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volume of a circular segment

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RIVetron

Mechanical
Apr 30, 1999
33
Can someone tell me how to calculate the volume of a circular segment? I have found the formulae to calculate the area, and length of arc, etc, pretty much everything except the volume.<br>
<br>
Any help would be appreciated<br>
<br>
Thanks<br>

 
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If you mean the volume of a spherical sector, formulae are in 'Burington', P.14, Art. 48. If that publication is not available in your library or elsewhere, send me your fax #. Click "Forummanager" in the 'Contact box'. One is on the front page, and elsewhere on the site
 
Hi, Jimbo,<br>
thanks for your reply, but I do not mean a spherical sector, I mean a circular segment, with depth/thickness, such as a rocker for a rockinghorse, etc.<br>
<br>
Paul<br>

 
If you calculate the area then multiply by the width/thickness, that should give you you're volume shouldn't it? Just like a cylinder - the area of the circle times the hieght of the cylinder is the volume.
 
SPG,<br>
I wondered about that also, at first I thought that was the way to do it. Maybe it is, it certainly makes sense. I just thought I'd put the question out to see if anyone has had to tackle that sort of problem before. - you know how sometimes if you look at a problem long enough you start to question the solution, or wonder if you've left something out. Just being careful.<br>
<br>
Thanks<br>
Paul<br>
<br>
<br>

 
Does anyone know a link to this publication?

'Burington', P.14, Art. 48.
 
Treat the cross section as a complete ring
and get the volume of the ring.
If you know the sector angle, the volume of
the sector would be the sector angle/360
and then time the volume of the ring.
 
Fire up your favorite CAD solid modeler, create your object and the have it compute the volume for you. The results will be the same as the others have suggested. Area x depth = volume.
 
You can integrate by parts, the circular segment subtended by an angle say, B, either side of the vertical axis of symmetry.

Given the equation of a circle as (x-x0)^2 + (y-y0)^2 = R^2 and setting the center of the circle to the origin, (x0,y0) = (0,0), the equation is simply:

A = INT (A/B) [sqrt(R^2 - x^2)], A & B = limits.

A = R^2[B - sinB cosB]

Clearly if you have a circular sector, R = outer radius, r = inner radius, then the difference between the two slices is the area required in between.

A = (R^2 - r^2) [B - sinB cosB]

This is very easily verified, for if the sector was a complete circle, B = pi by symmetry, then we get A = pi (R^2 - r^2), which is the area of the donut.

So if the circular sector is of thickness, t:

V = (r^2 - r^2) [B - sinB cosB] t

You may wish to convert to polar coordinates and retry, personally I find Cartesian to be a bit more challanging.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
 
If you know the length of arc, call it L. Then the volume is simply
L*w*t
where
w= width of segment
t= thickness
 
Please forgive me if this is way off. My math stopped in the 9th grade and about 8 years ago I decided to write a program for calculating the volume of metal required for weld joints here is the VB code I came up with

Function reinforcement(wHeight, wwidth)
If wwidth = 0 Then wwidth = 1
reinforcement = (wHeight / (6 * wwidth)) * ((3 * wHeight ^ 2) + (4 * wwidth ^ 2))
End Function


Using it always came within 3 decimal points of what some textbooks had for weight of weld reinforcement but I never had anyone check it out. Anyone with a formula more accurate that I may use that only requires the width of the segment and height from the chord? It literally took me weeks for the one above

Thanks



Any

Gerald Austin
Iuka, Mississippi
 
PipeWelder1999, if you are within three decimal points accuracy, then continue using it. For welding purposes, is it necessary to be measuring to thousands of an inch? Typically tape measures don't have that kind of accuracy, even the new fancy digital ones!

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
 
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