# Half-full frustum

• Dec 20th 2006, 11:01 AM
xh558
Half-full frustum
The volume of a frustum of a cone is found by:

V = pi/12 x h x (D^2 + d^2 + Dd)

Where h = height, D = major diameter, d = minor diameter

I have an upside down cone frustum (i.e. larger diameter at top) with known values for D, d and h.

The solid is half full of liquid (by volume). How do I find the height of the liquid?
• Dec 20th 2006, 11:58 AM
xh558
Someone made a post on this thread but when I refreshed it dissapeared.

Anyway, to whomever it was:

http://www.mathhelpforum.com/math-he...6e440fb6-1.png

Is when you use the radii. My equation is for the diameters.

• Dec 20th 2006, 12:11 PM
galactus
That's the reason I deleted. I seen my error.
• Dec 20th 2006, 12:16 PM
xh558
Oh. Thanks for helping anyway:)
• Dec 20th 2006, 12:38 PM
galactus
I was thinking, though. Maybe try the ratio:

$\frac{h}{D-d}=\frac{h_{2}}{d_{2}-d}$

Where the subscript 2 refers to the height and diameter at the halfway mark.

$h_{2}=\frac{h(d_{2}-d)}{D-d}$

Maybe this will point you in the right direction. I'll look at it some more. Someone else will be along.
• Dec 20th 2006, 01:37 PM
xh558
This is the problem. Whatever I do, I always end up with two unknowns:

http://www.mathhelpforum.com/math-he...414ed3c5-1.png

In the above, h2 (on the left) is what I am looking for but to find it I need to know d2 (i.e. the diameter at the height of the liquid when the cone is half-full).

I don't see any way of substituting this expression (or any rearrangement of it) in to the volume equation. Somehow, we need to eliminate one of the unknowns.

This is a tough one!
• Dec 20th 2006, 02:31 PM
galactus
Here's an idea. Maybe you can generalize it.

I will go ahead and use dimensions. Let D=2, d=1, h=3

This gives the total volume as $\frac{7{\pi}}{4}$

Half that is, of course, $\frac{7{\pi}}{8}$

Draw a diagram of the frustrum with d centered at the origin and laying on the x-axis with D at the top.

You can find an equation of the line that makes its side.

Using the dimensions I mentioned, we get an equation of $y=6x-3$

Let's revolve about the y-axis, so we'll solve for x and get $x=\frac{y+3}{6}$

${\pi}\int_{0}^{h}(\frac{y+3}{6})^{2}dy=\frac{h(h^{ 2}+9h+27){\pi}}{108}$

This must equal $\frac{7{\pi}}{8}$

$\frac{h(h^{2}+9h+27){\pi}}{108}=\frac{7{\pi}}{8}$

Solving for h we find $h\approx{1.95}$

Therefore, with this frustrum the liquid is 1.95 units high when it is half full.

I know, I know, there are easier ways. I just thought this was a fun enough way to go about it.

EDIT: I derived you a formula in terms of the given variables alone. I derived it using calculus and some technology, but it appears to work.

$h_{2}=\frac{((12VD+d^{3}h{\pi}-12Vd)h^{2}{\pi}^{2})^{\frac{1}{3}}}{{\pi}(D-d)}-\frac{dh}{D-d}$

If you try enter in d=1, D=2, h=3, and V=7Pi/8 you get the same as above. Which seems reasonable. Give it a try.
• Dec 21st 2006, 12:06 PM
earboth
Quote:

Originally Posted by xh558
The volume of a frustum of a cone is found by:

V = pi/12 x h x (D^2 + d^2 + Dd)

Where h = height, D = major diameter, d = minor diameter

...

Hello,

I can't offer you a complete solution, but maybe the following will help a little bit.

I've attached a sketch of the frustum.

You get the proportions:

$\frac{2x}{h_2}=\frac{2(D-d)}{h}$. With $d_2=d+x$ you'll get:

$d_2=d+\frac{D-d}{h} \cdot h_2$. Now you can substitute the variable d_2 by this term and you have an equation with only one unknown (h_2).

If you do that, the equation will become pretty ugly very soon. So I never tried to solve it.

Good luck!

EB
• Dec 22nd 2006, 02:09 AM
xh558
Galactacus & Earboth,

Thank you both for your kind help so far. I'm going to go away now and see if I can make this work.

I'll report back later!
• Dec 23rd 2006, 10:05 AM
galactus
Something else I thought of is to extend your frustrum on down to finish the cone it is cut from. Make two similar cones. One, the entire frustrum and extended portion, the other a similar cone made form the extended portion and that part of the frustrum up to h. The difference in their volumes is half the volume of the frustrum. Follow me?. I managed to do that, but it still turns into a cumbersome equation. See if you can get anywhere with that idea.

The point where the two lines intersect is the tip of the cone below. Since the smaller diameter of the frustrum lies on the x-axis, it'll have coordinates

$0,\frac{-dh}{D-d}$

The height of the entire, big cone is then $h+\frac{dh}{D-d}$

The height of the smaller cone comprised of the extended portion and the height of the liquid is $h_{2}+\frac{dh}{D-d}$

The difference in their volumes is half the frustrum volume:

$\frac{\pi}{3}(\frac{D}{2})^{2}(h+\frac{dh}{D-d})-\frac{\pi}{3}(\frac{d_{2}}{2})^{2}(h_{2}+\frac{dh} {D-d})=\frac{1}{24}{\pi}h(D^{2}+Dd+d^{2})$

Whew!!.

Just an idea. Maybe you can get inspiration or something. I remembered I have actually seen this method used for other frustrum problems. For instance:

"Coffee is poured at a uniform rate(say 20 cm^3/sec) into a frustrum-shaped cup. Given the larger and smaller diameters and height(say 4 and 2 and 6, respectively), how fast will the coffee level rise when the coffee is halfway up?."

Extending the frustrum to a cone is a good way to solve this.

If you extend the sides of the cup to complete the cone and let $V_{1}

$
be the volume of the portion added, then

$V=\frac{1}{3}{\pi}r^{2}h-V_{1}$

Similar triangles:

$\frac{r}{h}=\frac{4}{12}; \;\ r=\frac{1}{3}h$

$V=\frac{1}{3}{\pi}(\frac{h}{3})^{2}h-V_{1}$,

$\frac{dV}{dt}=\frac{1}{9}{\pi}h^{2}\frac{dh}{dt}, \;\ \frac{dh}{dt}=\frac{9}{{\pi}h^{2}}\cdot\frac{dV}{d t}$

$\frac{dh}{dt}=\frac{20}{9{\pi}} \;\ cm/sec$

I know I rambled on, but maybe you can get some ideas. I hope I didn't mess up somewhere.

Merry Christmas.