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Author Topic: Re-thinking spin dynamics for world's largest top  (Read 11793 times)

ta0

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Re: Re-thinking spin dynamics for world's largest top
« Reply #30 on: September 16, 2014, 11:00:30 AM »

The equation that gives the conversion efficiency for the yo-yo can be modified for our case. The result is:

e = 1 /[1+ r2 M/I]

where M is the mass of the loaded cart, I the moment of inertia of the top and r the radius of the spool axle.

Notice that as the weight increases, the efficiency decreases. The amount of energy going into rotation still grows but there is a diminishing gain. For a very large weight, the rotation is limited by the free-fall velocity from that height converted to spin velocity by r.

Unless a new top is made, I and r cannot be changed. But there is a way around this using a pulley. I'll post this on the main thread about the testing.

===============
Derivation:

The potential converts to kinetic and rotational energy:  EP = EK + ER
EK = 1/2 M v2 where v is the velocity
ER = 1/2 I w2 where w is the rotation rate (in radians/second)

The efficiency e would be:
e = ER / Ep = ER / (ER + EK) = 1/(1 + EK/ER) = 1/[1 + (v/w)2 (M / I)]

Equating the rope taken by the rail car to the rope given by the spool: v = r w
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jim in paris

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Re: Re-thinking spin dynamics for world's largest top
« Reply #31 on: September 16, 2014, 01:08:40 PM »

"Greek to me ", said the Manager///
"math and poetry have that in common:"
They talk about reality
In a funny manner


good calculations,partner 8)

jim
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ta0

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Re: Re-thinking spin dynamics for world's largest top
« Reply #32 on: September 20, 2014, 12:26:21 AM »

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 w3 R4 L Cmc

where d is the density of the fluid (in our case 1.2 kg/m3 for air), w is the rotation rate, R the radius, L the length of the cylinder and Cmc 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 Cmc as a function of the Reynolds number, Rer, 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:

Rer = d w R2/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 Rer 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 Cmc 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 R4 term produces an extra 1/5 factor:

P = 1/10 pi d w3 R4 H Cmc

I 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/m3 * (20 1/s)3 * (1.8 m)4 * 3 m * 0.01 = 950 W

or about 1 kW!  :o

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 w3 term decreases enough. If the rotation rate decreases to half, the power loss decreases by 8 (except of a small change in Cmc).
« Last Edit: July 21, 2016, 11:21:11 AM by ta0 »
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the Earl of Whirl

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Re: Re-thinking spin dynamics for world's largest top
« Reply #33 on: September 20, 2014, 08:44:34 AM »

Exactly what I was thinking!
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Jeremy McCreary

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Re: Re-thinking spin dynamics for world's largest top
« Reply #34 on: March 23, 2016, 03:25:08 PM »

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