Timeline for Pushing a convex cone and equidistants
Current License: CC BY-SA 4.0
10 events
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| May 15 at 22:36 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| May 15 at 22:24 | comment | added | Saúl RM | Maybe the problem with the verification is when you set $x=y=0$? It seems from the previous comment that you are only considering distances to the part of the boundary which is in the plane $x_2=0$ | |
| May 15 at 22:17 | comment | added | Saúl RM | Yes, it seemed strange to me that something like $\arctan(1/4)$ didn't appear somewhere in the argument. I didn't comment on that because the same idea still applies. E.g. the point $(0,0,\sqrt{17}/4)$ is at distance $<1$ of the point $(0,\sqrt{17}/8,\sqrt{17}/8)$, which is in the boundary of $K$ (In particular, I don't think it is in the boundary of $K_1$) | |
| May 15 at 22:15 | comment | added | Richard Montgomery |
Verification: The eqns for $K_t$ can be obtained from the unit inward pointing normals of $K$. These yield the following eqns for $K_t$: $$a(y+z) \ge t$$ $$a(-y+ z) \ge t$$ $$b (x + 4z) \ge t$$ $$b (-x + 4z) \ge t$$ with $a = 1/\sqrt{2}$ and $b = 1/\sqrt{17}$. Setting $x = y = 0$ and $t =1$ and the $\ge$' to $=$' we see that $(0,0,\sqrt{17}/4)$ is on the boundary of $K_1$.
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| May 15 at 22:14 | comment | added | Richard Montgomery | I also made an arithmetic error. Set $x_* = \sqrt{17}/4$. The triangle with half angle $\theta/2$ has, from your measurement specifications $\tan(\theta/2) = 4/1$ whereas I had set the tangent to $2/1$ by mistake. The correct triangle is thus a $1:4: \sqrt{17}$ right triangle and $x_*$ comes from this. | |
| May 15 at 20:21 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| May 15 at 20:10 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| May 15 at 20:05 | comment | added | Saúl RM | It seems the point $(0,0,x_*)$ is at distance $<1$ of the point $\left(0,\frac{x_*}{2},\frac{x_*}{2}\right)$, which is in the boundary of $K$. So $(0,0,x_*)$ cannot be in $K_1$, right? | |
| May 15 at 19:31 | comment | added | Richard Montgomery | Thanks! But the minimal value of $x_3$ on $K_1$ is not your $\sqrt{2}$, but rather is $x_* = \sqrt{5}/2 = 1.18..$ which is less than $\sqrt{2}$. To see this, form the sector obtained by intersecting the plane $x_2 = 0$ with your $K$. You get a sector whose angle $\theta$ satisfies $tan(\theta/2) = 2/1$ and so $sin(\theta/2) = 1/x_*$. The point $P_* = (0,0, x_*)$ is achieved by propagating the bounding rays of this sector in one unit. (Your $\sqrt{2}$ comes from the angle made by the orthogonal sector made by intersecting with the plane $x_1 = 0$.) I believe $K_1 = P_* + K$. | |
| May 15 at 16:16 | history | answered | Saúl RM | CC BY-SA 4.0 |