Questions on Delta V....

[Fizwalker]

Noobie here.... I gots a question! =P

I've been looking over the KSP Wiki and I noticed the section on Delta V requirements for various near Kerbin bodies (Mun and Minimus) from various altitudes of Kerbin orbits. Do these numbers reflect a certain mass of vessel or does the amount of Delta V needed remain the same for all vessels?

Thank you in advance!

Edited September 8, 2013 by Fizwalker
Question has been answered!

[Snarfster]

Mass is included in the number. A heavier rocket will simply need more fuel to reach the same delta V.

[M4ck]

It is constant regardless what you try to get there. It is a measure of how much do you need to "speed up" your ship in order to get somewhere else.

Think of it like of throwing a ball and a stone. If you want them to land in a particular spot, you would have to throw each of them with the same speed(well, the air resistance would be a bit different, but it can be omitted here). This is the dV from a stationary ball/rock to one that has just left your hand.

Of course, as the rock is heavier, it would be harder to throw it at the same speed as the lighter ball; you would need to apply more force to it. Just like a heavier ship needs to burn more fuel(as the engines need to apply more force to the vessel) to gain the same amount of speed as a lighter one. But after both of them are traveling at the same speed, they will continue on the same trajectory, no matter if it is a tiny probe or a huge mothership.

I hope it helps

Edited September 7, 2013 by M4ck

[Blue]

Delta-V is a measure of acceleration, i.e. "speed-change". Also if it's useful, ISP or Specific Impulse can be alternately termed "speed-change potential per amount of fuel".

[Raptor831]

If you really want to make your head spin about numbers, check out Nyrath's Atomic Rockets. It goes over the actual rocket science of this stuff, and it's great for getting a good understanding of this (especially combined with KSP play). Just be prepared for some math!

[Dispatcher]

[EDIT] You might try Wikipedia for more information. [/EDIT]

I second Raptor831's suggestion.

Edited September 7, 2013 by Dispatcher
Attempting to add to thread in a meaningful way.

[blizzy78]

I second Raptor831's suggestion.

Please don't. Posts containing "me too" or "I second this" don't add anything useful to a discussion.

[capi3101]

Snarfster answered the question well; the same delta-V is required, but a heavier payload will require more fuel in order to achieve that level of delta-V. Long post ahead.

Let's do some examples: assuming you're already in orbit of Kerbin, it takes 1,070 m/s of delta-V to perform a Mun transfer and orbit under optimal conditions (according to the delta-V map). Let's say we're using an LV-N Atomic Thruster (Isp 800) and that our payload is 1.75 tonnes. Add that to the 2.25 tonnes of the LV-N and you've got four tonnes of "dead mass". You work the Tsiokolvsky Rocket Equation (delta-V = ln (M/Md) * 9.81 * Isp) backwards (warning: higher intensity math ahead)...

1070 = ln(M+4/Md+4) * 9.81 * 800

ln(M+4/Md+4) = 0.13634

(M+4/Md+4) = e^0.13634 = 1.14607

M+4 = 1.14607Md + 4.58423

M = 1.14607Md + 0.58423

Youu then look for a fuel tank/combination of fuel tanks where the final requirement for M is less than what's actually there. The fueled mass of all the main tanks in the game (except for the Oscar-B, I think) is nine times that of the dry mass, so...

M = 9Md = 1.14607Md + 0.58423

7.85393Md = 0.58423

Md = 0.07438 - going through the dry mass of the tanks, that's slightly more than an FL-T100, so an FL-T200 should have enough fuel for the transfer.

And just to check...

M = 1.14607(*.125) + 0.58423 = 0.727, M=1.125 > 0.727, so yes an FL-T200 has enough fuel to make the burn.

M = 1.14607(*.0625) + 0.58423 = 0.6556, M=0.5625 < 0.6556, so no an FL-T100 doesn't provide enough fuel.

With me so far?

Now let's up the payload to 11.75 tonnes, bringing the dead mass to 14 tonnes. Same process:

1070 = ln(M+14/Md+14) * 9.81 * 800

ln(M+14/Md+14) = 0.13634

(M+14/Md+14) = e^0.13634 = 1.14607

M+14 = 1.14607Md + 16.04498

M = 9Md, 9Md = 1.14607Md + 2.04498

7.85393Md = 2.04498, Md = 0.26038 ... which is slightly more than the dry mass of an FL-T400 tank. Tack on an FL-T100 to the FL-T400 and you're good to go.

Same distance, same amount of delta-V, but the increase in payload required a lot more fuel...

So what if we were to go with a different engine, let's say an LV-909? 0.5 tonnes, but only 390 Isp? Well...let's go with the 1.75 tonne payload case. Our dead mass is 2.25 tonnes. Here we go again:

1070 = ln(9Md+2.25/Md+2.25) * 9.81 * 390

ln(9Md+2.25/Md+2.25) = 0.27967

(9Md+2.25/Md+2.25) = e^0.27967 = 1.322697

9Md+2.25 = 1.322697Md + 2.97607

7.6773Md = .72607

Md = 0.09457 ... again, an FL-T200 is more than enough while an FL-T100 is not enough.

Increase the payload by ten tonnes...

1070 = ln(9Md+12.25/Md+12.25) * 9.81 * 390

ln(9Md+12.25/Md+12.25) = 0.27967

(9Md+12.25/Md+12.25) = e^0.27967 = 1.322697

9Md + 12.25 = 1.322697Md + 16.2030

7.6773Md = 3.953

Md = 0.5149 ... an FL-T800 wouldn't be enough, but you'd have enough if you tacked on an FL-T100. That's still more fuel than you would need with the nuclear engine case.

Hopefully these help better illustrate the answer to the question; I definitely hope I haven't confused anybody.

[Specialist290]

If you really want to make your head spin about numbers, check out Nyrath's Atomic Rockets. It goes over the actual rocket science of this stuff, and it's great for getting a good understanding of this (especially combined with KSP play). Just be prepared for some math!

Agreed; I generally link to the "Mission," "Engines," and "Blast Off" subpages whenever the topic comes up. Nyrath is good at explaining these things in terms that anyone can understand.

Another resource I'd highly recommend (to add something a little new to the discussion) is WernherVonKerman's

. The "delta-v = movement points" analogy isn't perfect, but it works well enough to get the concept across, and he spends a little time going into further detail as the video progresses.

[Fizwalker]

Thank you all for your replies! What I have gotten from this is that the dV is constant and that to get to Minmus or the Mun I have to have a ship capable of developing that constant. Mass is irrelevant to that constant, in that it determines thrust and fuel needed to get there.

Capi, I think I am following you but I would like to confirm some variable definitions... (I'm an ex-soldier so bare with me....)

M= "Wet" Mass, i.e. the mass of the ship with fuel and consumables with the mass of the structure = Initial Mass

Md= Dry Mass. i.e. the mass of the structure with empty tanks. = Final Mass

The appropriate equation is dV= Effective Exhaust Velocity*ln(Initial Mass/Final Mass)

and Effective Exhaust Velocity can be written as Isp*Standard gravity, where Isp is the specific impulse expressed over a time period.

Did I get the variables correct?

[capi3101]

@Fizwalker: Yeah, that's correct. The Isp ratings of the various engines are listed over at the wiki and so I tend to think in those terms.

Bear in mind that standard gravity is treated as a constant and that it is 9.81 m/s^2 regardless of the planet/moon involved; there have been folks on these forums that have gotten confused about that in the past, which is the only reason I bring it up.