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[Aerobie]

Hello Iacopo,

I'm getting excited about the potential of "unweighting" a top with a magnet.  With careful positioning of the vertical gap, you should be able to reduce the "weight" on the tip to about 5% of actual weight.

If you try it, I suggest that you make a cluster of small steel parts at the top of the stem, rather than a single piece of steel.  that's to minimize eddy currents, which would introduce drag.  A bundle of insulated (varnished) steel wires would work.  You could also mount the magnet on a ball bearing.

At first, I worried about the weight of the steel at the top of the spindle raising the Cg.  But upon further thought, I don't think it's a worry because the magnet will be pulling most of the weight upward.

I'll bet you can run this top for days!

Alan

[Iacopo]

I'm getting excited about the potential of "unweighting" a top with a magnet.

Yes, you can eliminate great part of the tip friction with magnets.
I made this experiment, long time ago:



This also is interesting, it is not a spinning top but it gives an idea of how much long can spin a flywheel with almost no tip friction:

https://www.youtube.com/watch?v=_STnL0U9PyQ&list=UUD-YcF3YmwU9sxcwK4T6Byw&index=79

[Aerobie]

Hi Iacopo,

As we both know, aero drag is the major limitation on spin time.  You've conquered that with vacuum, leaving tip friction.  So the magnet combined with vacuum should be great. 

I repeat my suggestion to use a bundle of insulated steel wires, instead of sold metal, to minimize eddy currents - which consume energy.  In your video, the eddy currents would probably circulate in both the rotating and the stationary magnets.  If you use a stationary magnet, and a bundle of varnished soft steel wires at the top of the stem, that should prevent eddy currents.  But it might be necessary to use multiple small diameter stationary magnets, instead of one large one.

Alan

[Iacopo]

As we both know, aero drag is the major limitation on spin time.  You've conquered that with vacuum, leaving tip friction.  So the magnet combined with vacuum should be great. 

I agree. Maybe one day I will experiment with it.

I repeat my suggestion to use a bundle of insulated steel wires, instead of sold metal, to minimize eddy currents - which consume energy.  In your video, the eddy currents would probably circulate in both the rotating and the stationary magnets.  If you use a stationary magnet, and a bundle of varnished soft steel wires at the top of the stem, that should prevent eddy currents.  But it might be necessary to use multiple small diameter stationary magnets, instead of one large one.

I am wondering if there are magnets which are not electric conductors, for to avoid eddy currents in them, they should be the most suitable for this project.

Anyway, if magnets are centered in the rotation axis, eddy currents seem not to form.
See this experiment I made recently, and the comment of Ta0:

http://www.ta0.com/forum/index.php/topic,5248.msg56040.html#msg56040

[Jeremy McCreary]

It appears to me, that diameter is nearly irrelevant at higher speeds in air...
...when I compared two 48 gram tops, both on 3/8" balls, with diameters of 1" and 2", their decays in the range of 800 to 1,000 RPM were about equal. I also compared 1" and 1.25" tops of the same weight and ball diameter, but closer in configuration than the 2" above.  In the range of 1,000 to 1,400 RPM, the 1.25" top has slightly lower decay.

These are interesting and important observations. Here's one way to understand them...

Consider an idealized finger top consisting of (i) a tip and stem of negligible mass, and (ii) a solid, flat-faced cylindrical rotor of mass M, radius R, axial length (thickness) L, uniform density d_C, volume V, and axial moment of inertia (AMI) I₃. The subscript "C" points to the cylinder (as opposed to the surrounding air), and the "3" points to the spin axis. Then

V = pi L R²
M = d_C V
I₃ = ½ M R² = ½ pi d_C L R⁴

To estimate the aerodynamic component of the top's deceleration, I'll use the von Karman model (VKM) of the steady-state swirling laminar flow induced by the smooth, flat face of a cylinder spinning in otherwise still air. The VKM has been validated in countless experiments and applies whenever the dimensionless Reynolds number

Re = w R² / v_A < 300,000,

where R is the cylinder's radius in meters as before, w is the cylinder's angular speed in rad/s, and v_A is the kinematic viscosity of air. The subscript "A" here points to the air. At room temperature (20°C), v_A ~ 1.5e-5 m²/s. At and above the pressures Iacopo can reach, it varies little with pressure.

Since finger tops operate at Re values far below 300,000, the VKM applies. The aerodynamic braking torque (ABT) coming from both cylinder faces at once is then

Q_F = -b d_A v_A^0.5 w^1.5 R⁴,

where b = 1.93 is a dimensionless constant, and d_A is the density of air. The minus sign marks Q_F as a braking torque. The subscript "F" points to the cylinder faces (as opposed to the lateral surface) as the source of the ABT. At room temperature and sea-level pressure, d_A ~ 1.20 kg/m³.

