How to calculate the height of subdividing lines of a conical frustum? I am trying to figure out how to calculate the heights of subdividing measuring lines of an Erlenmeyer Flask. So if my flask is $100$ mL and has $r_1 = 64$ mm and $r_2 = 22$ mm, and it has measuring lines for $50$, $75$ and $100$ mL, then what are the heights for those volumes?
I tried to rearrange the equation for a conical frustum to the height h and replace h with other variables but I'm simply stuck.
Image for reference: https://www.glas-shop.com/images/104677-100ml-erlenmeyerkolben_1038.jpg
 A: The volume of a frustum with radiuses $R$, $r$, and height $h$ is
$$V= \frac{1}{3} \pi h \cdot \frac{R^3 - r^3}{R-r}$$
Let's see what is the volume of the frustum obtained by sectioning it at height $t \cdot h$, $0<t<1$. The base radius is still $R$,  the top one is $r_t \colon = (1-t)R + t r$, and the height is $t h$. We get
$$V_t = \frac{1}{3} \pi \cdot t h \cdot \frac{R^3 - r_t^3}{R- r_t}$$
Now $R- r_t= t(R-r)$ so we get
$$V_t = \frac{R^3- r_t^3}{R^3 - r^3} \cdot V$$
Therefore, if we cut the $t$ part of the height we get the following part of the full volume
$$\frac{V_t}{V} =\frac{1- (\frac{r_t}{R})^3}{1- (\frac{r}{R})^3}$$
Note that $\frac{r_t}{R} = \frac{(1-t)R+ t r}{R} = (1-(1-\rho)t)$, where $\rho = \frac{r}{R}\ $. In the end
$$\frac{V_t}{V}  = \frac{1- (1- (1-\rho) t)^3}{1- \rho^3}$$
Now the equation $\frac{V_t}{V} = s$ has solution
$$t = t(s) = \frac{1- (1- (1-\rho^3) s)^{\frac{1}{3}}}{1-\rho}$$
In our case $\rho = \frac{22}{64}$, and
$t(.5)\simeq .29820\ldots$, and $t(.75) = .52636\ldots$, that is: for $\frac{1}{2}$ we cut at about $\frac{1}{3}$, and for $\frac{3}{4}$ just over the middle.
