Volumn of cone with dip stick

timeout

I have done these a thousand times at college but for some reason I just can't seem to work it out today.:( I'm trying to write a program to do it but i need it on paper before i can do that. The idea is simple:

You have a cone shaped tank with a flat top and base,

   _
  / \
  ---  

Like that!
you want to put a dip stick in and read of the number of liters.

Volumn of a cone is 1/3 ([pi] r2 h) but don't think that works here unless it should be 1/3 ([pi] R1 R2 h)

[edit] diagram is not perfect

dudara

A truncated cone like that is called a frustum. More info here

timeout

Right so I now have the formula for the volumn of this frustum. How do I work out the height of a set volumn of water within the container?

dudara

Right, you have two unknowns there, the height (l) and the radius of the water at that height (R).

You do know the height of the frustum (h) and the radii of the bottom and top (R1 and R2 respectively).

What you need to do is work out an expression that relates the radius of the water(R) at any height (l) to the set parameters R1, R2 and h.

By using a Cartesian graph, you can show that R = l*(R2-R1)/h + R1.

Substitute this into the volume formula and solve for l.

If this isn't clear, I'll go through it more clearly.

cunnins4

could you not use integration? The volume of a section of a cone is the integral between the max and min (i.e. the height) of the section of pi times the radius squared.

So all you have to do is express the outer side of the cone as a function and revolve it around the x-axis.

b
V = | pi[f(y)]^2 dy (here i used | as the integral sign)
a

So if you're given the set volume of the water in the section, you can work backwards to find the height (b) if you set a=0.

(edit: the b and a should be above and below the | respectively, but the editor keeps moving them back.)

timeout

Yeah I was think i had only one unknown and then i realised i had two which has thrown me off. Thing is,i think, this is what i wanted. I mean 2 unkowns and just sub a value for one to get the other. Ie. I know that the total volume is 1,000 but i want the height of 500. I have 2 radius values so the one i want has to be between the two. So substitute the unknown radius with a value to determine the height. So I get:
500 = 1/3 pi h(100sq + 100 * 20sq+20)
=> h = 1/3 pi 500(100sq + 100 * 20sq+20)
right?

dudara

What I did above was express one unknown in terms of the other (R in terms of l, the unknown height). Then I substituted that into the formula (leaving only one unknown, l. All other variables are known). Then solve for the unknown variable.

timeout

right so :
Values:
wh - water hight
wr - water radius
r2 - top radius(5)
r1 - bottom radius(10)
h - total height(10)
v - volume of water(500)
V - volume of Frustum(1832)

the formula is :

v = 1/3 pi wh (r1^2 + r1 * wr^2 + wr)
=> 500 = 1/3 pi wh (r1sq + r1 * (wh*(R2-R1)/h + R1)sq + (wh*(R2-R1)/h + R1))

=> 500 = 1/3 pi wh(110 * ((wh*(-95)/10+10)sq + wh*(-95)/10+10)))

Correct???