[Guest]

The credit of this one goes entirely to Bonanova. I just made the figures.

"Given a wine glass [assume it's an inverted cone] with angle theta and known Height,

determine the Radius of the sphere which when inserted into the full glass will

displace the maximum volume of wine.

Clearly for a very small radius, the sphere can sink to the bottom of the glass

and displace its own small volume of wine.

And for very large radius, bottom surface of the sphere will approximate a

plane and not enter the glass at all, displacing a diminishingly small volume

of wine.

Somewhere in between, a maximum volume will be displaced.

Clearly the answer will be Radius = f[theta] x Height.

Determine f[theta]."

[Guest]

basic geometric problem.

sphere should be fully inside with upper circle of the cone acting as tangent to sphere.

(h-r)/r = tan (theta/2)

Edited April 3, 2009 by King Zeus

[bushindo]

Since the shortest line connecting the center of the sphere to the cone will have to meet the cone perpendicularly, the angle made by the shortest connecting line between the sphere center and the cone with the vertical line in the center is a constant with value (90 - theta/2 )

r = h cos( 90 - theta/2 )

The volume of any sphere satisfying the requirement of `resting on the wine cup' is

V = 4 pi ( h cos( 90 - theta/2 ) )^2

Here is where it gets messy. There are three different situations. Let H be the height of the cone and h be the height of the center of the sphere

h > H ( Less than half the sphere is inside the cone )

h < H - r ( The entire sphere lies inside the cone )

H - r < h < H (between 1/2 and the entire sphere lies inside the cone )

We can construct equations to determine the maximum volume for the displaced water for each scenario. My intuition tells me that its scenario 3. In that case, the displaced volume is equal to the volume of the sphere subtracting the volume made by the spherical cap above the cone. (see equation for volume of spherical cap here http://mathworld.wolfram.com/SphericalCap.html )

Volume = 4 pi ( r )^2 - (1/3) pi ( r + h - H )^2 ( 3r - ( r + h - H ) )

Substitute the exp​ression for h

Volume = 4 pi ( r )^2 - (1/3) pi ( r + r/cos( 90 - theta/2) - H )^2 ( 3r - ( r + r/cos( 90 - theta/2) - H ) )

Taking derivative with respect to r and setting it equal to 0, we'll get an exp​ression for the minimizing r under scenario 3. It'll be a quadratic equation with two answers, one of which probably will make no sense. The computation is rather messy, so maybe I'll do it later or someone else can do it. I'll have to get back to work on stuff I'm actually getting paid for.

Edited April 3, 2009 by bushindo

[bonanova]

No takers for the general solution?

Adopt the notation of the OP figures except let theta be half the angle shown.

Observe: for maximum displacement, the sphere cannot sit on top of the glass, as in the left figure [clearly decreasing its radius deepens its penetration] nor can it be fully submerged, as in the right figure [clearly increasing its radius displaces more fluid, at least until its top surface emerges from the fluid.] So the optimum case is shown by the middle figure.

We are given h, theta and r.

Define two more distances:

y = distance from apex to center of sphere - thus [y-r] is the distance from apex to botom of the sphere.

d = height of the submerged portion of the sphere: i.e. d = h - [y-r]

We need d to calculate the volume of displaced fluid, see below.

Looking at the triangle in the middle figure we see y = r/sin[theta], or

[1] y = r csc[theta] and thus

[2] d = h - fr ... where f=csc[theta]-1 depends only on theta.

The volume of a truncated sphere of height d is

[3] V = pi d²[3r - d]/3 - we seek to maximize this with respect to r.

The chain rule gives

[4] dV/dr = aV/ar + aV/ad dd/dr [where a denotes partial derivative]

[5] aV/ar = pi d² ... from [3]

[6] aV/ad = pi d[2r - d] ... from [3]

[7] dd/dr = -f ... from [2]

Substituting [5]-[7] into [4] and simplifying,

[8] dV/dr = pi d [d - f(2r - d)]

Since d is never 0, setting the derivative = 0 implies

[9] d = f(2r - d) Now substitute [2] to get rid of d, yielding: h(1-f) = fr(3+f). Solving for r,

[10] r_max = h [(1+f)/f]/f(3+f). Recall that f = csc[theta]-1.

