By the way, the torque is equal to the friction force times its lever (distance to the axis.)

Interesting. I have found this formula for torque:

Torque (newton meters) = moment of inertia (kg m^2) x angular acceleration (radians/sec^2)

The moment of inertia of the wooden top of this test is 0.000255 kg m^2.

The RPMs lost around 520 RPMs in one minute are 160.

160 RPMs are 16.75 radians/sec , so the average angular acceleration during that minute was 16.75 radians/sec^2.

The torque for air and tip friction on this top at 520 RPMs should be:

0.000255 x 16.75 = 0.0043 newton meters (rounded off)

Since one newton is 102 grams, 0.0043 newtons are 0.44 grams (rounded off).

So the total drag on this top at 520 RPMs could be thought as a force of 0.44 grams at one meter from the axis.

We have seen that about 3/4 of it, is air drag, and about 1/4 tip friction.

3/4 of 0.44 grams are 0.33 grams;

this torque acts at various distances from the axis, so it can't be calculated for a single distance from the axis, but, just to have an idea of the intensity of the torque, the radius of this top is 62 mm, and, at this distance from the axis the torque for air drag is 5.6 grams.

1/4 of 0.44 grams are 0.11 grams;

The diameter of the contact point seems to be one tenth of millimeter and not larger than so.

At 1/20 mm from the axis, 0.11 grams become 2200 grams, (more than two kg !).

I wonder if this data is correct... the top weighs 113 grams... is it possible to have more than two kilograms of friction force on its tip ? Certainly the tip is relatively sharp and the weight of the top is concentrated on that tiny contact point with a pressure of about 10 kg/square millimeter (113 grams on 1/10 millimeter diameter contact point)..

in fact it is no surprise if there is wear of the contact points. I don't know what to think about this.