# Calculating volume of a cone? Question

Posted:
edited January 2014
I have two cones. Each are identical in size and are right angle cones.

The diamter of the base is 52 feet and the height from base to tip is 26 feet.

The cones overlap in their respective centers. How do I calculate the total volume of the two cones without duplicating the volume of the overlap?

Yeah, I bet when you saw the thread title you thought I was a dumb a\$\$.

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Quote:

Originally posted by aplnub Yeah, I bet when you saw the thread title you thought I was a dumb a\$\$.

I thought maybe it was about ice cream.
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Overlap how?

Like

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Quote:

Originally posted by aplnub

I have two cones. Each are identical in size and are right angle cones.

The diamter of the base is 52 feet and the height from base to tip is 26 feet.

The cones overlap in their respective centers. How do I calculate the total volume of the two cones without duplicating the volume of the overlap?

Yeah, I bet when you saw the thread title you thought I was a dumb a\$\$.

Integrate.
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Quote:

Originally posted by curiousuburb

Overlap how?

Like

[code]

. ^ ^

. / ^ \\

. /A /\\ B\\

The top. Imagine two dunce caps on a desk in front of you (base down and tip up). Now slide them together.

I was hoping to dodge the calculus (I have been out of school way to long) but if all else fails I may.

Any other suggestions would be appreciated.

My AutoCad LT is not 3D. That may be a quick solution if any of you are 3d draftsmen.
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I have a test on this crap tomorrow.
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It's easier to think of the intersection as a plane cutting a wedge out of each cone. You need to figure out the distance this plane of intersection is from the cone centre (because they are 90 degrees, it's just half the radius) then work out the volume of the wedge and subtract two of those from the volume of the total volume of the cones.

You can calculate the wedge volume using integration on the cone volume equation over the radius from the point of intersection to the length of the radius.
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Originally posted by aplnub

I have two cones. Each are identical in size.....

MMM... Have a nice trip mate
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Do you mean something that looks like this?

acording to the software, this has a volume of 32759.2537 cubic feet.
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That's so cheating. Descartes* would be rolling over in his grave...

*Or whichever the dude was that figured out the volume of cones and integration and stuff
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Quote:

Originally posted by NielsD

Do you mean something that looks like this?

acording to the software, this has a volume of 32759.2537 cubic feet.

Exactly! Did you do that in MathCad?

What math program is the hottest thing going these days and what is the best thing going on the Mac platform?
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Quote:

Originally posted by sunilraman

That's so cheating. Descartes* would be rolling over in his grave...

*Or whichever the dude was that figured out the volume of cones and integration and stuff

Cheating is quicker

aplnub, I did this in Rhino 3D which, unfotunetly is not available for the Mac platform.. (so yes, I'm on a windows box right now..) (would that make me even more of a cheater? )
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Yeah, it's a tricky one. It actually requires triple integration to get the volume of the cone wedge:

http://mathforum.org/library/drmath/view/65406.html

Ouch.

Anyway, the equation for that is:

V = (1/3)*H*R^2*(T - 2*cos[T]*sin[T] + cos^3[T]*ln[sec(T)+tan(T)]).

where T is the angle between the intersection of the plane that I mentioned before and the line passing through the cone centers.

Since the distance of the plane intersection is just R/2, we can work out the other values with Pythagoras. We get:

cos(T) = 1/2

sin(T) = sqrt(3)/2

tan(T) = sqrt(3)

sec(T) = 2

T = pi/3

plugging them in, we get:

V = 2025.87939 cubic feet

The volume of the cone is 1/3*pi*R^2*h = 18405.5442

Therefore that shape = 2*cone - 2* wedge = 2*18405.5442 - 2*2025.87939

= 32759.3296 cubic feet.

But yeah Rhino3D is faster. Time to get an Intel Mac to run it under Windows maybe .
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I had this problem on a calc final way back when. Not sure if the measurement are the same but it was one of those that tied alot of the course work together. The professor helped out by giving similiar homework problems for these. Wish we had mathforum.org back in 78.

reg
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that sure is some equation...

Could never have come up with that even if I knew in what direciton to look.

Impressive

(Glad I have Rhino though )
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The below is from a post in Feb of 2006.

Anyway, the equation for that is:

V = (1/3)*H*R^2*(T - 2*cos[T]*sin[T] + cos^3[T]*ln[sec(T)+tan(T)]).

where T is the angle between the intersection of the plane that I mentioned before and the line passing through the cone centers.

Since the distance of the plane intersection is just R/2, we can work out the other values with Pythagoras. We get:

cos(T) = 1/2

sin(T) = sqrt(3)/2

tan(T) = sqrt(3)

sec(T) = 2

T = pi/3

I do not understand what you mean by "Tis the angle between the intersection of the plane and the line passing thru the cone center".  Can you explain? Better yet, can you provide a sketch that explains this?

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jman3423 wrote:
I do not understand what you mean by "T is the angle between the intersection of the plane and the line passing thru the cone center".  Can you explain? Better yet, can you provide a sketch that explains this?

Sure:

edit: more accurately, T is half the angle at the top of the triangle where the vertical line goes through, it just so happens that because it's all right angles in this example, T is the exact angle between the vertical and the line that goes through the cone centre. If it was any other setup, that wouldn't be the case.
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I have went to the Dr. Math site that you supplied the link for. I know that you took the equation from this site and modified it to reflect a set of cones that protrude into the center of each other.  But comparing the two, I just am not following.

I need to figure the storage volume for grain in a building that has multiple dumps at the top of the building.  Is there any way that you could give me the equation for would represent a height that changes, distance between dump points that changes and a varying angle of repose for the grain?  I realize the radius would be the result of the height and the angle.  Also, what about the distance from the axis of the cone to the plane?

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jman3423 wrote: »
I have went to the Dr. Math site that you supplied the link for. I know that you took the equation from this site and modified it to reflect a set of cones that protrude into the center of each other.  But comparing the two, I just am not following.

I need to figure the storage volume for grain in a building that has multiple dumps at the top of the building.  Is there any way that you could give me the equation for would represent a height that changes, distance between dump points that changes and a varying angle of repose for the grain?  I realize the radius would be the result of the height and the angle.  Also, what about the distance from the axis of the cone to the plane?

Ignore that last post (I've removed it so it doesn't cause confusion), I put the chord and T in the wrong place. The chord and angle T are what you see from above, not from the side. So imagine looking down on the cones from above. Where they intersect is where the chord of length C is created. It is at distance P from the centre of the cone.

The description was that T is the angle between the point where the plane intersects the cone at the top-left and a line drawn through the centres of both cones.

As far as this applies to the grain in the building, if it looks something like this:

What you'd do is work out the height of the grain pile and angle of repose. From this you can work out the cone radius: R = H1 x tan(R1)
Then you'd work out the distance the intersection point is from the cone centre = P. You can work this out knowing the radii of the intersecting cones and the separation of the centres.
With R and P, you can work out angle T and plug the values into the equation. Work out the volume of each grain cone and subtract all the overlapping wedges.
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Marvin, what software do you use to make these shapes? They look very nice!

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pb wrote: »
Marvin, what software do you use to make these shapes? They look very nice!

Just Photoshop but you should really use Illustrator or some other vector or CAD software so that you can do boolean vector shapes instead of rasterizing.