How to Math?

[EliasDanger]

So I want to learn the maths, and I know there are people here who would have fun solving a problem and explaining how to go about figuring it out. So, if anyone is up for it:

I have a tug powered by two 800 ISP Nukes. This tug currently has a mass of 21,255 kg with 1462 units of liquid fuel out of a possible 3200. I'm about to dock back with my LKO space station. The next mission for my tug is to take a new station into orbit around the mun. I'm shooting for a 75 km orbit. Now, I don't want to take anymore fuel with me than I have to, other wise, I'm just wasting fuel. So I figure, 2,000 delta-v worth of fuel is a safe amount to get to the mun, drop off the station, and come back to LKO. Now the space station that the tug will be hauling has a mass of 11.3 Tons. The question is, how much liquid fuel should my tug have to have to have (English, baby!) 2,000 delta-v when it docks with the new munar space station?

Edited October 12, 2015 by EliasDanger

[Reactordrone]

Assuming you don't use any fuel docking with the new station you'll need 1463 units of fuel onboard for the combined stack to have 2000m/s so pretty much what you've already got in there. Of course your delta-v will go up once you drop off the station.

ETA-For how to do it, you need to know the empty mass of your ship with each unit of fuel weighing 5kg, 1462 units weighs 7.31 tonnes so your empty mass is 13.945t. Add the weight of your station to that to get a combined mass of 25.245t. Rearranging the rocket equation ,2000=9.82x800xln(mass with fuel/empty mass) your mass ratio needs to be 1.29 so 32.56t / 25.245t which is 7.32 tonnes of fuel.

Edited October 12, 2015 by Reactordrone
added some maths

[Crown]

I'm just throwing some links in here. It takes a little bit of reading for this topic.

https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

http://wiki.kerbalspaceprogram.com/wiki/File:KerbinDeltaVMap.png

So basically, the rocket equation is all you need. What you calculate with it is the mass of fuel, which means you need to convert into units. The KSP Wiki has information about fuel density.

What you do is two calculations. One has the mass of tug and station, and the other has solely the tug's mass. So it's fuel for bringing the empty mass of tug and station to Mun orbit, plus the fuel bringing the tug back to LKO.

[EliasDanger]

Hell yeah, thanks you guys for the speedy and thorough response! It's going to be nice being able to calculate this stuff myself instead of over budgeting everything cuz I'm not sure.

[EliasDanger]

So, if I understand correctly....if I wanted instead to have 5,000 dv when I dock with my station, I would need 4,504.8 units of fuel, which would be more fuel than my ship is capable of handling, so I would know I'd need a bigger ship.

In fact, the most delta-v my ship would have it I filled the tanks with 3200 units would be 3,848.67 delta-v. That math is right, right?

Another question I have is, you don't add ISPs if you have multiple engines? So, if you have a payload that weighs say 35 tons. You want to get that into orbit, and you want at least 4,000 delta-v VAC to do it. You're going to use 1 mainsail and 4 skippers. How would you go about figuring out how much fuel to bring to do that without trial and error and kerbal engineer? Cuz I wanna be smarter than that.

Edited October 12, 2015 by EliasDanger

[KerikBalm]

Fuel is 0.005 tons per unit.

Your Tug with 1462 units has 7.31 tons of liquid fuel. Its total mass is 21.255 Its dry mass is thus 13.945.

The station's total mass is 11.3 tons. Tug+ station is 32.555.

Thus its dry mass is 32.555-7.31 = 25.245

The rocket equation says you'd have: 800*9.81* ln (32.555/25.245) = 1995.768 m/s of delta V

If you wanted to have 5,000 dV....

5000= 800*9.81* Ln ((25.245+x)/25.245)

0.6371= ln ((25.245+x)/25.245)

1.891= (25.245+x)/25.245

47.738= 25.245 + x

22.49 = x X is how many tons of fuel... divide by 0.005 ot get how many units

4,498 units of fuel... ok, the difference must be rounding errors...

5,000 dV in a single stage is A LOT for the kerbin system, I wouldn't both trying for that much.

Conservatively, you'd need 1,300 m/s to put something in orbit around Mun (using 900 instead of 860 for the ejection, 400 instead of 310 for insertion and circularization)

so to calculate how much fuel you'd have left when you drop off the station...

1300= 800*9.81* Ln (32.555/(32.555-X))

1.18= 32.555/(32.555-x)

38.42 - 1.18x= 32.555

1.18x= 5.865 -> x = 4.97 tons of fuel

So, after dropping off your station, you'd have 7.31-4.97 = 2.34 tons of fuel left.

It thus has 800*9.81*ln ((13.945+2.34) / 13.945) -> 1217 m/s

You can spend 1,300 m/s before dropping off your station, and the resulting loss in dry mass means you'd have 1217 m/s to get back to Kerbin... with proper aerobraking, you'd need less than 400 m/s

[Jovus]

It also helps to realize that, for the game, your maximum attainable wet/dry mass ratio is 9:1. (practically speaking less, since this is the mass ratio of a fuel tank. Yes, I'm ignoring Mk3 tanks with their holier-than-thou 9.3)

So using the rocket equation you can do a quick check of maximum possible delta-V for each engine in the game, which can help you with design decisions, though it is not the absolute optimization parameter.

[KerikBalm]

Yes, I'm ignoring Mk3 tanks with their holier-than-thou 9.3

Knock Knock

- Who is there?

The 1.0 release and part balancing

- "The 1.0 release and part balancing" Who

The 1.0 release and part balancing the mk3 tanks to an 8:1 ratio...

Yea... its a lame joke...