# SpaceX Body Flaps - Why so Much Power Required

I don't doubt the credibility of this particular source, but why would control surfaces, even at hypersonic speed (in rarefied air, mind you), and even with substantial windward surface, require 1.5 megawatts to actuate?

I consider this guy to be a pretty credible source...so I'm definitely not questioning his veracity.

I'd just like to better understand why so much power would be needed to move EACH body flap on the Starship.

• it's more about response time I guess. You don't need much energy but you need it fast hence the high power – user3528438 Jun 20 at 3:18
• The number is somewhat within the realm of reason. The Space Shuttle had three 135 horsepower auxiliary power units that were used to drive the hydraulics for the Shuttle's flaps. On conversion to metric that 3*135 horsepower is about 300 kilowatts. – David Hammen Jun 20 at 4:02

Well, an approach to this is to try and calculate how much power might be needed, and you can make a hack at this. Since I don't know how large the flaps are or various other critical parameters (and I am not sure this information is public) I can't actually calculate what the real power requirements are: rather this answer gives you the expression you need to approximately calculate the power requirements if you do know those values.

First of all there's an expression for the dynamic pressure of a body moving through a medium:

$$q = \frac{1}{2}\rho v^2$$

Here $$\rho$$ is the density of the medium and $$v$$ is the speed. This $$q$$ is the $$q$$ of 'max $$q$$', of course: it tells you how much pressure the atmosphere exerts per unit area you're trying to force through it at a given speed.

Then starting from this we can work out some more things:

\begin{align} q &= \frac{1}{2}\rho v^2 &&\text{pressure, as above}\\ F &= qA &&\text{force, A is area presented to the airstream}\\ &= \frac{1}{2}\rho v^2 A\\ E &= Fd&&\text{energy: d is the distance you're trying to move}\\ &= \frac{1}{2}\rho v^2 A d\\ P &= \frac{E}{t}&&\text{power: t is the time you move over}\\ &= \frac{1}{2}\rho v^2 A \frac{d}{t} \end{align}

OK, so this last expression tells you how much power you need if:

• the vehicle is moving at $$v$$;
• the air density is $$\rho$$;
• the flaps present an area of $$A$$ to the airstream;
• you want to move them a distance $$d$$, in time $$t$$.

The last one is complicated for hinged flaps, but we're just going to guess some numbers.

So, well, let's guess some numbers:

• $$v = 5\times 340\,\mathrm{ms^{-1}}$$ (mach 5);
• $$\rho = 0.1\times 1.2\,\mathrm{kg m^{-3}}$$ (a tenth of atmospheric density at sea level);
• $$A = 0.5\,\mathrm{m^2}$$ (effective area of flap);
• $$d = 0.5\,\mathrm{m}$$ (how far you want to move it);
• $$t = 0.25\,\mathrm{s}$$ (how long you want to move it in);

Well you can check the dimensions are right – this is a power – and the answer is that

$$P = 173400\,\mathrm{W}$$

So, OK, this is a fair bit lower than the quoted figure – it's around $$10\%$$ of it in fact. But I just made up plausible values for the numbers involved: I have no idea what any of them actually are, as I said above. If the flaps are required to work at higher speed, or in denser air, or more quickly, or they are larger, then you need more power to drive them.

Someone with actual numbers for the flaps rather than my invented ones should be able to get more reasonable values.

One apocryphal bit of information which may help is the comment by Musk that 'it's like moving the entire wing of an aircraft'. Well, how big are aircraft wings? Let's take the F-15: its wing area is $$56.5\,\mathrm{m^2}$$, so each wing is $$28.3\,\mathrm{m^2}$$. If we simply use that value for the size of the flaps in the above expression we need $$9.8\times 10^6\,\mathrm{W}$$! This is now much higher than the stated power.

Well, if we assume the flaps are deployed at an angle of about $$10^\circ$$ to the airflow then $$A = 28.3\,\mathrm{m^2}\sin 10^\circ \approx 4.9\,\mathrm{m^2}$$, and this gives a required power of $$1.7\times 10^6\,\mathrm{W}$$, which is about the stated power.

What's clear from this is that the figure of $$1.5\times 10^6\,\mathrm{W}$$ is certainly in the right area for the power you would need to drive flaps like this.

• curious downvote: perhaps whoever did so would like to comment so I can make the answer better? – tfb Jun 21 at 14:10
• +1 for science! Sometimes aerodynamic surfaces are moved at much higher frequencies, especially when used in unstable flight. I don't know if that's the case here, but if they are it might require higher peak power. Your answer also makes me wonder if they will implement something analogous to regenerative braking by generating power from movement in the opposite direction ;-) – uhoh Jun 21 at 23:20
• @uhoh: thanks. I was slightly frustrated by the silent downvote – I sometimes do that but only when I think the answer is completely unrecoverable, and I don't think mine was. I think if there's hope for the answer it's polite to comment if you're downvoting it. (also not trying to imply the silent downvote was you of course!) – tfb Jun 22 at 11:37

On a usual aircraft the wings which are redirecting the airflow (and carrying the weight of the aircraft) are mounted firmly to the body. Most control surfaces are already in the direction of the airflow and just redirect it a little.

If I understand the proposed Starship landing behavior correctly, those flaps can be almost perpendicular to the flow, so much larger forces would apply. Imagine if one of those big flaps was fixed to the ground at the hinges and you tried to use it as a windshield in a storm. The engines on the hinges would need to be pretty strong to lift it up against the storm.