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

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

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and your TI use different numeric integration algorithms to
approximate the length. I don't know which is better...

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'

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:

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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.

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

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

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

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 ..

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.

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.

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

-- 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

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