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I think that it is called 'packing efficency'. In euclidien 3D space the sphere minimizes the surface. With a fractal object you can maximize the surface without limit, I think.

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If you divide a volume by an area you get a length (as you have found), this length is physically just the length of a cylinder, using the XKCD example (you could use any n-sided prism) where the circle face has an area equal to the surface area (of your original shape). you can see this image that demonstrates it: NOTE: The scale between the sphere and ...

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let's consider some simple examples: a sphere, a cube, and a rectangular parallelepiped. Let's denote the radio of the volume to the surface area of a given object by $\ell$, then we have \begin{align} \ell(\mathrm{sphere}) &= \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{1}{3} R = \frac{1}{6}D \\ \ell(\mathrm{cube}) &= \frac{L^3}{6L^2} = ...

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For the case of a sphere the ratio you found is: $$\frac{V}{S} = \frac{ \frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3}$$ We can actually pass off the volume as being the integral of the surface area here. That's passable when you check the calculus. One approach is then to ask "what is a function divided by its derivative". This is really similar to ...

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The physical representation depends on the geometry of the system. In the case of a sphere, then we have the simple result $$\frac{V}{A} = \frac{(4\pi/3)R^3}{4\pi R^2} = \frac{R}{3}.$$ That is, the ratio is one-third of the radius. Now spheres are special in that they maximize this ratio. For example suppose you had a cube of side length $s$. Then $$... 0 With a short straightforward calculation, I came to this picture: That is, if the ellipse semi-major and semi-minor axes are given by vectors \pmb{a} and \pmb{b}, then the eigenverctors are proportional to \pmb{a}\pm i\pmb{b} (with maybe some complex factors), and their order would give the direction of rotation: from the ... 2 Since a worldline along the time axis on Minkowski diagram is at rest, it is more intuitive to measure angles from that axis instead, as then 'slope' is (space)/(time), i.e., a velocity. Then we have the trigonometric relationship:$$\frac{v}{c} = \tanh\alpha$$where Minkowski spacetime follows hyperbolic trigonometry because of the sign difference in the ... 2 Why do you stop your largest angle with ten 9s after the decimal point? If you added more of them, then you'd get a smaller bound for the velocity. And you keep adding 9s ad infinitum and you'll "eventually" reach 89.\bar{9}=90. So eventually, you'll see that the velocity could be arbitrarily small. This just means that the worldline can be vertical... and ... 2 A few thoughts about this. Let's first take it the other way around: determine if it would work with a given set of ellipses and then find for which mesh radius it would work. The general polar equation for a newtonian ellipse is$$r(\theta) = \frac{p}{1 + e\cos(\theta-\theta_0)}$$where p is the ellipse parameter, e the excentricity and \theta_0 ... 0 Ron Maimon is entirely correct when he says that GA is precisely Clifford algebra, as any book or paper using the phrase "Geometric Algebra" is sure to say. But I think he misses both the point of the question and the point of "GA". The question I'll paraphrase the question as: Is GA a good, pedagogical way to introduce the mathematical side of physics to ... 3 You are perfectly free to call the other angle \theta, so long as you are consistent. Then the area of interest would be written A \sin(\theta), and the other \theta marked in the diagram would have to be relabeled \pi/2-\theta. 1 Imagine taking the inclined surface to be close to horizontal. In this case, the angle between \mathbf A and the horizontal would be approximately 90^\circ, while the angle you have indicated in red would be approximately 0^\circ, so they can't possibly be the same. 2 Rather than looking at one orbit of Io, consider observing Io and Jupiter for around 200 days, starting when the Earth is exactly between the Sun and Jupiter, and ending when the Earth is opposite Jupiter, with the Sun in between. In the 200 days, Io will make around 110 orbits of Jupiter. But, importantly, the light from that last orbit of Io will need to ... 2 Actually, a hollow cube (assuming it's hollowed out symmetrically) completely filled with water/air just has its center of gravity at the intersection of the space diagonals while a solid cube doesn't generally have that. It only has that if the mass is distributed homogeneously. And in a hollow cube completely filled with air/water, the mass is distributed ... 3 What you call "centre of gravity" is more commonly called center of mass (COM). The general formula is not too hard:$$\mathbf{r}_{COM} = \frac{\sum \mathbf{r}_i m_i } {\sum m_i} where $\mathbf{r}_i$ are the position vectors of all individual masses $m_i$. In words: multiply a mass by its offset w.r.t. some coordinate system. Do this for all the ...

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