Ellipsoid surface area
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05-28-2021, 12:37 AM
Post: #1
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Ellipsoid surface area
semi-axis, a ≥ b ≥ c
cos(φ) = c/a m = (a²(b²-c²)) / (b²(a²-c²)) I = cos(φ)^2 * elliptf(φ,m) + sin(φ)^2 * ellipte(φ,m) → S = 2*pi*(c² + a*b/sin(φ)*I) HP Prime does not have elliptf(ϕ,m) and ellipte(ϕ,m), we might as well combine I, as 1 integral Let t = sin(ϕ)^2 = 1-(c/a)^2, s = √t = sin(ϕ): I = (1-t) * ∫(1/√((1-x²)*(1-m*x²)), x=0..s) + t * ∫(√((1-m*x²)/(1-x²)), x=0..s) I = ∫((1-m*t*x²) / √((1-x²)*(1-m*x²)), x=0..s) Code: #cas CAS> ellipsoid_area(1.1,1.2,4.7) → 54.6901240998 Example from The Surface Area Of An Ellipsoid, A. Dieckmann, Universität Bonn, July 2003 A good estimate with Thomsen formula, rel. err ≤ 1.061% CAS> ellipsoid_area_est(a,b,c) := 4*pi*mean(([a*b,a*c,b*c]).^p)^(1/p) | p = 1.6075 CAS> ellipsoid_area_est(1.1,1.2,4.7) → 54.4952199737 |
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Messages In This Thread |
Ellipsoid surface area - Albert Chan - 05-28-2021 12:37 AM
RE: Ellipsoid surface area - Albert Chan - 05-28-2021, 08:05 PM
RE: Ellipsoid surface area - Albert Chan - 05-30-2021, 06:04 PM
RE: Ellipsoid surface area - Albert Chan - 05-31-2021, 01:38 AM
RE: Ellipsoid surface area - Albert Chan - 08-05-2022, 04:40 PM
RE: Ellipsoid surface area - Albert Chan - 08-05-2022, 04:51 PM
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