How to calculate the height of subdividing lines of a conical frustum?

Question

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

Answer 1

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.

Answer 2

Given the radii of the top and bottom of the frustum ($r_1$ and $r_2$) and the height $h$, the formula for the volume of a conical frustum is as follows: $$V=\frac{\pi h}3(r_1^2+r_2^2+r_1r_2)$$ Before we start, it will be helpful to solve for $h$ in terms of $r_1$, $r_2$, and $V$: $$\begin{align}V&=\frac{\pi h}3(r_1^2+r_2^2+r_1r_2)\\ \frac{\pi}3h&=\frac V{r_1^2+r_2^2+r_1r_2}\\h&=\frac{3V}{(r_1^2+r_2^2+r_1r_2)\pi}\end{align}$$ I'm going to assume that $r_1=64$ mm and $r_2=22$ mm correspond to the frustum with volume $100$ mL. Converting $100$ mL to cubic millimeters, we get that $100$ mL $=100000$ mm.$^3$. Plugging this information into the formula, we get: $$h=\frac{3\cdot100000}{(64^2+22^2+64\cdot22)\pi}\approx\boxed{15.947389\text{ mm.}}$$

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Now that we have the height of $100$ mL, we can now compute the heights for the rest of the volumes. Let $h$ be the height of the conical frustum for $75$ mL = $75000$ mm.$^3$. Consider the following diagram below, which represents a cross section of the frustum going through the center of its circular base (excuse me for my horrible diagram; I'm not the best at drawing): Notice that $AB=r_2=22$ mm, and $CD=r_1=64$ mm. Using that, we can calculate that $CE=CD-AB=64-22=42$ mm. Also notice that $DH=GX=h$ and $BE=AD=15.947389$ mm., which if you recall, was the height for $100$ mL. Using the similar triangles $\triangle BEC\sim\triangle GXC$, we get the following equation: $$\begin{align}\frac{BE}{EC}&=\frac{GX}{XC}\\\frac{15.947389}{42}&=\frac{h}{XC}\\XC&=\frac{42h}{15.947389}\end{align}$$ With that, we can calculate the top radius of the $75$ mL. frustum, which is $GH=DX=CD-XC=64-\frac{42h}{15.947389}$. Also realize that the bottom radius is always the same: $64$ mL. Plugging all of this into our height formula, we get: $$h=\frac{3\cdot75000}{\left(64^2+\left(64+\frac{42h}{15.947389}\right)^2-64\left(64-\frac{42h}{15.947389}\right)\right)\pi}$$ Plugging this equation into Wolfram Alpha gives that $h\approx\boxed{8.39411\text{ mm.}}$ mm. for $75$ mL.

For $50$ mL., simply plug in $V=50000$ mm.$^3$ into the equation instead of $75000$ mm.$^3$ to get: $$h=\frac{3\cdot50000}{\left(64^2+\left(64+\frac{42h}{15.947389}\right)^2-64\left(64-\frac{42h}{15.947389}\right)\right)\pi}$$ Plugging this equation into Wolfram Alpha gives that $h\approx\boxed{6.31217\text{ mm.}}$ for $50$ mL.