Relation between volume of cone and frustum given find height
Answers
The volume of a frustum is derived from that of a cone Ah/3=d2hπ/12Ah/3=d2hπ/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(D−d)/H where DD is the larger and dd is the smaller diameter and hh is the height of the frustum. This gives the that the height of the tip is Hd/(D−d)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)/Hd+h(D−d)/H. Putting these together we get:
V=((d+h(D−d)H)2(h+HdD−d)−d2HdD−d)π12
V=((d+h(D−d)H)2(h+HdD−d)−d2HdD−d)π12
Where dd is the smaller diameter, DD is the larger, HH is the height of the frustum and hh is the depth of the liquid. The factor π/4π/4 is because of circular base and could be changed if you desire another shape of the base.
If you want to solve for hh you may expand the expression:
V=((d+h(D−d)H)2(h+HdD−d)−d2HdD−d)π12=((d2+2dh(D−d)H+(h(D−d)H)2)(h+HdD−d)−d2HdD−d)π12=(hd2+h22d(D−d)H+h3(D−d)2H2+Hd3D−d+2hd2+h2(D−d)dH−Hd3D−d)π12=(h3(D−d)2H2+3h2(D−d)dH+3hd2)π12=(D−d)2H2(h3+3h2Hd(D−d)+3h(Hd(D−d))2+(Hd(D−d))3−(Hd(D−d))3)π12=(D−d)2H2((h+Hd(D−d))3−(Hd(D−d))3)π12
V=((d+h(D−d)H)2(h+HdD−d)−d2HdD−d)π12=((d2+2dh(D−d)H+(h(D−d)H)2)(h+HdD−d)−d2HdD−d)π12=(hd2+h22d(D−d)H+h3(D−d)2H2+Hd3D−d+2hd2+h2(D−d)dH−Hd3D−d)π12=(h3(D−d)2H2+3h2(D−d)dH+3hd2)π12=(D−d)2H2(h3+3h2Hd(D−d)+3h(Hd(D−d))2+(Hd(D−d))3−(Hd(D−d))3)π12=(D−d)2H2((h+Hd(D−d))3−(Hd(D−d))3)π12
Which can be readily solved for hh:
h=V12πH2(D−d)2+(Hd(D−d))3−−−−−−−−−−−−−−−−−−−−−−−−√3−Hd(D−d)
h=V12πH2(D−d)2+(Hd(D−d))33−Hd(D−d)