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Community Tip - Visit the PTCooler (the community lounge) to get to know your fellow community members and check out some of Dale's Friday Humor posts! X

Mathcad Community Challenge March 2024 - Perimeter of an Ellipse

DaveMartin
16-Pearl

Mathcad Community Challenge March 2024 - Perimeter of an Ellipse

I am shocked that I have walked the surface of this planet for over 5 decades without realizing that there is no exact closed-form solution for the perimeter of an ellipse. In school, we learn the perimeter of shapes like triangles, squares, rectangles, circles, parallelograms, and so on, but I feel like schools and teachers conveniently skipped over ellipses.

(An ellipse, in case it’s been a while, is the set of points where the sum of the distances from two points – the foci – is a constant.)

 

DaveMartin_0-1709255800267.png

 

Your challenge, should you choose to accept it, is to correct that oversight. Create a Mathcad worksheet that:

  • Derives, depicts, or shows one (or more) of the various approximation formulas / methods for the perimeter of an ellipse.
  • Create a calculator whereby someone can change the values of the semi-major and semi-minor axis lengths in order to find the perimeter.
  • Create a 3D plot of the perimeter as a function of the ellipse semi-major and semi-minor axis lengths.

Although there is no exact solution for the perimeter of an ellipse, this is a fairly well-documented problem. Therefore, your documentation in your worksheet is key! This worksheet should be able to stand on its own and be understood by someone with a basic knowledge of calculus (since integrals and infinite series are involved).

Good luck and have fun!

 

Find the Mathcad Community Challenge Guidelines here!

 

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com
35 REPLIES 35

Hi,

File enclosed Prime Express 9.0

Cheers

Terry


@terryhendicott wrote:

Hi,

File enclosed Prime Express 9.0

Cheers

Terry


Ellipse-Prime.png

I once won such an argument a long time ago - I showed two ellipses (Mean) for which there is an exact formula for the perimeter. One of them is a circle, and the second - see the picture.
A more challenging and interesting problem is the Cassini Oval (gmean) and the Cayley Oval (hmean)!

Ellipse-4c.png

PS

“The race course was a large three-mile ring of the form of an ellipse..”

Leo Tolstoy. "Anna Karenina"

Ellipse-Perimeter.png

More interest plot

зз.png

My attempt attached (excludes part 3 as 3D -plotting not available in Express).

 

(As I couldn't do the 3D plot in Prime Express, I decided to do it in M15 - see below.

The plot is really boring!).

 

Ellipse3D.png

 

Alan

 

 

DJF
16-Pearl
16-Pearl
(To:DaveMartin)

Every problem deserves a numeric solution.  I'll leave the complicated derivations for the experts.  This was good enough for Archimedes so good enough for me too.  Bit too much to make the 3D plot at this time.

 

Prime 7 attached.

Hi,

Am enjoying the posts so far so want to add a quick from 1st principles approach:

Capture.JPG

Hi,

Just could not resist it.  A page of matrices and voila 3D plot in express.

Capture.JPG.

Cheers

Terry

DaveMartin
16-Pearl
(To:DJF)

DJF,

In the blog write-up, how should I refer to you in the blog write-up?

Thanks, Dave

 

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com

This is one I couldn't let go.  This question is easily solved using polar coordinates:

AndrewK_0-1709936677308.png

 

Calculation speed can be increased using the ellipse's symmetry by limiting the integral between 0 and pi/2 and multiply that value by 4.  With a similar technique surface, moment's of inertia and so on are easily calculated.

 

All this is pretty basic, really.  The fun starts when adding parameters to the equation, for instance making Superellipses and even (far) beyond.  In the following article I made use of Mathcad Prime to illustrate the above: https://www.athena-publishing.com/series/atmps/issbg-22/articles/260/view 

 

Enjoy!

Bert Beirinckx

@Bert

EllipseForm.png

Alan

 

Werner_E
25-Diamond I
(To:AlanStevens)


@AlanStevens wrote:

@Bert

EllipseForm.png

Alan

 


 

"... which is exactly  ..."

exactly??  The perimeter of a circle can't be calculated exactly, either. 😞

The closed form for the circle circumference is actually a deception, because the infinite series is hidden in the constant pi, which we have invented for the  ratio perimeter: (2*r) for the ellipse with a=b=r.

So there is no real difference between the perimeter of a general ellipse and a circle other than the pi approximation key being available on every pocket calculator...

 

I, too, enjoyed Bert's article.

BTW, I nice video on the topic of the ellipse perimeter can be seen here on Stand-Up Maths channel: Why is there no equation for the perimeter of an ellipse

I would say the circumference of a circle is exactly 2pi.r.  Of course, that doesn't mean we can get an exact numerical value.  However, let's not worry too much about semantic interpretations!  Perhaps it would have been better if I'd replaced the words "is exactly"  by  " has the closed form solution".

