Calculating Measurement Uncertainty for Conical Pendulum Experiment

[Vegeta]

The following question is about an experiment with the conical pendulum.
I have measured the length l = 43\,cm the radius r = 10\,cm, I have 3 measurement of the period with the same radius, where the measurements are

T= [1.288, 1.285, 1.301] s

The uncertainties of the measurements are \Delta l = \pm 1\,cm, \Delta r = \pm 1\,cm and \Delta T = \pm 0.02\,s.
I want to calculate the uncertainty in the measurement og g, when

g_i=4\pi^2\frac{\sqrt{l^2-r^2}}{T_i^2}

Can I calculate the uncertainty \Delta g by

\Delta g = 4\pi^2\frac{\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}}{(T_i-\Delta T)^2} - g_i

Where the expression 4\pi^2\left(\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}\right)/(T_i-\Delta T)^2 is the worst case scenario of the measuring g. Is that correct?
If that is how I can calculate the uncertainty in g, is the relative uncertainty then

\frac{\Delta g}{g_i} \qquad \mathrm{or} \qquad \frac{\Delta g}{\overline{g}}

Where \overline{g} is the mean value. Which one is the correct one? The first expression has a relative uncertainty for each measurement.

 

[HallsofIvy]

Yes, calculating the value for the largest possible and then smallest possible values of l, r, and T gives the possible error for the function.

The relative uncertainty is \frac{\Delta g}{g}.

 

[Vegeta]

Well in the equation

\Delta g_i = 4\pi^2\frac{\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}}{(T_i-\Delta T)^2} - g_i

There is acctually an uncertainty of g, for every measurement \Delta g_i. Should I say

\frac{\Delta g_i}{g_i} \qquad \mathrm{or} \qquad \frac{mean(\Delta g_i)}{mean(g_i)}?

Where in the last equation I only have one value for the relative error. I am not quite certain of what to choose?