Volume of a bead with square hole- Program approach?

[Albert Chan]

(11-05-2023 04:16 PM)Albert Chan Wrote:  

(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)

To show hole volume is over-estimated, we check sign of errors

lua> taylor(hv2 - b^2 * √(1 - (2b^2)/3), b, 10)

- b^6/45 - 23*b^8/945 - 143*b^10/5670 + b^12*order_size(b)

Negative under-estimated error → RMS formula over-estimated hole volume.

Quote: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.

height of rect cap (both of them), h = 1-k      // 1 is diameter, k is corner height

mean([1,1,k]) = 1 - h/3 = (1-h) + 2/3*h

MEAN formula under-estimated hole volume, if rect cap mean height > 2/3*h

XCas> ratio := (hv2/(b*b)-k) / (1-k)
XCas> taylor(ratio, b, 8, polynom)

2/3 + 4/45*b^2 + 5/63*b^4 + 23/252*b^6 + 1997/16632*b^8

Small hole, h and b^2 about the same size: k = 1-h = sqrt(1-2*b^2) ≈ 1-b^2
Ratio converge faster (smaller coefficients), if we use variable h.

XCas> taylor(ratio(b=sqrt((1-(1-h)^2)/2)), h, 6, polynom)

2/3 + 4/45*h + 11/315*h^2 + 1/84*h^3 + 25/8316*h^4



Compare mean height ratio with spherical cap.

V = pi*h/6 * (3*a^2 + h^2)
A = pi*a^2

ratio = V/A/h = 1/2 + (h/a)^2/6 > 1/2