Calculating fuel for Eve launch to orbit

[Ninzoris]

Having tried to launch from Eve with a lander than can easily reach orbit from kerbin only to see it apogee at 20km was a bit disappointing. So this time I would like to actually calculate how much fuel I need to reach orbit with out getting at that number with trail and error.

If I know the dry mass of my lander, twr, engine thrust and specific impulse, surface gravity and atmospheric drag. What formula would I use to calculate how much fuel I need to lift the craft to 120km orbit?

Thank you

Edited July 4, 2013 by purpletarget

[HoY]

10,000 meters per second. Might I recommend kerbal engineer? It can show your individual stages thrust And deltaV relative to eve(or any other body)

It just takes the manual math out of rocket building.

[K^2]

If you want to quickly estimate the Delta-V to make orbit from sea level from any planet using information from Wiki, use this:

Delta-V = 4*g*H/v_t + v_e/1.41 ± v_s.

Here g is surface gravity, H is scale height, v_t is terminal velocity at sea level, and v_e is the escape velocity. v_s is sedereal rotational velocity, and you subtract it for launches done East or add it if you launch West. Launches in other directions will fall somewhere in between. All of the numbers should be in meters, meters/second, or meters/second² as appropriate. (That's the way they are listed in Wiki as well.)

For Eve, this works out to 11,434 ± 54.6m/s. If you ask people who have done ascent from sea level, they would tell you 11,500m/s, so this is a very good estimate. Of course, you can get to orbit with as little as 8,500m/s if you start from a tall mountain, and that's what people usually end up doing. (You can use the same exact formula, substituting v_t for appropriate altitude. For highest point on Eve, 6,450m above sea level, v_t is 92.6m/s, so you get 8,476 ± 54.6m/s. You can compute v_t for any height using v_t = v₀ exp(h/(2H)), where v₀ is terminal velocity for sea level, and h is the starting height.)

As another point of comparison, for Kerbin, the formula gives 4,265 ± 174.5m/s, and again, this is very close to an absolute minimum you need to establish orbit.

[Mr Shifty]

Very nice, K². Thanks for the formulas; hadn't seen those before.

[K^2]

I haven't seen them anywhere either. Can't promise that nobody done this before, but I've only derived these yesterday as byproduct of my attempt to derive optimal ascent theory.

[numerobis]

Sweet! I'd love to see a writeup of that derivation.

[K^2]

Well, the short of it is that even with exponentially thinning atmo, the vertical part of the ascent should be done at v_t for appropriate altitude. So the altitude obeys a differential equation, h'(t) = v₀ exp(h(t)/(2H)). This DE can be solved to obtain h(t) = - 2H ln((2H-v₀t)/(2H)). In other words, ascent "to infinity" takes finite time equal to t = 2H/v₀. Naturally, this is only true if your TWR diverges to infinity, but even if TWR remains near the optimal value of 2, time to exit atmo is going to be only slightly longer. In vacuum, TWR of 2 gives you acceleration of 2g. So corresponding delta-V is 2gt.

Now, this underestimates fuel needed for vertical climb, but it also doesn't take into account gravity turn. The fact that two compensate is purely empirical at this point, but it might actually be related to how optimal gravity turn should be performed. I'm still working on optimizing the turn, so I can't say anything for sure yet.

So all in all, very simple formula that is pretty rough but works reasonably well for all of the current planets. The fun part was actually proving that terminal velocity climb is optimal even when atmosphere gets thinner with altitude. I did not honestly expect that outcome. I know everyone has been using this as a rule of thumb, but as far as I know, it's based on derivation for constant density.

P.S. I do plan to make a larger write-up on everything I learned about effects of atmosphere on launches, landings, and aerobraking, but I still have a few loose ends. Hopefully, I'll get them sorted soon enough, and it will probably be a link to a PDF in the Science Lab section when it's done.

Edited July 3, 2013 by K^2

[Ninzoris]

This is perfect, thank you everyone.

[Beanspoon]

Question solved, can't delete.

Edited July 28, 2014 by Beanspoon

[Beanspoon]

Please ignore post. I posted something, realised my mistake and tried to take it down but I can only edit, not delete :/

[Pecan]

I should point out that the answer has given the required deltaV, not the required fuel - which is what the OP originally asked.

To obtain a solution to that reverse the Tsiolkovsky rocket equation, now you know your engines' ISP, g, dry mass and deltaV.

[Oafman]

Well, the short of it is that even with exponentially thinning atmo, the vertical part of the ascent should be done at v_t for appropriate altitude. So the altitude obeys a differential equation, h'(t) = v₀ exp(h(t)/(2H)). This DE can be solved to obtain h(t) = - 2H ln((2H-v₀t)/(2H)). In other words, ascent "to infinity

Riiiiight. Maybe I will just stick with cheaty Mechjeb