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Maths Problem - Partial Ellipse 1st Moment of Area

Maths Problem - Partial Ellipse 1st Moment of Area

Maths Problem - Partial Ellipse 1st Moment of Area

(OP)
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

RE: Maths Problem - Partial Ellipse 1st Moment of Area

The first order momentum with respect to an axis is given by the product of the area of the surface and the distance between the centre of mass of that surface and the axis.

In your case you might want to calculate the first order momentum with respect of y-axis (being equal to zero the momentum with respect to x-axis), which
will be:

Aellipse*Xc = PI*a*b*c
 

RE: Maths Problem - Partial Ellipse 1st Moment of Area

(OP)
Thanks for the reply but the issue I have is I am performing the calculation to determine the centre of mass of the section where x>0.

Also, since it is just part of the ellipse the area is not PI*a*b but I've got that equation nailed.

Mike

RE: Maths Problem - Partial Ellipse 1st Moment of Area

the math looks right to me, but how does it pick up the -ve y value ?  i mean you're using dA = ydx, but you want both the -ve and +ve values of y ?  or 2*+ve y ??

what is the sense if c > a ?

RE: Maths Problem - Partial Ellipse 1st Moment of Area

(OP)
Thanks rb1957,

You're right, I need to double the result to account for the -ve of y, I'd overlooked that.  Unfortunately the error is not fully accounted by this.  

In fact I'm having difficulty in getting a workable integration. Since writing the original post I found an error in my excel sheet and having corrected the error there is one term that now can't be solved (asin of a value > 1).  Going to keep working on that.

I'm only interested in the result when c<a.

 

RE: Maths Problem - Partial Ellipse 1st Moment of Area

To calculate the centre of mass (Yc,Xc) you don't need x*f(x)dx, but rather:

Yc = ∫ xdxdy / ∫ dxdy

Xc = ∫ ydxdy / ∫ dxdy

Being

y= SQRT[1-(x-c)^2/a^2]

dy = - [(x-c)/ SQRT[1-(x-c)^2/a^2]]dx

RE: Maths Problem - Partial Ellipse 1st Moment of Area

(OP)
Thanks ione,

Haven't you missed b^2 from y = ?  

I get:

y=sqrt[b^2-b^2(x-c)^2/a^2]

assuming I'm right dy is then:

dy=-[b^2(x-c)/[a*sqrt(b^2-b^2(x-c)^2/a^2)]]dx



IRstuff - it's part of a design optimisation so need to keep the symbols for the design process.  If I just needed one value I'd just use the CAD output.

RE: Maths Problem - Partial Ellipse 1st Moment of Area

mike0709lum,

You are right, I've forgotten "b"

dy = -(b/a)*[(x-c)/SQRT[a^2-(x-c)^2]

Maybe for the integration a parameterization could be convenient:

(x-c) = a*cos(t)
y = b*sin(t)

Since you have to deal with just a portion of an ellipse it's up to you to establish the integration range for t
 

RE: Maths Problem - Partial Ellipse 1st Moment of Area

what if you integrate by hand ?  as the sum of slices in excel ??

RE: Maths Problem - Partial Ellipse 1st Moment of Area

First off, the "c" is an offset to the fundamental problem.

So you first center  the ellipse at  0,0 and  performing the integral of

x*f(x)dx between the limits of -c and a to get the first moment of area ; then divide by the area which is
Integral of
f(x)dx , limits -c,a to get the CM,after which  you add c. That,s all.
[For convenience you make the substitution
sin@=x/a]

I got for int x*f(x) dx  limits -c,a

a^2bCos^3(arcsin(-c/a))/3

int f(x)dx, lim -c,a comes to
int abcos^2@d@ limits arcsin(-c/a), pi/2
=ab(pi/2+arcsin(c/a) + .25*ab(sin2(arcsinc/a)

RE: Maths Problem - Partial Ellipse 1st Moment of Area

zeke, that looks like it's for a 1/2 ellipse, maybe a full one ... i thiught the OP was after the general expression 'cause he was interested in only a portion of an ellipse.  

 

RE: Maths Problem - Partial Ellipse 1st Moment of Area

"zeke, that looks like it's for a 1/2 ellipse, maybe a full one ... i thiught the OP was after the general expression 'cause he was interested in only a portion of an ellipse."

Yes, it's for the the upper half of the the truncated centered ellipse between -c and a whose x value of CM is the same as the full ellipse.

It obviously doesn't matter what his reference coordinates are
and using his, unnecessarily complicates the math.
 

RE: Maths Problem - Partial Ellipse 1st Moment of Area

Zekeman,

Not to jump into this again, but standing to the OP

Quote (mike0709lum (Mechanical) 7 Oct 10 6:39):

...the issue I have is I am performing the calculation to determine the centre of mass of the section where x>0.

What's the need to calculate the integral of x*f(x)dx?

Was I mistaken when I wrote:

[quote ione (Mechanical) 7 Oct 10 9:02]

Yc = ∫ xdxdy / ∫ dxdy

Xc = ∫ ydxdy / ∫ dxdy

[quote/]?
 

RE: Maths Problem - Partial Ellipse 1st Moment of Area

"What's the need to calculate the integral of x*f(x)dx?

Was I mistaken when I wrote:

[quote ione (Mechanical) 7 Oct 10 9:02]

Yc = ∫ xdxdy / ∫ dxdy

Xc = ∫ ydxdy / ∫ dxdy

[quote/]?

Double integrals are  messy and in this case unnecessary..

You're technically right, but since the problem due to symmetry will yield a y value of the CM to be 0, why complicate it. As I show, the integration is fairly straightforward if you make the trig substitutions.

RE: Maths Problem - Partial Ellipse 1st Moment of Area

zekeman,

I agree with you that necessarily yc=0, I also agree on parameterisation and I had suggested it in one of my previous post (7 Oct 10 11:42 ), but I still do not understand  the need to calculate the integral of x*f(x)dx: this is not in agreement with the definition of center of mass coordinates (via first order moment)

RE: Maths Problem - Partial Ellipse 1st Moment of Area

Same as your definition.
I think the confusion is in f(x)

Int x*dA=  int x*dxdy
Int dA=int dxdy and
dA=ydx  since  y=y(x)=f(x)

int x*f(x)dx and
xCM=int xdA/int dA

xCM=int x*ydx/int ydx


Definition of x position of CM

for the x axis symmetric area

BTW, how do you make the "integral" sign?

RE: Maths Problem - Partial Ellipse 1st Moment of Area

Thanks zekeman

Further to the integral sign I have made it in a word doc and then paste copy.

RE: Maths Problem - Partial Ellipse 1st Moment of Area

Thanks Ione for the integral tip.

You learn all sorts of things on forums.

RE: Maths Problem - Partial Ellipse 1st Moment of Area

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/
 

RE: Maths Problem - Partial Ellipse 1st Moment of Area

(OP)
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

RE: Maths Problem - Partial Ellipse 1st Moment of Area

(OP)
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

RE: Maths Problem - Partial Ellipse 1st Moment of Area

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

RE: Maths Problem - Partial Ellipse 1st Moment of Area

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/
 

RE: Maths Problem - Partial Ellipse 1st Moment of Area

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)

RE: Maths Problem - Partial Ellipse 1st Moment of Area

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

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