Algebra question: volume of spherically capped frustum

Hi Folks,

I spent about an hour trying to figure this out, and ended up with not much more than two pages of equations of ever-increasing length. If someone wants to help, I would appreciate it. I really need to go back to school, my skills have gotten way too rusty.

Lets say I want to make a frustum and cover one end with a sperical cap. The radius of the base of the spherical cap is equal to the radius of the end of the frustum it sits on. The radius of the sphere is equal to the height of the sperical cap plus the height of the frustum (i.e. the center of the sphere is the center of the other end of the frustum).

I have the volume of the entire thing, the total height (i.e. the radius of the sphere), and the radius of the uncapped end of the frustum. With that information, there should exist an equation that gives the radius of the capped end of the frustum (which is the radius of the base of the spherical cap), but I can’t find it. That’s the question: what is that equation?

Check this page.

Yeah, I looked at that page when I was trying to figure it out for myself. It features not the equation I desire, but it was quality source material for the Wikipedia entry I wrote on the subject.

There is such an equation, but it’s sufficiently complex and sufficiently late at night that I can’t easily provide a solution to it. Let’s call the radius of curvature of the spherical cap q; the radius of the uncapped end of the frustrum R; the radius of the capped end of the frustum r; and the height that the frustum would have without the cap h. My (hopefully error-free) calculations show that

V = pi h/3 (R[sup]2[/sup] + r R + 2 r[sup]2[/sup] + 2 h[sup]2[/sup])

By Pythagoras’ theorem, h = sqrt(q[sup]2[/sup] - r[sup]2[/sup]), so this can be written as

V = pi/3 sqrt(q[sup]2[/sup] - r[sup]2[/sup]) (R[sup]2[/sup] + r R + 2 q[sup]2[/sup])

Now if I understand your question correctly, you have V, q, and R, and you want to solve for r. The problem is that this is effectively a fourth-degree polynomial in q; squaring both sides,

(q[sup]2[/sup] - r[sup]2[/sup]) (r R + R[sup]2[/sup] + 2 q[sup]2[/sup])[sup]2[/sup] = 9 V[sup]2[/sup]/pi[sup]2[/sup]

Now, quartic (fourth-order) polynomials are soluble, but there isn’t a simple formula one can write down for the solution as there is for, say, quadratic polynomials. It’s possible that there’s a qay to simplify this, or that your problem is one of the special cases mentioned in the link, but I wouldn’t bank on it.

If, on the other hand, you know h instead of q, then you have a quadratic equation (the first one I wrote down above) in r and you should be able to easily find the solution.

It’s kind of late for me too.
You could try visiting my site’s geometry index
and see if there’s a calculator that has what you need.
(Yes, it has frustum and spherical cap calculators as well as their associated formulas).

Yes it is late. How about this instead?

Yes, I’d been using your frustum calculator previously; working with it partially led to me asking this question. You’re actually missing a ) on your frustum page.

I did do a google search before asking, so I’m pretty sure that there isn’t a calculator anywhere for this specific question. If I get an answer, though, I’ll make one.

Thanks, MikeS, I’m glad I’m not just missing some obvious solution.

Your comment has been noted and corrected.


Partially defines the, spherical cap, SC.

Completes definition of SC.
The conical frustrum, CF, is partially defined by it’s height and upper radius.
Neither the lower radius or the angle necessary to define the CFi s given.

You must determine the volume, V, of the solid assembly in terms of one variable.

Now it’s up to you!

Is this of some practical value or do you delight in mm that you can’t finish?

I’m not quite sure how to take your reply, springears.

As I said in the OP, “I have the volume of the entire thing, the total height (i.e. the radius of the sphere), and the radius of the uncapped end of the frustum”.


I do have an application for it, yes. I wouldn’t go to this much effort just for giggles.

It might help you to visualize your shape differently – consider the frustum to be a cone from which the upper portion has been subtracted twice (i.e., once to make it a frustum, and once to make it an isosceles triangle swept around a vertical). Then imagine the cap not as a flat-faced “slice” of a sphere, but as a conical solid angle taken from the sphere (a spherical cone).

