- #1

tjosan

- 33

- 2

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!

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!