Volume of a bead with square hole- Program approach?
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06-16-2020, 12:35 PM
(This post was last modified: 11-05-2023 03:59 PM by Albert Chan.)
Post: #22
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RE: Volume of a bead with square hole- Program approach?
(06-11-2020 06:53 PM)Albert Chan Wrote: Solve with A=0.1, B=0.1, we have hole = 0.00996658845699, about 2% of sphere For small square hole, we can estimate with simple formula 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> hv2(b = 0.1) // just to confirm square hole formula → 0.00996658845699 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) Since taylor series all have even powers, let dimensionless x = b^2 XCas> pade(x - x^2/3 - 7*x^3/90, x, 3,3) → (17*x^2-30*x)/(7*x-30) XCas> r(x) := x*(17*x-30)/(7*x-30) XCas> r(0.1^2) // rough estimate for small 0.1×0.1 hole → 0.00996658870698 |
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