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Can someone explicate the stability of the truncated cone shape of the reentry vehicle of a spaceship when moving in the atmosphere with its bottom facing forward? It seems counterintuitive that it should be, since the most naive static force consideration would suggest the opposite. This naive consideration is best described by the setup of this erroneous answer. I did the force analysis in the comment below that answer and pointed out the error in the conclusion.

There are several questions here,here and here considering the cone shape of the reentry vehicle. However, even though the answer to the first question is most closely related to my concern, it does not address the aerodynamic stability.

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    $\begingroup$ The angle of the cone and the position of the center of mass may influence the aerodynamic stability and the orientation in the atmosphere. $\endgroup$ – Uwe Jan 29 at 20:41
  • $\begingroup$ @Uwe: Of course. The question is how. $\endgroup$ – Hans Jan 29 at 21:43
  • $\begingroup$ Great question, could you tell more about this naive static force consideration ? $\endgroup$ – qq jkztd Jan 30 at 13:42
  • $\begingroup$ The Wikipedia page en.wikipedia.org/wiki/Atmospheric_entry#Entry_vehicle_shapes referenced in the answer to space.stackexchange.com/questions/11975/… discusses this at a high level. A spherical section with a cone is statically stable when the center of gravity is closer to the section than the center of curvature, and generates some lift. The basic shape was chosen because it was amenable to closed-form computations. $\endgroup$ – antlersoft Jan 30 at 19:21
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    $\begingroup$ @Hans found this $\endgroup$ – qq jkztd Feb 3 at 20:11
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We assume the following.

  1. The force exerted by the air on a surface is pure pressure thus normal to the surface without friction. The pressure increases with respect to the magnitude of the surface normal component of the incident air flow velocity.

  2. The surface of the capsule is axially symmetric. Label the intersection of the symmetric axis and the surface (bottom) facing the incoming airflow $B$. The inward normal vector $\vec n$ of any infinitesimal surface patch either intersects the axis at point $N$ some finite distance from $B$ or $\vec n$ parallels the axis. The center of mass of the capsule $C$ locates between $B$ and $N$.

The capsule will achieve aerodynamic stability.

Before presenting the proof of this proposition, I give a plausible toy model of this air flow pressure function. The realistic function will surely be more complicated. Suppose an air column of an infinitesimal cross section area $dA$ collide with a facet with its normal vector forming an angle $\theta\in\big[0,\frac\pi2\big]$ with the air flow direction vector. The air bounces off the facet completely elastically. The momentum change (all in the normal direction of the facet) per unit time is then $2\rho v^2\cos\theta dA$, where $\rho$ is the density of the air flow and $v$ the speed of it. The area upon which this momentum change occurs is $\frac{dA}{\cos\theta}$. Divide the first quantity by the second, we get the pressure $p(\theta):=2\rho v^2\cos^2\theta$. Now the early arriving particles bounce off of the surface normally and collide completely elastically with the late arriving particles and bounce back towards the surface again. By symmetry, the average particle velocity near the surface vanishes in the surface normal direction but its component tangent to the surface remains. Macroscopically, the fluid on average as a whole moves along the tangent of the surface.

Proof:

1) 2-dimension.

Let us formulate the problem formally. Let $s\in[-s_0,s_0],\,s_0>0$ measure the distance, with sign, from the intersection of the symmetry axis with the surface. Denote the unit inward normal vector at $s$ by $\hat n(s)$. Let $\theta(s)$ be the angle from $\hat n(0)$ to $\hat n(s)$ with counterclockwise direction as the positive direction for the angle. $\theta(-s)=-\theta(s)$ by the axial symmetry. Let the angle from $\hat n(s=0)$ to the incoming airflow direction be $\theta_a$ also with counterclockwise direction as the positive direction. Place the curve $(x(s),y(s))$ in the Cartesian coordinate such that $(x(s=0)=0,y(s=0)=0)$ and the center of mass be located at $(x=0,y=y_c)$. We have $(x(-s),y(-s))=(-x(s),y(s))$. Let $p(\beta)$ be the pressure as a function of the angle $\beta$ with respect to the incoming air flow. The torque at each curve with respect to $(0,y_c)$ is $l(s)p(\theta_a-\theta(s))$ where $l(s)\hat z = \big((x(s),y(s))-(0,y_c)\big)\times \hat n(s)$.

Without loss of generality we assume $\theta_a>0$. Otherwise we can just reflect the coordinate with respect to the $y$ axis and get back the same problem because of the axial symmetry.

The total torque is \begin{align} T&:=\int_{-s_0}^{s_0}l(s)p(\theta_a-\theta(s))ds \\ &=\int_0^{s_0}l(s)\big(p(\theta_a-\theta(s))-p(\theta_a+\theta(s))\big)\,ds \end{align} as $l(-s)=-l(s)$ by the axial symmetry of the curve. Stability is achieved if $T>0$. We have $l(s)>0,\,\forall s>0$ since $C$ is between $(0,0)$ and $(0,y_c)$. $p(\theta_a-\theta(s))>p(\theta_a+\theta(s))$, since $|\theta_a-\theta(s)|<\theta_a+\theta(s),\ \forall \theta_a>0,\, \theta(s)>0,\, s>0$, and the fact that $p(u)>p(v),\,\forall |u|<|v|$. Therefore $T>0$.

QED

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – called2voyage Feb 15 at 13:43
  • $\begingroup$ My down-vote is because the reduction of the fields of sub-, trans-, super- and trans-sonic aerodynamics, and reentry gas effects to a one line assumption means this answer bears very little relevance to the real world. Beyond that, the proof boils down to the well understood statement "an object will be aerodynamically stable if the centre of pressure is behind the centre of mass" $\endgroup$ – JCRM Feb 15 at 16:46

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