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Math trivia: a bead placed on the table (hole to the sky), volume = volume of sphere with diameter = height
(You can think of a sphere is a bead with no hole, ... Neat)

Here is a prove using messy integrals

Simpson's rule for bead volume is simpler:
let R = radius of sphere, r = radius of the hole, height of bead = 2 h, thus h² = R²- r²

Half-Bead volume = h/6 * (top_area + 4 * middle_area + bottom_area)

Bead volume = 2 Half-Bead
= h/3 (0 + 4 Pi ( (√(R²- (h/2)²))² - r² ) + Pi (R²- r²))
= Pi/3 h (4 R² - h² - 4 r² + h²)
= 4/3 Pi h^3

Here is a beautiful prove with no math, just visualization

Find a gem of science/math stuff link: https://www.lockhaven.edu/~dsimanek/home.htm

Physics Q&A: https://www.lockhaven.edu/~dsimanek/puzz...nswers.htm

#1 Normally, a bridge is build on top of river, with ships passing beneath it.
-> But there is water bridge, with ships run on top of it ...

#66 holey sphere, about volume of bead. Solution by Martin Gardner ...

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I were trying to get a volume of bead with square hole (side 2 r).
Its volume must be less than bead with round-hole, radius r.

My first approximation for removed square holes volume
~ 4/Pi * volume of removed sphere round-hole
= 4/Pi * (4/3 * Pi * (R^3 - h^3))
= 16/3 (R^3 - h^3), where h = sqrt(R^2 - r^2)

Volume of square-holed bead ~ 4/3 Pi R^3 - 16/3 (R^3 - h^3)

For relative small r, is that close ?

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(09-09-2018 09:22 PM)Zaphod Wrote: [ -> ]I like the sites sponsor:


https://www.lockhaven.edu/~dsimanek/ideal/ideal.htm

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Sounds like the ideal sponsor.

(Post 279)

To maximize the holes of bead, the hole (no caps) can reach 1/sqrt(3) ~ 58% sphere volume !

http://datagenetics.com/blog/july22014/index.html

If the problem get flipped, maximize size of sphere inside cyclinder, it is even higher = 2/3 ~ 67%

http://datagenetics.com/blog/july32014/index.html