Editing Thrust (section)

===Thrust to power===
The power needed to generate thrust and the force of the thrust can be related in a [[non-linear]] way. In general, <math>\mathbf{P}^2 \propto \mathbf{T}^3</math>. The proportionality constant varies, and can be solved for a uniform flow, where <math>v_\infty</math> is the incoming air velocity, <math>v_d</math> is the velocity at the actuator disc, and <math>v_f</math> is the final exit velocity:

:<math>\frac{\mathrm{d}m}{\mathrm{d}t} = \rho A {v}</math>

:<math>\mathbf{T} = \frac{\mathrm{d}m}{\mathrm{d}t} \left (v_f - v_\infty \right ), \frac{\mathrm{d}m}{\mathrm{d}t} = \rho A v_d</math>

:<math>\mathbf{P} = \frac{1}{2} \frac{\mathrm{d}m}{\mathrm{d}t} (v_f^2 - v_\infty^2), \mathbf{P} = \mathbf{T}v_d</math>

Solving for the velocity at the disc, <math>v_d</math>, we then have:
:<math>v_d = \frac{1}{2}(v_f + v_\infty)</math>

When incoming air is accelerated from a standstill – for example when hovering – then <math>v_\infty = 0</math>, and we can find:

:<math>\mathbf{T} = \frac{1}{2} \rho A {v_f}^2, \mathbf{P} = \frac{1}{4} \rho A {v_f}^3</math>

From here we can see the <math>\mathbf{P}^2 \propto \mathbf{T}^3</math> relationship, finding:

:<math>\mathbf{P}^2 = \frac{\mathbf{T}^3}{2 \rho A}</math>

The inverse of the proportionality constant, the "efficiency" of an otherwise-perfect thruster, is proportional to the area of the cross section of the propelled volume of fluid (<math>A</math>) and the density of the fluid (<math>\rho</math>). This helps to explain why moving through water is easier and why aircraft have much larger propellers than watercraft.