I know there are Math Wizards here

I was trying to calculate the circumference of an elipse in equations, so I
entered this formula
"D3@Sketch3" = 2*(pi)*Sqr(("D2@Sketch3"^2 + "D1@Sketch3"^2)/2)
Where D2 is half the diameter of the major axis and D1 is Half the minor
diameter. On the elipse that I have
D2 = 1.0922455 and D1 = .75 The equation figured this to be 5.8644 but
Solidworks says the length of the Elipse is 5.557 Then on my TI-86 it comes
out to 5.88661 which is different yet. .002 is almost acceptable.
that is almost .307 inches off what is the deal. Did I use the wrong
formula or should I call the VAR.
Corey Scheich
Reply to
Corey Scheich
Loading thread data ...
Funny, I just came across this a few days ago, while enhancing our SolidSketch add-in to make it place points at regular intervals along an ellipse... Actually, there is no exact formula for the circumference of an ellipse (yours is a ~rough~ approximation, if I may say... ;-) See
formatting link
and your TI use different numeric integration algorithms to approximate the length. I don't know which is better...
Reply to
Philippe Guglielmetti
Take an ellips with D1=1 and D2=100. My feeling says that the circumference should be pretty close to 200, right?
So now find a formula that gets you there ;-).
Harry 'math is my grandmothers middle name. Mine is exxxxxaggeration'
Reply to
Harry Kroonen
Corey,
As Phillipe says your formula is an approximation - there is no exact closed form solution.
I entered your formula and values into SolidWorks 2004 and got 5.8866107. That agrees closely with your TI, suggesting that it uses the same approximation you do.
The following site lists several improved approximations:
formatting link
They state that for modest eccentricity it is hard to beat the second Ramanujan method:
a - Major axis radius b - Minor axis radius e - eccentricity = Sqr(1 - b^2 / a^2) h = (a - b)^2 / (a + b)^2
approx circumference = pi (a + b) [ 1 + 3 h / (10 + Sqr(4 - 3 h) ) ]
(By modest eccentricity they mean b should not be very small compared with a - your example certainly qualifies.)
I added an extra angle dimension to my sketch - D4, to take the place of h, and then replaced your equation with the following two equations:
"D4@Sketch3" = (("D1@Sketch3"-"D2@Sketch3")/("D1@Sketch3"+"D2@Sketch3"))^2 "D3@Sketch3"= pi * ("D1@Sketch3"+"D2@Sketch3") * (1 + 3 * "D4@Sketch3" / ( 10 + Sqr( 4 - 3 * "D4@Sketch3" ) ))
The above equations gave a value of 5.837601 compared with SolidWorks measure 5.8378270, which looks like an improvement.
I used an intermediate dimension for h (D4) because I didn't know how to create intermediate variables not associated with dimensions. (I guess this is not possible?) It is important to use an angular rather than a length dimension, as h is a ratio and so should be unitless.
I can email you my sample part if you like.
Reply to
Paul Delhanty
Machinery's Handbook says the formula should be:
"D3@Sketch3" = (pi)*Sqr(2*("D2@Sketch3"^2 + "D1@Sketch3"^2)) Where D2 is *all* the diameter of the major axis and D1 is *all* the minor axis.
Verified w/ D2 = 1.0922455 and D1 = .75 and result was 5.88661 in SWX 2001+.
Ray
Reply to
Ray McHugh
Thank you all I ended up going to the site Pillippe suggested and read all the interesting equations for this and selected an easy semi acurate one seems to come out quite a bit closer to what SW calculates.
(pi) * ( 3 * ("D2@Sketch3"+"D1@Sketch3") - Sqr(( 3 * "D2@Sketch3" + "D1@Sketch3") * ("D2@Sketch3" + 3 * "D1@Sketch3")))
Again D2 is half the long axis and D1 is half the short one
Thanks again Corey Scheich
Reply to
Corey Scheich
Hi Corey.
Since the "Radius" is constantly changing, You are definitely in the "zone of calculus" and possibly even Trigulus or even Caligulus.
I was able to lift these two approximations from a trusted source -
Perimeter = PI * SQRT [ 2*(A^2 + B^2) ]
&
Perimeter = PI * SQRT [ 2*(A^2 + B^2) - ( (a-b)^2/2.2 ) ]
These might help.
Otherwise let's get our table of integrals & derivatives out and start scratching our heads.
Regards,
SMA
Reply to
Sean-Michael Adams
I believe the approximation formula is:
"D3@Sketch3" = 2*(pi)*Sqr(((("D2@Sketch3"/2)^2) + (("D1@Sketch3"/2)^2))) if D2 and D1 are diameters.
Reply to
David
Yogi Berra! I didnt know you used Solidworks! Welcome to the NG!
Reply to
rocheey
I beg to differ, sorry. Fairly easy with a spot of calculus IIRC if you know the equation .... And I think you can get the focal points from the major & minor axes .. and it may be stored in the part database as a general conic too. I don't know SW's database formats ..
Reply to
Cliff Huprich
Guglielmetti"
Cliff,
According to
formatting link
the 'simplest form' is the following definite integral:
pi/2 4 a Integral Sqr[ 1 - e^2 (sin(t))^2 ]dt 0
a - Major axis radius b - Minor axis radius e - eccentricity = Sqr(1 - b^2 / a^2)
I think that the issue is that the integral is not solvable as a closed form in terms of functions available in SolidWorks equations. It is true that the definite integral itself is just another function to be approximated on a digital computer, just like Sqr, Sin, Cos etc. However, for those functions supported by SolidWorks equations, the approximation is performed natively by SolidWorks to the limit of accuracy representable by double precision arithmetic.
Reply to
Paul Delhanty
Guglielmetti"
Cliff,
According to
formatting link
the 'simplest form' is the following definite integral:
pi/2 4 a Integral Sqr[ 1 - e^2 (sin(t))^2 ]dt 0
a - Major axis radius b - Minor axis radius e - eccentricity = Sqr(1 - b^2 / a^2)
I think that the issue is that the integral is not solvable as a closed form in terms of functions available in SolidWorks equations. It is true that the definite integral itself is just another function to be approximated on a digital computer, just like Sqr, Sin, Cos etc. However, for those functions supported by SolidWorks equations, the approximation is performed natively by SolidWorks to the limit of accuracy representable by double precision arithmetic.
Reply to
Paul Delhanty
Actually there is someone out there who has developed a formula for the circumference of an ellipse. He has published a book called Circular Elliptics.
The blurb for the book reads:
Relates the ellipse to the circle in ways you never dreamed of, offers a proof for an equation for the circumference, and the surface area of an ellipsoid of revolution.
If anyone is interested check out
formatting link
ciao
Reply to
Peter B. Olson
For math answers consult a mathematician.
Goto
formatting link
-- from MathWorld
equation (69) and (70):
perimeter = pi*(a+b)*(1 + h/4 + h^2/64 + h^3/256 + ...)
where h = ((a-b)/(a+b))^2
you should be able to get a decent precision.
Joe
Reply to
joe(usenews)
yeah i got the same as you.... not sure
Reply to
Anonymous

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.