Delta-V Calculation Question

[Endersmens]

So, say I don't wanna find out the ships wet and dry mass, and use all that to calculate exact delta-v. Say I just want a quick, rough delta-v estimation. Could I just take the average acceleration, and multiply it by the seconds of acceleration?

For example, say i have a ship that I know starts accelerating at 1g, and burns out at 2g. And I know it gives 60 seconds of thrust before burn out. Could I take (1.5*9.8m/s)*60=882? Would 882 be a valid rough Delta-V Estimation for this ship?

[The_Duck]

Yes, dV = (average acceleration due to thrust) * (burn time). This is an exact equation. However note that your vessel can accelerate for other reasons than thrust, for example gravity and drag. You have to factor these out before you compute the dV in this way. That is, the acceleration you need to use is the average acceleration in 0g with no air resistance.

[Endersmens]

Yes, dV = (average acceleration due to thrust) * (burn time). This is an exact equation. However note that your vessel can accelerate for other reasons than thrust, for example gravity and drag. You have to factor these out before you compute the dV in this way. That is, the acceleration you need to use is the average acceleration in 0g with no air resistance.

Alright, cool.Well like. People say returning to orbit on the Mun takes ~550 m/s. If this equation came up to 560 m/s, that would be enough to get to orbit right? That's why i need the delta-v, for transfers and such. I shouldn't have to factor in gravity for this, since I will be orbiting, therefor at zero-g right?

[Taki117]

Alright, cool.Well like. People say returning to orbit on the Mun takes ~550 m/s. If this equation came up to 560 m/s, that would be enough to get to orbit right? That's why i need the delta-v, for transfers and such. I shouldn't have to factor in gravity for this, since I will be orbiting, therefor at zero-g right?

Not quite true. You are still under the effects of gravity, 9.81m/s to be precise. the reason you are "weightless" in space is that you're moving horizontally faster than you are moving vertically. Basically to maintain orbit you have to be moving sideways at at least 9.82m/s as you fall 9.81 m/s.

Keep in mind that scenario is for an infinitesimally small gravitational body. Generally orbital velocity is much faster because you have to miss the ground.

Maybe this XKCD will help explain it better.

[Red Iron Crown]

If you know the beginning and ending acceleration, you can also run the rocket equation using the ratio of those instead of the masses:

dV= Isp * 9.82 * ln(a1/a0)

Where a0 is the starting acceleration and a1 is the final acceleration. Calculating it this way works for burns at less than full throttle, which extend burn time but don't increase dV.

[Endersmens]

Not quite true. You are still under the effects of gravity, 9.81m/s to be precise. the reason you are "weightless" in space is that you're moving horizontally faster than you are moving vertically. Basically to maintain orbit you have to be moving sideways at at least 9.82m/s as you fall 9.81 m/s.

Keep in mind that scenario is for an infinitesimally small gravitational body. Generally orbital velocity is much faster because you have to miss the ground.

Maybe this XKCD will help explain it better.

Well, I understand that. What i'm saying is, if the delta-v to orbit on Mun is 550, and say the equation I suggested comes out to 551. With near perfect Piloting, the craft will make orbit, no?

If you know the beginning and ending acceleration, you can also run the rocket equation using the ratio of those instead of the masses:

dV= Isp * 9.82 * ln(a1/a0)

Where a0 is the starting acceleration and a1 is the final acceleration. Calculating it this way works for burns at less than full throttle, which extend burn time but don't increase dV.

Ah. What is In?

Also, the equation I used also works for burns at less than full throttle, the burn time would increase, but the acceleration would decrease rationally with the increase of burn time, from the full throttle equation. So it would still balance out. The only limitation is the burn simulation has to be run at a set throttle. Although, with your equation, your starting throttle would have to match your ending throttle to get proportional a1 and a0 measures. However, the throttle in between does not matter, as it does in my equation. (note: I'm not claiming ownership of the equation, although i came up with it on my own, I'm just using it as a way to specify which equation)

[Red Iron Crown]

Well, I understand that. What i'm saying is, if the delta-v to orbit on Mun is 550, and say the equation I suggested comes out to 551. With near perfect Piloting, the craft will make orbit, no?

In theory, yes. In practice, no. Maneuver nodes and most dV maps assume an idealized instantaneous burn which is impossible, any time spent burning away from the moment increases the dV required to change the orbit in the desired way. Normally not that big a deal unless your TWR is really low, but a 0.2% margin of error would likely get eaten by it.

Ah. What is In?

The natural logarithm, it's LN not IN. Available on any scientific calculator or spreadsheet, difficult to calculate in your head.

Also, the equation I used also works for burns at less than full throttle, the burn time would increase, but the acceleration would decrease rationally with the increase of burn time, from the full throttle equation. So it would still balance out. The only limitation is the burn simulation has to be run at a set throttle. Although, with your equation, your starting throttle would have to match your ending throttle to get proportional a1 and a0 measures. However, the throttle in between does not matter, as it does in my equation. (note: I'm not claiming ownership of the equation, although i came up with it on my own, I'm just using it as a way to specify which equation)

That sounds right to me.

