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 ?
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.