# Volume Equations

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#### soho54

Does anyone have the equations for the volume of a hyperbolic, and conical horn on hand.

I'm taking about the horns volume, not one of the chamber volumes.

Thanks.

#### tb46

Hi soho54,

For cones I have always used the approximation: volume = (area1+area2)/2 * height.

Here is a reference, this can get difficult pretty fast:

Solid of Revolution -- from Wolfram MathWorld

Hope this helps,
Regards,

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#### wakibaki

Volume of a cone: 1/3 * pi * r^2 * h

I'm trying to lay my hands on a book that has the hyperbolic.

w

#### wakibaki

What is the equation of the hyperbolic?

w

#### soho54

There should be a direct way to do it. I'm trying not do to it with the slice method tb46 linked to.

This is the equation for the volume of an Exponential Horn
V=(St/k)*(e^(k*x)-1)

It is very easy to use as I already have St, K, and x from the cross-sectional area equations. Mo=St*e^(k*x)

The parabolic I can't remember right now, but it is similarly easy.

Conical cross-sectional area is:
Mo=St*((x+Xo)/Xo)^2

Hyperbolic cross-sectional area is:
Mo=St*((COSH(x/Xo)+(m*SINH(x/Xo)))^2

#### wakibaki

In the general form, volume of a rotation, its: the integral from a->b of pi()*(y^2)dx

You substitute for y^2 and perform the integration.

For a sphere, for example, the equation of the boundary curve is x^2 + y^2 = r^2

Re-arranging: y^2 = r^2 -x^2

Substituting, the volume of a hemisphere is: the integral from 0 -> r of pi()*(r^2 - x^2)

Which evaluates to [0 -> r] [pi()*(r^2*x - x^3/3)]

=pi()*(r^3-[r^3/3])

= 2/3 pi()*r^3

therefore volume of a sphere = 4/3 pi()*r^3

You just need the equation of your hyperbolic in terms of y, then you can work out y^2 and substitute, integrate.

w

#### Ron E

There should be a direct way to do it. I'm trying not do to it with the slice method tb46 linked to.

Conical cross-sectional area is:
Mo=St*((x+Xo)/Xo)^2

Hyperbolic cross-sectional area is:
Mo=St*((COSH(x/Xo)+(m*SINH(x/Xo)))^2

Look for the theorems of Pappus Guldinus - if you can do an integral and find the centroid, you are all set.

#### soho54

Conical horn volume
V =(1/3)*PI()*x*((r1^2)+(r2^2)+(r1*r2))

r1=sqrt(St/PI())
r2=sqrt(Mo/PI())
x=distance
St=throat Sd
Mo=mouth Sd

#### tb46

Hi soho54,

That looks correct, naturally I made a typo in Post #2, obviously it's divide by 3 not 2.

Hope you find an easy solution for the expo/hypex horns.

Regards,

#### wakibaki

Conical horn volume
V =(1/3)*PI()*x*((r1^2)+(r2^2)+(r1*r2))

r1=sqrt(St/PI())
r2=sqrt(Mo/PI())
x=distance
St=throat Sd
Mo=mouth Sd

No, that just can't be right.

The formula is obtained by calculating the volume of the cone as it would be if extended to a point (the original formula I gave), and subtracting the volume contained by the cone which is 'missing'

(1/3 * pi * r1^2 * h1)-(1/3 * pi * r2^2 * h2)

where r2 is the radius of the throat, r1 is the radius of the mouth, h1 is the distance from the mouth to the point of the cone as it would be if extended to a point and h2 is the distance from the throat to the point of the cone.

By inspection it is obvious that the second formula,

(1/3 * pi * r1^2 * h1)-(1/3 * pi * r2^2 * h2)

rearranged,

Can never contain this term

(r1*r2)

which is in the formula you have given. The formula you have given, moreover has no provision for calculating the value of h1 or h2.