If we define g[theta] = [(1+f)/f]/f(3+f) we have the maximum displaced fluid when the sphere has radius

[11] r_max = g[theta] x h, which has the form anticipated in the OP.

It's possible that g[theta] simplifies with trigonometric identities.

I haven't verified this with calculations yet.

Maybe someone would like to do that.

[bonanova]

basic geometric problem.

sphere should be fully inside with upper circle of the cone acting as tangent to sphere.

(h-r)/r = tan (theta/2)

take a closer look at your triangle ... (h-r) looks like the hypotenuse; so (h-r)/r is cosecant[theta/2].

So r_max = h/(csc[theta/2] + 1), which agrees with the previous post for the case of the largest submerged sphere.

What still remains to be shown is what radius displaces the maximum volume of fluid.

[Guest]

No takers for the general solution?

Maybe someone would like to do that.

Adopt the notation of the OP figures except let theta be half the angle shown.

Observe: for maximum displacement, the sphere cannot sit on top of the glass, as in the left figure [clearly decreasing its radius deepens its penetration] nor can it be fully submerged, as in the right figure [clearly increasing its radius displaces more fluid, at least until its top surface emerges from the fluid.] So the optimum case is shown by the middle figure.

We are given h, theta and r.

Define two more distances:

y = distance from apex to center of sphere - thus [y-r] is the distance from apex to botom of the sphere.

d = height of the submerged portion of the sphere: i.e. d = h - [y-r]

We need d to calculate the volume of displaced fluid, see below.

Looking at the triangle in the middle figure we see y = r/sin[theta], or

[1] y = r csc[theta] and thus

[2] d = h - fr ... where f=csc[theta]-1 depends only on theta.

The volume of a truncated sphere of height d is

[3] V = pi d²[3r - d]/3 - we seek to maximize this with respect to r.

The chain rule gives

[4] dV/dr = aV/ar + aV/ad dd/dr [where a denotes partial derivative]

[5] aV/ar = pi d² ... from [3]

[6] aV/ad = pi d[2r - d] ... from [3]

[7] dd/dr = -f ... from [2]

Substituting [5]-[7] into [4] and simplifying,

[8] dV/dr = pi d [d - f(2r - d)]

Since d is never 0, setting the derivative = 0 implies

[9] d = f(2r - d) Now substitute [2] to get rid of d, yielding: h(1-f) = fr(3+f). Solving for r,

[10] r_max = h [(1+f)/f]/f(3+f). Recall that f = csc[theta]-1.

If we define g[theta] = [(1+f)/f]/f(3+f) we have the maximum displaced fluid when the sphere has radius

[11] r_max = g[theta] x h, which has the form anticipated in the OP.

It's possible that g[theta] simplifies with trigonometric identities.

I haven't verified this with calculations yet.

This is incorrect: [4] dV/dr = aV/ar + aV/ad dd/dr. The correct form should be dV/dr=dV/dd*dd/dr.

You are also forgetting that there is a product of two functions that depend on r. So in [3] you need to apply:

dV/dr=(pi/3)*[2d*dd/dr*(3r-d) +2d²*(3-dd/dr).

I am working on it....to verify the rest

[Guest]

This is incorrect: [4] dV/dr = aV/ar + aV/ad dd/dr. The correct form should be dV/dr=dV/dd*dd/dr.

You are also forgetting that there is a product of two functions that depend on r. So in [3] you need to apply:

dV/dr=(pi/3)*[2d*dd/dr*(3r-d) +2d²*(3-dd/dr).

I am working on it....to verify the rest

Well it seems I miss understood your notation in the chain rule.

Rmax=h*(f+1)/(3+f)

Verified using H=12. Angle=30 degrees.-> Rmax=6