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

  • $\begingroup$ Well, using a Conical Frustrum Calculator that I have found online, the height seems to be about $15.94739$ mm for $100$ mL. $\endgroup$
    – Aiden Chow
    Jun 28, 2022 at 22:29
  • $\begingroup$ You need some extra information. Is $r_1$ the radius where the volume is 100mL? If not, you need to know the height where $r_1$ is measured (the height of the frustum part of the glass). $\endgroup$
    – Andrei
    Jun 28, 2022 at 23:08
  • $\begingroup$ @AidenChow well yeah. But how do I get the heights for 50 and 75 mL while not changing the geometry of the frustum? $\endgroup$
    – J.Doe
    Jun 28, 2022 at 23:28
  • $\begingroup$ Well, given the height for the $100$ mL, and let's say the height of the $75$ mL is $h$, you can then use similar triangles to find the radii of the new frustum in terms of $h$. Then plug that into the formula for the volume of a conical frustum and solve for $h$. Repeat for $50$ mL and you have your heights. $\endgroup$
    – Aiden Chow
    Jun 28, 2022 at 23:38
  • $\begingroup$ Can you please just type a full answer with an example? $\endgroup$
    – J.Doe
    Jun 28, 2022 at 23:45

2 Answers 2


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.

  • $\begingroup$ Some good design for the Erlenmeyer Flask for estimates by eye. $\endgroup$
    – orangeskid
    Jun 29, 2022 at 2:28

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: $$h=\frac{3\cdot100000}{(64^2+22^2+64\cdot22)\pi}\approx\boxed{15.947389\text{ mm.}}$$

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.

  • $\begingroup$ I just realised why the volume is off. The 64 mm and 22 mm are the diameters, not the radii. The specs should be correct though since they are from the ISO norms for those flasks. $\endgroup$
    – J.Doe
    Jun 29, 2022 at 19:42

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