Calculation of energy requirement per drive

[Elodin]

I really like the complexity of the game and how it is necessary to understand and learn the details in order to make good decisions. While i was on the endeavour to understand how drives and power plants interact, i stumbled over something that i think is a miscalculation:

The drives all have power requirements in terms of Watts that are internally calculated via thrust * efficiency * thrustCap. For the example of the Protium Inertial Torch x1 this makes 275,000 N * 0.85 * 60 m/s = 1.4 Terrawatt. Since we are talking about the power that the reactor has to provide, we are talking about the power before efficiency losses occur. Or in other terms, the reactor of the example should have to provide 275,000 N * 60 m/s = 1.65 Terrawatt of which the drive can only turn 85 % into thrust.

[Morti]

for me, the best answer is there :
https://www.reddit.com/r/TerraInvicta/comments/y32akq/updated_ship_drives_chart_with_all_ship_parts_and/

[ulzgoroth]

Why do you think that's what the calculation is? I don't think that's what the calculation is.

Your correction would also be wrong, though.

The exhaust power of the drive is, I'm pretty sure, 1/2 thrust * exhaust velocity (should be * thrust cap, but I think in the example I've looked at it isn't?). That power is what the drive does, efficiency doesn't come into it. The input power should then be that value divided by the efficiency.

If you check these numbers against Zeta Helion I think they line up. I haven't done the math for a lot of drives because the power requirement stat is only displayed in-game so it's annoying to extract.

[Banelord]

i dont think much, i only use zeta helions and dont think ive seen any better

[Elodin]

Originally posted by ulzgoroth:

Why do you think that's what the calculation is? I don't think that's what the calculation is.

Your correction would also be wrong, though.

The exhaust power of the drive is, I'm pretty sure, 1/2 thrust * exhaust velocity (should be * thrust cap, but I think in the example I've looked at it isn't?). That power is what the drive does, efficiency doesn't come into it. The input power should then be that value divided by the efficiency.

If you check these numbers against Zeta Helion I think they line up. I haven't done the math for a lot of drives because the power requirement stat is only displayed in-game so it's annoying to extract.

I've extracted the numbers from the game files, where there are more values than you are shown ingame. My calculation of the given values lines up perfectly with every ingame drive i could check it against. I'm no rocket scientist - so i can't or won't say if this math is physically correct. But this is how the game calculates it. And with the background i do have, i can say that efficiency has to be taken into account here or can be thrown away as a value entirely.

To understand, its important to know what efficiency is (which might not be common knowledge, so i'll explain in short): Its the rate at which the drive can convert its input into the output we want.
In the example of the protium inertial torch, the drive operates with a power of 1.4 Terrawatt. So thats our output - but what is the input? Thats the power the drive gets from the reactor. But if the drive operates with less than 100% efficiency, which is true in our case because efficiency is only 0.85 or 85%, than what we put in needs to be greater than what we get out. Or in other words: if the reactor provides 1.65 Terrawatt of power, but the drive can only convert 85% of that into what we want, then the drive will output 1,6 TW * 0.85 = 1.4 TW.

[lemurs2]

And, honestly, the efficiencies in game are overrated a great deal. We can only provide, today, 400W per kg to a drive like the VASIMR but that drive, like all electrics has better performance with higher power input.
We think we can in the very near future reach 1000W per kg which would move to about a 50% eff.
To reach 70% eff you would need 6000W per kg, which we have no idea how to output.