Volume of Cone Frustrum

[larry cottrill]

It is sometimes desired to find the volume of one of our pulsejet "cones" - really, the frustrum of a right circular cone. Here is (I think) a simple method, based on the fact that all we usually know (from some drawing) are the two end plane diameters and the length (i.e. the on-axis distance) between them. This method uses two convenient "shortcuts":
- The rate-of-change R of the cone diameter with distance from the apex
- The ratio Q of the volume of the major cone (the cone bounded by the larger end diameter and the apex) to the volume of the minor cone (the cone bounded by the smaller end diameter and the apex)
These may be immediately determined thus:
Assume D1 = the larger diameter, D2 = the smaller diameter, and L = the length. Then,
R = ( D1 - D2 ) / L
and
Q = ( D2 / D1 )^3 (where '^3' means 'raised to the third power' or 'cubed')

We can determine the height of the major cone immediately:
L1 = D1 / R

Next, the major cone volume:
V1 = D1^2 x L1 x pi / 12

And finally, and very simply, the frustrum volume by subtracting the minor cone volume (which is simply V1 x Q):
V = V1 - ( V1 x Q )

Example: Find the volume of the Lady Anne Boleyn chamber cone, where D1 = 64mm, D2 = 29mm and L = 171mm. The volume is as follows:

R = ( 64 - 29 ) / 171 = 35 / 171 = 0.2047
Q = ( 29 / 64 )^3 = .4531^3 = 0.0930
L1 = 64 / 0.2047 = 312.69
V1 = 64^2 x 312.69 x 3.1416 / 12 = 335308 cu.mm
V = 335308 - ( 335308 x 0.0930 ) = 304124 cu.mm or 304.124 cc

You may challenge or verify at will.

L Cottrill

[WebPilot]

Larry,

The formulation used for the volume of a frustum is:

V=1/3×pi×h×(r1²+r1×r2+r2²)

where
h is your L and
r1 is your D1/2 and
r2 is your D2/2 and
r2² is r2×r2

Your formula should turn into this one.

Does it?

I get 304107 mm³ in a spreadsheet so we're close.

[Irvine.J]

See, heres where your both confused,
Taking that reading, then divide by the standard amount of beer in a stubbie, (375ml) = 0.810952 Standard beers.
Now, knowing that in a common XXXX (Four X) brand beer, is 1.4 standard drinks, the volume of your cone is now 1.08 standard drinks... I think... I can't remember, pass me another XXXX. Then we'll fire up solidworks and check with that, cause its much easier thhhheeeeennnn woorrkkkkinggnnsssn outsya selfs ehhhh...yarrrghghg.
Man, it seems to be bushweek on the forum talking about gobbled up cats and stuff WOOT!

[Irvine.J]

I get 304106.954265 as well, to the last digit of my calculamchator. Which is 304ml or so using webpilots formulae there.

[tufty]

http://mathworld.wolfram.com/ConicalFrustum.html

Wolfram is the place for stuff like this (WebPilot has it right, by the way)

[larry cottrill]

Larry,

The formulation used for the volume of a frustum is:

V=1/3×pi×h×(r1²+r1×r2+r2²)

where
h is your L and
r1 is your D1/2 and
r2 is your D2/2 and
r2² is r2×r2

Your formula should turn into this one.

Does it?

I think it has to. My division by 12 comes from recognizing that the volume of a cone is exactly 1/3 the volume of the cylinder it will fit, and that you have to divide by 4 if you're squaring the diameters rather than the radii. The hard part would be working back out of my factor Q.

I get 304107 mm³ in a spreadsheet so we're close.

Well, 304124 / 304107 = 1.0000559... Yes, 6 thousandths of one percent is indeed pretty "close" ;-)

L Cottrill

[larry cottrill]

http://mathworld.wolfram.com/ConicalFrustum.html

Wolfram is the place for stuff like this (WebPilot has it right, by the way)

Of course. But anyody can go look something up. I wanted to derive it in a way that made sense to me. The crucial element for me was realizing that if we "grow" a series of cones with a particular included angle working out from the apex, the volume will simply increase with the cube of the height. (It follows similarly that the cone surface area and the area of the plane circular end will each grow with the square of the height.) From this, the factor Q was derived intuitively as the key to the method.

The proof of its validity is its close agreement with what the formula produces. I agree that the formula is probably more straightforward for purposes of calculation. It totally obscures the geometric nature of the problem, however, which is made very clear in my method, once the meaning of factor Q is understood.

The factor Q can be used in another amusing way: It reveals how little is lost as we cut off the top section of a cone. For example, if D2 / D1 = 2, meaning that we are cutting off the cone exactly at mid-height, we are removing a mere 1/8 of the volume. The bottom half (frustrum) that's left is 7/8 of the entire volume of the original cone, from a cut halfway up! This result is (to me) fairly non-intuitive, and shows an example of how difficult it is to estimate volumes of even simple three-dimensional forms by visual inspection. I can practically NEVER guess the right-sized Tupperware thingy to hold leftovers.

L Cottrill

[tufty]

http://mathworld.wolfram.com/ConicalFrustum.html

Wolfram is the place for stuff like this (WebPilot has it right, by the way)

Of course. But anyody can go look something up. I wanted to derive it in a way that made sense to me. The crucial element for me was realizing that if we "grow" a series of cones with a particular included angle working out from the apex, the volume will simply increase with the cube of the height. (It follows similarly that the cone surface area and the area of the plane circular end will each grow with the square of the height.) From this, the factor Q was derived intuitively as the key to the method.

The proof of its validity is its close agreement with what the formula produces. I agree that the formula is probably more straightforward for purposes of calculation. It totally obscures the geometric nature of the problem, however, which is made very clear in my method, once the meaning of factor Q is understood.

Just using the formula is, of course, mere brainless application of acquired knowledge, and to be avoided. I thoroughly enjoy working stuff out myself (a few weeks back, I did a bunch of derivation to prove that not only can you write a 3d rendering engine without using a single call to trig functions, but that it will be faster to do it that way).

No, the simplified formula itself doesn't cover the geometric nature, but the derivation of it does...

Simon

[WebPilot]

I am sure some of you can derive this formula, but for
those of you that are having troubles, I am posting how I
accomplished it.

You only need to know a little geometry and trig.

(c.t.)

[Anders Troberg]

a few weeks back, I did a bunch of derivation to prove that not only can you write a 3d rendering engine without using a single call to trig functions, but that it will be faster to do it that way

Yep, you shouldn't need anything more complicated than an occasional division if you stick to using triangles of the same proportions (unsure about the English term, "likformiga trianglar" for the Scandinavians here).

I'm not sure if it will work if you start rendering reflections and refractions, though, such as if you try to emulate viewing through a lens.

[Al Belli]

Hi,

I use this calculator which uses the classic formula.

If You enter the three values, you get the frustum volume.
If You enter only one diameter and length you get the volume of a cone.
If the diameters You enter are the same value, You get the volume of a cylinder.

[Bent]

I turned it into a gui for those without ms-exel.

[tufty]

What about those of us without MS-*anything*?

/me ducks and runs

[Al Belli]

Hi,

For those MS less, download a 60 day free trial from MS.

http://us20.trymicrosoftoffice.com/defa ... ?nd_loc=us

Or a free Excel viewer from MS.

http://www.microsoft.com/downloads/deta ... edbd03aaf0

Al Belli

[Jim Berquist]

Try Open Office! Free, Does the stuff, Exel, campatible, word campatible. and open source!


http://www.openoffice.org/


jim