Calculating Major & Minor Radius of an Ellipse

**AdamW** · October 9th 2011, 04:20 AM

Afternoon all,

Firstly apologies if this is posted in the wrong section. I believe I have but if not please feel free to move.

As the title suggests I am trying to calculate the major and minor radius of an ellipse.

See the attached image for estimated measurements. Note the image is not to scale - it is rough sketch.

The shaded section is a solid part that I have measured from. I believe the curve to be an ellipse, hence calculating the 2 radius.

As can be seen from the image, the curve is not a complete quarter of an ellipse. The curve ends at an approximate angle of 40degrees, where the part then continues vertically.

I have measured the distance from the start of the curve to the 40degree point as 15mm.

The distance between the horizontal surface at the top and the point at which the curve ends (at 40degrees) is measured at 5mm.

I am trying to model this part in CAD software hence require the major radius (a in image) and minor radius (b in image).

I will state that maths is not my strong point, but I have tried to find a solution online, with no luck.

So, firstly is is possible to calculate the major & minor radius from the dimensions given.

If so, how?!

If more information is required, and its possible for me to measure, I can do, just let me know.

Many thanks in advance and appreciate any help that can be given.

**Opalg** · October 9th 2011, 08:35 AM

This problem has quite a nice solution. The ellipse has equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , and the slope of the tangent at the point (x,y) is given by $\frac{dy}{dx} = -\frac{xb^2}{ya^2}.$ You want the point $(15, b-5)$ to lie on the ellipse, and for the slope there to be $\tan(-40^\circ)\approx -0.839.$

I prefer doing algebra to arithmetic, so I'll write $t = \tan(40^\circ)$ and then substitute in the numerical value when we get to the end of the calculation.

The condition for the point $(15, b-5)$ to lie on the ellipse is $\frac{225}{a^2} + \frac{(b-5)^2}{b^2} = 1.$

The condition for the slope there to be –t is $\frac{15b^2}{(b-5)a^2} = t.$

Solve the second of those equations for a, $a^2 = \frac{15b^2}{t(b-5)}$ , and substitute that into the first equation:

$\frac{225t(b-5)}{15b^2} + \frac{(b-5)^2}{b^2} = 1.$

After a bit of simplfying and cancellation, that boils down to $\boxed{b = \frac{15t-5}{3t-2}}$ .

Now you can plug in the value of t, to get $b\approx 14.67$ . Then substitute that into the equation for $a^2$ to find that $a\approx 19.44.$

**AdamW** · October 9th 2011, 10:41 AM

Hi Opalg.

Thank you very much for the prompt response.

I have created the ellipse using the dimensions calculated and it seems to fit just fine!

Once again thanks for the help, much appreciated.

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