I Calculate new height of truncated cone

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To calculate the new height of a truncated cone after removing water, the volume formula for a truncated cone is used, but adjustments must be made for the new upper radius. The ratio between the original height and the difference in radii remains constant, allowing for the derivation of a new height equation. An alternative approach suggests expressing the volume as a function of height, simplifying the calculations. The relationship between the cone's dimensions leads to a cubic equation that can be solved for the new upper radius, which then helps determine the new height. The discussion emphasizes the complexity of the calculations involved in determining the new dimensions of the truncated cone.
tjosan
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Hi,

Suppose you have a truncated cone filled water with the lower radius being R, and upper r (R>r), and the height is H.

R, r and H is known so the volume, V, can be calculated using V=1/3*pi*H*(R^2+R*r+r^2). Now suppose you remove some water so that you end up with a lower volume, V1.

The water surface will now have a radius of r1, and the height will be h. The overall shape of the cone will remain the same though, its just that the surface has moved down.

How can I calculate the new height? I cannot wrap my head around this. First I just used the new volume in the formula above and solved for H, but then I realized the upper radius isn't the same anymore, so that wont work.

I attached an image to illustrate.

Thanks!
 

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think about how the radius of the cone changes with height.
 
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Dr Transport said:
think about how the radius of the cone changes with height.
I think I solved it.

The ratio, C, between the H and R-r must remain the same for the new cone (because the angle is the same, tan (angle) = constant) , so C = H/(R-r) = H1/(R-r1) => H1=C*(R-r1) [1]

V1=1/3*pi*H1*(R^2+R*r1+r1^2) = 1/3*pi*C*(R-r1)*(R^2+R*r1+r1^2), where r1 is the new upper radius, V1 is the new volume. Solve for r1 and then use equation [1] to solve for H1.
 
tjosan said:
The ratio, C, between the H and R-r must remain the same for the new cone (because the angle is the same, tan (angle) = constant) , so C = H/(R-r) = H1/(R-r1) => H1=C*(R-r1) [1]

V1=1/3*pi*H1*(R^2+R*r1+r1^2) = 1/3*pi*C*(R-r1)*(R^2+R*r1+r1^2), where r1 is the new upper radius, V1 is the new volume. Solve for r1 and then use equation [1] to solve for H1.
Your method should work (I haven’t tried it) but it looks like you will end-up having to solve a really messy cubic equation in ##r_1##.

Here are some hints for an alternative approach.

With conventional notation, the volume of a cone is ##V(r, h) = \frac 13 \pi r^2 h##. The difficulty here is that ##V## is a function of 2 variables, ##r## and ##h##. In your question, the cone angle is effectively given; this gives a simple relationship between ##r## and ##h##. You should be able to show that ##V(h) = kh^3## where ##k## is a constant. ##V(h) ## is now a sinple function of the single variable ##h##.

(You have enough information to find ##k## and the height of the ‘full’ cone in terms of the given data.)

The required frustum is the part of a full cone which remains after the cone’s ‘tip’ (itself a cone) is removed. ##V_{frustum} = V_{full cone}~-~V_{tip}##.

Using the above gives a more manageable way to find '##h_1##'. But it’s still a bit messy.

(Note. We prefer LaTeX for equations here. The link to a guide is https://www.physicsforums.com/help/latexhelp/). This is the link shown at the bottom left of the edit window.)
 
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