ahh if only i could remember calculus 2. integration would be so useful now. :D

you can use double integrals.
http://personal.inet.fi/koti/ollip/p...tical_cone.png

so let's set the bottom left of the cone to be at (0,0,0)

measure the veritcal height of the water. so say it's 4cm high. the upper bound in the z direction is z = 0. the lower bound, we see is defined by the equation of a circle. so say ur cone has a radius of 5. this means in terms of z, the lower bound is z = sqrt(25-x^2)

now let's do the y direction. say the base of the cone is at y = 0. the distance to which the water extends to the right is say, 4. now u have ur y boundaries.

somebody should definately check my maths.

now for x. find the distance the water extends into and towards you, (in terms of the picture above). so from the radius of the base. say the water extends 2 cm forwards and therefore 2cm towards you. you now have ur x boundaries.



you can integrate
http://hsinmao.googlepages.com/Picture4.png

you have two cones so multiply ur answer by 2. and ure done.

[π(2xr - x2)Δx + π x 2Δx] × 2r = 4π r2x Δx .

obviously

This is the first thing that crossed my mind but I think the area of the base is not 25% covered with 2 cm. But if I did calculate the correct area would this calculation give me the correct answer?

Can anyone confirm this?

Bump....

Help calculating a volume of a horizontal cone
 

Hi,

I need help calculating amount of liquid in a horizontal conical object (tank with conical ends). The cone looks like this: > and the tank would be something like:

I know the radius of the base and the height of the cone + the height of the liquid level. How can I calculate the amount of liquid on the conical part of the tank?

This is too hard for me so any help is greatly appreciated http://www.discussworldissues.com/fo...ies/laugh1.gif

BTW, sorry for the really poor presentation. I can draw a picture tomorrow if you'll need one...

Edit: Damn, title does not change in thread list...

http://en.wikipedia.org/wiki/Volume

A cone (circular-based pyramid): http://upload.wikimedia.org/math/2/4...4ef86f49de.png r = radius of circle at base, h = distance from base to tip

Just calculate the volume of the cylinder. ((pi) * (radius)^2)*height of cylinder. Then find the volume of the cylinders on each side, which is (1/3)*(((pi) * (radius)^2)*height of cone)

I think both of you were trying to calculate the volume when the cone is full. I'm sure I didn't explain the thing correctly.

Here's a picture (yeah, I suck in Gimp too)
http://personal.inet.fi/koti/ollip/p...tical_cone.png
I'm trying to find out how much liquid is at the bottom when I know the height of the liquid level...

Thanks for help so far...

One possible method (although you might not like it!): the cone has a circular base, so the water level will mark out a segment of that circular base. The area of the segment will decrease as you move along the cone - i.e. going from the largest part of the purple bit in your picture to the end of the point.

Now you can find the area of the segment via:

http://upload.wikimedia.org/math/a/3...aa2b5b6d5e.png

and you can get the angle via the height of the water:

http://upload.wikimedia.org/math/1/c...849d6bfb76.png

Use trigonometry to find the horizontal length of the purple bit and then integrate the segment area equation over this length to find the volume of the water. There's probably a much simpler way of doing this but it's first thing that popped into my pre-morning-coffee mind http://www.discussworldissues.com/fo...ies/smile1.gif.

then you simply have half a cone with the base of the water level

Neeyik: http://www.discussworldissues.com/fo...ies/blush1.gif Too damn hard for a dummy like me

I don't think so.

Pic from bellow...
http://personal.inet.fi/koti/ollip/pics/from_bellow.png

If you put 2 of those together the base will not be circular anymore.

If you know how high the liquid sits you can just find the proportion of the diameter and multiply it by the equation I posted. So if its an 8 cm tall tank and the water is 2 cm high, just multiply the entire volume of the tank by .25.