I'll put here what I did to try to estimate the order of magnitude of the air drag on the top. Feel free to skip this post, it is not the usual fare of the forum (Nerdy alert).
What I did was based completely on the book
Rotating Flow by Peter Childs, in particular on section 6.2, Rotating Cylinder Flow, which is available online
here. Now, a spintop is a cone not a cylinder. I assumed I could treat it as a pile of short cylinders, although I am sure this is not completely true. I bet a rotating cone, contrary to a cylinder, produces some axial flow, probably from tip to base due to centrifugal force. Also, I will only consider the top between a radius of 30 cm and 1.8 m as the narrower tip will not contribute much drag. Hopefully these assumptions won't throw off the results too much.
According to the book, the power lost to drag of a cylinder that is only rotating is:
P = 1/2 pi d w
3 R
4 L C
mcwhere d is the density of the fluid (in our case 1.2 kg/m
3 for air), w is the rotation rate, R the radius, L the length of the cylinder and C
mc is a coefficient that is measured experimentally.
Note that the power lost depends on the third power of the rotation rate and on the fourth power of the radius. It is very difficult to get a large top to spin fast!
The book gives you C
mc as a function of the Reynolds number, Re
r, based on extensive experiments (from the 1940's).
The Reynolds number determines, for example, if the flow is laminar or turbulent. The equation to calculate the rotational Reynolds number is:
Re
r = d w R
2/u
where u is the dynamic viscosity of the fluid (0.000018 kg/m s for air).
For our peak w = 195 RPM it gives 4,370,000 at 1.8 m and 119,000 at 0.3 m. Above Re
r of a few thousand the flow is completely turbulent and below about 60 completely laminar. So we are very much in turbulent regime. By the way, for a standard top, 2 inch in diameter spinning at 3000 RPM the Reynolds number (at the rim) using the same formula is 13,000, so also turbulent or perhaps mixed.
The C
mc only varies from 0.0069 to 0.014 between those values, so we'll take the average of 0.01.
We can divide the cone into many little cylinders or just use the usual integration trick to convert the equation to that of a cone of height H. The R
4 term produces an extra 1/5 factor:
P = 1/10 pi d w
3 R
4 H C
mcI am not sure if Joe included the bar extensions in the 3.88 m height of the top, but I'll assume H to be 3 m. We now have all the data to calculate the power dissipation.
P = 1/10 * 3.1416 * 1.2 kg/m
3 * (20 1/s)
3 * (1.8 m)
4 * 3 m * 0.01 = 950 W
or about 1 kW!
Considering that 500 W over 30 seconds is an energy loss of 15,000 Joules and the top spinning at 193 RPM has an energy of 57,800 Joules, this is pretty significant. This will slow down the top fast until the w
3 term decreases enough. If the rotation rate decreases to half, the power loss decreases by 8 (except of a small change in C
mc).