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
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
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
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
what is the sense if c > a ?
RE: Maths Problem - Partial Ellipse 1st Moment of Area
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
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
TTFN
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RE: Maths Problem - Partial Ellipse 1st Moment of Area
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
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
RE: Maths Problem - Partial Ellipse 1st Moment of Area
Cheers
Greg Locock
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RE: Maths Problem - Partial Ellipse 1st Moment of Area
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
RE: Maths Problem - Partial Ellipse 1st Moment of Area
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
Not to jump into this again, but standing to the OP
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
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
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
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
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
RE: Maths Problem - Partial Ellipse 1st Moment of Area
You learn all sorts of things on forums.
RE: Maths Problem - Partial Ellipse 1st Moment of Area
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://
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
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
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
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
still an interesting result ... ellipse CG is independent of a/b
RE: Maths Problem - Partial Ellipse 1st Moment of Area
htt
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
"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