# Volume of a point inside a truncated cone

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• Nov 5th 2012, 04:41 AM
nucci93
Volume of a point inside a truncated cone
Hey guys I want to use a 12oz cup to measure things with, but I can't figure out how I would go about doing this. The measurements are roughly 1.1 inch radius at base, 1.6 inch radius on top and 3.8 inch height.

For example, if I wanted to fill it with 4oz of water, how would I calculate what height to fill it at to get this?

Thanks!
• Nov 5th 2012, 04:52 AM
fkf
Re: Volume of a point inside a truncated cone
You want to use a 12 oz cup to measure things with. The measurements are 1.1 inch at base, 1.6 inch radius and 3.7 inch height.

The formula you're searching for says
Volume = pi/3*h(R^2+R*r+r^2)
where R is the larger radius
• Nov 5th 2012, 05:00 AM
nucci93
Re: Volume of a point inside a truncated cone
I know that, but let's say I want to fill it with 4.5oz of water or something. How would I know what height to fill it to? (without cutting it, etc.)
• Nov 5th 2012, 05:14 AM
fkf
Re: Volume of a point inside a truncated cone
Quote:

Originally Posted by nucci93
I know that, but let's say I want to fill it with 4.5oz of water or something. How would I know what height to fill it to? (without cutting it, etc.)

Since I'm from Europe I gotta love the metric system ;) Use the given formula and solve for the unknown variables. I'll walk you through this:

The formula says
Volume = pi/3*h(R^2+R*r+r^2)

We also know that the relationship between the height and the larger radius (top) is
h/R = 3.7/1.6 <=> h = 3.7R/1.6

In this case we then have
Volume = pi/3*h(R^2+R*r+r^2) => 4.5 = pi/3*h(R^2+R*1.1+1.1^2)
You can solve this equation by first using the above given relationship between R and h into this equation. This will give you only h as unknown to solve for.
• Nov 5th 2012, 05:40 AM
skeeter
Re: Volume of a point inside a truncated cone
Quote:

Originally Posted by nucci93
I know that, but let's say I want to fill it with 4.5oz of water or something. How would I know what height to fill it to? (without cutting it, etc.)

in addition to converting ounces to cubic inches, you would need to use concepts from calculus (volumes of rotation) to solve this problem.

$R$ = upper radius of the glass

$r$ = lower radius

$H$ = glass height

$h$ = height to fill for desired volume

$V$ = desired volume in cubic inches

$V = \pi \left[\frac{(R-r)^2 \cdot h^3}{3H^2} + \frac{r(R-r) \cdot h^2}{H} + r^2 \cdot h\right]$

convert your desired volume to cubic inches, substitute in the radii and height measurements of the glass ... solve for $h$.