How can I calculate fuel/thrust/weight ration in this game?

[TheLunaticist]

How can I calculate how much fuel/thrust/weight I need to get into orbit and from there to other planets? (Calculating how much I need to get any ship into orbit would be alot more useful but if you can explain both that would also be nice)

[Chibbity]

That's called Delta V which the game displays, and you can compare it to a KSP Delta v map on Google.

[FourGreenFields]

Once in orbit, delta V is all you have to care about.

When taking off from planets with an atmo, thrust-to-weight does matter, but there are still rules of thumb for how much delta V you need. Without atmo, thrust to weight greater 1 is required, beyond that just a pretty constant amount of delta V.

[Randox]

The game will do it for you. When you are building a rocket in the assembly building, in the staging list on the right side will be a blue tab under some/all of the stage numbers (any stage that has engines and fuel) with a number in it. That number, in m/s, is your dV for that stage. Under the staging list in an orange highlight is the total dV for that craft. Clicking on the blue highlighted numbers opens a detailed listing window, which by default includes TWR for the start of that stage (TWR increases as the fuel is burned and ship mass decreases).

At the bottom right side of the screen are a few buttons, one of which is dV (it uses the greek letter for delta, a triangle). That button gives you options to change the reference body. So dV for example is pretty much always measured using the vacuum value (all dV maps use this version). For TWR, you'll want to use the relevent reference body (to make sure you can take off from or safely land on a specific moon or planet). If you're on Kerbin or deep space, use Kerbin as the reference, since this value is in g's (a TWR of 2 will accerlerate at 9.81 m/s^2).

[Jupiter3927]

TWR is basically your acceleration when fighting against gravity in terms of the gravity of whatever body you're referring to.
Kerbin's gravity can be rounded up to 10 m/s/s so any rocket you make will need at least 10 m/s/s of acceleration (1 TWR) to get off the launchpad.
Having it at least 15 m/s/s or 1.5 TWR works a lot better.

Acceleration can be calculated by dividing mass by thrust so M/F=a.

[andylaugel]

Click on the orange parts of the staging diagram, like so.
https://steamcommunity.com/sharedfiles/filedetails/?id=1848144224

[andylaugel]

Alternatively, there is the manual method.

dV (in m/s) = 9.81 * engine Isp * Ln( starting spacecraft mass / final spacecraft mass )

So that previous post had a spacecraft with a mass of 6.95 t, or 6,950 kg. Liquid fuel and oxidier mass 5 kg/unit. This craft has 440 Liquid Fuel and 360 Oxidizer, so the total fuel masses 4,000 kg or 4t. The "Terrier" engine has an Isp of 345. So 9.81 * 345 * Ln( 6.95 / ( 6.95 - 4 )) gives us 2,900 m/s dV. The screenshot says it is only 2,899 m/s, but that is only 1 m/s dV off--well within the margin of error.

As a bonus exercise, we can compute how much dV we have with this craft based on monopropellant, once all the Liq/Ox fuel is used up. Monopropellant masses 4 kg/unit. The RCS thrusters have an Isp of 240 in a vacuum. So with 86.52 mono propellant left, we have 293.8 m/s dV to spare.

TWR is easier. To find our acceleration in m/s/s, we take our total thrust (60 kN for a Terrier), and divide it by our mass (6.95t). So this craft accelerates at 8.63 m/s/s. Then we divide that by the world's surface gravity to determine TWR. So on Kerbin, it has a TWR of 0.88--not enough to lift itself off the ground. But on the Mun with its 1.6606 m/s/s surface gravity, it has a TWR of 5.2. ... Okay, that is a little less than what the KSP read-out shows, but that just leads me to suspect it takes altitude into account. The mun's pull at higher altitudes is less, making TWR's higher.

[andylaugel]

I found myself with a practical math problem in KSP, and thought I'd share.

The Question

How much fuel do I need to reserve on my mining ship for a safe landing on the Mun?

The idea is that my mining ship will refuel itself on the Mun's surface, dock with a ship in orbit, convert any remaining ore into fuel, transfer some fuel to the other vessel, undock, and land again.

Given

When I gathered my data, my mining ship was safely on the Mun's surface. Going to the map screen and clicking on that 'i' icon on the right side, I found it massed 56.04 tons. Going to Resources a few icons above that, I found it had 1491 units of Liquid Fuel, 1733 units of Oxidizer, and 600 units of ore.

I also know that liquid fuel and oxidizer both mass 5 kg/unit, and ore masses 10 kg/unit. The mining ship uses "Poodle" engines with an Isp of 350. 1 metric ton is 1,000 kg. Poodle engines use Liquid fuel & oxidizer in a 9:11 ratio.

I want 1,000 m/s of dV to de-orbit and land. (This is based on altitude, the Mun's gravity, and how much of a safety margin I (the player) desire.)

Finding the Solution

I need to know how much my mining vessel masses without fuel or ore. That's:
"Dry" ship mass = 56,040 kg - 5(1733 oxidizer) - 5(1491 liquid fuel) - 10(600 ore) = 33,920 kg = 33.92 t.

Now, let's plug this into the dV equation.

dV = g * Isp * Ln( initial mass / dry mass )
1,000 = 9.81 * 350 * Ln( initial mass / dry mass )
1,000 / ( 9.81 * 350 ) = Ln( initial mass / dry mass )
0.291 = Ln( initial mass / dry mass )
e^0.291 = e^(Ln( initial mass / dry mass ))
1.338 = initial mass / dry mass
1.338 = ( dry mass + fuel mass ) / dry mass
1.338 = ( 33.92t + fuel mass ) / 33.92t
1.338 * 33.92t = 33.92t + fuel mass
45.39t - 33.92t = fuel mass
11.47t = fuel mass

Now we need to convert fuel from total mass into units of liquid fuel and oxidizer.
liquid fuel (in units) = total fuel mass (in kg) * (9 units / 20 total equivalent units) / 4 kg per unit
oxidizer (in units) = total fuel mass (in kg) * (11 units / 20 total equivalent units) / 4 kg per unit

The Solution

I need to reserve at least 1,290 units of liquid fuel & 1,576 units of oxidizer to ensure my mining ship has 1,000 m/s dV desired to safely land on the Mun. That also assumes all the heavy ore aboard has been consumed.