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

 

[ione]

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

 

[mike0709lum]

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

 

[rb1957]

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 ?

 

[mike0709lum]

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.

 

[ione]

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

 

[IRstuff]

It looks like you're trying to do a symbolic solution? Why not just numerically integrate?

TTFN

FAQ731-376

 

[mike0709lum]

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.

 

[ione]

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

 

[rb1957]

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

 

[GregLocock]

I can't help feeling that an annotated sketch might be worth a thousand posts.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[zekeman]

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)

 

[rb1957]

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.

 

[zekeman]

"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.

 

[ione]

Zekeman,

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

…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:

Yc = ? xdxdy / ? dxdy

Xc = ? ydxdy / ? dxdy

[quote/]?

 

[zekeman]

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

Was I mistaken when I wrote:

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.

 

[ione]

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)

 

[zekeman]

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?

 

[ione]

Thanks zekeman

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

 

[Occupant]

Just to muddle the waters some, the C.G. can also be found using an equivalent circle segment as per attached.