We'll come back to this important formula for the facial ABT Q_F in a moment. But first we have to assess how Q_F compares to Q_L, the ABT coming from the cylinder's lateral surface.

Although the facial flows over the cylinder remain laminar up to Re ~ 300,000, the lateral flow goes turbulent at Re ~ 64, well below typical finger top values. No simple formula for Q_L exists in the turbulent regime, but a reasonably accurate iterative formula depending only on Re and the cylinder's length L can be implemented in a spreadsheet. When you do that for Re and L values typical of our finger tops, you find that the facial ABT Q_F is generally 10 to 100 times larger than Q_L.

Hence, it's reasonable to ignore the lateral ABT and use Q ~ Q_F, where Q is the total ABT acting on the cylinder.

From the rotational version of Newton's 2nd Law, the aerodynamic angular deceleration a due to Q is

a = Q / I₃ ~ Q_F / I₃ = -(2 b / pi) (d_A / d_C) v_A^0.5 w^1.5 / L

Things to notice here...
(1) The minus sign marks a as a deceleration. The units are rad/s². To convert a to RPM/min, multiply it by pi / 1800 ~ 0.0017.
(2) The cylinder's radius R cancels out completely at all relevant speeds, in keeping with your observations.
(3) The deceleration depends on w^1.5, not on w².
(4) The deceleration varies inversely with the cylinder's density d_C. In our finger tops, the density ratio d_A / d_C is on the order of 0.001 to 0.0001.
(5) The inverse relationship between a and L reflects the fact that the cylinder's AMI is proportional to L via M and V.

Note that the aerodynamic angular deceleration a here does not include the angular deceleration due to tip processes. The latter becomes increasingly important as the speed-time curve flattens out during the final phase of spin-down. But in the early phase where the speed-time curve is steep, the above expression for a should describe most of the deceleration.

Alan: Does (5) above square with your observations?

At a given RPM the surface velocity at the perimeter of a top equals diameter squared...

Here we part company. The tangential speed u of any material point at radius r on or within a spinning cylinder is u = w r, not u = w r². For a point on the lateral surface of a cylinder of diameter D = 2 R,

u = ½ w D

The correct expressions follow directly from the fact that the circumference of a circle or cylinder is proportional to its radius or diameter. The units alone tell you that u at the lateral surface can't be proportional to either R² or D².

...aero drag is proportional to velocity squared.

You keep saying this. It's certainly true of some solid objects translating through the air at certain speeds without significant spin. But a large body of experimental data shows that it does not apply to objects spinning in air without significant translation. Our tops are in the latter category.

[Aerobie]

Hi Jeremy,
Great post.  I love your use of math.
I agree with it. 
I have some references that I'd like to review about drag coefficient, but I agree that it's less than velocity squared at low Reynold's numbers.
I'm busy at a trade show for most of the coming week, back to tops after that.
My latest record for small tops is over 23 minutes for a 40 gram top with a single twirl.
Best,
Alan

[Aerobie]

Hi Jeremy,
Back from a trade show, I have more time.
I have corrected my reply #11 in this thread, in agreement (I think) with your comments. 
That left decay rate is proportional to D^2.   Please re-read #11 and tell me if you agree.
Best,
Alan

[Jeremy McCreary]

I have corrected my reply #11 in this thread, in agreement (I think) with your comments. 
That left decay rate is proportional to D^2.   Please re-read #11 and tell me if you agree.

Sorry, we're still a good distance apart.

[Aerobie]

Tell us more.  Where do we still depart?

Alan

[Jeremy McCreary]

Tell us more.  Where do we still depart?

To be perfectly honest, I feel like we're talking past each other. But I'll give it another go...

At a given RPM the surface velocity at the perimeter of a top is proportional to diameter...

Agree now.

... and aero drag (at very low Reynold's numbers) is proportional to velocity.  So the aero drag at the perimeter would be proportional to D^2.

For starters, not sure exactly what you mean by "drag" here. Guessing it's braking torque per unit lateral surface area from the next quote. That's not what engineers usually mean by "drag". When using a widely accepted term in a special way, best to say so up front to avoid misunderstandings.

Second, your tops aren't operating at "very low" Reynolds numbers (Re). In fluid dynamics, "very low" generally means Re << 1 and almost always means Re < 10. (See White's classic text Viscous Fluid Flow.)

But in your best-case scenario (1" top at 200 RPM), Re ~ 225 -- practically speaking, far above the "very low" range. For your 2" top at 1,000 RPM (the kind of speed actually under discussion here), Re ~ 4,500.

