Definitions:
r[sub]1[/sub] = radius of the base of the frustum (unknown)
r[sub]2[/sub] = radius of the top of the frustum (known)
R = radius of the sphere (known)
h = height of the frustum, R[sup]2[/sup] = h[sup]2[/sup] + r[sub]1[/sub][sup]2[/sup], by Pythagorean Theorem, so h is known
V = volume of the capped frustum (known)
The volume can then be expressed as:
V = (Pi/3)(h(r[sub]1[/sub][sup]2[/sup] + r[sub]2[/sub]r[sub]1[/sub]) + 2R[sup]2/sup)
Note that this is just quadratic in r[sub]1[/sub], which is easy to solve (though the actual expression is a bit complex). Since the coefficient of the linear and quadratic terms are positive definite, only the + root of the quadratic formula need be considered.
As a check, let’s look at the limiting behavior in the cases h=0 and h=R. In the first case, the volume reduces to (2Pi/3)R[sup]3[/sup], which is just the volume of the hemisphere of radius R. In the latter case (a cone with a zero radius cap), the volume reduces to (Pi/3)hr[sub]1[/sub][sup]2[/sup], which is also just what we expect. This isn’t a proof, but it’s at least a decent sanity check.
State the problem mathematically.
The Volume of the whole thing:
V = f(x) + f(r). where x is an variable defined by a geometrical solid or envelope and r is some constant which can’t be defined in terms or x.
When r can be defined in terms of x then the solution follows, messy though it may be.
ASSUMING that you mean a RIGHT ANGLE FRUSTRUM (as linked in your OP), this reduces to a simple two dimensional problem, if you take a cross-section through the axis-- i.e. an isosceles right triangle (sides at 45° angle to the axis) of base B, intersecting a semicircle of radius R.
Please excuse my wordiness, if I had suitable webspace at the moment, a diagram would clarify it in an instant.
R = the radius of the sphere; a known quantity it is also the radius of the circle in our 2-D cross-section B = the base radius of the frustrum; a known quantity it is also the base of the intact cone, or the triangle in our figure
a = x = the radius of the base of the cap
h = y = the height of the frustrum
Place the center of the base at the origin (0,0). Since the figure is mirror symmetric about the Y-axis, we’ll take the right side (1st quadrant)
For all points on the circle: x²+y² = R²
For all points on the triangle: y = B - X The height of a right angle cone is equal to the radius of its base
Therefore R²= x² + (B-x)²
Since B and R are known constants, we can solve numerically or algebraically for X (the desired radius).
The solution for a frustrum of any other given angle is solved the same way, except you’d use y = B-(tan Θ/2)x for the triangle, yielding:
R²= x² + (B-mx)² [where m=tan(Θ/2), a constant. Tan (90°/2) just happens to be 1]
You could also solve this using trigonometry or polar coordinates in 2 dimensions
No, in my original formula q was the “radius of curvature of the spherical cap”, which means that if the “centre” of the sphere would be on the base of the frustum, then q is the total height of frustum + spherical cap.
To add to my earlier answer: in order to obtain the shape described, sqrt(2) R ≤ B < R
(where B is the base radius of the cone, and R is the radius of the sphere)
If B<R the right angle cone would fit entirely inside the hemisphere. If B>R sqrt(2), the sphere would fit entirely inside the right angle cone. A frustrum only exists in he range specified, but it is only unique at B=sqrt(2) R, where the cone is externally tangent to the hemisphere (a single ring of contact). Elsewhere in the range, the side of the cone intersects the hemisphere at two different circles.
Perhaps this is why the total volume of the frustroid is needed: to distinguish between the two possible cases for a given allowable value of R and B. I presume by inspection, but have not bothered to prove, that the two possible frustroids will have differnt volumes. There may be one or more special cases where they are equal.
The parts are identified by color. Blue is the base of the frustum, a known value. Green are the walls of the frustum. Black is the radius of the sphere, a known value. Red is the spherical cap. Orange is the base of the spherical cap and the top of the frustum, the value we are trying to find. Also known is the volume of the entire thing.
