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

trigonometryvolume

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!

Best Answer

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 $g$: $$\begin{align*} V&=\frac{1}{3}\pi\left[\left(\frac{r}{2}\right)^2(h+g)-\left(\frac{d}{2}\right)^2g\right]\\\\ &=\frac{1}{12}\pi\left[r^2h+(r^2-d^2)g\right]\\\\ &=\frac{1}{12}\pi\left[r^2h+(r^2-d^2)\left(\frac{-dH}{D-d}\right)\right]\\\\ &=\frac{1}{12}\pi\left[\left(d+\left(\frac{D-d}{H}\right)h\right)^2h+\left(\left(d+\left(\frac{D-d}{H}\right)h\right)^2-d^2\right)\left(\frac{-dH}{D-d}\right)\right] \end{align*}$$

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