# The "reverse rocket equation" explained

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Here's my own example of figuring out how much fuel to put on a Munar probe. I first calculated what my ballpark delta-v for a 10000m orbit would be, assuming that the orbital velocity is ~ the delta-v for achieving that orbit on an atmosphereless planet.

`Vorbit = sqrt{((Mass of Mun)(Gravitational Constant))/(Radius of orbit around center of Mun)}`

About 557 m/s. This probe is expected to land and launch, so twice this is a minimum. I then used my highly dubious math skills to find my fuel ratio needed when using a Terrier engine.

`(Mwet/Mdry)=e^{delta-v/(Isp*9.8)} ORPercent of initial total mass that is fuel = 1 - e^-(delta-v/(ISP*9.8)`

About 29% of my final total mass should be fuel for my probe. So, my probe was 1.2 tons empty, so I made it ~ 1.55 tons full. And sure enough, this was just enough fuel to take it on a round trip for a munar landing and back to the mothership. (I would have preferred a bit more fuel for comforts sake, but it worked out.)

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All,

Since the release of KSP1.0, the g0 figure is now no longer approx. 9.82. It has been corrected back to it's proper value of 9.8066.

Also, check out Meithan's excellent charts here. They replace Tavert's Julia code and are updated for KSP 1.02.

Best,

-Slashy

Edited by GoSlash27

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A late late update!

I have integrated thrust requirements in my application of the reverse rocket equation, but haven't talked about that here.

Designing a stage with a minimum acceleration requirement involves a couple additional steps, but it's all pretty logical.

First, you figure out how much mass one of your proposed engines can lift at the desired acceleration rate.

Next, you figure out how much of this theoretical rocket needs to be fuel in order to achieve your DV budget.

Given the required fuel mass (and thus tank mass) and the mass of 1 engine, you figure how much payload one of these engines could handle in this mission.

Finally, you scale this to your actual payload requirement and proceed as usual.

It works like this:

I wish to design an upper stage for a 20t payload. It needs to generate 1800 m/sec DV at no less than .7g acceleration. I will simultaneously check all engines, but for this exercise I will look at the Terrier.

Specs: Isp= 345s T= 60kN M=0.5t

Since this is an upper stage, I will assume vacuum figures.

How much mass can a Terrier lift at .7g?

M=T/(g0*a)

where

M=theoretical rocket mass in tonnes

g0= local surface gravity (9.81 m/sec^2 in this case)

a= acceleration requirement

60/(9.81*.7)= 8.74 tonnes

How much of this theoretical rocket needs to be fuel in order to achieve 1800 m/sec DV?

Rwd= e^(DV/9.81Isp)

Rwd= 2.718^(1800/(9.81*345) = 1.70

Fuel fraction= (Rwd-1)/Rwd

Fuel fraction= (1.70-1)/1.70 = .412. 41.2% of our theoretical rocket is fuel.

How much of our theoretical rocket is tanks to hold the fuel?

Fuel fraction/ Rfe= tank fraction

.412/9 = .0458. 4.6% of our theoretical rocket is tankage. Adding them together yields 41.2+4.6= 45.8% of our theoretical rocket is fuel and tankage.

Multiplying it by the mass of our theoretical rocket,

.458*8.74= 4.00 tonnes of our theoretical rocket is fuel and tanks.

How much payload could one of these engines handle in this requirement?

Total mass- fuel and tank mass- engine mass= payload mass.

8.74t - 4.58t - .5t = 3.66t payload.

This needs to be scaled up accordingly to meet our 20 tonne payload requirement, so we need (rounding up to the nearest whole number of engines) 20/3.66= 6 Terriers to do this job.

Given this figure, we proceed normally.

Best,

-Slashy

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Speaking of rearranging equations... You can rearrange the TWR equation to give you the kN needed on the pad for a given rocket mass for a given TWR. While this isn't nearly as sophisticated as the previous examples it can give you a quick idea of what you need for acceptable launch conditions.

TWR = (kN / (mass in kg * g0)) * 1000 (You multiply by 1000 to translate thrust in kN to N.)

Rearranging this provides you with the following formula:

kN = (TWR / 1000) * g0 * mass in kg

You can shorthand this for a certain body, Kerbin for instance, by precalculating in your desired values. Let's say I'm looking for an on-pad TWR of 1.3 on Kerbin (typical for any of my stock launchers):

kN = (1.3 / 1000) * 9.81 * mass in kg

kN = 0.012753 * mass in kg

If I have a 40 ton rocket I would need 0.012753 * 40000 = 510.12 kN to achieve my desired TWR of 1.3 on Kerbin. A single Skipper would fit the bill.

Also cool, once you've boiled down the shorthand you can rearrange again to see that a Skipper at 1 atm, with a thrust of 568.75, could lift 568.75 / 0.012753 = 44597.34964 kg, or 44.5 tons from Kerbin with a TWR of 1.3.

Remember that these results must include the mass of the engine, so that Skipper is actually only lifting 41.5 tons because it masses 3 tons.

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How do I use the rocket equation for a specific destination requiring a certain delta-v if I know nothing but the dV and the mass of my payload?

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[quote name='MAFman']How do I use the rocket equation for a specific destination requiring a certain delta-v if I know nothing but the dV and the mass of my payload?[/QUOTE]

1. Pick an engine (cluster); then you know Isp too.
2. Use the rocket equation to work out how much fuel you need.
3. If the resulting vehicle doesn't have the TWR required, start again (specifically for launch vehicles, but also prevents horrendous orbital burn-times).

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On 11/21/2015, 5:20:14, MAFman said:

How do I use the rocket equation for a specific destination requiring a certain delta-v if I know nothing but the dV and the mass of my payload?

MAFman,

The way I do it is as described above.

1) figure out how much mass a single engine can handle by converting it's thrust into weight at my desired acceleration.

2) Run the reverse rocket equation to model the single engine example

3) scale up the single engine example to match my desired payload.

Best,

-Slashy

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On 12/14/2014 at 1:19 AM, GoSlash27 said:

NOTE:

g0 has been corrected to 9.81 in KSP's engine as of version 1.0. Please disregard the "9.82" stuff from these earlier equations.

Small alteration needed. From the changelog for 1.2:

```Now always use g0 = 9.80665 and G (big G) = 6.67408e-11 for gravitational constants.
```

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