Rocket design and maths.

[LN400]

First, let me just say this is not about cost efficiency or dv efficiency as such. The topics are:

How to design a rocket based on pre-considered performance demands in terms of TWR and dv.

How to calculate payload mass for a given engine or cluster, given the demands for TWR and dv.

This topic is all about the maths that go into calculating this before hitting the VAB, for no other reason than the fun of it.

The main topic though is:

If you have any maths that is useful in building, that goes beyond the rocket equation, and that might be new to someone, then please share.

 

I have just started out looking into this. First I wanted to look at what limitations a given engine has in terms of performance. In particular, I wanted to know what the maximum payload for a given engine is, when considering performance requirements. For my example, I have a 909 driving an upper stage + payload on top.

The specifications for the 909 (vaccuum) are:

Thrust max: 60 kN (T)

I_sp: 345s

Mass: 0.5t

 

For my upper stage, I decide on:

Required dv: 2000 m/s (dv)

Required TWR: 1.8 (t)

 

From T, one can calculate maximum total mass (wet mass W) for the stage + payload that will give the required TWR.

W = T / (t * 9.80665) = 3.3990540...

I will use W = 3.399.

The rocket equation says

dv = I_sp * 9.81 ln(W/D) (equation 1)

where D is the dry mass. Furthermore, we have that

W = m_payload + m_engine + m_{fuel tank} + m_fuel (equation 2)

D = m_payload + m_engine + m_{fuel tank} (equation 3)

 

The rocket equation can then be rewritten

dv = I_sp * 9.81 ln [1 + m_fuel / (m_payload + m_engine + m_{fuel tank})] (equation 4)

 

We set

r = dv / (I_sp * 9.81)

and solve the equation with respect to the ratio W/D (R):

R = e^r (equation 5)

Substituting R with 1 + m_fuel / (m_payload + m_engine + m_{fuel tank}) we can rearrange and solve this equation with respect to m_fuel

m_fuel = (m_payload + m_engine + m_{fuel tank})(e^r - 1) (equation 6)

 

In the stock game, the standard tanks have a standard ratio between the mass of a tank and the mass of fuel in a full tank. The m_{fuel tank} of the 4.5t tank is 0.5t, or 1/9 the mass of a full tank. The m_fuel is 8/9 the mass of a full tank. The mass of the tanks increase/decrease linearly as you go from one size to another. If we call the 4.5t tank a 1/1 tank, then each smaller step in size is half the size of the bigger one so that in stock game, the 1.25m tanks come in these sizes.

1/1, 1/2, 1/4 and 1/8 with the masses of 0.5t * 1/1, 0.5t * 1/2, 0.5t * 1/4 and 0.5t * 1/8.

In my game, I also have access to a 1/16 tank (mod is MRS) and my calculations are based on that. So, this tank has a mass of 0.5 * 1/16 = 0.03125t.

One can not have anything other than whole tanks, so the total mass of 1/16 tanks becomes:

m_\fuel tank = n * 0.03125t (equation 7)

where n is a positive integer.

 

Each tank's fuel level can be adjusted in 10% increments. For the 1/1 tank, the total mass of fuel is 4t. For the 1/16 tank the fuel mass is therefor 0.25t. 10% increments means an increment in mass of 0.025t. This is the smallest amount one can adjust the fuel level, using standard series tanks (Oscar-B and ROUND-8 have a different ratio). This means the total mass of fuel must be an integer k times 0.025t.

m_\fuel  = k * 0.025 (equation 8).

Back to equation 6, we can now rewrite it as

k = (m_payload + m_engine + m_{fuel tank})(e^r - 1) / 0.025 (equation 9)

 

The total wet mass W is the sum of payload mass, fuel mass, engine mass and tank mass (ref equation 2). That gives us that the payload mass

m_payload = W - m_engine - n * 0.03125 - k * 0.025 (equation 10)

Substituting for m_payload in equation 9 and cancelling out adding and subtracting terms, we get

k = (W - k * 0.025)(e^r - 1) / 0.025 (equation 11)

Solving equation 11 with respect to k gives

k = W(e^r - 1) / (0.025 e^r) (equation 12)

k is then rounded up to nearest positive integer.

 

Since a tank can hold no more than 100%, for k = 10, this means that for a k from 1 to 10 requires 1 tank, k from 11 to 20 requires 2 tanks, k from 21 to 30 requires 3 tanks and so on. One can therefor use the value of k to determine n in equation 7 .

k from 1 to 10 -> n = 1

k from 11 to 20 -> n = 2

etc

With k and n known, we can then solve equation 10 to find the payload mass that will give us the required dv and TWR.

 

Conclusion:

For the 909, I found that

k = 61 from equation 12.

n = 7 from equation 7.

W = 3.399

Payload mass: 1.15525t rounded to 1.155t from equation 10.

Since n is the number of 1/16 tanks, I then substituted 4 of them with a 1/4 tank and 2 with a 1/8 tank. The remaining 1/16 tank had 10% fuel in it.

These calculations were done before building. The result of the build was:

dv = 2015 m/s

TWR = 1.80.

 

These steps can be used when you want to look at the capabilities of a given engine or if you simply want to nerd out and design a rocket taylored for a particular payload. Most of the time though, you know the payload and the challenge is to find the right engine and the right amount of fuel that will give you the required dv and TWR.

Edited September 16, 2016 by LN400

[GoSlash27]

 

Hooray for spreadsheets!

-Slashy

[Kerbart]

19 minutes ago, GoSlash27 said:

Hooray for spreadsheets!

-Slashy

I'm more headed towards Python scripts these days. But hooray for automated calculations.

May I also congratulate you on being friendly on the eyes in your post, and using e^x+1 exponential notation instead of the ugly e^(x+1) clutch? We have superscript, why not use it after all, right?

[OhioBob]

Here's the method I use...