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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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Proposition: 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 and is zero when the surface normal component becomes negative.

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 achieves 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.

However, interestingly, two and a half months after I posted this answer, I happened upon the theory of hypersonic aerodynamics that surprisingly endorsed fully the following derivation as the correct computation for the pressure of hypersonic (Mach 3-5) airflow on an largely axial symmetric body with blunt surface geometry. c.f. equations (11-2) and (11-3) of chapter 11 on the hypersonic aerodynamics of W. H. Mason's lecture on configuration aerodynamics. Search for "Newtonian Impact Theory" in this accompanying PPT to that chapter.

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. Alternatively we can assume the complete inelastic collision of the air molecule with the surface, so that the momentum normal to the surface completely dissipates only the tangential component is unmolested so the air molecules after the collision move parallel along the surface. In this case, it is clear $p(\theta):=\rho v^2\cos^2\theta$ which is half of the previous value as the surface normal momentum transferred is half of that in the elastic case. In the case of fractional elastic collision, the $p(\theta):=(1+\alpha)\rho v^2\cos^2\theta$ where $\alpha\in[0,1]$ is the coefficient of collision elasticity.

Moreover, the part of the object surface that is in the "shadow" of the incoming airflow will remain untouched by the airflow and thus experience no pressure.


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, needing to account for only the surface facing the incoming airflow, \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, by Assumption 2), the center of mass $C$ located at $(0,y_c)$ is between $N$ (at the origin of the coordinate $(0,0)$) and $B$. $p(\theta_a-\theta(s))>p(\theta_a+\theta(s))$, 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
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    $\begingroup$ 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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    $\begingroup$ @JCRM: You are right that the airflow, particularly the pressure it exerts on the surface of a body behaves differently in the regimes of sub-, trans-, super and hyper-sonic speeds. Fortunately, the assumption of my answer turns out to be correct in the hypersonic regime. The proposition, mathematically correct in the first place, is now very relevant to the geometry to achieve aerodynamic stability for the initial hypersonic reentry flight into the atmosphere. Please check out the edited second paragraph of the preamble to my proof. It contains the reference to the hypersonic aerodynamics. $\endgroup$ – Hans Apr 30 at 8:54
  • $\begingroup$ The linked document does not "endorsed fully " your derivation, it talks about it being an approximation. section 11.16 and figure 11.20 partly explain why this is useless in a real-world scenario. $\endgroup$ – JCRM Sep 27 at 20:59
  • $\begingroup$ @JCRM: The linked equations are exactly the same as what I have derived. To what extent does the reference not "endorsed fully" my derivation? Your complaint of it being an approximation per se is a meaningless truism since no mathematical formula describing natural phenomena is not an approximation. Do you have one example that is an exception? There is no section 11.16. There is Section 11.6. How exactly and which part of that section and figure 11.20 explain Equation (11-3) is useless? Please write out the specific details of your logical derivation rather than traffic in vagueness. $\endgroup$ – Hans Sep 28 at 0:53

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