Talk:Ellipse

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Mistake concerning figure related with polar form of ellipse?

hi,

concerning formula : $r(\theta )={\frac {a(1-e^{2})}{1+e\cos \theta }}$ I think that the figure related on the right is wrong, $\theta$ comes from perihelion and not aphelion. r is minimum when $\theta =2\pi$ or =0 and maximum when $\theta =\pi$. Here, r is minimum when $\theta =\pi$. — Preceding unsigned comment added by 84.99.70.139 (talk) 17:04, 1 June 2012 (UTC)

Hi,

I don't know if this is actually your problem, but when I tried to use this formula, I thought that 'e' referred to the constant e, when it actually refers to eccentricity. Once I corrected that, the formula started producing ellipses instead of hyperbolas. Also, it would be worth mentioning that vertical ellipses should swap sine for cosine.130.85.56.84 (talk) 05:36, 28 October 2012 (UTC)

General polar form

The section General polar form gives without citation a general formula. See this math reference desk discussion of 1 Oct. 2012 about whether the formula could be correct. In particular, it seems to me that the radical should have a plus minus sign before it. Otherwise, when the origin is outside the ellipse and the angle theta is such that the ray cuts through the ellipse twice (entering and exiting), the formula fails to give two solutions. Duoduoduo (talk) 15:16, 1 October 2012 (UTC)

Reason for reverting edit

I'm reverting the formula added today for the circumference, because the only citation given is http://ellipse-circumference3.blogspot.com/ , which is not a WP:reliable source according to Wikipedia standards. And that blog itself gives no other reference. Wikipedia requires that the formula be sourced to a refereed publication. Duoduoduo (talk) 20:53, 2 October 2012 (UTC) Actually MrOllie beat me to it! Duoduoduo (talk) 20:55, 2 October 2012 (UTC)

A formula of such a great breakthrough does not require a reference other than the formula as standing on its own. Show me one simplistic, one step formula capable of producing 8 significant digits, without the use of the Hoelder mean, and without using pi. Why reference a site that is not used as a blog as a blog? Was there any blogging on it? No! So why was the word "blog" mentioned"? If a person had a site named "mountainspot" would it really trip a person up if it wasn't about mountains? Get past the needless block and inspect the formula to see the tremendous contribution. If a person suddenly came up with an explicit formula, I suppose there will be haters of the contribution for wanting some encyclopedia to first publish it. Amazing. Please ask a mathematician about this - one who is versed in it, such as Michon or Cantrell. This amazing formula needs to be out in the open to be shared throughout the world to simulate investigation of the form used. Note that the error function listed on the site verifies the quality.Numbertruth (talk) 03:45, 3 October 2012 (UTC)

No, even a formula for a 'great breakthrough' requires a source that meets wikipedia's guidelines. This reference is a self published site, so it does not qualify. Since this is a very important formula, no doubt the mathematics community will notice it quite rapidly and write about it in a few peer reviewed journals, we can afford to wait until then. - MrOllie (talk) 10:57, 3 October 2012 (UTC)

Well, so it is, but some things just won't make it into journals when the one coming up with the material is not part of the status quo.

Look at these three new forms I derived using the artithmetic-geometric mean:

http://mathforum.org/kb/thread.jspa?threadID=2422080&messageID=7942499#7942499 C = 4*(pi*(e-1)*e*(e+1)*(d/dx)[agm(1,(1-x=e)/(1+x=e))]-pi*(e-1)*agm(1,(1-e)/(1+e)))/(2*(e+1)*agm(1,(1-e)/(1+e))^2)

http://mathforum.org/kb/thread.jspa?threadID=2420827&messageID=7937451#7937451 C=8e^2(1-e^2)*d/dk(K(k=e^2))+2(1-e^2)*pi/(AGM(1,sqrt(1-e^2)))

http://mathforum.org/kb/thread.jspa?threadID=2428401&messageID=8048879#8048879 C = 4*E((1-x)^2/(1+x)^2)= (2pi)*((((d/dx)[AGM(x,x^2)]|x=Q)*(1-x)*x^2)/agm(x,x^2)^2 + (2x^2-x)/agm(x,x^2))

Where Q = (1-e)/(1+e) = (1-sqrt(1-b^2))/(1-sqrt(1-b^2)); e= eccentricity, and b = length of minor axis.

These formulas make it easy for anyone to compute the ellipse circumference since the agm is simplistic and converges rapidly, thus so little effort would be required. The derivatives of the agm function require a good numeric processor, but other than that, these add to the theory on ellipses. As can be noticed, since I am not of the group that wants to reward those with credentials, progress will be slow. I saw this even when I was inventing in the sciences - it simply produced jealousies and then I get shut out. I hope this changes as humans need to look beyond certification to assess quality.

I will wait, even if I am long gone as this is what may happen.

