Delta-V Problems

[DunaManiac]

Hi everyone this is KerbalXcer. Today I want to talk about delta-v. For some reason I cant get it about delta-v. I've done all the calculations, but then I try to calculate the final equation, It just doesn't work. Like I watched this video, and he went so fast, and I thought "Okay I have to multiply it all together." So I multiplied 345*9.807*0.141 (the example in the video) and I calculated it, they got 1,400 m/s DV, But I got 477. Can someone help me with this? Thank you

[Gaarst]

Where da video at?

Edit: anyway, delta V is pretty straightforward to calculate. It's given by: dV = g0 * Isp * ln(m1/m0)

• g0 is the standard gravitational parameter, it's 9.81 regardless of the body you're standing on
• Isp is the specific impulse
• m1 is the wet mass of your stages: this stage and all stages above fueled
• m0 is the dry mass of your stage: current stage empty from all fuel, all stages above fueled
• ln is a function: the natural logarithm

Edited February 2, 2017 by Gaarst

[Bill Phil]

I'm sorry, but I can't help without the video...

[Shpaget]

Well, 345*9.807*0.141 is not 1400, so you either missunderstood the guy, or he made a mistake.

[0111narwhalz]

Perhaps an understanding of the why will lead to an understanding of the what.

Tsiolkovsky's Rocket Equation is, as seen above, g_0 * I_sp * ln(m_0 / m_1), where

g_0 is the standard gravitational parameter,

I_sp is specific impulse,

ln(X) is the natural logarithm of X, and

m_0 and m_1 are the wet and dry masses, respectively.

Interestingly, specific impulse times g is exhaust velocity. Thus, an equivalent form is V_e * ln(m_0 / m_1). I will base this monologue on this form.

So, why is exhaust velocity important? Well, a change in velocity (/\ V) requires a change in momentum. However, momentum is conserved in a closed system. Thus, we must open the system. This is done by expelling mass. The velocity of the expelled mass determines how much mass carries how much momentum. How about the mass ratio and logarithm? Well, it's obviously important to know the mass if you want to know the acceleration, but that doesn't matter right now. What matters is how much total change in momentum (and thus velocity) you can do. And that's about ratios*. Now, why the logarithm? Well, as you chuck mass, you'll find yourself less massive. The change of velocity for the same change in momentum is greater because you're lighter. The logarithm models this.

*Take a moment to prove to yourself that it's all about mass ratios by letting two rockets burn themselves out. One shall be as two of the other, attached nose-to-nose but with one engine off. Burn both to completion and compare their /\ V. See figure below.

R1 »=>

R2 »=><=

[Streetwind]

9 hours ago, KerbalXcer said:

So I multiplied 345*9.807*0.141 (the example in the video) and I calculated it, they got 1,400 m/s DV, But I got 477.

The third number, where you have 0.141, is supposed to be the logarithm of a mass fraction. And for that purpose, it looks pretty small. In order to get this number, you'd have to have a rocket stage that is only 13% fuel. That's really, really miniscule. Like, that's the equivalent of mounting eight FL-T100 tanks and then using only one of them. I would say it's even somewhat difficult to accidentally get such a low mass fraction in KSP - just adding one single tank generally gives you more. You'd have to intentionally go out of your way to find a tank that's small enough to end up with so little.

Thus, are you sure you understood that youtuber correctly? Are you sure the number you should be multiplying here is in fact 0.141? If the youtuber got a result of 1400, then what he actually multiplied was 0.414, not 0.141. Did you just accidentally swap the numbers around?