Comments on: The Cold Equations

By: TB

TB — Sat, 18 May 2013 19:52:55 +0000

Another good reference:

Rocket Propulsion Elements. Not cheap, which is why mine is still a 4th edition.

By: ER

ER — Sat, 18 May 2013 18:47:22 +0000

My example is meant to be assembled and launched from deep space, with no gravity, no air resistance, and an environment where things like thrust/weight ratios are irrelevant (you just burn for a longer period). And of course, it doesn’t have to stop when it gets there, or turn around and come home, and stop again when it gets back.

The Starflight Handbook has an excellent discussion of how the mass ratio varies from stage to stage and how it should be optimized for overall performance in a staged rocket launched from earth; and how factors like fuel density must be taken into account. I’m learning a lot.

I’m aware of those factors, I was just trying to show how even the the ideal case was so damned hard.

By: TB

TB — Sat, 18 May 2013 18:24:23 +0000

Then comes the hard part.

Once you get past the theoreticals, you deal with things like multi-stage calculations, tank weights and materials, propellant mix ratio issues, drag issues from vehicle size, development costs of the technology, development times, propellant costs and handling issues, modifying ideal trajectories for things like Max Q, and dozens of other things that pull you away from the “ideal” numbers.

Somewhere in the middle of that cloud of multiple variables is a workable and cost-effective design.

By: ER

ER — Sat, 18 May 2013 17:45:25 +0000

So, how do we know the most energy efficient fuel load for a rocket is 3.92 times the payload mass? And how do we know that efficiency is 65%, and the final velocity is 1.6 times the exhaust velocity at that fuel load?

Review:

Vf = Ve ln [ (M1+M2) / M2 ]

Where Vf is the payload’s final velocity when the propellant is all used up
M1 is the mass of the propellant
M2 is the mass of the payload
and Ve is the exhaust velocity.

First we normalize the mass of the empty rocket (M2) and set it equal to 1. (One rocket, or one payload. No cheating here, it’s like saying the Enterprise has a mass of exactly “one Enterprise”). When we solve for M1 it will be in terms of M2.

When the rocket runs out of fuel, it will have a speed Vf.

Its kinetic energy at that time is by the vis-viva equation: KE = 1/2(M2)Vf**2 or KE = 1/2 Vf**2

The amount of energy in the fuel is FE = C M1, where Capital C is just a constant that depends on what kind of propulsion system we have.

We shall define the Efficiency of our fuel usage as KE/FE. We want to pick an amount of fuel M1 which will maximize the Efficiency.

Since we have the equations for KE and FE, we know Efficiency = (1/2 Vf**2) / (C M1)

But we know from the rocket equation that Vf = Ve ln [ (M1+M2) / M2 ] or Vf =Ve ln [ (M1+1) / 1 ]

So substituting into the Efficiency equation

KE/FE = 1/2 ( [ Ve ln (M1+1)] **2 ) / ( C M1)

We clean up a little by collecting the 1/2, Ve**2 and 1/C constants and lumping them together and calling them Z = Ve**2/2C. Regardless of which values we wind up with for the Efficiency and M1, none of these guys change, they are all constant engineering parameters peculiar to your propulsion system.

KE/FE = Z(ln(M1+1))**2 / M1

This is an equation giving the Efficiency in terms of the Fuel Mass. Plug in any fuel mass you want and it gives you its Efficiency.

You can use calculus, take the derivative and set it to 0, and then solve for M1, or you can solve it brute force by substituting different values for M1 until the Efficiency is maximized.

at M1 = 3.92, KE/FE =0.64761 Z and Vf = Ve ln (3.92 + 1) = 1.593 Ve