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A crystal goblet contains water to the brim. It has a conical shape, with a 5-inch vertical height and a 30-degree angle at the base. I mischievously want to spill as much of the water as possible. To accomplish this I have a supply of spherical lead weights of any radius I desire. Which single lead weight shall I drop into the goblet?

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This is a solution if the sides are 30 degrees away from vertical. There is another way to go about solving it that makes it more tractable if you have an arbitrary angle away from vertical, but I found this result particularly interesting. Unfortunately spoilers are still not working at the moment and figures can't be hidden by making them white, but at least the text will be hidden as white so you can click here and press Ctrl-A (or whatever the equivalent on a Mac would be) if you want to make it visible.

 

Spoiler

 

566baebda2b96_gobletfigure.jpg.f8b4644b5

From this figure, R/h = sin(30) so h = R/sin(30) = 2R.

Since the goblet height is 5, s = 5-h = 5-2R.

To find the submerged volume, think of the sphere as a stack of horizontal circles, and take the integral of the circles' area as you go from the surface of the water in the goblet to the bottom of the sphere.

Let x be the variable of integration and go from s to -R (from the height of the topmost circle relative to the center of the sphere, to the bottom of the sphere). At any distance x from the center of the sphere, the radius of the circle (r, not to be confused with the radius of the sphere R) is given by r2 + x2 = R2 (consider the figure where r is shown when x=s). So r = sqrt(R2-x2). And the area of the circle is of course pi*r2. So the total volume submerged is the integral

566baf106a910_gobletformula1.jpg.467cc36

That function (the displaced volume) will have a derivative d/dR of zero at its maximum, so take the derivative

566baf44a960e_gobletformula2.jpg.548523e

That must equal zero, so divide all terms by 2 pi to get

2 R2 – 15 R + 25 = 0

And by the quadratic formula, this equals zero when

R = [15 +/- sqrt(152 – 4 * 2 * 25)] / 4

R = [15 +/- sqrt(25)] / 4

R = 2.5 or 5

The solution at R=5 would not make physical sense because h = 2R = 10 would mean that the sphere is completely out of the water and no longer displacing anything (hence the derivative equaling zero, I guess).

At R=2.5, the submerged volume (based on that first formula solving the integral) would be 98.17 – 65.45 + 0 = 32.72. And perhaps surprisingly, R=2.5 would put the center of the sphere right at the surface of the water since h = 2R.

 

 

Edited by plainglazed
added spoiler tags and unwhited text
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You did not need to solve the integral. The submerged volume is a spherical cap, the volume of which is pi*h2*(3r-h)/3, where h is the height of the sphere from the intersecting plane. In your diagram, this is (R+s). For the more general case this is 5 - (r/sin(t) - r), or (1-1/sin(t))r + 5. Now, after some review of the derivative, which I was barely introduced to some 14 years ago, I have a complete solution for the question as posted.

 

f(r) = pi/3 * h2 * (3r-h)

h = (1-1/sin(15)*r + 5; letting n be the constant value (1-1/sin(15)), we have h=nr+5, then:

f(r) = pi/3 * (nr+5)2 * (3r-nr-5)

f(r) = pi/3 * ((3n2-n3)r3 + (30n-15n2)r2 + (75-75n)r - 125)

f'(r) = pi/3 * (3(3n2-n3)r2 + 2(30n-15n2)r + (75-75n))

f'(r) = pi * ((3n2-n3)r2 + (20n-10n2)r + (25-25n))

(3n2-n3)r2 + (20n-10n2)r + (25-25n) = 0

The radical cleans up nicely, giving us (-b + 10n)/2a and (-b - 10n)/2a

(-b + 10n)/2a evaluates to 1.74599, which is 5 * sin(15) / (1- sin(15)), where the sphere just touches the surface of the water, giving a submerged volume of 0.

(-b - 10n)/2a evaluates to 5*sin(15)/(1+sin(15)+2*sin2(15)) = 1.150465939351, giving a submerged volume of 5.3177342075568337849950 cubic inches.

