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I need to calculate the volume of a shape which is kind of like a truncated cone but has an elliptical base

eg. Minor axis a: 5 Major axis b: 15

and a second more round elliptical top

eg. Minor axis c: 1 Major axis d: 2

The 'vertex' ellipse is more round than the base which is more elliptical.

The height of the shape would be

h = 10

I've been looking for a formula for this but can't seem to find it ...

Can You help ?

Example

Edit. The shape is simple enough 2 cocentric ellipses on parallel planes form the top and bottom of the shape. The walls of the shape resemble those of a truncated cone. Basically I need the formula for a truncated elliptical cone but the ratios of minor to major axis on the ellipses describing the top and bottom are different.

Please help :)

Thanks a million :)

KReiser
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Wanton
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  • This doesn't specify the solid completely. What do the cross sections look like? Can you give a formula? – saulspatz May 12 '19 at 17:52
  • Are the axes of the top ellipse projected on to the axes of the bottom ellipse, or is the cone "twisted". Is the center of the top ellipse projected on to the center of the bottom ellipse. – MasB May 12 '19 at 18:07
  • Hi, thx for the comments. The ellipses do not twist or anything just to ellipses on parallel planes 10 units apart. One ellipse lofted to the other. – Wanton May 12 '19 at 18:15
  • Do You really require a formula ? My maths has not been in use for X years :) – Wanton May 12 '19 at 18:15
  • This is not a cone, because bases are not similar between them. – Intelligenti pauca May 12 '19 at 18:29
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    The problem with this question is that there is not necessarily a default way of defining the shape of the cross sections. What is it that defines their shapes? Is it (as I assumed in my answer) a linear dependency of the length of the axes on the height? Or do you want the convex hull of the top and bottom ellipses? You can also connect each point of the bottom ellipse with the corresponding point on the top ellipse with a line, but then you have to define what "corresponding" means. Same angle when projected on the plane of the bottom ellipse? Same parameter, using the usual parametrization? – Reinhard Meier May 12 '19 at 18:41
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    Why should this question be closed ? The asker is reactive, the question is natural and moreover puzzling by the fact that there is no unique answer. – Jean Marie May 12 '19 at 22:18

1 Answers1

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I assume that each cross section is an ellipse and that the length of the major axis and the length of the minor axis vary linearly with the height of the cross section above the bottom.

Let $a_0$ be the length of semimajor axis at the bottom, $b_0$ the length of the semiminor axis at the bottom, $a_1$ be the length of semimajor axis at the top, $b_1$ the length of the semiminor axis at the top. Let $a(z)$ be the length of the semimajor axis at the height $z$ above the bottom and $b(z)$ the length of the semiminor axis at the height $z$ above the bottom. Then $$ a(z) = \frac{z}{h}(a_1-a_0)+a_0 \\ b(z) = \frac{z}{h}(b_1-b_0)+b_0 $$ And the volume is $$ V=\int\limits_0^h \pi\,a(z)\,b(z)\,dz $$ After some lengthy calculations, I got $$ V = \pi\,h\,\frac{2a_0b_0 + 2a_1b_1 +a_0b_1+a_1b_0}{6} $$ If you use the lengths of the complete axes instead of the semi axes, the formula is $$ V = \pi\,h\,\frac{2A_0B_0 + 2A_1B_1 +A_0B_1+A_1B_0}{24} $$

Reinhard Meier
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  • Thx a million for Your answer ... I calculated a simple version of Your formula for a circle radius 1m (eg. ellipse Ao= 1m, Bo= 1m) lofted 1m to an ellipse Ai=2m, Bi=1m.

    The Answer using Your formula is : V = pi *9/6 = 4.7123 m3

    When I simulated this shape in my CAD program and calculate the Volume in the program (Autocad Fusion 360) it gives me 1.178m3 ?

    I think we may have mis-understood each other :|

     – Wanton May 12 '19 at 19:08
  • Note that I have used the lengths of the _semi_axes. – Reinhard Meier May 12 '19 at 19:13
  • I have added the formula for the use with the complete axes. – Reinhard Meier May 12 '19 at 19:29
  • My bad ... Sorry for that :)

    You're 100% correct ... :)

     – Wanton May 12 '19 at 19:38
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    That is correct (+1). However it is worth to underline that such a surface won't (in general) be a cone, since the directrices won't met at a single point, nor in general it is a conical wedge, since the directrices don't meet along a line, and (always for general values of the parameters) it might not be developable. All characteristics that were not specified by the OP. – G Cab May 12 '19 at 22:41
  • In particular the picture "hints" to that the surface is made by bending a sheet, i.e should be developable. If that is the case, then refer to this post – G Cab May 12 '19 at 22:49