 2000 if you don't account for gravity losses reaching orbital velocity...)

[modwizcode]

I'm not quite sure I understand that last part. I really do want to make an actual calculation of it or at least figure out how it would be approximated in the first place.

[modwizcode]

Seems like my second message just got deleted. I really want to find the correct calculation and/or find the correct way to measure it through trial and error. Please explain the last paragraph of your post

[Shna_na]

The precise measurement of the minimum amount of Delta-V required to achieve low Kerbin orbit from an altitude of 0m is 4550m/s.

[EtherDragon]

@modwizcode

A couple things to consider in your calculations.

1. While gravity isn't constant from the surface to orbit - it's effectively the same.

You really can simplify this part of your math because in essence the difference between gravity on the surface and at 80km is very similar.

2. While the FORCE of gravity is (basically) constant - its affect on your flight path isn't.

As you ascend, you get closer and closer to an orbit, and gravity acts directly against you less and less. In fact, once you have your ascent phase done (with a ballistic trajectory at 80km apoapsis), you no longer need to figure in gravity.

So, as you start off on the ground, you apply 100% of gravity as "lost" delta-V. As you pitch over and begin to gain momentum, that percentage gets less and less, eventually down to zero (at MECO).

3. Gravity doesn't add to the Delta-V of your circularization burns.

During circularization, you don't need to calculate gravity into the Dv at all - at that point it's just like any other orbital maneuver. It just happens that you are raising your Periapsis from inside Kerbin to above it's atmosphere. So calculate this part like any other orbital maneuver to change orbital parameters.

[modwizcode]

Anyone care to add the exact place that someone is getting that 4500? It's super common, an exact sequence of steps basically taken to derive it would be great.

[Specialist290]

I think the "4500 m/s" figure was arrived at through multiple field tests, not derived analytically. At least, I haven't been able to pinpoint a source where someone actually laid out the calculations.

[Shna_na]

Here are my sources for 4550m/s:

My post detailing Delta V for planets etc.

Jellycubes' Map

[UmbralRaptor]

4500 m/s is just a rule of thumb. There's no analytic way to get the required amount when there's an atmosphere in the way, so it's best to use MechJeb's ÃŽâ€V recorder, or setup a numeric approximation.

I have done the calculation for the transfer ðÂݴv added to the ðÂݴv required to combat gravity due to potential energy. This produces a higher number than what's possible because the gravity ðÂݴv will decrease as the craft rises.

However since air resistance isn't factored into the equation I was using it should probably balance out.

However for a 80km circular orbit I get about ðÂݴv = 5932.1614 as the total ðÂݴv.

I get transfer ðÂݴv of 2278.9316 and a gravity ðÂݴv of 3653.2298.

transfer ðÂݴv = 2278.9316 km/s²

gravity ðÂݴv = 3653.2298 km/s²

total ðÂݴv = 5932.1614 km/s²

It's worth noting that rockets are more about changing your momentum than your energy. Also, going straight up, then sideways is notably less efficient that 2 sideways burns, even if you have infinite TWR(!) That said, I'm rather confused about the units (shouldn't they be m/s, rather than km/s²?), and how the gravity figure showed up. (If you hit that speed at 0 km without any atmosphere, you'd be on an escape trajectory)

Here are my sources for 4550m/s:

My post detailing Delta V for planets etc.

Jellycubes' Map

http://wiki.kerbalspaceprogram.com/w/images/7/73/KerbinDeltaVMap.png

Edited April 1, 2014 by UmbralRaptor

[purpletarget]

There's no simple solution to the problem, but it is possible to calculate within a reasonable approximation through iterative calculations if you are handy with a spreadsheet or similar tool for such calculations. If you look here, at ep's 1-2A, 1-2B, and 1-3, you'll see the various equations and steps needed to pre-calculate the numbers. For mine, I just do a 1 second time lapse...not overly accurate but does get a reasonable approximation. The exact number does vary from craft to craft depending on the TWR and various other factors that can affect gravity and atmospheric losses.