Volume of a bead with square hole- Program approach?

[Albert Chan]

(06-16-2020 12:35 PM)Albert Chan Wrote:  For unit diameter (d=1), with square hole, b=a

XCas> k := sqrt(1-2*b*b)
XCas> hv2 := b*b*k/3 - atan(b*b/k)/3 + b*(1-b*b/3)*atan(b/k)
XCas> taylor(hv2,b,10)
→ b^2 - b^4/3 + -7*b^6/90 + -3*b^8/70 + -83*b^10/2520 + b^12*order_size(b)

hv2 ≈ b^2 * √(1 - (2b^2)/3) = b^2 * √(1 - (1-k^2)/3)

Remove b=a requirement: hv2 ≈ a*b * √(1 - (a^2+b^2)/3)
Remove d=1 requirement: hv2 ≈ a*b * √(d^2 - (a^2+b^2)/3)

It is not as accurate hv2 pade approximation, but likely good enough
Previous example, d = 1, a = b = 0.1

lua> d, a, b = 1, 0.1, 0.1
lua> a*b * sqrt(d*d - (a*a+b*b)/3)
0.009966610925150703

Hole volume with 5 good digits, over-estimated.

Last term (estimated average height) is equivalent to RMS(d, d, sqrt(d*d-a*a-b*b))
Repalce RMS with arithemetic mean, we get the under-estimated hole volume.

lua> a*b * (2*d + sqrt(d*d-a*a-b*b)) / 3
0.00996649831220389

This is not as good as RMS version, does not reduce computation, thus not recommended.