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I've seen several tutorials on rocket stability where one can empirically obtain the center of pressure by cutting out a 2D imprint of the rocket and balancing the 2D shape as seen in the pictures below:

2D cutout of a rocket

Balancing a 2D imprint to determine center of pressure

What is the physical/mathematical justification for this, otherwise, extremely simple method for determining center of pressure? What are the underlying assumptions behid this method? In what cases is it applicable? When it is applicable, what order error should I expect from this method? In what cases is it not applicable?

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  • $\begingroup$ I think this assumes that the CP is the same as the CG... $\endgroup$
    – Ron Beyer
    Sep 18, 2019 at 22:32
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    $\begingroup$ @RonBeyer: The cg can be different on the vehicle compared with the 2d cut-out, since weight in the 3D vehicle can be distributed unevenly. $\endgroup$
    – Paul
    Sep 18, 2019 at 22:39
  • $\begingroup$ In physics this is known as argument from symmetry $\endgroup$
    – crasic
    Sep 19, 2019 at 0:41

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I couldn't justify this method more simply than NASA does:

For model rockets, the magnitude of the pressure variation is quite small. If we assume that the pressure is nearly constant, finding the average location of the pressure times the area distribution reduces to finding just the average location of the projected area distribution.

So the pressure, constant everywhere, "cancels out" leaving you with only area.

And the simplification from a 3-D shape to its 2-D projected area is justified by the shape's relative simplicity and symmetry. This wouldn't work with a grand piano.

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  • $\begingroup$ Is axisymmetry a prerequisite? $\endgroup$
    – Paul
    Sep 18, 2019 at 23:16
  • $\begingroup$ Reducing axisymmetry makes the projected area depend more on the direction of projection, thus increasing the chance of a poor choice and thus the error of estimation. Even in this question's illustration, the fins introduce asymmetry: if you instead projected along the fins' X instead of their +, then they'd contribute less, and the estimated CP would sit closer to the nose. This method also can't notice if you have six (or a dozen!) fins instead of four. $\endgroup$
    – Camille Goudeseune
    Sep 18, 2019 at 23:38
  • $\begingroup$ This makes me wonder how much of a deviation from true CP is accumulated with each additional fin. I’m guessing there is no heuristic n-fin correction to this method, is there? $\endgroup$
    – Paul
    Sep 19, 2019 at 1:39
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    $\begingroup$ When a fin's angle of attack is near zero, the aerodynamic pressure (aka lift) on it is much lower than when it's correcting for a gust (or actively steering). Each fin adds drag even at zero AoA, so we prefer fewer; 3 or 4 suffice for stability or steering; so it's unlikely that anyone's bothered to estimate a heuristic for larger n. $\endgroup$
    – Camille Goudeseune
    Sep 19, 2019 at 3:43

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