Calculate needed fuel fraction

[twotoes02]

Hey gang, Two Toes here.

I was hoping to get some help with a bit of rocket science I've tripped up on. I'm trying to calculate the mass fraction of fuel a rocket will need to carry based on the needed delta v and a few other variables.

Let me show you what I have so far:

This derivation provides the delta v available for a single stage rocket knowing g, the specific impulse of the engine, the starting mass (mo), the burn out mass (mbo) and the thrust to weight ratio R.

What I'd like to do (and this is where I'm having trouble) is redefine the equation in terms of mbo. With this I can determine the mass fuel fraction (mff) of the vessel knowing the thrust, specific impulse, delta v and starting mass. The issue as you can see is that one of the mbo's is located within a natural log.

Does anyone know of another derivation that allows me to calculate the burn out mass with these (or more) variables? Other suggestions are more than welcome of course.

Thanks!

Edited May 11, 2013 by twotoes02

[Jason Patterson]

Do you have a question? The first part of the equation is Tsiolkovsky's rocket equation.

Is the second part your attempt to include TWR? I'm kind of confused about its purpose. If you're trying to calculate an effective delta-v by subtracting out part of the entire quantity to compensate for gravity drag, this won't work. You'd have to know how long the engines were burning upward and the characteristics of the gravity turn (assuming there is one) and all of that sort of thing. Additionally, as the rocket moves upward, its TWR changes because the force of gravity decreases. On several bodies you'd also have an atmosphere to deal with.

[nyrath]

twotoes02, the opposite of the natural log of a number is called e^x or natural antilog.

For example:

deltaV = (g*Isp) * ln[massRatio]
deltaV / (g*Isp)  = ln[massRatio]
massRatio = e^(deltaV / (g*Isp))

You sort of confused me because I'm used to seeing the letter R to symbolize Mass Ratio, not thrust to weight ratio.

Now, below, Mp is the mass of the propellant, Me is the mass of the rest of the rocket, deltaV is delta V, Isp is specific Impulse

MassRatio = (Mp/Me) + 1
deltaV = (g*Isp) * ln((Mp/Me) + 1)
deltaV / (g*Isp) = ln((Mp/Me) + 1)
e^(deltaV / (g*Isp)) = (Mp/Me) + 1
e^(deltaV / (g*Isp)) - 1 = Mp / Me
(e^(deltaV / (g*Isp)) - 1) * Me = Mp

so the last equation will give you the mass of propellant, given the other variables.

The total initial mass is Me + Mp, divide the mass of the propellant by that to get the propellant fraction.

Edited May 11, 2013 by nyrath

[twotoes02]

Wow fast responses, that's awesome!

I do apologize for confusing you, its most likely in my clunky communication skills...

Jason if I may,

Let's assume the rocket is already in space and I'm using this derivation to determine the delta v needed for an orbital maneuver. I derived this equation from William Tyrrell Thomson's "Introduction to Space Dynamics" equation (8.1-10) on page 242 of the second edition (I believe its the second edition...). Now knowing this, if you believe TWR is not needed under these circumstances then we can certinally modify the equation or find another derivation that will serve my purpose better.

And it seems that I may have not posted my question clearly. Please allow me to restate the question. Is it possible to rewrite the equation to solve for mbo knowing delta v, and the other variables? Or is there another derivation that would better suit my purpose. However, it seems that nyrath may have answered my question.

nyrath, yes I certainly do understand the natural log and its opposite, the problem here is that one iteration of mbo is within a natural log and the other is not.

I am a little confused unfortunately, you stated that R represents the mass ratio where the texts I have at hand show R as the thrust to weight ratio. Written as R = T/(m*g) where T is the thrust of the rocket. However it seems you have provided me another equation that will better suit me and my need! I do thank you good sir, please allow me to try this out and I will let you know if this is the answer I am looking for.

Thank you for your time. I'll see if this answer is what I'm looking for.

[tavert]

Edit: Ninja'd by OP. Might look for that book, maybe the relevant page can be previewed in Google books or something. The first term in the equation is Tsiolkovsky's rocket equation which most of us here are familiar with. We're just not sure where the second term came from.

Follow nyrath's math (though natural antilog is a silly name for what I always just call the exponential function). One thing is you may not know your empty mass Me because you don't know how many fuel tanks you need to add and they add dry mass. If you use fuel tanks as big or bigger than the FL-T200 (so none of the small Oscar-B or Round-8 tanks), then all the tanks in KSP have the same propellant fraction of 8/9. So if you have a final payload of mass Ml, then you can add tanks such that:

Mt = Mp/8, where Mt = dry tank mass, Mp = propellant mass

Me = Ml + Mt = Ml + Mp/8, where Me = empty mass, Ml = payload (non-fuel-tank) mass

(e^(deltaV / (g*Isp)) - 1) * (Ml + Mp/8) = Mp

(e^(deltaV / (g*Isp)) - 1) * Ml = Mp * (1 - (e^(deltaV / (g*Isp)) - 1)/8)

(e^(deltaV / (g*Isp)) - 1) * Ml / (1 - (e^(deltaV / (g*Isp)) - 1)/8) = Mp

Edited May 11, 2013 by tavert

[nyrath]

though natural antilog is a silly name for what I always just call the exponential function

Yes, it is true that is its real name. But when trying to explain it is not intuitively obvious that the exponential function is the inverse of natural logarithm. Which is what you have to do when solving for Mp

[purpletarget]

Let's assume the rocket is already in space and I'm using this derivation to determine the delta v needed for an orbital maneuver. I derived this equation from William Tyrrell Thomson's "Introduction to Space Dynamics" equation (8.1-10) on page 242 of the second edition (I believe its the second edition...). Now knowing this, if you believe TWR is not needed under these circumstances then we can certinally modify the equation or find another derivation that will serve my purpose better.

The page you mention isn't available for preview, but 241 is. So looking at the initial steps of what I think was leading to your OP formula, the equation isn't the Delta-V proper, but rather the max vertical velocity. So the first part is the Rocket Equation, and everything after the Minus is trying to account for Gravity Drag because the equation assumes vertical flight from Earth. It's an equation dealing in performance more than characteristics.

So, in terms of figuring out your actual Rocket Delta-V, particularly for orbital maneuvers, just use the Rocket equation. ( dV = Isp*g0*ln(m1/m2) ) and drop the second part after the Minus.

After that, rearranging for your unknown mass ratio should be pretty straight forward.

[Jason Patterson]

I'm not 100% certain, but I don't believe that that function, delta-v(Mbo), has an inverse. I think that's what you're after, but it just doesn't look to be possible without resorting to infinite series.

[tavert]

I'm not 100% certain, but I don't believe that that function, delta-v(Mbo), has an inverse. I think that's what you're after, but it just doesn't look to be possible without resorting to infinite series.

You can express it in terms of the Lambert W function (also known as product log). Named after the same guy as the orbital transfer Lambert problem, funny enough.

http://www.wolframalpha.com/input/?i=solve+a+%3D+b+ln%28x%29+%2B+c%2Fx%2C+for+x