Power Points, Megawatts and Spaceships

[Anthony]

Another way of calculating this is to work out reaction drives. The power requirement of a perfectly efficient electric thruster is 0.5 * exhaust velocity * thrust. For 1 G and 1 mps/tank, that works out to 8890N/ton * 32,180m/s * 0.5 = 143MW/ton. Ion drives give 0.0005G and 3 mps/tank, so that's 215 kW/ton. Mass driver drives give 0.01G and 0.3 mps/tank, so that's 429 kW/ton.

[David L Pulver]

Originally Posted by Anthony

Another way of calculating this is to work out reaction drives. The power requirement of a perfectly efficient electric thruster is 0.5 * exhaust velocity * thrust. For 1 G and 1 mps/tank, that works out to 8890N/ton * 32,180m/s * 0.5 = 143MW/ton. Ion drives give 0.0005G and 3 mps/tank, so that's 215 kW/ton. Mass driver drives give 0.01G and 0.3 mps/tank, so that's 429 kW/ton.

That seems reasonable. I do remember the electric drives were fudged to make them about 5-10x as efficient so as to have mass-driver powered manned ships that were actually worth taking. For realism reduce G thrust by about 1/5 (allowing some efficiency losses) for ion and by about 1/10 for mass driver.

In short, the intent is that the beam weapons are broadly correct, but the electric drives are unrealistic.