# Does the minimum power required for level flight depend on CD0?

I came across this video few days back on Edx 16.101 youtube channel.

At 7.00 min of the video, the presenter derived an equation for minimum flight power which does not depend on Cd0.

$$P_\min = \frac{4}{3\pi e AR} \sqrt{\frac{2W^3CL}{\rho S}}$$

I don't have a doubt on the derivation, but this result seems to be very contradictory for me.

For example, let's take 2 identical aircrafts, one with fuselage airbrakes deployed and one without. (Identical CL, AR, $\rho$, S, CL and e but different Cd0) According to this expression the power requered cannot change. But,

1. For same CL, V is going to be the same
2. With airbrakes deployed, aircraft 1 drag is much higher than the aircraft 2
3. Power required (D*V) should be higher for aircraft 1.

May be I am missing something simple. Why is this apparent contradiction?

• I think what you're missing is that the CL will not remain the same between the two aircraft. – Porcupine911 Feb 29 '16 at 5:29
• To expand on what @Porcupine911 said, two identical aircraft, one with flaps up, one with flaps down, have different CL. – Simon Feb 29 '16 at 8:13
• Hi Porcupine, It cannot be the case, as clearly the Pmin is a function of CL. CL is given to evaluate the Pmin. one possible answer is that for different Cd0, 2 aircrafts has 2 speeds. but for same CL even that cannot be the case. Anyway thanks for the input. appreciate it .. – Guest Feb 29 '16 at 8:31

The whole expression is only valid for the optimum polar point where the ratio $\frac{c^2_D}{c^3_L}$ reaches its maximum. If you deflect spoilers, this optimum will shift to a higher lift coefficient, so both aircraft will not fly at the same lift coefficient or the same speed any more! Always keep in mind that in this special condition the lift coefficient can also be expressed as $$c_L=\sqrt{3\cdot\pi\cdot AR\cdot e\cdot c_{D0}}$$ Just plug that back into the equation and the minimum power expression will make a whole lot more sense: $$P_{min}=\frac{4}{3\cdot\pi\cdot AR\cdot e}\cdot\sqrt{\frac{2\cdot W^3\cdot\sqrt{3\cdot\pi\cdot AR\cdot e\cdot c_{D0}}}{\rho\cdot S}}$$ This is still the same equation, but now it should be obvious that opening the spoilers will require more power.