Taking x as the distance from throat to mouth since h1 = x + h2 we have:

(1/3 * pi * r1^2 * [x+h2])-(1/3 * pi * r2^2 * h2)=

1/3 *pi * [(r1^2 * [x+h2])-(r2^2 * h2)]

(x+h2)/h2=r1/r2 (similar triangles)
x+h2=h2*r1/r2

so in terms of h2 we have

1/3 *pi * [(r1^2 * [h2*r1/r2])-(r2^2 * h2)]
=1/3 *pi * h2 * [(r1^2 * [r1/r2])-(r2^2)]
=1/3 *pi * h2 * [(r1^3/r2)-(r2^2)]

but

h2=h1-x
h1=h2*r1/r2

therefore

h2=h2*(r1/r2)-x
x=[(h2*r1)/r2]-h2
x=h2[(r1/r2)-1]
h2=x/[(r1/r2)-1]

hence in terms of x

1/3* pi * x/[(r1/r2)-1] * [(r1^3/r2)-(r2^2)]

w

It's a bit late. I'll look at it again in the morning

#### soho54

No, that just can't be right.
It works out right against everything I have checked it with.

Looking back over tb46's link I found =(1/3)*PI()*x*((r1^2)+(r1*r2)+(r2^2)), which I missed the first time scanning for the hyp bits. Due to the mix and match style of addition and multiplication, it works out the same as mine.

All three give the same answer.

Now, can you pull the hyp one off? I never got it to fly.

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#### wakibaki

All three give the same answer.

Ah, so they do.

Do you have the formula for the hyperbola from which the horn is generated in the form y=f(x)?

w

#### soho54

No, still working on that. The fact that I remember "zero" calculus doesn't help.

I have only ever dealt with the hyperbolic flare using the equation I added before. M or T is all I'm used to thinking about.

#### tb46

Hi,

I'm not even going to try to correct my mistakes anymore divide by 2 divide, by 3 what's the difference? (volume = (area1 +area2)/2 x height). I'll try to attach two pages from the VNR Concise Encyclopedia of Mathematics. Kepler's rule may be what you're looking for.

Regards,

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#### David McBean

Does anyone have the equation for the volume of a hyperbolic horn on hand.

Hi soho54,

Given S1, S2, L12 and T, the volume of a hyperbolic-exponential horn can be calculated as follows:

A = 1 + T
B = 1 - T
C = -2 * Sqrt(S2 / S1)

m = Ln((-C + Sqrt(C ^ 2 - 4 * A * B)) / (2 * A)) / L12

V1 = A ^ 2 / (2 * m) - B ^ 2 / (2 * m)

V2 = 2 * A * B * L12
V2 = V2 + A ^ 2 * Exp(2 * m * L12) / (2 * m)
V2 = V2 + B ^ 2 * Exp(-2 * m * L12) / (-2 * m)

Volume = S1 / 4 * (V2 - V1)

I derived the expressions for Hornresp some years ago .

Hope this helps.

Kind regards,

David

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#### soho54

Hope this helps.

Kind regards,

David
Yes, thank you. Already put it into Excel.

Of course, now I am going to have to borrow some text books to see how you did it.

#### wakibaki

Cool.

I'd have got there, but it would have taken a while.

And another while after that, probably...

w

#### soho54

I'll try to attach two pages from the VNR Concise Encyclopedia of Mathematics. Kepler's rule may be what you're looking for.
I tried that one before. It doesn't work when the length is several times the radii.

Thanks for the help.

#### soho54

I'd have got there, but it would have taken a while.

And another while after that, probably...

w
Add a couple more whiles for me. I was getting the first V2 result through another terribly convoluted method, and had no idea what to do next.

#### David McBean

Given S1, S2, L12 and T, the volume of a hyperbolic-exponential horn can be calculated as follows:

The expressions can be further simplified if required:

A = 1 + T
B = 1 - T
C = 2 * Sqrt(S2 / S1)

m = 2 * Ln((C + Sqrt(C ^ 2 - 4 * A * B)) / (2 * A)) / L12

V1 = (A ^ 2 - B ^ 2) / m
V2 = 2 * A * B * L12 + (A ^ 2 * Exp(m * L12) - B ^ 2 * Exp(-m * L12)) / m

Volume = S1 * (V2 - V1) / 4

Kind regards,

David

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