I have noticed errors and some parts needing clarification regarding elliptical integrals on wiki but I'll leave it to the experts to spot.  :) You may delete the earlier discussion regarding the pade variant or if you give me permission, I will do it. I don't want to do anything here I am not suppose to do. 184.100.17.31 (talk) 07:44, 22 December 2012 (UTC)Numbertruth Numbertruth (talk) 00:27, 11 January 2013 (UTC) A formula I just came up with today is simplistic, having a maximum relative error an entire magnitude lower than Ramanujan's second formula: C~a*[((1-sqrt(2)/2) + (sqrt(2)/2) * (b/a)^(-0.454))^(2-2pi)](2pi-4)-4 Maximum absolute relative error: ~3.8E-5 (~3.4E-5 with -0.454012 as inner power) Note: b<>0

The above could be written as the following to see the direct decomposition into the elliptical integral of the second kind (C=4aE(1-x^2)): C~4a[1+(pi/2-1)/(((1-sqrt(2)/2)+(sqrt(2)/2)*(b/a)^(-0.454))^(2pi-2))]

This certainly is noteworthy to be inserted on the main page. Waiting around for this being shown in a popular publication shouldn't be the major point in preventing the world from seeing this. Numbertruth (talk) 05:46, 18 January 2013 (UTC)Numbertruth (talk) 19:35, 19 January 2013 (UTC)

I just edited the circumference section adding Bessel's series (which converges much more rapidly than the series in e). I also removed an elaborate series which is unlikely to be of much practical use. For people who need to compute the circumference accurately for very eccentric ellipses, I included the necessary AGM method (following the paper by Carlson). cffk (talk) 18:25, 10 April 2013 (UTC)

Ellipse as Conic Section

Several pages on Wikipedia including this page and the page Conic Sections claim that ellipses are one type of conic section. My idea of such a conic section would result in a curve like an ellipse but with one end with a smaller radius, or 'pointier' than the other. This is backed up by an edit in this talk pages archives which suggests that an ellipse being a conic section is fallacious and the true conic section would be an egg-shape. If this is true, could this be clarified? Otherwise, am I (and the commenter in the archives) just wrong? 124.177.190.63 (talk) 12:57, 19 November 2012 (UTC)

We could sit down and figure out how wide an actual conic section is at chosen points along its length. If we did, we'd find it was symmetrical, like the ellipse it is. Tim Zukas (talk) 19:50, 19 November 2012 (UTC)

Turns out if we choose an example conic section it's easy to show it's symmetricral.

On graph paper draw the line y = x and the line y = -x; the two lines above the x-axis are a cross-section of a cone with its vertex at the origin and its axis along the y-axis.

Draw a line from (-1,1) to (2,2). That's the cross-section of the plane that's perpendicular to the paper and sections the cone. Equation of that line is y = (x + 4)/3.

Call the width of the conic section w (measured perpendicular to the paper). Then (w/2)² = y² - x²  ; substitute (x +4)/3 for y and we find that (as a function of x) w² is an upside-down parabola, symmetrical about x = 1/2 as it should be. Tim Zukas (talk) 02:46, 21 November 2012 (UTC)

Being Ben BernankGrinch of the Federal Reserve stole yet another Christmas from me, I had time to improve upon the Blankenhorn-Ramanujan ellipse circumference formula. This formula provides exact end-points and with a maximum relative error of several magnitudes better than the Ramanujan formula. Some day, the Wiki editors will recognize the improvement, but I am not holding my breath for it.

C~pi(a+b)*(1+3h/(10+sqrt(4-3h)) +(1.5*h^6-0.5*h^12)/((11pi/(44-14pi))+24100(1-h)))

Where a and b are the half length axes and h = (a-b)^2/(a+b)^2

Referencing http://i39.photobucket.com/albums/e191/toomers/ell1.jpg, replace the h^5.20114 with 1.5*h^6-0.5*h^12, and the 19176 with 24100.

See: http://ellipse-circumference.blogspot.com

BEING THIS IS FACTUAL, as anyone can check out and I can provide a spreadsheet to anyone interested this should be able to be showing on wiki's site for people to have easy access to. When there is no doubt about the veracity, there should not be power mongers preventing publication. Numbertruth (talk) 00:41, 26 December 2012 (UTC)

It would be nice for the general public to see an equivalent form of the Ramanujan II formula without the Hoelder mean:

C~pi(a+b)[1+((x-1)^2(10(x+1)-sqrt(x^2+14x+1)))/((x+1)(33x^2+62x+33))], where x=b/a

Anyone here should be able to verify this. If the power structure here still doesn't want this shown to the public, well, I made my appeal.Numbertruth (talk) 08:41, 9 January 2013 (UTC)

an approximation to an ellipse with a circle

I simply didn't understand what exactly was meant here. Is the approximate ellipse which is really a circle is to be drawn with 2 compasses or with one? Perhaps a picture, or more precise details, or a reference would resolve the lack of understanding. — Preceding unsigned comment added by 31.44.140.246 (talk) 19:29, 19 December 2012 (UTC)

Numerical and linear eccentricity

Unfortunately, nowadays there is a confusion in notation. Traditionally the linear eccentricity should be denoted with the latin e, and the numerical eccentricity should be denoted with the greek ε, where e=εa. Even on this page there is an obvious confusion in notation.Theodore Yoda (talk) 15:29, 25 March 2013 (UTC)

Polar angle

Parametric form in canonical position, missing angles in drawing

I agree with the archived comment requesting labels on the diagramme http://en.wikipedia.org/wiki/File:Parametric_ellipse.gif. —DIV (138.194.10.62 (talk) 06:54, 12 May 2013 (UTC))

Accuracy of formula

Mathworld seems to have a different formula as their #58 compared to Ellipse#Parametric_form_in_canonical_position. I think WP is correct, but am puzzled at the discrepancy. —DIV (138.194.10.62 (talk) 06:55, 12 May 2013 (UTC))