4 ms·
Thoughts about Pi
- agjacobson 11y agoYeah but the radius of the moon's orbit is sure interesting.
- agjacobson 11y agoDiameters might appear in drawings. But they NEVER appear in physics. (The distance between particles determines forces. Diameters are a boring parameter about the size of something.) Machine tool programming is mostly radii in the paths, and diameters in the setup parameters.
- avmich 11y ago> Diameters might appear in drawings. But they NEVER appear in physics. Even as a characteristic size in, say, Reynolds number definition?
- sebastos 11y agoThe statement about the diameter being the interesting parameter is kind of exactly wrong. This is true for all the same reasons Tau is a better constant than Pi. For example, how does one derive the area of a circle? The most natural way is to integrate with respect to radius. When using Tau and radius, you get A = (1/2)Taur^2, which is nice since it looks a lot like the integral of, say, momentum, which gives you kinetic energy. If you use Pi or diameter or both, then weird factors of 2 start to creep in. It's also much more natural to think about defining the radial extent at each angle, rather than the diameter, especially if you're dealing with something other than a circle that has changing radius.
- gaze 11y agoTau is not happening.
- justinlardinois 11y agoStop trying to make tau happen. It's not going to happen.
- delish 11y agoI'll echo what gp said to help make tau happen: tau : the circumference (perimeter) of a circle. (1/2)tau^2 : the area of a circle. Beautiful. To go from one to the other, you integrate or take a derivative. Like gp said, there's no weird factor of two. Just the power rule in action.
- drdeca 11y agoYou seem to have left out the r in both of those expressions. Tou r. .5 tou r^2 .
- cmiller1 11y agoNope, all circles have a perimeter of 19.7392088022.
- Someone 11y ago(1/2)tau^2? I assume that's not what you meant to write, and just mention: 2pi r : the circumference (perimeter) of a circle. pi r^2: the area of a circle. Beautiful. To go from one to the other, you integrate or take a derivative. There's no weird factor of 1/2. Just the power rule in action.
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- alimw 11y agoYour conjecture as to the surface area of a hypersphere embedded in 4 dimensions (a "3-sphere") is not borne out :) The answer is in fact d^3 pi^2 / 4. See https://en.wikipedia.org/wiki/N-sphere#Volume_and_surface_area https://en.wikipedia.org/wiki/N-sphere#Volume_and_surface_ar...