Graduations of volume on the side of a cone

Question

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!

Answer 1

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.

Answer 2

The total volume of the cone is $\displaystyle V=\frac 13 \pi r^2h$.

Since the radius is uneasy to measure, I'll trust the figure for $V$ much.

[ trusting $r$ give $V=1211\ l$ while trusting $V$ gives $r=103.8cm$ ].

Then we have the trig formulas $\begin{cases} \sin(\theta)=\frac rl \\ \cos(\theta)=\frac hl \\ \tan(\theta)=\frac rh\end{cases}\qquad\bbox[5px,border:2px solid]{\displaystyle \theta=\arctan\bigg(\sqrt{\frac{3V}{\pi h^3}}\bigg)}$

If I calculate $L$ from $h,V$ I got $L=149.8\ cm$ for $t=43.88°=0.7658656680$

This is the interior slant, but because of the metal thickness of the tank it is slightly different, let's just multiply by $\sigma=\frac {150.3}{149.8}\simeq 1.003$ and we should be good.

Now that we have our angle, we are going to calculate the graduations for $10\ l$ intervals.

$\displaystyle v=\frac 13\pi \tilde r^2\tilde h=\frac 13 \pi\times l^2\sin(\theta)^2\times l\cos(\theta)\iff\bbox[5px,border:2px solid]{\displaystyle l=\sqrt[3]{\frac{3v}{\pi\sin(\theta)^2\cos(\theta)}}}$

I assumed here $V$ and $h$ are the main measures and calculated everything from it.

Don't forget to use correct units for calculations : $V=1.220\ m^3$ and $H=1.08\ m$.

If you want to calculate for $v$ in liters and $l$ in centimeters do :

$v=0\cdots 1220\quad:\quad\displaystyle l=100\sqrt[3]{\frac{3v}{1000\pi\sin(\theta)^2\cos(\theta)}}\times \sigma$

I advise you to pour known quantities of liquid in the tank (like 50 l, 200l) and measure $l$ empirically and by the formula and adjust $\sigma$ so that they match.

Ideally $\sigma$ should be $1$, but to tuning your graduations a little bit couldn't hurt before making them permanent.

Here is my samples with $\sigma=1.003$ for every $40$ liters

0       0.00
40     48.10
80     60.60
120    69.37
160    76.35
200    82.25
240    87.40
280    92.01
320    96.20
360   100.05
400   103.63
440   106.97
480   110.12
520   113.10
560   115.93
600   118.62
640   121.20
680   123.68
720   126.06
760   128.35
800   130.56
840   132.70
880   134.78
920   136.79
960   138.74
1000  140.64
1040  142.49
1080  144.30
1120  146.06
1160  147.78
1200  149.46

Answer 3

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.)