Vtyjkyuk

I need to measure the height liquid would come up to in a jug I have. I need to determin how far up the jug the liquid would go (similar to markings on a measuring jug).

Using the volume and lower radius and slant angles to Im looking for a formula I could use. (After translating ml in mmcubed to make it possible)

As an example, if i know the dimensions of the vessel, I would want to know how far up 200ml of liquid would go.

Thanks in advance people!

Ok, so call $r$ the radius of the basis, $theta$ the angle between the vertical axis of the cone and the surface.
$$colorredh=fracrtan theta-sqrt[3]fracr^3tan^3 theta-frac3Vpi tan^2 theta.$$
Of course you have to plug in the measures with the right units, for example: $r$ in $mm$ and $V$ in $mm^3$.

EXAMPLE: you have that $r=45mm$, the graph below for computing $tan theta$ remains valid, so $tan theta= frac6.5110$. Let's say you want $V=200ml=200000mm^3$, the formula gives

$$h=frac45cdot 1106.5-sqrt[3]frac45^3cdot 6.5^3110^3-frac3cdot 200000cdot 110^2picdot 6.5^2=sim 32.8 mm$$

EDIT: Since $tan theta$ can be difficulte to find, this is an easier way you can calculate it: here is a section of the truncated cone, you have to measure the lenghts $a$ and $b$. Then
$$tan theta =fracab$$

Once you have found $tan theta$, it should be easy to compute $tan^3 theta$ with a calculator. The quantity $r$ can be measured directly and $V$ is known.

Assume the container has shape of a truncated cone .. $r$ is minimum radius at bottom, $R$ at top, the cone has a flare up with $alpha$, $h$ is height reached. Using well known relation of truncated cone volume we work up the algebra, defining $H,a^3$ as

$$ V = fracpi h3 ( R^2 + r^2 + R r ), quad tan alpha = fracR-rh
,quad H = h tan alpha,, , a^3= fracV tan alphapi tag1 $$

When $R$ is eliminated to make a cubic in $H$,

$$ H^3/3 + H^2 r + H r^2 - a^3 = 0 tag2 $$

whose only real root is found by Mathematica as

$$ H = (3 a^3 +r^3)^frac13 - r $$