Volume of an Elliptical Cone
An elliptic cone is a cone whose base is an ellipse. If you take a flexible circle-base cone and gently squeeze the sides, you will make a shape that resembles and elliptical cone. To find the volume of such a cone, you need to know its height and the lengths of the semi-major and semi-minor axes of the ellipse at the base. Here is the formula with several examples.
Elliptical Cone Volume Formula
If the height is H and the semi-major and semi-minor axis lengths are A and B respectively, then the volume of the cone is given by the formula
V = (π/3)*A*B*H = (H/3)(πAB)
In other words, the volume is one-third of the height (H/3) times the area of the elliptical base (πAB). This is consistent with the volume formulas for circular cones and pyramids of any base shape. Their volumes are all H/3 times the area of the base.
As a very simple example, suppose a cone has a height of 7 inches and an elliptical base whose major axis is 10 inches and whose minor axis is 6 inches. This means H = 7, A = 10/2 = 5 and B = 6/2 = 3. The volume is, therefore
(π/3)*7*5*3 = 35π ≈ 109.956 cubic inches.
The semi-major and semi-minor axis lengths of an elliptical cone are 8.61 cm and 14.07 cm. If the longest side length of the cone is 23.95 cm, what is the height and the volume of the cone?
For this problem, we need to apply the Pythagorean Theorem to a right triangle whose leg lengths are 14.07 cm and H cm, and whose hypotenuse is 23.95 cm. This gives us a height of
sqrt(23.95^2 - 14.07^2) ≈ 19.3814 cm
The volume of the cone is then
(π/3) * 8.61 * 14.07 * 19.3814 ≈ 2458.73 cm^3.
Flat Pattern of an Elliptical Cone
Unlike a circular cone whose flat pattern is a circular arc, an elliptic cone produces a more complicated curve when it is cut along a side and unrolled. The flat pattern of an elliptical cone is shown above. The curve of an unfolded elliptic cone cannot be expressed in terms of elementary functions whether you use rectangular coordinates or polar coordinates.
The volume an elliptical cone is 359.77 and the height of the cone is the arithmetic mean of the semi-major and semi-minor axis lengths. If the length of elliptic cone base is 1.8 times the width, what are the dimensions of the cone?
Letting A be the length of the semi-major axis, we have A = 1.8B and H = (A+B)/2 = 1.4B. Using the volume formula for elliptic cones we get the equation
359.77 = (1.4H/3)(1.8πB^2)
359.77 = 0.84πB^3
B^3 = 359.77/(0.84π)
B = [359.77/(0.84π)]^(1/3)
B ≈ 5.1467
Using the other relations gives us A = 9.2641 and H = 7.2054.
Example 4: Elliptic Cone Frustum
For any base shape, the volume of a frustum is (H/3)(X + sqrt(XY) + Y) where X and Y are the areas of the top and bottom faces. A truncated elliptic cone or elliptic conical frustum is no different. Suppose an elliptic conical frustum has a base whose width and length are 10 and 16 inches, while the top face has a width and length of 5 and 8 inches. If the height of the frustum is 17 inches, then the volume is
(17/3)[160π + sqrt(160*40*π^2) + 40π]
= (17/3)[160π + 80π + 40π]