Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations MintJulep on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
volume of a circular segment
- Thread starter RIVetron
- Start date
- Staff
- #2
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
- Thread starter
- #3
- Thread starter
- #5
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>
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>
diamondjim
Mechanical
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.
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.
Mandrake22
Mechanical
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
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
pipewelder1999
Industrial
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
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
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
- Status
Similar threads
- Question
- Locked
- Question
- Question
- Question
