# Height of a part volume in a cone

Hi, I know this is high school math but i feel kinda stupid right now so I am asking it here:

Given is a truncated cone (upside down if relevant). Given is the volume as well as the upper and lower radius.

Now the cone got filled with a given volume. How do i determine the height of the filled liquid?

As you see in the picture, there is a cone (not a triangle). Given is r1w, r2w, and VW and VA

With that at least I am able to calculate everything else, but not hA. Can someone of you please help me?

If $$\alpha$$ is the half-aperture angle of a cone, $$r$$ its base radius and $$h$$ its height, then $$h=r\cot\alpha$$ and the volume of the cone can be written as $$V={\pi\over3}r^3\cot\alpha$$.
Let then $$r_A$$ be the radius of the surface of the liquid. We have: $$V_W={\pi\over3}\cot\alpha(r_{2W}^3-r_{1W}^3), \quad V_A={\pi\over3}\cot\alpha(r_{A}^3-r_{1W}^3).$$ From the first equation we get $$\cot\alpha={3V_W\over{\pi}(r_{2W}^3-r_{1W}^3)}$$ and substituting that into the second equation we can solve for $$r_A$$: $$r_A=\root3\of{{V_A\over V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}$$ and finally: $$h_A=(r_A-r_{1W})\cot\alpha={3V_W\over{\pi}(r_{2W}^3-r_{1W}^3)} \left(\root3\of{{V_A\over V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}-r_{1W}\right).$$
• Thank you so far. I now get how you got to $r_A$ . But the next line is not really clear to me. You said $h=r\cot\alpha$, so why did you said in the last equation $h_A=(r_A-r_{1W})\cot\alpha$? Applying the volume formula of a a conial frustum to $V_A$: $${V_A={1\over3}\pi(r_A^2+r_Ar_{1W}+r_{1W}^2)*h_A}$$ Solve for $h_A$: $${h_A = {V_A\over{1\over3}\pi(r_A^2+r_Ar_{1w}+r_{1w}^2)}}$$ Now comparing your $h_A$ solution to mine, there are differences. Where is my mistake? – Frosty Jan 30 at 9:08
• Height $h$ in $h=r\cot\alpha$ is referred to the whole cone, i.e. it is the height from the cone vertex to the base. The height of a frustum is the difference of the heights of two cones. – Aretino Jan 30 at 9:54
• Sorry for the weird formula. I typed it in many other mathjax-previews and they were shown without failours. I went wrong. Maybe now its going to work. I got a Volume formula for a truncated cone and solved it for $h_A$: $h_A={V_A\over{1\over3}\pi(r_A^2+r_A*r_{1w}+r_{1w}^2)}$. – Frosty Jan 30 at 12:08
• Now if i type i try your take your $r_A$ and $h_A$ i get other results as with my $h_A$ calculation. – Frosty Jan 30 at 12:13