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From thin airfoil theory, at subsonic speeds, the center of pressure of any uncambered airfoil is at quarter chord. For sounding rockets, however, the center of pressure is generally assumed to be located at the centroid of the projected area. Note that model sounding rockets generally stay subsonic.

What is the theoretical derivation for this? For this question, one can ignore the effect of fins and restrict to the axisymmetric slender bodies.

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  • $\begingroup$ Just FYI, you're more likely to get answers on questions like this at Space.SE $\endgroup$ – Machavity Sep 20 at 2:59
  • $\begingroup$ Isn’t the aerodynamic center at the quarter cord and the location of center of pressure changes with AoA at the airfoil. aviation.stackexchange.com/a/19399 $\endgroup$ – Kolom Sep 21 at 9:53
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    $\begingroup$ @kolom For symmetric thin airfoil it stays at quarter chord $\endgroup$ – Jimmy Sep 21 at 15:51
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This answer may not be what you are looking for, but I think the issue is we aren't really talking about the rocket as a "wing" here since these rockets fly ballistically. I think the term CP in the NASA article would be better expressed as "neutral point", the point at which aerodynamic moments are in balance when the rocket is not aligned with the airflow, taking into account the body and all of its appendages.

So this would include the fins as part of the total lateral area subject to lateral air loads when the rocket yaws while flying ballistically, and to be stable, the rocket's center of mass needs to be forward of the neutral point, so that when it yaws, the corrective moment, acting at the aerodynamic neutral point, is created about the center of mass to bring it back into alignment with the airflow.

The term CP suggests that the rocket is flying like a lifting body wing, with a constant angle of attack and lift being generated to support its weight. But as a ballistic object, its AOA is always ideally 0 and what lateral aerodynamic forces are being created are strictly to keep the rocket's tail lined up behind its nose as it travels its ballistic path.

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  • $\begingroup$ Restricting the discussion to the body only, just as in aircraft, the definition of neutral point is identical to aerodynamic center. That is, the point where the pitching moment is invariant with flow incidence. If your CG is ahead of the AC, and the Cm at AC is zero at zero incidence, then you have stabilized flight at zero incidence. The difference is, it is mentioned by various sources that CP is relatively constant for rockets (perhaps only for small incidences), so CP and AC coincide and there is no differentiation. $\endgroup$ – Jimmy Sep 19 at 19:15
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Thanks for the link, we may need to break out the old NACA films and do some refreshment for the benefit of NASA.

After correctly explaining the center of pressure would the sum of pressure of the nose, fuselage, and tail, they proceed to the "simplified" version of "hang a 2 dimensional cutout from a string". I think I'd rather try my office fan out as a crude wind tunnel!

The rocket pictured in the link has large fins, which as cruciform "horizontal and vertical" stabilizers give the rocket directional stability. Thrust(lift) is from the rocket engine. It path is straight up, it does not use any part of its body for lift, nor does it have a wing. The center of drag (the rocket is symmetrical) is right in the middle of the nose looking directly at the FRONT of the rocket.

The fins give it directional stability, and moving CG towards the nose more so. But remember, directional stability from fins is AERODYNAMIC, so we put it on a launch rod so it has speed before it flies free. (F-16s launching Sidewinders at 400 knots do not have this issue)

Now, as in aircraft, let's move on to side forces. Here is where center of pressure analysis becomes applicable. A crosswind will weathervane a rocket just like an airplane. So now you decide how much weather vaning you want, trading a little directional stability for less weather vaning, or like the US Army did with their Nike Hercules, putting fins on the second stage too. This gives resistance to gyrations of the fuselage (caused by a lack of fins and imperfect thrust vector), and it worked because the Hercules was guided. This forgotten "cold war" relic may have been the best airplane the Army ever made.

But for a non-guided directionally stable model rocket there will be some weather vaning in a cross wind. A possible solution would be thrust vectoring, or add some finnage to the nose to reduce it to acceptable levels (and remaining directionally stable).

Cutting out a 2 dimensional shape (of consistent thickness) ignores the effects of wind on a larger width tail area, though any error would be in favor of stability. It is surprising that NASA would not encourage young modelers to understand aerodynamic stability. Firstly, explain "centroid" as the center of AREA. Next explain for flat plate stabilizing effects (as compared to airfoils), that is where the Center of Pressure is because .....

Finally, scaring people off by saying it is all complicated calculus?? No, a simple side force wind tunnel and a string works fine. Set the rocket upright and apply aerodynamic side force with a fan. Attach the string to one point and see if it tips over. Keep trying until you find CP.

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    $\begingroup$ So why is the center of pressure of a thin body located at its hydrostatic center? You haven't explained any of it, which is my question. $\endgroup$ – Jimmy Sep 20 at 2:08
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    $\begingroup$ Sorry, I meant centroid. $\endgroup$ – Jimmy Sep 20 at 2:14
  • $\begingroup$ We may be looking at flat plate lifting characteristics. Since a thin airfoil may lose top lift after a few degrees AOA and behave like a flat plate (or a stalled airfoil), CP would move to the center. $\endgroup$ – Robert DiGiovanni Sep 20 at 21:34

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