# Simple dipstick problem

by pat666
Tags: dipstick, simple
 P: 709 1. The problem statement, all variables and given/known data I have a problem involving a dipstick, a cone and a cylinder. Basically I want to make a dipstick for a cone and cylinder which shows the volume of remaining fluid in the containers. 2. Relevant equations 3. The attempt at a solution for the cone V=pi r^2 *h/3 now I cant find how to relate depth to volume? please help
 P: 709 I just remembered to say that what I'm really really having a problem with is finding the correct relationship between height and radius. Thanks
 P: 51 Well, you could start by focusing on the cylinder. If the volume of the cylinder is volume = pi * radius^2 * height Can you rewrite the equation so that you solve for height? If you can do that you know that a certain height on the dipstick will correspond to a certain volume of the cylinder. Then you can figure out how the cone will add on to that.
P: 709

## Simple dipstick problem

the cylinder is on its side so its not that simple I don't think??? sorry I forgot to say that in the OP.
 P: 51 In that case you need to rewrite the volume equation for the cylinder in terms of the diameter of the cylinder. Remember that radius = 1/2 the diameter. That way you'll know the volume that corresponds to the "diameter height" that you have written on the dipstick. Is the cone attached to the cylinder or is it a separate question?
 P: 709 the cone is a separate question. so with the diametre V=pi*1/4 D^2 *h then D=sqrt(4V/pi) but that will only work until 1/2 way won't it?
 P: 51 You know what, I think I actually made a mistake. I think you're actually going to have to use some trigonometry to solve this. I'll have to think about this a bit.
 HW Helper P: 3,327 Alright sure, and yes it does make the problem easier Ok so I'm guessing that the dimensions of the cone are known constants. Let's give it a base radius R and perpendicular height H, so the volume of the cone is $$V=\frac{\pi R^2 H}{3}$$ Now let's fill this cone up a bit with water, the water level will be at a height h. Now, just look at the part of the cone that isn't filled with water - it's a smaller cone with height H-h. Let the radius of this cone be r. So the volume of the water is now simply $$V=\frac{\pi R^2 H}{3}-\frac{\pi r^2 (H-h)}{3}$$ Ok so what constants do we know in this formula? We know R, H, h (because this will be observed on the dipstick) and we don't know r. Well r is pretty easy to find, just take a side-view of the cone and look at one side of the cone. It will be a right triangle with height H and length R. There will be another similar but smaller right triangle within it with height H-h and length r. Since these triangles are similar, their dimensions are proportional to each other. Can you take it from here?