Surface Area of an Oblique Cone first created 11/11/07 - last modified 11/11/07 Page Author: Ty Harness
Sorry still unfinished

Finding the surface area of a right cone is straight forward but finding the surface area of an oblique cone is not easy. Cantrell [1] proposes a formula for finding the surface area of an oblique cone but it's not a simple formula to resolve:



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Figure 1 An oblique cone with the following properties: r = 2, h = 3, d=1


Cantrell's Solution S = 2*r*Sqrt[s]*(EllipticE[m]-EllipticK[m]+(1-n)*EllipticPi[n,m])

s = Sqrt[ (h^2+ (r-d)^2) * (h^2+(r+d)^2) ]

m = 1/2*(1 - (h^2+r^2-d^2)/s)

n = 1/2*(1 - (h^2+r^2+d^2)/s)

So using some test parameters (say) those shown in figure 1:

r = 2, h = 3, d=1 then

s = 13.41640786, m = 0.0527864, n = -0.02174919

It's possible to calculate the 3 elliptical integrals from tables but very much easier using functions from the wolfram.com[2] web site.

EllipticE[m] = 1.5921645, EllipticK[m] = 1.5498573, EllipticPi[n,m]) = 1.5750129

and finally the surface area

S = 22.95811858 [units^2]



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Figure 2 - CONET Sheet Metal Application: r = 2, h = 3, d=1




The cone transformer CONET software divides the base circle in to n divisions. The higher n becomes the more accurate the solution to the area. The software uses Heron's formula for finding the area of the triangle as shown in red in figure 3:

$A = sqrt(s(s-a)(s-b)(s-c))$ where s is the semi-perimeter $s = 1/2 (a+b+c) $ and a,b,c are side lengths of the triangle. Finding the true lengths and pattern triangulation onto a flat plane is described at : stormaths.htm

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Figure 3 - Summation of triangles


h=5, r=3, d=2 h=1, r=3, d=4 r=2, h=3, d=1
Cantrell, D.W. [1]56.15133.858 22.958
CONET56.13433.84322.950
Table 1 - Results from Cantrell and ConeT - ConeT is only an approximation suitable for general sheet metal work
References
[1] http://mathforum.org/library/drmath/view/55017.html

[2] http://functions.wolfram.com/

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