# How is thrust per unit power measured?

Typically, thrust efficiency is measured in specific impulse, as thrust per fuel consumption rate:

$$I_\text{sp} = \frac{F_\text{thrust}}{\dot{m}_\text{fuel}} \quad \text{or} \quad I_\text{sp} = \frac{F_\text{thrust}}{\dot{m}_\text{fuel}} g_0$$

However, this only really applies to methods of thrust that depend on fuel mass. In an electric airplane, for instance, you'd really want to measure something more like this:

$$I_\text{sp} \stackrel{!}{=} \frac{F_\text{thrust}}{P}$$

I've never seen thrust efficiency measured in $s/m$ before, and some time searching online for such a unit hasn't really been fruitful. What's the power-unit equivalent of specific impulse?

• en.m.wikipedia.org/wiki/Ion_thruster can help you. – Yuzuriha Inori Apr 4 '18 at 9:17
• @YuzurihaInori you would think so, but it was the first place I looked. There isn't actually anything there about the power-unit equivalent of specific impulse – Tal Apr 4 '18 at 9:44
• adsabs.harvard.edu/full/1997ESASP.398..131K Did you check this? – Yuzuriha Inori Apr 4 '18 at 9:48
• It occurs to me that the inverse of the quantity you are discussing has units m/s and could reasonably be called “effective exhaust velocity”. Low exhaust velocities do require much less power per thrust, that is why ion thrusters use so much energy for so little force (high exhaust velocity). Don’t know if that is useful to you at all, just thought I would throw it out here. – Duncan Harris May 23 '18 at 3:46
• @DuncanHarris Is that because here the momentum resulting from the "effective exhaust velocity" is equivalent to the momentum imparted to the vessel, which grows proportionally with velocity ($mv$), while higher exhaust velocity requires higher power and therefore higher kinetic energy, which grows with the square of velocity ($m\frac{v^2}{2}$)? – Tal Feb 1 '19 at 0:31