# How to know the flattening factor for a ellipse?

I want to know how can I get the flattening factor for a ellipse by knowing its semi-major and semi-minor axes ?

Actually I tried this formula:

$f=\left(\frac{a}{b}-1\right)$

While $f$ is the flattening factor, $a is the semi\,major\,axes$, $b is the semi\,minor\,axes$. I think it's true because when I try it with circle it gives me $\left(\,f=0.0\,0\right)$. but I don't have any source for this formula and I'm not sure if it's true !

So if any one know what is the exact formula for finding the flattening factor for a ellipse ?

• Wikipedia gives the flattening factor as $1-\frac ba$, which would be zero for a circle. That shouldn't be surprising, as a circle isn't flattened. Jul 3, 2013 at 21:59

The flattening factor is given by $\;f = 1 - \cfrac ba$.

A closely related term you might be interested in is the eccentricity of an ellipse, usually denoted $e$ or $\varepsilon$. Eccentricity in general represents ratio of the distance between the two foci, $2h$, to the length of the major axis, $2a$: $$e = \dfrac{2h}{2a} = \dfrac ha$$

where the distance between a focus and the center is given by $\;h = \sqrt{a^2-b^2}.$

In fact, we can represent eccentricity in terms of $a, b$: $$e=\varepsilon=\sqrt{\frac{a^2-b^2}{a^2}} =\sqrt{1-\left(\frac{b}{a}\right)^2} = \frac ha$$

For an ellipse, the eccentricity is $0 < e < 1$. Eccentricity is zero when the foci coincide with the center point, i.e, the figure is a circle. As eccentricity increases, the shape gets more elongated (stretched/flattened): the closer to $1$ it gets, the flatter the ellipse.

• The OP is using $f=1-\frac ba$ as the flattening, and you are using $f=ae$. This might be a bit confusing.
– robjohn
Jul 3, 2013 at 22:13
• Thanks a lot :) Jul 3, 2013 at 22:32
• You're welcome, Mohammad! Jul 3, 2013 at 22:34
• @amWhy: Due to grammar and formatting ugliness, I edited my post today about six times. I hate doing it, but I'd rather it be correct than deal with the ambiguities of how bumping works. Jul 4, 2013 at 0:48
• Hello, @Babak! $\large \overset{\small\star\star}{\smile}$ Jul 6, 2013 at 14:25