Suppose I have a physical coffee mug that is in the shape of a frustum. I can find the physical dimensions like smallest and largest radius and the height of the frustum. My question is till what height should I fill my cup to get a specific volume of water in it. For example I want 150 ml of water inside my cup so what height do I fill it till.The radius of the liquid at this level is also not know by me.
A,B,C are all known.
X is unknown.
Cup |Drawing
[Math] Relation between height and volume of frustum
geometryvolume
Best Answer
The volume of a frustum is derived from that of a cone $Ah/3 = d^2h\pi/12$.
The frustum is just a cone that you have removed the tip from and the volume is the difference of the volume of the cone and the tip. Now you probably don't have the height of the cone and the tip so you will have to use the fact that the diameter varies linearily along the axis. The slope is $(D-d)/H$ where $D$ is the larger and $d$ is the smaller diameter and $h$ is the height of the frustum. This gives the that the height of the tip is $Hd/(D-d)$. So you have to calculate the difference of the cone formed by the liquid and the tip and the tip itself. What we need is to find the base diameter of that cone which is $d + h(D-d)/H$. Putting these together we get:
$$V = \left(\left(d+{h(D-d)\over H}\right)^2\left(h+{Hd\over D-d}\right)-d^2{Hd\over D-d}\right){\pi\over 12}$$
Where $d$ is the smaller diameter, $D$ is the larger, $H$ is the height of the frustum and $h$ is the depth of the liquid. The factor $\pi/4$ is because of circular base and could be changed if you desire another shape of the base.
If you want to solve for $h$ you may expand the expression:
$$V = \left(\left(d+{h(D-d)\over H}\right)^2\left(h+{Hd\over D-d}\right)-d^2{Hd\over D-d}\right){\pi\over 12}\\ = \left(\left(d^2+2d{h(D-d)\over H}+\left({h(D-d)\over H}\right)^2\right)\left(h+{Hd\over D-d}\right)-d^2{Hd\over D-d}\right){\pi\over 12} \\ = \left( hd^2 + h^2{2d(D-d)\over H} + h^3{(D-d)^2\over H^2} + {Hd^3\over D-d} + 2hd^2 + h^2{(D-d)d\over H} - {Hd^3\over D-d} \right){\pi\over 12} \\ = \left( h^3{(D-d)^2\over H^2} + 3h^2{(D-d)d\over H} + 3hd^2 \right){\pi\over 12} \\ = {(D-d)^2\over H^2}\left( h^3 + 3h^2{Hd\over (D-d)} + 3h \left({Hd\over (D-d)}\right)^2 + \left({Hd\over (D-d)}\right)^3 - \left({Hd\over (D-d)}\right)^3 \right){\pi\over 12} \\ = {(D-d)^2\over H^2}\left(\left( h + {Hd\over (D-d)}\right)^3- \left({Hd\over (D-d)}\right)^3\right){\pi\over 12} $$
Which can be readily solved for $h$:
$$h = \sqrt[3]{ V{12\over\pi}{H^2\over(D-d)^2} + \left({Hd\over (D-d)}\right)^3} - {Hd\over (D-d)}$$