So your volume will be the volume of a spherical cone (SC) plus a frustum (F), minus the volume of the cone they share (CC, the “common cone”). Let’s define this in terms of radii R1 and R2 (the frustum’s greater and lesser radii) and H (height of the solid and also the radius of the spherical cap).

V[sub]total[/sub] = V[sub]SC[/sub] + V[sub]F[/sub] - V[sub]CC[/sub]

To compute V[sub]SC[/sub] we need to know the height of just the spherical cap (which we’ll call “h” when we isolate it). From Pythagoras, the frustum height h2 = H[sup]2[/sup] - R2[sup]2[/sup] .

Therefore the cap height h = H - h2 = (H - H[sup]2[/sup] + R2[sup]2[/sup])

Knowing this, we can calculate V[sub]SC[/sub]:
V[sub]SC[/sub] = [sup]2[/sup]/[sub]3[/sub] * pi * H[sup]2[/sup] * h
= [sup]2[/sup]/[sub]3[/sub] * pi * (H[sup]3[/sup] - H[sup]4[/sup] + R2[sup]2[/sup]*H[sup]2[/sup])

…which is ugly but do-able.

V[sub]F[/sub] = [sup]1[/sup]/[sub]3[/sub] * pi * (H[sup]2[/sup] - R2[sup]2[/sup]) * (R1[sup]2[/sup] + R1*R2 + R2[sup]2[/sup])

…which I’m not going to expand out because typing all these "sub"s and "sup"s is killing me.

Lastly, you subtract away V[sub]CC[/sub]:

V[sub]CC[/sub] = [sup]1[/sup]/[sub]3[/sub] * pi * R2[sup]2[/sup] * (H[sup]2[/sup] - R2[sup]2[/sup])

So, in terms of only the variables you specified, we have

V[sub]total[/sub] = { [sup]2[/sup]/[sub]3[/sub] * pi * (H[sup]3[/sup] - H[sup]4[/sup] + R2[sup]2[/sup]H[sup]2[/sup])} + {[sup]1[/sup]/[sub]3[/sub] * pi * (H[sup]2[/sup] - R2[sup]2[/sup]) * (R1[sup]2[/sup] + R1R2 + R2[sup]2[/sup])} - {[sup]1[/sup]/[sub]3[/sub] * pi * R2[sup]2[/sup] * (H[sup]2[/sup] - R2[sup]2[/sup])}

…unless of course I’m totally wrong.

…or didn’t read the OP thoroughly enough. :smack:

This is a problem in Solid Geometry to be solved by algebra.

Look at the problem this way.
The Volume, V, is known at say 120 cu. in.

ASSUME the volume of the geometric volume is: a + b^2 + c^3.
Setting: V = 120 = a + b^2 + c^3 you can assume values for a and b and solve for c.
However if the volume can be expressed in terms of only one variable the solution is greatly simplified.

The OP complete defines the spherical cap.
For the conical frustum the OP omits the critical third ‘dimension,’ i.e. radius of the base, or slant height, or cone angle. Without this dimension the problem is not completely defined and no solution is possible.

This is essentially what I did to get the formula in my post above.

You mean sqrt(H[sup]2[/sup] - R2[sup]2[/sup]), right? I think this is the reason that we didn’t get the same answer.

Check the OP again — he says that he has the radius of the base.

He is still missing a piece of information, from what is given there are an infinite number of solutions. Think of a solid that fits the determined base radius and height. It will have a particular volume. If we then change the angle of the side of the frustum we can change the height of the frustum itself (and consequently the radius of the top of the frustum and radius of the sphere from which the spherical cap comes) to get back to the same volume. For example, nothing that you have stated so far forces the spherical cap to be on the large or small base of the frustum.

So, you need to pick one of the following to define: Are of the top of the frustum, radius of the top of the frustum, angle of the side of the frustum, volume of either frustum or spherical cap alone, height of either frustum or spherical cap alone.

Otherwise you could add a constraint that the angle of the side of the frustum matches the angle of the spherical cap where the two intersect, thereby making a smooth connection without an edge.

I will now find the equation for the radius of the capped end of the frustum if you do this, check back soon.

