Timeline for Mapping a cube to a sphere
Current License: CC BY-SA 4.0
42 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2023 at 10:54 | comment | added | Harry van Langen | Without iteration and large enough p (or test for singularity) the worst quality I get is 95%. This is already a major improvement over Nowell and brute contraction. I tried the simplest iteration I could come up with just to see if it would give further improvement (which it does). | |
| Oct 13, 2023 at 9:41 | comment | added | Harry van Langen | $\lim\limits_{p \to \infty} x^p = \begin{cases}0&for\ \left\vert \text{} x\right\vert <1\\ 1&for\ \left\vert \text{} x\right\vert =1\end{cases}$ so it is a (clunky) substitute for the simple test and (at least in the limit) the quality at the central line and edges is not affected. | |
| Oct 12, 2023 at 16:33 | comment | added | HelloGoodbye | You can perhaps improve the quality of the interior further, but I still don't understand the motivation for making the method iterative? If two iterations lead to better quality than one, what is your intuition for why that works? | |
| Oct 12, 2023 at 16:25 | comment | added | HelloGoodbye | I wasn't talking about small $p$; I was talking about the case when $p\to\infty$. (For small $p$, you should reasonably get suboptimal positions after one iteration.) | |
| Oct 12, 2023 at 15:18 | comment | added | Harry van Langen | Yes that is correct for small p. It distracts from the central question though - can the mapping in the "interior" be improved further? | |
| Oct 12, 2023 at 15:00 | comment | added | HelloGoodbye | You say that the quality of the central coordinate line and edges cannot be improved. However, my point was that I think the quality should get worse after you perform additional iterations, since after one iteration, you should have obtained optimal quality for those lines, and additional iterations will move the points on those lines from their optimal positions. I'm still wondering if I'm missing something. It it possible that the quality will improve for some locations, but I at least think that the quality should get worse for the central and corner lines. Isn't that correct? | |
| Oct 12, 2023 at 14:56 | comment | added | HelloGoodbye | Why is it important that the expression is on a closed form? | |
| Oct 12, 2023 at 11:08 | comment | added | Harry van Langen | Similar to Nowell, I wanted to come up with closed form expressions of the form $\underline{x}_{sphere} =\underline{f} \left( \underline{x} \right)$. Hence the idea to include the power term to deal with the singularity. The central coordinate line and edges remain straight under the transformation $\underline{x}_{c} =\underline{g} \left( \underline{x} \right)$ and the "quality" cannot be improved. However, elsewhere the iteration helps a bit. Since I was so close to achieving 100% quality throughout, I wondered if there was an alternative (closed-form) expression that would achieve this. | |
| Oct 11, 2023 at 19:47 | comment | added | HelloGoodbye | And instead of using $x_c=\sqrt{x^p+y^2+z^2} \tan\left(x \arctan\left(\dfrac1{\sqrt{x^p+y^2+z^2} } \right)\right)$, why don't you just use $x_c=\cases{\sqrt{y^2+z^2} \tan\left(x \arctan\left(\dfrac1{\sqrt{y^2+z^2} } \right)\right), & |x| < 1\\x, & |x| = 1 }$? | |
| Oct 11, 2023 at 19:44 | comment | added | HelloGoodbye | I see. But I don't understand the rationale for making the method iterative. According to the reasoning on your blog, if we consider the line on the cube where $y=0$ and $z=1$, shouldn't the "optimal" image point on the cube be $x_c=\sqrt{y^2+z^2} \tan\left(x \arctan\left(\dfrac1{\sqrt{y^2+z^2}}\right)\right)=\tan\left(x \dfrac{\pi}{4}\right)$ and $y_c=y=0$ and $z_c=z=1$? Because this would be achieved in the very first iteration, in the limit when $p\to\infty$, and if you iterate further on that, you're just going to move away from that ideal point, right? Or am I missing something? | |
| Oct 9, 2023 at 15:07 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor formatting
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| Oct 9, 2023 at 13:25 | answer | added | user1277628 | timeline score: 2 | |
| Oct 9, 2023 at 6:39 | comment | added | Harry van Langen | It is explained here: hvlanalysis.blogspot.com/2023/05/… | |
| Oct 5, 2023 at 17:51 | comment | added | HelloGoodbye | Interesting formula / algorithm. How did you derive it? Also, it seems to me that the larger $p$ is, the better the quality becomes. Is that so? If so, have you considered deriving a formula for the limiting case for which $p\to\infty$? | |
| May 22, 2023 at 8:46 | history | edited | Harry van Langen | CC BY-SA 4.0 |
Removed picture duplication
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| May 22, 2023 at 8:34 | history | edited | Harry van Langen | CC BY-SA 4.0 |
Further development