The first three drawings all have the same diameter frustum base, but different diameters for the base of the spherical cap. The first is a normal one. The second and third show the limits of volume and radius that I mentioned earlier: the second having the cap base radius equal to the radius of the sphere, producing a frustum of height 0 and ending up with a hemishere; the third having the cap base radius equal to 0, producing a frustum with a height equal to the radius of the sphere and ending up with a cone.
The fourth drawing demonstrates that the base of the frustum can be larger than the radius of the sphere, violating KP’s addendum.
Now the volume, V, is a numerical value, i.e. a constant, C1.
The bottom radius of the frustum, r, is a numerical value, i.e. a constant, C2.
Since the OP defines the object as a sperical cap and frustum of a cone joined at the base circle of the cap and the top of the frustum it is reasonable to assume that it looks somewhat like and old fashioned spittoon with a ‘spherica’ bowl’ joined to a flaring insloping rim. If we knew the values of C2 andC2 we could make some reasonable assumptions regarding the values of R, and perhaps home in on final dimensions. As far as I can see there is no difinitive relationship between the cap and the frustum other that the overall height is R, the height of the Cap is, h, and the height of the frustum is, R - h. All of which result in very messy algebra and nothing useful to the present. I have doubts that an algebraic solution will be forthcoming.
It is an algebra question in that the inputs and output would be algebraic. I have no desire or intention to limit the scope of the methodology used to arrive at the answer. Using calculus does seem to be the easiest approach at this point, but I’d be just as happy if it used topology and quantum mechanics if the resulting answer worked.
I’m going to stick with notation that’s used in the original Dr. Math links.
Let r[sub]1[/sub] = radius of the spherical cone (also the radius of the closed end of the frustum), r[sub]2[/sub] = radius of the open end of the frustum, R = radius of the sphere and total height of the object, h = height of the spherical cone, and H = height of the frustum. The volume of the object is
V = ([symbol]p[/symbol]/6)(R-H)(3r[sub]1[/sub][sup]2[/sup]+(R-H)[sup]2[/sup])+([symbol]p[/symbol]/3)H(r[sub]1[/sub][sup]2[/sup]+r[sub]1[/sub]r[sub]2[/sub]+r[sub]2[/sub][sup]2[/sup]).
Since H[sup]2[/sup] = R[sup]2[/sup]-r[sub]1[/sub][sup]2[/sup], we can write an expression for V that is completely in terms of R, r[sub]1[/sub], and r[sub]2[/sub]. After doing so (and I did!), you will have a pretty ugly formula. From this formula, you then want to solve for r[sub]1[/sub] in terms of V, R, and r[sub]2[/sub].
Unfortunately, I really don’t think that you will be able to find a close-form solution for r[sub]1[/sub] that is in terms of the known values V, R, and r[sub]2[/sub]. That’s the bad news. The good news is that it’s a really trivial matter to iterate to a solution for r[sub]1[/sub]. I coded the formula into Excel and used Solver to get an answer for r[sub]1[/sub] for input values of V, R, and r[sub]2[/sub]. So, if that’s your objective, that would be an easy way to do it. You can’t always have nice close-form formulas for everything, and in practice, such formulas are not really necessary because of the computational power at your disposal.
Where R is the radius of the sphere (total height), V is the total volume, r[sub]2[/sub] is the radius of the uncapped end of the frustum and r[sub]1[/sub] is the radius of the cap.
This does not seem to be symbolically solvable (using any of the methods that MikeS linked to). Pretty easy to solve numerically though. Since there does not seem to be repeated roots, Newton’s method should converge supra-linearly…
“nivlac” has provided a relatively simple solution by a practical iteration.
The “algebraic solution” then appears to have been a red herring or sidetrack from the real objective of finding the dimensions of the envelope meeting your dimensional criteria and description.
You may want to double-check this because I don’t think it works. Not wanting to do the algebra (since I don’t think it’s particularly helpful in this case to solving the problem), but wanting a cross-check for what I previously posted, I plugged in a known solution: V = 1501, R = 10, r[sub]2[/sub] = 5, r[sub]1[/sub] = 9 (with precision to about 4 decimal places). The resulting heights of the spherical cone and frustum are 5.641146 and 4.358854, respectively. These inputs do not satisfy your polynomial equation (not even close). Perhaps you just erred in transcribing your solution.