 

Edited by Logophobic
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Let me point out that we don't need two separate formula for Iceberg 1 and Iceberg 2. The formula given for Iceberg 2 is valid for Iceberg 1. As for Vsw decreasing for Iceberg 2, this is only true when θ <= 30 (base angle not greater than 60.)

Spoiler

First, an error in my solution: 5*sin(15)/(1+sin(15)+2*sin2(15)) = 1.150465939351
Should have been: 5*sin(15)/(1+sin(15)-2*sin2(15)) = 1.150465939351

For the general solution:

From bonanova's solution: (g > 0; 0 < θ < 90)
R1 = g sin θ / (1 + sin θ)
R2 = g sin θ
R3 = g sin θ / cos2 θ

From my solution:
n = (1-1/sin θ), g = height of cone
f(r) = π/3 * ((3n2-n3)r3 + (6gn-3gn2)r2 + (3g2-3g2n)r - g3);  R1 <= r <= R3
f'(r) = π * ((3n2-n3)r2 + (4gn-2gn2)r + (g2-g2n))

(3n2-n3)r2 + (4gn-2gn2)r + (g2-g2n) = 0

a=n2(3-n); b=2gn(2-n); c=g2(1-n)

b2-4ac = 4g2n2 = (2gn)2 

(-b + 2gn)/2a = g sin θ / (1 - sin θ) > R3 and therefore not a solution

(-b - 2gn)/2a = g sin θ / (1 + sin θ - 2 sinθ) is the only solution

r = R2 when sin θ = 2 sinθ; that is when sin θ = 1/2, θ = 30

 

 

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Assuming we can't just break the goblet and spill it all...

I get approx 0.606 in which will displace approx 0.908 cubic in of water.

Oops...I had an extra parenthesis somewhere.

So what I really got was a radius of approx 0.643 in and displacement of approx 1.02 cubic inches.

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After further consideration, a larger sphere, while not displacing its total volume, could possible displace a greater volume than that which I gave at first.

A sphere with radius 5 * Sin(15) = approx 1.294 inches would displace half of its volume, approx 4.539 cubic inches.

This is less than my first answer, but a test case with radius 1.2 inches would displace greater than 5.2 cubic inches. I have not calculated the exact volume for this case, nor do I expect that a radius of 1.2 inches is the final solution.

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On 12/5/2015 at 2:28 PM, Logophobic said:

After further consideration, a larger sphere, while not displacing its total volume, could possible displace a greater volume than that which I gave at first.

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I agree with your calculations so far. But a different radius will displace more volume.

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Clarification: when you say the goblet has a 30 degree angle at the base, do you mean it forms a V and there are 30 degrees between the \ and / of the V, or do you mean that the sides are 30 degrees away from vertical?

If the former (like I'm currently running with), the math gets quite hairy. If the latter, it should be much more tractable.

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Logophobic found R1 and R2 and found the radius of greatest displacement with numerical methods.

plasmid rederived the formula for a sperical cap and got a closed form solution for the 60-degree goblet.

Logophobic then found the closed form solution for the 30-degree goblet.

Apologies to plasmid for not clarifying the issue of goblet angle.

Since Logophobic was the first solver, I'll give his post the mark.

I came across this puzzle some years ago and decided it would be fun to solve it again. I didn't remember what angle was given for the goblet so I used a fairly narrow one. As plasmid points out, 60 degrees might have made it an easier task.

Here's my solution of the puzzle

Spoiler

 

This problem is simple to state and mildly challenging to solve. We have a conical goblet full to the brim with water. We also have an infinite supply of lead spheres of any desired size. We impishly desire to spill as much water as possible by dropping one of the spheres into the goblet. Which one shall we use?

Small spheres displace their entire-but-small volume, while very large spheres barely dent the water’s surface. Somewhere in between, the spill is maximized.

To decide on the optimal sphere we say the goblet has height g, an angle 2θ at the base; its rim radius r = g tan θ. The sphere diameter is R. If the sphere contacts the inside of the goblet (not just the rim) then its center has a height of y = R /sin θ. If the sphere is only partially submerged we call the radius of the circle that intersects the water’s surface a.  