 

I also liked the Stand-Up Maths video.

 

Alan

 

Werner_E
25-Diamond I
(To:AlanStevens)


@AlanStevens wrote:

I would say the circumference of a circle is exactly 2pi.r.  Of course, that doesn't mean we can get an exact numerical value.  However, let's not worry too much about semantic interpretations!  Perhaps it would have been better if I'd replaced the words "is exactly"  by  " has the closed form solution".

 

I also liked the Stand-Up Maths video.

 

Alan

 


What I meant was that 2* pi *r just looks like a closed form solution but it actually isn't. It just looks that way because we decided to choose a simple looking greek letter as an abbreviation for he involved infinite sum. That way I guess we could also derive "closed form" solutions for the perimeter of an ellipse with any other ratio a:b as well.

Val Ochkov has asked me to post this picture:

pir.png

 

Val says he has just been banned from the forum!  I hope that's just a temporary glitch as he's been a great supporter of Mathcad and the forum for years.

 

Alan

Werner_E
25-Diamond I
(To:AlanStevens)

I guess and hope that it is not a ban on the part of the forum operators, but on the part of the Russian rulers.
If this is true, the use of a VPN could probably help Val  to overcome this hurdle.
If not, maybe the admins can say something about it -> @admin , @AndrewK , @Jaime_Lee 



@Werner_E wrote:

I guess and hope that it is not a ban on the part of the forum operators, but on the part of the Russian rulers.
If this is true, the use of a VPN could probably help Val  to overcome this hurdle.
If not, maybe the admins can say something about it -> @admin , @AndrewK , @Jaime_Lee 


Seems to have been a forum ban (Val sent me the following image):

ban.png

 

I assume this was an automated system response.  Perhaps an actual PTC person could overturn it?

 

Alan

AndrewK
Community Manager
(To:AlanStevens)

Thanks for the additional information @AlanStevens we are looking into this now.

A friendly hello to the puzzle group,
I only have MC14 available. Therefore I couldn't read all the posts. Therefore, here is just a reference to a very good approximation function for the circumference of the ellipse - it is Ramanujan's formula. I don't know of a better one from the literature. I have attached them, but I cannot provide the desired 3D plot. I'm asking for help here.

Hi,

Your function uu() has two parameters so producing a 3D plot is simple with the matrix() function.

Cheers

Terry

I use MC14 and therefore cannot load your file. Would you please reply in the appropriate file format? I haven't created a 3D plot yet. They were mostly vicious equations that I solved using equivalent minimum problems.

Hi

Mathcad 12 version you should be able to open in Mathcad 14

Thank you. It helped 🙂


@terryhendicott wrote:

Hi,

Your function uu() has two parameters so producing a 3D plot is simple with the matrix() function.

Cheers

Terry


I would strongly vote against using the matrix function to do the job! It may give a decent result here but you can't use it to provide intermediate values (non-integer arguments) or negative arguments.

The way to go sure is using the CreateMesh() function which is exactly made for this.

We could also plot the function uu directly but in this case it will fail because the function uu is written in a way so that it yields non-real values for negative input arguments. And because the standard interval for a 3D Quickplot is from -5 to 5, this makes the whole plot fail.
But if we force the argument values to be in the desired range, plotting uu directly works as well.
Nonetheless I sure would vote for using CreateMesh when creating 3D plots of functions in 2 variables.

Werner_E_0-1710092992830.png

BTW, here is a different approach allowing for negative input values, too (even they may not make much sense in the context of the ellipses here)

Werner_E_2-1710093881644.png

 

Can someone derive an equation for the contour lines ??

 

 

 

@AlfredFlaßhaar 

You should be able to open the xmcd file posted by Terry initially with your MC14!?

Yes, now it runs. Do not know why it was in opposite before. At my age, nothing surprises me anymore 😉 .

ppal
17-Peridot
(To:DaveMartin)

 

Using Ramanujan Approximation.

ppal_1-1710113706155.png

ppal_0-1710116304140.png

From MARK B. VILLARINO

 

192_05.pdf (emis.de)

 

Nice plots already done - see    AlanStevens 

Reminder to everyone that we're now in the last one-third of March! Great participation thus far.

I manage the Creo and PTC Mathcad YouTube channels for PTC, as well as all PTC Mathcad marketing in general.

This isn't really a response to the challenge.  The attached sheet poses a slightly different question.

 

I started out to answer the challenge; seemed easy to simply integrate along the curve and easier to do it in polar coordinates:

Fred_Kohlhepp_0-1711713089477.png

For the ellipse posed in the challenge:  

Fred_Kohlhepp_1-1711713164090.png

Then I looked at the other responses. . . .

And went back and checked, did the same thing in cartesian coordinates:

Fred_Kohlhepp_2-1711713385372.png

So why won't polar work?

 

 

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