0

Can you please help me with the problem of computing the lateral area and volume of a frustrum but the base and top ellipses are arbitrary 2 ellipses.

My problem is: if a1,b1 are the semi- major and minor axes of the base ellipse, the a2, and b2 are semi- major and minor axes of the top ellipse, a2 can be the minor semi-axis of the top ellipse, and b2 can be the major semi-axis, i.e., the major axes at the bottom and top can be orhogonal.

I have found obviously the solutions everywhere on the web, also here, for the case where the base and top ellipses are similar in shape (that is the top ellipse is just a scaled ellipse of the base ellipse). So the Thales ratios between a1,a2,b1,b2 with height H are all known. Easy to compute the frustrum lateral area and volume as the difference of two elliptical cones. But here this does not work.

Is there analytical solution to this?

Thank you so much for reading. You are my last hope.

Best regards, Zuheyr

zuheyr alsalihi
  • 5
  • 3
  • There can be several different lateral surfaces fit to the same couple of bases: you should point out how the lateral surface is constructed. – Intelligenti pauca Aug 02 '21 at 08:11
  • @intelligenti pauca: Lateral surface I meant is the sides, say the lateral surface area of a cylinder is 2PiRadius. – zuheyr alsalihi Aug 02 '21 at 11:08
  • Of course, but how is that surface constructed? Perhaps, joining those points on top and bottom base corresponding to the same azimuthal angle? And is the line joining the centers of the ellipses perpendicular to both bases? – Intelligenti pauca Aug 02 '21 at 13:21
  • Yes, perpendicular. I would presume that the lateral surface is constructed that way, linear between the top and the bottom ellipses at the the same azimuthal angle. Thank you for your help. – zuheyr alsalihi Aug 02 '21 at 17:25

1 Answers1

0

A point $P_1$ on the base ellipse with azimuth $\theta$ has coordinates: $$ P_1=\left( {b_1\cos\theta\over\sqrt{1-e_1^2\cos^2\theta}}, {b_1\sin\theta\over\sqrt{1-e_1^2\cos^2\theta}}, 0 \right) $$ and the corresponding point $P_2$ on the top ellipse is then: $$ P_2=\left( {b_2\cos\theta\over\sqrt{1-e_2^2\cos^2\theta}}, {b_2\sin\theta\over\sqrt{1-e_2^2\cos^2\theta}}, H \right) $$ where $e_1$ and $e_2$ are the eccentricities: $$ e_1^2=1-{b_1^2\over a_1^2},\quad e_2^2=1-{b_2^2\over a_2^2}. $$ A generic point on segment $P_1P_2$ can be expressed as $$P=(1-t)\cdot P_1+t\cdot P_2, \quad\text{with}\quad t\in[0,1]. $$ This is then a parametrization of the lateral surface and the standard way to compute its area is: $$ area=\int_0^{2\pi}\int_0^1 \left|{\partial P\over\partial t}\times {\partial P\over\partial \theta}\right|\,dtd\theta. $$ For the volume, integrating in cylindrical coordinates gives: $$ volume={H\over2}\int_0^{2\pi}\int_0^1 \left(P_x^2+P_y^2\right)\,dtd\theta. $$

enter image description here

Intelligenti pauca
  • 55,765
  • Thank you so much. I do not know if I will be able to carry out the integrations but I will try. Px and Py are partial derivatives I think. H is the distance between the centers of the ellipses. – zuheyr alsalihi Aug 04 '21 at 10:19
  • The integration cannot give a result in terms of elementary functions, at least for surface area (that is true even in the case of an elliptic cylinder). $P_x$ and $P_y$ are the cartesian components of $P$, not partial derivatives. – Intelligenti pauca Aug 04 '21 at 10:22
  • Thank you so much. I do not know if I will be able to carry out the integrations but I will try. Px and Py are partial derivatives I think. H is the distance between the centers of the ellipses. Thank you so much for the beautiful figure which shows curved lateral surfaces but the connections are straight lines as your analysis correctly uses this information. I think the easiest solution for me is to numerically integrate your equations, I am engineer unfortunately, not mathematician. A very sincere thank to you!! Best regards. – zuheyr alsalihi Aug 04 '21 at 10:35
  • Sorry while I think and edit, the comment is disabled. Million thanks. I know the result will involve complete elliptic integrals of the 2nd kind because of the perimeter of the ellipse, but the genius Ramanujan formulas for the perimeter are perfect. . I do not know how general and correct is the intuitive answer for the surface area: an average slant multiplied by the sum of the perimeters of the base and top ellipses? I think the easiest solution for me is to numerically integrate your equations, I am engineer unfortunately, not mathematician. A very sincere thank to you!! Best regards. – zuheyr alsalihi Aug 04 '21 at 11:14
  • Sum of the perimeters*average slant/2. – zuheyr alsalihi Aug 04 '21 at 13:53
  • @zuheyralsalihi That formula might give a good approximation. As I wrote in a comment above, $$ and $$ are the cartesian components of $$, not partial derivatives. – Intelligenti pauca Aug 04 '21 at 14:16
  • Thank you again. Yes Px and Py I understood but while I was thinking, writing the comment, the comment time delay was expired. So when i reposted them, they were no longer actual. Many many thanks to you. Best regards. – zuheyr alsalihi Aug 04 '21 at 17:25
  • As I do not have enough reputation I cannot close the thread I think, neither can I vote your answer. I accepted as correct solution and I wrote a progam based on what you have suggested and as I mentioned, I just compute an average slant, that's it for the lateral surface, The perimeters are computed either complete elliptic integral of the 2nd kind or with Ramanujan. Volume numerical integration using local axes at z as you give and from: https://math.stackexchange.com/questions/3223528/volume-of-a-truncated-elliptical-cone-with-2-different-elliptical-semi-axis-on-e. THX – zuheyr alsalihi Aug 06 '21 at 08:11