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| May 22, 2023 at 3:54 | comment | added | Harry van Langen | That is for the 9x9x9 grid and p=50. However, for this grid any even p>=16 will achieve better than 96.0 quality. The best I can achieve for this grid is 96.1 for p>=24. For a finer grid, a larger p is required to deal with deteriorating quality near the edges of the cube. | |
| May 22, 2023 at 0:35 | comment | added | Oscar Lanzi | When you claim your quality is $96\%$, is that for the $p=50$ example you gave? Did you seek to optimize $p$? | |
| May 21, 2023 at 16:59 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| May 21, 2023 at 11:58 | history | edited | Harry van Langen | CC BY-SA 4.0 |
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| May 15, 2023 at 11:22 | history | edited | Harry van Langen | CC BY-SA 4.0 |
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| May 13, 2023 at 7:42 | comment | added | Harry van Langen | Sam, I want to make the grid points on a “great circle” (in the meaning discussed with Timothy) equidistant. At the moment some “great circles” are worse than others. | |
| May 13, 2023 at 7:25 | comment | added | Sam Nead | Do you care more about length distortion or "ratios of length distortion"? Also, how do you define "most distorted great circle"? (If it is in terms of Q, then your exposition is a tiny bit circular... no pun intended.) | |
| May 13, 2023 at 3:53 | comment | added | Harry van Langen | Thanks Daniel, I have updated the description slightly. | |
| May 13, 2023 at 3:50 | history | edited | Harry van Langen | CC BY-SA 4.0 |
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| May 12, 2023 at 22:55 | comment | added | Daniel Asimov | The description still looks like 8 points to me. Maybe you want {(x,y,z) ∊ ℝ^3 | max{|x|, |y|, |z|} = 1}. | |
| May 12, 2023 at 12:52 | comment | added | Harry van Langen | Hi Timothy, you are right. That’s what I meant. And the quality then becomes the uniformity of the map of those squares (or what I call coordinate lines in my response to Mark). | |
| May 12, 2023 at 12:40 | comment | added | Timothy Chow | @HarryvL I don't understand your definition of $Q$. What do you mean by a "great circle"? Here's my guess. We intersect the surface of the cube with a plane that is parallel to one of the faces of the cube; this gives us a square $S$ on the surface of the cube that runs all the way around the cube. Then we map $S$ to the sphere, and the image of $S$ on the surface of the sphere we call a "great circle." Is this correct? Note that this is not what is usually meant by a great circle on the surface of a sphere. | |
| May 12, 2023 at 11:59 | comment | added | Harry van Langen | Thanks Mark, but is there a solution in the context of my limited definition of quality? It merely requires the coordinate lines on the cube to be stretched/compressed uniformly. | |
| May 12, 2023 at 11:58 | history | edited | YCor | CC BY-SA 4.0 |
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| May 12, 2023 at 11:55 | comment | added | Harry van Langen | Thanks Alexandre. I changed the description | |
| May 12, 2023 at 11:44 | history | edited | Harry van Langen | CC BY-SA 4.0 |
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| May 12, 2023 at 11:36 | comment | added | Alexandre Eremenko | Your equation "of unit cube" defines 8 points, not the unit cube. | |
| May 12, 2023 at 11:20 | comment | added | Mark McClure | The question of mapping a portion of a sphere to the plane has been well studied due to its application to map projection. In that context, it's well known that no such map can represent all lengths in their correct proportions. My favorite text in this field is Portraits of the Earth, where this result is attributed to Euler. It was likely known in antiquity, as well though. | |
| May 12, 2023 at 10:42 | history | edited | Harry van Langen | CC BY-SA 4.0 |
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| May 12, 2023 at 9:53 | history | edited | Harry van Langen | CC BY-SA 4.0 |
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| May 12, 2023 at 9:43 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Updated according to the presentation in [this answer](https://math.stackexchange.com/a/4696995/317063)
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| May 12, 2023 at 9:04 | review | Close votes | |||
| May 19, 2023 at 3:07 | |||||
| May 12, 2023 at 8:57 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Embedded images, formatted and a bit Math Jaxed
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| May 12, 2023 at 8:42 | history | edited | YCor |
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| S May 12, 2023 at 8:38 | review | First questions | |||
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| S May 12, 2023 at 8:38 | history | asked | Harry van Langen | CC BY-SA 4.0 |