# How do you calculate fuel usage ratio for rocket acceleration vs deceleration using a fixed total amount of fuel?

If a rocket is accelerating and then decelerating to stop at a destination (not necessarily continuous acceleration), and you want to use a given total amount of fuel, how do you calculate the amount of fuel to use on the acceleration burn vs the deceleration burn?

I get how if I have a desired peak velocity I can work backwards from the destination using the Tsiolkovsky rocket equation and get the required fuel to complete the trip, but in my example I want to use all of the given fuel and derive the delta v. Since I'm applying the equation in two parts, that leaves both mass change and velocity change unknown, and I don't know how to deal with that.

An example using this calculator

If I have this rocket:

specific impulse=300, dry mass=1000, fuel mass=100000

and I use half the fuel (50,000 kg) on an acceleration burn I'll end up going 2010.25 m/s

Because the fuel mass is reduced, I only need to use 25,247.5 kg of fuel slowing down, leaving me with unused fuel.

There is a ratio of fuel usage on acceleration vs deceleration that will consume all of the given fuel, but I can't find or figure out how to calculate it, can someone help me out?

• Note that, if you've only got 40 km/s of delta-V to play with, orbital mechanics, orbital mechanics is going to be a dominating factor in what speed you have when you arrive at any intended destination within the Solar System. Sep 2 at 11:57

The general answer is $$fuel\;fraction= 1-\frac{\sqrt{r} -1}{r-1}$$ where the fuel fraction is the fraction of the mass of fuel used in the first burn over the total fuel mass, and $$r$$ is the fuel mass ratio of the whole craft,( $$r = \frac{m_0}{m_1}$$ where $$m_0$$ is the total initial mass of the craft and $$m_1$$ is the dry mass of the craft)

My derivation:

$$m_0$$ = The initial mass of the craft. Payload + full load of fuel.

$$m_1$$ = The final mass of the craft. Just the payload, all fuel is expended.

$$m_{1/2}$$ = The mass of the craft at the halfway point

In order to accelerate to a given velocity and decelerate back down to zero, you must expend half your delta-V in each burn.

This can be expressed in terms of the rocket equation as follows: $$v_e ln \left( \frac{m_0}{m_1} \right) = 2v_e ln \left( \frac{m_0}{m_{1/2}} \right)$$ Or in plain english, the full burn has twice the delta V as the halfway burn. By doing some cancellation and exponentiating both sides, we end up with: $$\left( \frac{m_0}{m_1} \right) = \left( \frac{m_0}{m_{1/2}} \right)^2$$ This is a useful rule of thumb to remember in general. If you want to double the delta-V of a craft, you need to square the mass ratio. If your rocket is 1:10 payload : fuel and you want to double the delta-V, you need to make the ratio 1:100.

The quantity we are after to solve the problem is not $$m_{1/2}$$, the total mass after the first burn, it is the fuel fraction $$\frac{m_{1/2}-m_1}{m_0-m_1}$$, which is the mass of the fuel left after the first burn divided by the total mass of fuel on the craft.

From rearranging the previous equation, we find that $$m_{1/2} = \sqrt{m_0m_1}$$ and so we plug that into the fuel fraction to get: $$fuel \:fraction =\frac{\sqrt{m_0m_1}-m_1}{m_0-m_1}$$ and then we let $$r = \frac{m_0}{m_1}$$ and do some algebra that I am too lazy to reproduce in mathjax here to end up with: $$remaining \;fuel\;fraction = \frac{\sqrt{r} -1}{r-1}$$ which of course means that $$spent \;fuel\;fraction = 1-\frac{\sqrt{r} -1}{r-1}$$

Let's do some sanity checks. If your craft is the size of the Titanic and its propulsion system consists of two firecrackers. Each firecracker is a 'burn' in one direction. Since the overall mass hardly changes between burns, the expended fuel fraction is approximately 0.5.

If your craft is mostly fuel, you will have a high $$r$$ value. Since most of the delta-V comes from the last bits of fuel in the tank, you have to burn a huge proportion of the fuel to get to halfway. As $$r$$ approaches infinity, the fuel fraction for the first burn approaches 1.

To answer your original question with the numbers you supplied, r = 101, so therefore the fraction of fuel you need to burn to the halfway point is 90.95%

• This is cool. Nice graph. Sep 2 at 1:40
• @OrganicMarble Classic interstellar rocket physics, where you don't have to worry (so much) about gravity wells and proper motions and stuff. It's where I cut my teeth on rocket stuff, even before Kerbal. Sep 3 at 0:13

Interesting problem. I'll use your numbers, divided by 1000. The initial mass is 1+100=101 (kg, tons whatever) and final or empty mass is 1. The velocity change is the same magnitude. So, we will let $$M_0$$ be initial mass, burn propellant down to mass $$M_1$$, then the slowdown starts at mass $$M_1$$ and all burned is mass $$M_e$$.

The rocket equation is then:

$$v=c\ln(M_0/M_1) = c\ln(M_1/M_e)$$ g Just to define everything, $$c$$ is exhaust velocity ($$g$$ times specific impulse) and the fractions after $$\ln$$ are the mass ratios. The mass ratios have to be equal, so $$M_0\cdot M_e=M_1^2$$.

Choose$$M_1= \sqrt(101 \cdot 1)= 10.05$$. Fuel burn = 101-10.05=90.95. Mass ratio is 101/10.05=10.04975, rounds to 10.05. Second burn mass ration = 10.05/1=10.05.

• just noting for posterity that i believe this answer is correct, but i found Ingolifs's to be more informative so i'm accepting that one. thank you for your help W H G. Sep 1 at 17:54