5
$\begingroup$

Ellipticity seems to have many definitions. So far, I'm aware of three. Are there other definitions of ellipticity that you know of, and if so, where did you encounter them?

The three definitions that I'm listing all have that $a \geq b$.

Definition 1: $$ \varepsilon = \frac{a-b}{a} = 1 - \frac{b}{a}. $$ This definition comes from my "Dictionary of Physics and Mathematics" from McGraw-Hill.

Definition 2: $$ \varepsilon = \sqrt{\frac{a^2-b^2}{a^2}} = \sqrt{1 - \frac{b^2}{a^2}}. $$ This definition comes from http://mathworld.wolfram.com/Ellipticity.html. As a side note, the Wolfram page on ellipticity also defines something called flattening that is equivalent to my definition 1.

Definition 3: $$ \varepsilon = \frac{a^2-b^2}{a^2+b^2}. $$

I've done a light search for information on definition three, but I have not turned up anything in the literature.

$\endgroup$
5
$\begingroup$

There are several definitions of ellipticity (also called eccentricity), though they aren't different things quite so much as different measurements of divergence from perfect circularity:

  • Your first definition is the first (or "primary") flattening f, which is equal to the versine of the angle [ref. 1];
  • Your second definition is the first (or "primary") eccentricity ε, which is equal to the sine of the angle [ref. 1];
  • Your third definition is the third eccentricity ε'', the trigonometric identity of which is ambiguous [ref. 1].

Within the context above, there are two other measures of flattening and one other of eccentricity [ref. 1]:

  • The second flattening, f ' = (a-b)/b, is the exsecant of the angle;
  • The third flattening, f '' = (a-b)/(a+b), is the havertan (half of the versed tangent) of the angle;
  • The second eccentricity, ε' = (a2-b2)/b2, is the tangent of the angle.

There exists at least one extension of these two sets (a rather hard-to-find item, it seems), a fourth degree of flattening and of eccentricity, respectively [ref. 2, pp. 2-3]:

  • This fourth flattening, μ0 = (a2-b2)/b2,
    • note that this would be f ''', in the established Western convention;
  • This fourth eccentricity, e0 = $\surd$((a2-b2)/ab),
    • note that this would be ε''', in the established Western convention.

The first 3 flattenings follow a general pattern of 2sin2(θ/2)/[denom.], wherein the denominator in each case is {1, 1-numer., 2-numer.}; I haven't yet checked to see if the 4th flattening follows suit with {3-numer.}. The first 3 eccentricities follow a rather similar pattern of sin2(θ)/[denom.], wherein the denominator in each case is also {1, 1-numer., 2-numer.}. I have not yet checked to see if the denominators continue in like manner, with {4-numer., ..., n-numer.,...}.

You might also note that the first four denominators are {a, b, (a+b)*0.5, ab0.5}. Whether this pattern continues to a fifth degree via tetration of 0.5, much less higher orders of hyperoperation (of 0.5, I expect), I don't yet know.

If I find data regarding the denominators' pattern, and/or that of the increasing hyperoperation, then I will return with the update(s).


References:

  1. http://wiki.gis.com/wiki/index.php/Angular_eccentricity
  2. https://zenodo.org/record/32854/files/ganshin69.pdf
    • "Geometry of the Earth ellipsoid", Vladimir Nikolaevich Gan'shin, Moscow, 1967
$\endgroup$
  • $\begingroup$ Don't forget the definition of ellipticity commonly used in the laser industry; $e=a/b$ $\endgroup$ – Jim Mar 12 at 13:51
  • $\begingroup$ @Jim , it's been a while since I last dug into this; are you referring to circularity (diametric ratio, aspect ratio, ratio of the conjugate to transverse radii, angular eccentricity cosine): wiki.gis.com/wiki/index.php/Angular_eccentricity#Circularity ? $\endgroup$ – Charles Rockafellor Mar 17 at 14:55
  • 1
    $\begingroup$ Seems to be the same thing. No idea if that's where everyone got it. I just assumed they took ellipticity and tried to make it as simple as possible. Either way, in laser optics, that's how they define ellipticity. I don't make the standards, I just work here $\endgroup$ – Jim Mar 18 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.