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Dec 31, 2022 at 0:04 answer added Woody timeline score: 0
Aug 29, 2020 at 0:36 vote accept Hans
Sep 28, 2019 at 6:00 history tweeted twitter.com/StackPhysics/status/1177825436149133312
Mar 31, 2019 at 2:01 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Feb 7, 2019 at 2:10 comment added Hans @VladimirKalitvianski: I have solved the problem and written up an answer below. One can see from my solution that the geometry of the bottom surface is crucial. For example, had the bottom been concave, even if it is wider than the top and the center of mass close to the bottom, as you stated above, the bottom orientation will not be stable. Check my solution out.
Feb 7, 2019 at 1:34 answer added Hans timeline score: 1
Feb 1, 2019 at 7:27 history edited Hans CC BY-SA 4.0
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Jan 9, 2019 at 2:24 comment added Hans @VladimirKalitvianski: As a matter of fact, Mike Dunlavey's answer while being wrong, shows exactly why this is intricate and somewhat counter-intuitive.
Jan 8, 2019 at 15:16 history edited Hans CC BY-SA 4.0
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Jan 8, 2019 at 14:57 answer added Mike Dunlavey timeline score: -1
Jan 8, 2019 at 6:27 comment added Hans @MikeDunlavey: My question is exactly the detailed geometry of why and how "if it gets out of line it will tend to self-correct".
Jan 7, 2019 at 19:58 comment added Mike Dunlavey It depends on where the center of mass is. If the center of mass is close to the center of the heat shield, it will tend to line up with the heat shield pointing forward. The heat shield is not planar but dished, so if it gets out of line it will tend to self-correct.
Jan 7, 2019 at 11:19 comment added Hans @VladimirKalitvianski: We can just set the reference frame coordinate on the center of the mass of the cone and consider the air pressure. It has nothing to do with the gravity except to locate the center of the mass. If the axis of the cone is perpendicular to the velocity, I agree that the air pressure torque pushes the vertex away. However, once the angle between the axis and the velocity is smaller than the that between the axis and the side of the cone, the geometry is not so obvious to me that the air pressure torque would decrease the angle between the axis and the velocity.
Jan 7, 2019 at 10:16 comment added Vladimir Kalitvianski Because if it is not well oriented, there is a net torque from the narrow part of the cone. As the bottom is heavier (I guess), its attraction to the Earth is higher than that of the narrow part, but the hydraulic resistance is proportional to the size or the size squared, not as the force (size in the third), so there is inequality of forces and torques that orients the cone bottom downwards.
Jan 7, 2019 at 10:12 answer added kalle timeline score: 0
Jan 7, 2019 at 10:02 history edited Hans CC BY-SA 4.0
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Jan 7, 2019 at 9:58 comment added Hans @PM2Ring: There is no active rocket firing to control the orientation.
Jan 7, 2019 at 9:57 comment added Hans @VladimirKalitvianski: If the right side up orientation is also unstable, object will just chaotically tumble. Why does it seem to regress to this orientation? Most importantly, the question asks for the aerodynamics of any of the orientation, stable or unstable.
Jan 7, 2019 at 9:46 comment added Vladimir Kalitvianski The cone upside down fall is not stable, that is why.
Jan 7, 2019 at 9:42 comment added PM 2Ring Because it's bottom-heavy? But that's also true of a cone of uniform density. My guess is that the primary concern is thermal, not aerodynamic, per se.
Jan 7, 2019 at 9:30 history asked Hans CC BY-SA 4.0