setting I_{1} = I_{3} gives h = r/2

I find this an interesting formula, which gives an immediate idea of the proportions of a top with the two moments of inertia equal.

I deduce from it that h is 0.707 times the radius of gyration in tops with I_{1} = I_{3}.

True, but keep in mind that h = r/2 assumes the mass of the top is in the shape of a thin disk. Your Nr 30 top is more of a torus shape (beautiful top by the way) Using your mass of 0.119 Kg and radius of 0.02875 m and assuming a thin disk:

I

_{3} = 1/2(.119)(.02875)

^{2} = 4.9x10

^{-5} Kg m

^{2}you give I

_{3} = 6.7x10

^{-5 }Kg m

^{2}which makes since because more mass is on the outside.

By the way, how did you get this value of I

_{3}?

Using your value of 140 rpm for the toppling down speed and using the critical speed formula I get

I

_{1} = 3.1x10

^{-5} Kg m

^{2} for your Nr 30 top.

Not sure how accurate this is but it's probably in the ball park.

Edit:

For a symmetric mass distribution, its impossible to have I

_{1} < I

_{3}/2. If h > 0 and or thickness along the symmetry axis then I

_{1} > I

_{3}/2. So this calculation is not accurate.

Did you measure or calulate I

_{1} for your Nr 30 top?