Increasing accuracy of solution of transcendental equation

℡╲_俬逩灬. 提交于 2019-11-26 11:24:51

If I understand this correctly, you're trying to infer (but not measure) the radius r0 of the tube from measurements for y and a.

Applying the usual error propagation to your formula for r0, one obtains (an estimate for) the error of the resulting r0. In the limit of small angles (applicable here, since a(t) is limited to 20 degrees), this gives roughly (using the small-angle approximation for the trigonometic functions)

dr0^2 ~= dy^2 + z0^2 (pi*da/180)^2

Thus, in the case of r0 much smaller than z0, the relative error on r0 is always much larger than the relative errors of y and z0*sin(a). This is already clear from your graph: the measured quantities depend only weakly on r0.

In other words, this is not a clever way to determine the radius r0. There is not much you can do about this fundamental limitation (except you can increase the range of angle a). Making many measurements (the usual method to beat down noise/errors) presumably won't help, because these measurements aren't independent of each other due to the internal workings of your machine. So, the only help would be more accurate measurements.

To analyse the situation, I recommend to make plots/figures of, say, the inferred r0 as function of y or of y as function of a for fixed r0.

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