Figuring the volume of a partially filled cone without the radius of the material inside the cone

Question

Say I have a cone. For simplicities sake, this cone is at the bottom of a storage silo. Is has a flat bottom, flat top and angled sides. I know the height of the material inside the cone but don't know the angle of the cone nor do I know exactly how to figure that out. Ultimately I need to figure out the formula used to calculate the occupied volume of this cone (I think it's referred to as a frustum?). I'm positive some trig is involved but my days of high school Pre-calc with Trig are gone away (and just when I needed them, shucks).

|<------ D ------>|
-------------------  =================
\                /              ^
 \              /               |
  \ |<-- r ->| /                |
   \==========/      ========   H
    \xxxxxxxx/         h ^      |
     \xxxxxx/            V      V
      ------         =================
     |< d >|

I know H. I know h. I know D. I know d. I do NOT know r and I do not know the occupied volume (the x's above). My boss has tasked me with designing a spreadsheet that you punch in the values known and it outputs the volume. Can anyone assist?

Thanks!

Answer 1

Assuming that the cone is symmetric (i.e., it "points straight down"), then the fact that the cone goes from radius $d/2$ to radius $D/2$ over a distance of $H$ tells you that the slope of the side (viewed horizontally) is $\frac{D-d}{2H}$. Thus, $$\frac{r}{2}=\frac{d}{2}+\left(\frac{D-d}{2H}\right)h.$$ Similarly, the "height" at which the cone would come to its apex (let's call it $g$) can be obtained by solving for $$0=\frac{d}{2}+\left(\frac{D-d}{2H}\right)g,$$ which gives us $$g=\frac{-dH}{D-d}.$$ Now we find the volume of the frustrum by taking the volume of the cone with radius $r/2$ and height $h+g$, and subtracting the volume of the cone with radius $d/2$ and height $g$: $$\begin{align*} V&=\frac{1}{3}\pi\left[\left(\frac{r}{2}\right)^2(h+g)-\left(\frac{d}{2}\right)^2g\right]\\\\ &=\frac{1}{12}\pi\left[r^2h+(r^2-d^2)g\right]\\\\ &=\frac{1}{12}\pi\left[r^2h+(r^2-d^2)\left(\frac{-dH}{D-d}\right)\right]\\\\ &=\frac{1}{12}\pi\left[\left(d+\left(\frac{D-d}{H}\right)h\right)^2h+\left(\left(d+\left(\frac{D-d}{H}\right)h\right)^2-d^2\right)\left(\frac{-dH}{D-d}\right)\right] \end{align*}$$

Answer 2

Do either of you now have a spreadsheet to do these calculations? Seems a LOT of time to spend to reinvent the wheel.

Actually, this is a variation on the theme. We're mixing 2 different liquids, at 2:1 ratio, and all we have is 5-gal buckets. It's 13" tall, with radius at the bottom (Rb) = 5", and radius at the top (Rt) = 5.75".

The question becomes: In order to keep the 2:1 as precise as possible, and because this gets used several times/day for significantly varying volumes, the shop wants a table that says (effectively), "If I pour in ingredient A to the 3 1/4" line, to what ht. do I pour in ingredient B to have a precise 2:1 mix?

If is was a true cylinder, this is a no-brainer, but with that slope...

Normally, I'd just slog through and build my own, but I'm on a bit of a time crunch here.

Mark