566857908164e_sphereingoblet.jpg.c98c745

We can distinguish four cases, in order of increasing sphere radius, (See Figure, where only the dark green portion of the sphere is submerged.)

  1.           Submarine: sphere is totally submerged.                               R < R1 = g sin θ / (1 + sin θ).
  2.         Iceberg 1: equator is below the rim.                                R1 < R < R2 = g sin θ.
  3.         Iceberg 2: equator is above the rim; still tangent.           R2 < R < R3 = g sin θ /cos2θ.
  4.           Squatter: sphere rests on the rim.                                  R3 < R.

The volume of displaced water can be found from two formulas: sphere and spherical cap:

Sphere:

Vs = (4π/3) R 3.

Spherical cap:

A cap is delineated by the plane of the water’s surface and occurs for all cases but the Submarine. It has height h and radius a. Its volume Vsc is determined by any two of (R, h, a) as follows:

Vsc [h, R] = (πh 2/3) (3Rh)
Vsc [h, a] = (πh /6) (3a 2 + h 2)
Vsc [h (R, a), R]      when R and a are known, find h (R, a) = R – (R 2a 2)½.

Volume Vsw of spilled water:

  1.       Submarine:         Vsw  = Vs                                         sphere (Vsw  increases with R)
  2.       Iceberg 1:            Vsw  = VsVsc [(h = R + y - g), R]    sphere – top cap (region of maximum Vsw )
  3.       Iceberg 2:            Vsw = Vsc [(h = R - y + g), R]            bottom cap (Vsw  decreases with R)
  4.       Squatter:             Vsw = Vsc [h (R, a = r), R]                bottom cap (Vsw  decreases with R)

These equations show that Vsw increases with R in case 1 (submarine) and decreases with R in cases 3 and 4 (iceberg 2 and squatter.) Maximum displacement thus occurs in case 2 (iceberg 1) at the radius where the growth rates of the sphere and the top cap are equal. An analytical solution is thus achievable.

Before doing this, let’s look at a plot of Vsw vs R for a specific case.

Specific case: g=5, θ =15o:

ball_in_goblet.jpg.3d60b87b112db12030dc1

Here r = 1.3397; R1 = 1.028; R2 = 1.2941; R3 = 1.387.  Vsc is plotted vs R showing a maximum spilled volume of 5.3177 for a sphere radius of 1.150, about midway in the Iceberg 1 region, as expected. The volume of that sphere is 6.3706; its cap volume is 1.0529.

Interestingly, the displaced volume is nearly the same value for the radius of the largest submarine and the radius for which the equator is rim-high (R1 and R2). This results from the choice of base angle. The ratio of these radii is (1 + sin θ) which for 15o is 1.259 which, cubed, is 1.995, which, halved, is nearly 1.

The regions are colored blue (submarine), red (iceberg 1), green (iceberg 2) and gray (squatter.)

 

 

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How did you get that Vsw decreases with R in the Iceberg 2 regime? It's true for the case of theta = 15, but I don't think it's true in general -- in the extreme case of a goblet with theta = 89.999 degrees (where the goblet is very close to a flat plane) I'd imagine a sphere with a large radius would get more underwater than a sphere with its center below the surface.

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1 hour ago, plasmid said:

How did you get that Vsw decreases with R in the Iceberg 2 regime? It's true for the case of theta = 15, but I don't think it's true in general -- in the extreme case of a goblet with theta = 89.999 degrees (where the goblet is very close to a flat plane) I'd imagine a sphere with a large radius would get more underwater than a sphere with its center below the surface.

Good point. You're right that my statement refers to the case at hand. How general it is now is clearly questionable. I had in mind to calculate  R2 and Rmax vs θ for a unit height cone, but left it undone for some personal matters at home. I imagine one or the other of the iceberg regions always contains Rmax, and for one particular value of θRmax would just equal to R2.

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