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Determine distance to Centre of Gravity of shape from 2 intersecting circles 4

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Judge

Mechanical
Oct 12, 2002
34
Hello. How can I determine the X distance to the centre of gravity of the area within the red arcs? See attachment. I need formula(e) to put in an excel spreadsheet.
Conditions:
Circles have different radii.
Centre of bigger circle is (0,0).
The included angle will always be less than 180 degrees.
Centres of both circles lie on the X axis.
Any suggestions would be much appreciated.
Thank you
Judge.


 
 http://files.engineering.com/getfile.aspx?folder=db8d9af6-3a1e-4f27-a07a-df22b5884a2f&file=Intersection.JPG
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(1)[ ] Note that "Crescent = Sector2 + Quadrilateral - Sector1".
(2)[ ] Note also that all four component shapes share a common axis of symmetry.
(3)[ ] Apply the parallel axis theorem multiple times, guided by the above "equation".
 
See attached spreadsheet (which gives the same answer as robyeng's formula).

I took the segment area and centroid equations from my Section Properties spreadsheet, which you can download from:

If you are using Excel 2013 I have just discovered that Microsoft have messed up the drawing function (again) by reverting to the conventions they used in 2003, so the link below might work better for the graphics:




Doug Jenkins
Interactive Design Services
 
By the way, it's quite interesting that the two overlapping segments always have exactly the same first moment of area about their own arc centres.

I'm sure there must be a use for that somewhere.

Doug Jenkins
Interactive Design Services
 
The answer turns out to be quite simple.
Defining d as the distance of the two center points, A1 as the area of the segment in the larger circle, A2 as the area of the segment in the smaller circle, xG=A2d/(A2-A1)
where, of course: A1=R2(α-sinαcosα), A2=r2(β-sinβcosβ), cosβ=(R2-r2-d2)/2dr, sinα=(r/R)sinβ

prex
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If you are using Excel 2013 I have just discovered that Microsoft have messed up the drawing function (again) by reverting to the conventions they used in 2003, so the link below might work better for the graphics:

Forget that bit, I must have been using an old version.

The attached spreadsheet allows groups of shapes to be combined, which allows the centroid calculation to be done automatically:
1. Calculate the angle from the arc centres to the top of the common chord for the two circles (Theta).
2. On the DefShapes sheet of the attached spreadsheet select Circular Segment and enter the R and Theta for the smaller circle.
3. Scroll down and in the Group properties table enter the X offset for the smaller circle, and set Elastic Modulus to 1. All other columns are zero, and the other rows should be blank. Click the Create New Group Button.
4. Go back to the main table and enter the R and Theta for the larger circle.
5. In the Group properties table set the offset to zero and Elastic Modulus to -1 and click the Add shapes to group button.
6. The Group properties results updates, showing the same centroid value (Xc) as calculated by the other spreadsheet.
7. The resulting shape can be plotted on the Coords_Group sheet. The group properties calculated on that sheet are not exact because they are based on curves made up of short straight lines (but it's pretty close).


Doug Jenkins
Interactive Design Services
 
IDS & Prex.

It seems you each had the pivotal insight that, in Doug's words, "the two overlapping segments always have exactly the same first moment of area about their own arc centres".[ ] It is this insight that leads to Prex's comment that "the answer turns out to be quite simple".[ ] Great work, leading to a very elegant result.[ ] I was expecting reams of algebra, so didn't have the guts to wade in.
 
My first spreadsheet actually calculated the positions of the centroids of the two segments; I didn't pick up that you could avoid that.

I have now uploaded a revised version using Prex's simplified formula.

Also I did a search for centroid of a lune and found some Fortran code which claims to do that but actually returns the centroid of a circle segment!

Doug Jenkins
Interactive Design Services
 
Thank you all for your replies.

RobyEngIT.
Your formulae worked perfectly. To confirm the results, I drew a constrained sketch in NX. Tried a range of values, & compared the resultant XG value to that in the spreadsheet. Perfect every time. Many thanks.

Prex.
I revised my sheet with your simplified formula. Nice work. Thank you.

Doug.
Thanks mate. Awesome spreadsheet. I have added it to my arsenal.

Many thanks.
Judge.


 
"If you're a glutton for punishment . . ."

Hmmm, I think that's why we all got into this business in the first place [bigsmile] LOL

It is better to have enough ideas for some of them to be wrong, than to be always right by having no ideas at all.
 
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