An inverted cone of radius a and height h is filled with water. A sphere made of a material denser than water is placed in the cone. Find the radius of the sphere, r, that will displace the largest volume of water.
Kenny M. raises the interesting question as to whether this problem might be solved using the analogous two dimensional problem: a circle fitted into an isosceles triangle (one inverted, lacking a base, thus having a "rim") and finding the radius circle that "displaces" the maximum area (rather than volume) below the rim. But, is the optimal 2D radius also the optimal 3D radius?
In my first post in this thread, I argued against this idea, pointing out that a circle, when rotated about its diameter into a sphere, displaces relatively more volume with its outer parts than its inner parts. Therefore, maximizing displaced area is a different maximization from maximizing displaced volume. But, I have also argued the need to match the circular and conical shapes as closely as possible, so the circle case perhaps applies. Anyway, let's find out by getting the exact 2D answer.
In a previous post, I showed that the exact 3D answer was:
R_sphere = a c h / (c^2 + ac - 2 a^2) with c = sqrt (a^2 + h^2)
In the 2D case, again the triangle height is h and its half-base is a. The circle radius is R and we will use the variable y for the position of the rim level while the circle's center is at (0, 0). Note, when the center of the circle is above the rim, y is negative. We will find the formula for the "displaced" area as a function of R and y, and then express y in terms of R, h, and a. Finally, we will try to maximize the displaced area A by finding R where dA/dR = 0
We have derived the position of the rim with respect to the circle center already here. What we called z2 in the link above is now called "y":
y = h - R/a sqrt(h^a+a^2)
Note: y may be positive (Case 2) or negative (Case 3).
The area of the circle (a segment for Case 2 or a circle less a segment for Case 3) that is immersed in the triangle is given by:
A = pi R^2/2 + R^2 arcsin(y/R) + y sqrt(R^2 - y^2),
with y=h-(R/a) sqrt(h^2+a^2)
This may be derived using geometry identities in a masterful way here or via a plodding Cartesian-coordinate integration here.
The only caveat is that our "y" a the displacement from the circle center to the rim and as such, may be positive or negative, while in the linked proofs, the positive displacement case portrayed. The negative case has the same form as the positive, which follows easily from the linked results.
Substituting-in gives the displaced area formula vs. a and h as shown here. But the derivative aA/dR(R,a,h), also shown, is god-awful to behold and does not lead easily to an analytic form for dAdR = 0.
Substituting-in a power series for arcsin (after differentiating) does not help, even as an approximation. (maybe try before differentiating?)
However, tracking the area displaced and its derivate shows that indeed the 3D and 2D problems give the peak displacement of volume or area respectively at very different radii. (The maximized radii for the two 2D example runs verify Kenny M's results as well. All of this can be seen in two new tables for the two test cases: a=3, h=6 here and a=2, h=8 here with comparisons between area and volume highlighted.
One way the results are confirmed is that the boundary between Case 2 and Case 3 occurs when the Sphere/Circle has its diameter on the rim, and the fractional immersed volume/area is 0.5. This comes out correctly in the simulations in all cases.
Finally, I tested Kenny's thought that when the ratio of a to h is large (a wide opening angle) the optimization moved to the 3rd Case (where the sphere center was above the rim but the sphere's contact with the cone was still below the rim. I verified this as well with a single example. For a cone with a=4 and h = 1, Case 3 runs from R=0.97 to R=16.5. Both the sphere and he cone are maximized in this regime: Sphere at R=11.05 and circle at R=7.96, shown here.
Note that the 2D solution always seems to be smaller than the 3D solution for the same a and h? This is due, as noted, to the volume being weighted more to the outer portions or the sphere, and so shallower displacements by larger spheres are better than smaller sphere that can sink in lower. This is not true for circles.
Closing, I must say that I found it odd that the 2D case, while simpler to specify than the 3D case, was was intractable (for me) analytically. Maybe there is a better set of variables to use? Also, I note that there are analogs to the isosceles triangles found in this solution to the volume of an intersection between spheres.
Edited on October 3, 2019, 1:03 am
Comparing the 3D problem with the 2D analog