The flows around the lateral surface of a cylinder at such Re values are fully turbulent, and that makes them fundamentally different from the creeping laminar flows observed at very low Re < 1.

To be more specific, you appear to be invoking a Stokes-type creeping laminar flow, but it just doesn't apply. Stokes flow is a very specific solution of the Navier-Stokes equation (NSE) applying only to a non-spinning sphere translating through a viscous fluid at Re < 1. As I pointed out last time, we're talking about a non-translating cylinder spinning in a viscous fluid at Re > 225. Apples and oranges. See White again.

To complicate matters further, the Stokes flow solution gives the total resistance (a force) acting on the sphere's CM. That's not the same thing as a torque per unit area acting on a small part of the sphere's surface. In fact, there is no net torque in the Stokes context.

Bottom line: You can't just lift an NSE solution from one context and expect it to hold in other contexts with very different geometries and relative fluid-body motions. Fluid dynamics is way too complicated for that.

Then the surface area of the perimeter increases with diameter, so we have aero drag at the perimeter proportional to D^3.   Furthermore, the perimeter's braking torque at the axis is proportional to D.

For aspect (diameter/length) ratios typical of real tops, my sources and calculations indicate that the bulk of the aerodynamic braking torque acting on a spinning non-translating cylinder comes from the faces, not the lateral surface.

But you're still talking about the lateral surface and ignoring the faces. Apples and oranges again.

Then the surface area of the perimeter increases with diameter, so we have aero drag at the perimeter proportional to D^3.   Furthermore, the perimeter's braking torque at the axis is proportional to D.  So we have total braking torque = D^4.

Yes, the braking torque coming from the faces is proportional to D^4, but not for the reasons your gave. The torque coming from the lateral surface is a different matter altogether and minor at best.

I was careful to point out last time that there's no simple plug-in formula for the aerodynamic braking torque coming from the lateral surface of a non-translating cylinder spinning in the turbulent regime at Re > 64. This comes straight from Childs' textbook, chapter 6, available free online.

If Childs couldn't come up with a simple plug-in formula here, surely none of us can.

This is only countered by inertia, which equals diameter squared.

I think you meant to say "is proportional to" rather than "equals" here. When you take into account the dependence of cylinder mass on diameter -- as you must in this context, and as I did -- the axial moment of inertia (the one that counts here) is proportional to D^4, not D^2.

This leads to D^4 / D^2 = D^2 as a decay factor for diameter.  That is larger decays faster.

I got D^0 for the "decay factor for diameter", not D^2. So we don't even agree on a final result for the total dependence of aerodynamic braking torque on cylinder diameter.

But then how could we given all the differences in assumptions and approaches? The devil's in the details here, and there's no way around that.

[ta0]

I haven't tried to follow this thread, but I have a comment.

My sources and calculations indicate that the bulk of the aerodynamic braking torque acting on a spinning non-translating cylinder comes from the faces, not the lateral surface. But you're still talking about the lateral surface and ignoring the faces. Apples and oranges again.

You must be assuming a maximum aspect ratio. If you elongate the cylinder enough, the drag of the lateral surface has to become greater than the constant drag from the faces at some point.

The lateral faces are further from the axis of rotation than any point of the faces. I guess the explanation for a greater drag from the faces has to do with the "centrifuge effect" the faces would have on the air (ejecting it in the radial direction).

[Jeremy McCreary]

My sources and calculations indicate that the bulk of the aerodynamic braking torque acting on a spinning non-translating cylinder comes from the faces, not the lateral surface. But you're still talking about the lateral surface and ignoring the faces. Apples and oranges again.

You must be assuming a maximum aspect ratio. If you elongate the cylinder enough, the drag of the lateral surface has to become greater than the constant drag from the faces at some point.
The lateral faces are further from the axis of rotation than any point of the faces. I guess the explanation for a greater drag from the faces has to do with the "centrifuge effect" the faces would have on the air (ejecting it in the radial direction).

Yes, I'm assuming an aspect ratio typical of a real top (axial length no more than a few times radius). A spinning pencil-like cylinder might well see more torque from the lateral surface than the combined faces, but the aspect ratios of real tops are more like those of soup cans, hockey pucks, or pancakes. I went back and made the aspect ratio assumption explicit just to be clear.

Spot on about the centrifuge effect. If the faces weren't pumping air radially, the facial torques might be less than the lateral torque. In fluid dynamics, the spinning disk problem is also referred to as the "von Karman pump".

Iacopo visualized the pumping action for us in the beautiful video below, and I can readily feel it with a hand next to any of my larger tops with more or less cylindrical rotors.