Maths Problem - Partial Ellipse 1st Moment of Area 1

[mike0709lum]

Hi,

Looking for some help on a maths problem, I'm a bit rusty and can't see what I'm doing wrong.

I'm trying to calculate the first moment of area of a portion of an ellipse which is not centered at 0,0.

The ellipse is described by:

(x-c)^2/a^2 + y^2/b^2 = 1 , this means the centre is at (c,0)

I know I need the integral of xdA with x between 0 and (a+c) but the answer I get is wrong (I know this from CAD comparison).

What I did was rejigged the ellipse equation to y=f(x) then used integral tables to evaluate the integral of x*f(x)dx:

I(x*sqrt(dx^2+ex+f))dx

where:
d=-b^2/a^2
e=2cb^2/a^2
f=b^2-b^2*c^2/a^2

The equation resulting from the solving of the integral is quite long so before I confuse everyone with that, have I got something fundamental wrong here?

Cheers,
Mike

 

[zekeman]

Thanks Ione for the integral tip.

You learn all sorts of things on forums.

 

[IDS]

Occupant beat me to it, the position of the centroid of a segment of an ellipse is the same regardless of the a/b ratio of the ellipse, so we can use the formula for the centroid of a circular segment, which is:
Xc =(2*R/3)*((sin(Theta)^3)/(Theta-sin(Theta)*cos(Theta)))

where:
The origin is at the centre of the circle
Theta = Cos^-1(C/R)
C is x ordinate of limit of the segment.

I have attached a screenshot (in ellipse.zip) of a section properties calculator that can be downloaded from:

http://newtonexcelbach.wordpress.com/2008/03/25/section-properties-of-defined-shapes-spreadsheet/

An alternative approach is to do a numerical integration of:
=((b^2/a^2)*(a^2-((x)^2)))^0.5*x

The second file in the attached zip file shows a screenshot of a spreadsheet to do that calculation, that can be downloaded from:

http://newtonexcelbach.wordpress.com/2010/09/11/eval-xls-and-the-alglib-integration-functions/

The results for the position of the centroid are identical for the circular segment (R=10) and the elliptical segment (a=10, b = 5) in the numerical integration example.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[mike0709lum]

Thanks Zekeman (& ione),

Worked a treat. I hadn't realised that centering the ellipse would work but thinking about the area and centroid relationships I see that it makes sense. Useful fact to bear in mind and simplifies the problem greatly.


Cheers,
Mike

 

[mike0709lum]

IDS,

Had a bit of a play with your integration xcel file, really useful, I'm sure others will find it useful also. Thanks for sharing that.

Cheers,
Mike

 

[rb1957]

a tought i had was to transform the ellipse into a circle ... y' = y*a/b ... but i wasn't sure how this'd affect the integration.

still an interesting result ... ellipse CG is independent of a/b

 

[IDS]

Mike - you might be interested in the latest post on my blog:

http://newtonexcelbach.wordpress.com/2010/10/10/faster-integration-with-the-tanh-sinh-method/

which is about an alternative numerical integration method (tanh-sinh quadrature). I have also added the tanh-sinh function to the Eval spreadsheet, with the ellipse area and first moment of area as an example:

http://interactiveds.com.au/software/Eval.zip

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[zekeman]

rb1057,

"still an interesting result ... ellipse CG is independent of a/b "

Actually, independent of "b" so the circle of radius "a" has the same CM.

This is obvious from the CM eq

Xcm= int(x*ydx)/Int(ydx)

for the circle, y1=a/b*y

If you substitute into Xcm eq , the a/b drops out and you get

int(xy1dx) / int(y1dx)

 

[moon161]

If the offset complicates it, couldn't you use the parallel axis theorem?