Graduations of volume on the side of a cone I am trying to put graduated volume markings (every 10 liters) on the side of a cone.
Specifically this is the conical section of a wine tank. I know the dimensions of the whole cone (h: 108cm, r: 103.5cm, l: 150.3cm, V: 1220L) but I am having trouble figuring out the volumes.
Because the tank is stainless steel, I can't see through or measure the r and h of the marks on the way up. Is the slant height l also proportional in the same way that it is proportional to r and h? I believe it is, however it has been quite a while since I have done math of this sort (pro tip, kids, listen to your Mom when she says you darn will use math in your life).
Just wanted to get the input from experts on this. Thanks!
 A: Here is something that may help. If you have a conical frustum (basically a cone with the top sliced off) then the formula for the volume is
$$V=\frac{\pi}{3}h(R^2+Rr+r^2)$$
Where $h$ is the height, $R$ is the radius of the base, and $r$ is the radius of the top. You can use this to solve your problem in the following way:
If you have that the volume of the cone is $1220L$ and, for example, you want to find the height of the water that would fill the cone halfway, then you would solve for $h$ in the equation
$$\frac{1220}{2}=h(103.5^2+103.5r_h+r_h^2)$$
where $r_h$ is the radius as a function of the height. In your case, we can use similar triangles to find
$$r_h=103.5-\frac{103.5}{108}h$$
and so you need to solve for (or approximate) $h$ as a solution of the cubic
$$\frac{1220}{2}=h\bigg(103.5^2+103.5\bigg(103.5-\frac{103.5}{108}h\bigg)+\bigg(103.5-\frac{103.5}{108}h\bigg)^2\bigg)$$
Does this make sense? If you need another example, just ask.
A: Think of the cone as situated vertex-down with its axis vertical, with slant neight $\ell$ (in cm) measured from the vertex. As you suspected, neglecting the bit of cylinder near the vertex, the volume held by the tank from the vertex to a slant height $\ell$ is proportional to $\ell^{3}$, so that
$$
V = k\ell^{3} = \frac{1220}{(150.3)^{3}} \ell^{3} \text{ liters}
$$
at a slant depth of $\ell$ cm.
The gradation mark at volume $V$ liters should therefore be at slant height
$$
\ell = 150.3 \times \sqrt[3]{\frac{V}{1220}} \text{ cm}
$$
from the vertex. To find the required positions, let $i$ run from $1$ to $122$ in the formula
$$
\ell_{i} = 150.3 \times \sqrt[3]{\frac{i}{122}} \text{ cm}.
$$
(This assumes the last mark, $\ell_{122}$, is at slant height $150.3$ cm, i.e., the tank is brimming.)
