It matters more for ion and fission engines because you're not generating the power to accelerate the exhaust as a function of burning the propellant. It might matter to them if efficiency is less than one, meaning the cooling system of the engine will need to handle some watts soaking into the engine walls and components. It's not really an everyday calculation for rocket engines, since chemical rocket engines mostly address power in the exhaust as a side issue. It allows you to answer, 'How many Watts need to be applied to your propellant for an observed thrust and exhaust velocity?' The equation I've been using is the jet power equation.

Nope, I gave the perfect efficiency value. For example, if the engines were only 50% efficient then the equation.

Any decrease in efficiency means the engines will pump more heat into the atmosphere. Quote from: cray on 07 August 2016, 16:09:44 I gave the value seen in the game based on fuel consumption (kilograms per fuel point), movement rules (0.5G per thrust point), and vehicle stats (3,600 metric tons). Construction rules aren't set up to build combatant spacecraft with 50 to 90% of their mass dedicated to fuel, and aerospace movement rules would change if thrust dropped significantly. The way to reduce energy release is to change the rules for fuel consumption or thrust, but those would significantly impact BT spacecraft design. Some of the energy will be emitted in the exhaust and some will be emitted in the cooling system, but it's still going to increase from 44.7 kilotons per second to something larger as efficiency decreases. Power = 0.5 x 10,584,000 m/s x 35,309,970 N / 0.5ДЄnd then you get twice the energy output as compared to the perfectly efficient engines. I gave the value seen in the game based on fuel consumption (kilograms per fuel point), movement rules (0.5G per thrust point), and vehicle stats (3,600 metric tons).

40k drop pod reentry effect