Simple dipstick problem

1. pat666

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?

2. pat666

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

3. ebits21

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.

4. pat666

709
the cylinder is on its side so its not that simple I don't think??? sorry I forgot to say that in the OP.

5. ebits21

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?

6. pat666

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?

7. ebits21

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.

709

9. ebits21

51
Well, I can say this much. If you look at the cylindrical tank from one of the ends, you would have a circle with the liquid coming up a certain height of the circle.

Therefore the problem becomes a question of finding the area of the segment of the circle below the chord marked by the top of the liquid. This is then multiplied by the length of the cylinder to find the volume.

Since it's a circle, you can draw a line from the center of the circle to the ends of the chord that marks the height of the liquid. Bisecting this, you get two right angle triangles with a hypotenuse equal to the radius, and a height equal to the radius minus the height of the liquid. The angle between these two is then used to figure things out...

Unfortunately... I'm kind of rusty on this so I'm afraid I can't be of much more help. Maybe someone else can take it from here?

10. pat666

709
What you are saying at the start makes perfect sense to me and hopefully someone can help me further with the math.

Thanks

11. ebits21

51
I thought I would mention another thing.

If you take the area of the "pie slice" that is marked by the lines drawn to the top of the chord (including the area that is swept out to the bottom of the tank), and then you subtract the area of the triangles above the liquid, that should give you the area of the liquid in the circle.

12. ebits21

51
I found this on wikipedia... which might help http://en.wikipedia.org/wiki/Circular_segment

You need to rewrite the area formula in terms of the height of the liquid, and I think that R = h + (r-h) gives you a way to do that.

13. Mentallic

3,701
I'm willing to give this question a go, but I can't seem to concentrate at the moment, probably because of my headache. If anyone could give me a jump start with what we're trying to solve, I'll give a lending hand to finish the problem off.

14. pat666

709
Hey Mentallic, any/all help is appreciated. Were trying to make a dipstick for a sideways cylinder and a cone. Currently I'm having trouble relating the height of the liquid on the circle to the area of that part.

Thanks

15. Mentallic

3,701
Ok since you said in post #6 that the cylinder and cone are separate questions, I'm going to assume that you want a relation between the height and volume of liquid present in a cone on its side (like a wheel) and the cone will be laying down which way exactly?

First of all, do you know about and are you supposed to use integration?

16. pat666

709
Cone is upright, cylinder is on its side. I know integration but this is for an 11th grader I'm helping, I seem to have forgotten everything from then. No integration for him yet though. It is the cone question that I need help with as there is ample information on the cylinder on the net.

Thanks and hopefully the fact that its upright (ice cream cone ways) makes this simpler and sorry for not stating this directly.

17. Mentallic

3,701
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?

18. pat666

709
I should be able tne thing that I'm not totally sure about is the "cone" that isn't filled with water, is this a cone, or at least does its volume follow the standard volume formula for a cone??

19. pat666

709
also now that I'm thinking about it I also can't find r, sorry I just didn't realize how much of this I've forgotten. Fairly embarrassing that I can't do yr 11 math anymore!!!

Thanks

20. Mentallic

3,701
Of course it's a cone. It has the same apex (point) as the larger cone but a shorter height and thus shorter radius. The cones are similar and both have flat bottoms (because water settles due to gravity of course ).

So we have two similar right triangles, the larger one with height H and length R, and the other one with height H-h and length r. Since they're similar, the proportions of each of their sides are in the same ratio. For example, a circle of radius 1 is similar to a circle of radius 2 such that their radii are a ratio of 1:2 and their circumferences are also a ratio of 1:2.
This means you can create the equality $$\frac{R}{r}=\frac{H}{H-h}$$