No, the radius of the sphere does not change, it’s one of the givens. The radius of the base of the cap can change, and the two heights corespondingly, but there is only one shape that has the given volume. The shape could range from a hemisphere (where the height of the frustum is 0) to a cone (where the height of the cap is 0) which does limit the range of volumes, but as long as the given volume is within that range, there should only be one solution for the cap base radius/upper frustum radius.

I’ll post a drawing this evening, perhaps that will help get everyone on the same page.

Mike, your SQRT is well-placed; that’s what I get for not double-checking. Yes, we arrive at the same answer for the volume equation. Solving it for R2 is going to be a pain.

I know it’s probably not going to help, but I want to re-cast the problem in a different way: imagine a cone intersecting a sphere. Where the cone’s pointy end emerges from the sphere, there is a circle of radius “r” (R2 in my parlance above - the variable to be solved), and where the cone’s wide end emerges there’s a larger circle of radius “R” (depending how we define the problem, this also qualifies as an unknown measurement!). Furthermore, if you cut the sphere in half parallel to these circles and look at the exposed great circle, there’s a circle inside where the cone’s midsection is exposed - that known radius is the R1 from above. And of course, the sphere is defined by its radius “h” (or “q” if you must).

Consider the following trivial cases:
R2 = q : the frustum/sphere is just a hemisphere; the volume is strictly a function of q or R2 and R1 is irrelevant.
R2 = 0 : the frustum/sphere has no spherical end-cap and is just a cone; volume is a function of q and R1.

Beacuse R2 can’t be less than zero or greater than q, we should be able to graph all of the possible volumes V against R2 (limited from 0 to q) for any known value-pair of R1 and q. A horizontal line drawn across the graph will intersect all valid solutions for R2.


Damn me for not previewing; however, I disagree with this statement of uniqueness – couldn’t there be two frusta of opposite tapers (one tapering down to the cap, one flaring up to the cap) that had the same volume?

OK, I was wrong, I failed to take into a constraint that came in the form of the sqrt you guys were using.

Unfortunately, aside from the squared typo, I think there is another reason you two don’t get the same. In MikeS’s original equation he seems to be using the same h for the height of the spherical cap and the height of the frustum, is that correct?

I get:

V[sub]scap[/sub] = pi/6 * (3*r[sup]2[/sup]+H[sup]2[/sup]) * h[sub]2[/sub]

V[sub]f[/sub] = pi/3 * (R[sup]2[/sup]+rR+r[sup]2[/sup]) * h[sub]1[/sub]

h[sub]1[/sub] + h[sub]2[/sub] = H

V = pi/6 ( (2R[sup]2[/sup]+2rR-r[sup]2[/sup]-H[sup]2[/sup]) * h[sub]1[/sub] +3r[sup]2[/sup]H+H[sup]3[/sup] )

h[sub]1[/sub][sup]2[/sup] + r[sup]2[/sup] = H[sup]2[/sup]

V = pi/6 ( (2R[sup]2[/sup]+2rR-r[sup]2[/sup]-H[sup]2[/sup]) * sqrt(H[sup]2[/sup] - r[sup]2[/sup]) +3r[sup]2[/sup]H+H[sup]3[/sup] )

(6V/pi - 3*r[sup]2[/sup]H - H[sup]3[/sup])[sup]2[/sup] = (2R[sup]2[/sup]+2rR-r[sup]2[/sup]-H[sup]2[/sup])[sup]2[/sup] * (H[sup]2[/sup] - r[sup]2[/sup])

r[sup]6[/sup] - 5Rr[sup]5[/sup] + 10H[sup]2[/sup]r[sup]4[/sup] + 8R[sup]3[/sup]r[sup]3[/sup] + (5H[sup]4[/sup]-4R[sup]2[/sup]H[sup]2[/sup]+4R[sup]4[/sup]-36VH/pi)r[sup]2[/sup] - (8R[sup]3[/sup]H[sup]3[/sup]-4RH[sup]4[/sup])r + (4R[sup]4[/sup]H[sup]2[/sup]+4R[sup]2[/sup]H[sup]4[/sup]-12VH[sup]3[/sup]/pi+36V[sup]2[/sup]/pi[sup]2[/sup])

Good luck solving that.