Edited August 27, 2014 by Red Iron Crown

[FleshJeb]

EDIT: Nevermind, RIC covered it.

[Endersmens]

In theory, yes. In practice, no. Maneuver nodes and most dV maps assume an idealized instantaneous burn which is impossible, any time spent burning away from the moment increases the dV required to change the orbit in the desired way. Normally not that big a deal unless your TWR is really low, but a 0.2% margin of error would likely get eaten by it.

The natural logarithm, it's LN not IN. Available on any scientific calculator or spreadsheet, difficult to calculate in your head.

That sounds right to me.

Ah, thanks for clearing all that up. (and about the munar orbit thing, i understand about maneuver nodes assuming instantanious and perfect burns and all that, that's why i said "near perfect piloting") I only wanted to know this so I could make rough delta-v assessments of my munar landers transfer stages and all that in stock. (no i'm not a stockaholic, I'm actually a modaholic, but I keep one copy of KSP stock, and vow to play through the career in every version stock first, before doing it with mods. This only includes updates that change career mode, of course.)

And about all that math stuff, yeah I love math. But lets test it for fun to make sure my logic is correct. Say we have a craft with an average acceleration of exactly 1 g at full throttle. And the burn duration is exactly 10 seconds. Using the equation I used, it would be 1*9.81m/s*10= 98.1 m/s. Now, say The burn is done at 90 percent throttle. The average acceleration would be 0.9g. Since the throttle got decreased 10 percent, due to a rule in math that i can't remember, the burn time would have to increase by 10 percent. so, 11 seconds burn time. 0.9*9.81m/s*11=

Well crap. It didn't work, it came out to 97.1. Done with .08 and 12, it comes out to 94.2. Can someone help? Or explain what i'm doing wrong?

[Red Iron Crown]

Check your burn time calculation, should be ~11.111s at 90%.

0.9*9.81*11.111=98.1

[Endersmens]

But...how do you get 11.111? .1 percent of 10 is 1. We decreased the throttle by 10 percent. Shouldn't the acceleration decrease by 10 percent? And the burn time increase by 10 percent?

[Jouni]

The average acceleration is a bit messy to calculate. Your (a0 + a1) / 2 overestimates it, leading to too optimistic delta-v figures.

Let's start from one interpretation of specific impulse. Assuming that the thrust of the engine is F, then Isp in seconds is the time the engine uses to burn amount F/g of fuel. Therefore the burn time for m0–m1 fuel is t = Isp * g * (m0–m1) / F. Because m0 = F/a0 and m1 = F/a1, the burn time is

t = Isp * g * (1/a0 – 1/a1).

Let's plug that into Red Iron Crown's version of rocket equation:

dv = Isp * g * ln(a1/a0) = t * ln(a1/a0) / (1/a0 – 1/a1) = t * a0 * a1 * ln(a1/a0) / (a1–a0).

The a0 * a1 * ln(a1/a0) / (a1–a0) part is the average acceleration.

[Endersmens]

oh.

Weeeelllllllllll then ima go plug it all in and see it work! Thanks for the help!

[Endersmens]

Ok, so for a craft whos starting acceleration is 1g and ending is 2g, the average acceleration is 1.39. (Rounded) now how would my equation work? as asked in my second to last post.

[Yasmy]

It is not true that the average acceleration is (initial + final acceleration)/2. Your formula might work fairly well for some rockets, but not for all. This is because the acceleration is not a linear function of time.

Assumptions: constant thrust (F), Isp, and fuel consumption (dm/dt = c).

Let the burn time be T, the initial mass be m0 and the final mass be m1 = m0 - c * T.

a(t) = F / m = F / (m0 - c * t)

The average of a(t) is a_avg = 1/T * (the integral of a(t) from 0 to T). You should not be surprised to discover this gives the rocket equation:

a_avg = 1/T (F/c ln(m0/(m0 - c*T)) = F/(m0-m1)/T * ln(m0/m1) = Isp/T * ln(m0/m1)

Then, like The_Duck said, dV = average acceleration times time: dV = Isp * ln(m0/m1).

So when will your formula work well? For small mass loss: (m0-m1) << m0. Then ln(m0/m1) is approximately linear in mass loss: ln(m0/m1) ~ (m0-m1)/m0. And since mass loss is linear in time at constant thrust, for small mass loss, the average acceleration is also approximately linear.

TLDR: If you use a small amount of your total mass, dV ~ (final - initial acceleration)/2 * time.

But if you use a bunch of your mass, you have to use the rocket equation.

Edit: ninjad!

Edited August 27, 2014 by Yasmy
factor of 1/2

[Red Iron Crown]

But...how do you get 11.111? .1 percent of 10 is 1. We decreased the throttle by 10 percent. Shouldn't the acceleration decrease by 10 percent? And the burn time increase by 10 percent?

The burn time for the 90% burn is calculated as 10s/.9. Easier to see this is correct for a 50% thrust burn, it doesn't burn 50% longer, it burns 100% longer.

I totally missed the incorrect average acceleration calculation though, Jouni and Yasmy know their stuff.