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[Jeremy McCreary]

============ Nerd alert! Extreme glaze-over danger ahead! ============

Iacopo: Many thanks for the raw data. I took the liberty of graphing and analyzing it in Excel. Sorry for the fuzzy screen-grab below.



The plot puts time t in seconds on the horizontal axis and angular speed w in rad/sec on a logarithmic vertical axis so as to render purely exponential trends as straight lines. Linear trends appear as curves strongly concave toward the origin. The curve on the right represents the vacuum data.

Each of the tan boxes reports the result of a regression analysis (curve-fitting) on a portion of the data. I ran 3 fits on each data set: An exponential fit at the highest speeds and a linear and exponential fit at the lowest.

Vacuum data: From t = 0 to t = 1,680 sec, decay is purely exponential with a decay constant of k_VHE = -0.00056 /sec and a near-perfect correlation (R_VHE² = 0.99817). The "VHE" subscript identifies the fit as "vacuum, high-speed, exponential."

From t = 2,540 sec to the end of the series at t = 3,840 sec, the vacuum data best fits a linear decay with a decay constant of k_VLL = -0.0272 /sec and a near-perfect correlation (R_VLL² = 0.99823). The "VLL" subscript here stands for "vacuum, low-speed, linear."

However, this same tail in the vacuum decay curve also fits an exponential decay with k_VLE = -0.00466 /sec and R_VLE² = 0.97465.

Air data: From t = 0 to t = 1,800 sec, decay is purely exponential with a decay constant of k_AHE = -0.00115 /sec and a near-perfect correlation (R_AHE² = 0.99928). The "AHE" subscript identifies the fit as "air, high-speed, exponential."

From t = 2,340 sec to the end of the series at t = 2,580 sec, the air data best fits a purely linear with a decay constant of k_ALL = -0.0264 /sec and a near-perfect correlation (R_ALL² = 0.99852). The "ALL" subscript identifies the fit as "air, low-speed, linear."

However, this same tail in the air decay curve also fits an exponential decay with k_ALE = -0.00466 /sec and R_ALE² = 0.98142.

Observations: At least 4 observations to note here:

(i) Your top spins down through 3 distinct decay regimes whether or not air is present: (a) Highly exponential decay at the highest speeds, (c) highly linear decay at the lowest speeds, and (b) an apparent combination of the two at intermediate speeds.

(ii) The fact that k_AHE is about twice k_VHE points to a significant boost in high-speed exponential decay due to aerodynamic drag.

(iii) The near-equality of k_VLL and k_ALL, and of k_ALE and k_VLE, suggests that the same dissipative processes control decay at the lowest speeds in both vacuum and air.

(iv) The low-speed exponential fits to the vacuum and air data aren't quite as tight as the linear fits, but they're still pretty darned good. (R² > 0.974 in all cases.) This isn't a big surprise, though. The farther one goes out on the tail of any truly exponential decay curve, the more linear it looks -- even to a regression analysis.

I find observation (i) rather counter-intuitive on the vacuum side.

Preliminary discussion: Overall, the results bear out Alan's statement that spin decay is exponential at the highest speeds and linear at the lowest -- at least in your top.

What really goes on at a top's tip-substrate "contact patch" is far from settled in the physics literature, but the 2 main candidate processes are sliding friction and rolling resistance. Since both generally show little speed dependence, one would expect that the braking torque due to either one or both would be largely independent of speed. The linear decays at the lowest speeds in air and vacuum are consistent with that expectation, but the exponential decay at the highest speeds in vacuum is strongly contradictory.

Observation (i) suggests at least 3 possibilities, not mutually exclusive: (a) For some reason, sliding friction, rolling resistance, or both, are strongly speed-dependent at high speed in your test setup. (b) Other speed-dependent contact patch processes dominate high-speed dissipation in a vacuum. (c) The vacuum was imperfect.

One last point: Now that we've established exponential decay in the high-speed regime, we have to let go of the idea that spinning tops are subject to braking torques that grow with the square of angular speed. Exponential spin decay requires a total braking torque that's linear in angular speed, not quadratic. Linear decay, on the other hand, requires a braking torque independent of speed. I'll back up these statements in a separate math post.

[Jeremy McCreary]

I have not read this thread, but I saw the title, and then within minutes I saw this as a recommended video on my You Tube feed...

Neff: Thanks for sharing this great discussion of tip material issues, Neff.

Iacopo: Now that you've had more time to evaluate tip and substrate materials, how do your observations stack up against those in the video?

[ta0]

Jeremy, I think Alan meant that the (instantaneous) energy loss is a linear function of spin for tip friction and a quadratic function of spin for air drag. Both will give exponential decays of spin with respect to time.

EDIT: If energy loss is proportional to spin rate (because friction torque is constant), the decrease in spin is linear with time. If energy loss is proportional to the square of spin rate, the decrease of spin is exponential. As Alan said.

[ta0]

The very low spin behavior may have to do with the difference between static and dynamic friction (compounded by the fact that there is oil there). The surfaces are probably "sticking". That is a very complicated behavior but not of much interest for our tops. The higher speed data is what I would have expected and it is very clean.

[Jeremy McCreary]

Jeremy, I think Alan meant that the (instantaneous) energy loss is a linear function of spin for tip friction and a quadratic function of spin for air drag. Both will give exponential decays of spin with respect to time.

Thanks, ta0. Yes, spin decay will be exponential if the power loss (kinetic energy decay rate) is quadratic in speed for any reason. If that's what Alan meant by "drag", then he's right on this point, but quadratic power loss comes from a braking torque that's linear in speed.

.... Aerodynamic drag is proportional to velocity squared....

I'm used to the fluid dynamic definition of "drag", which considers it a force or torque. (See https://en.wikipedia.org/wiki/Drag_(physics).) By this definition, drag isn't a power loss per se but will certainly result in one.

And if drag in the fluid dynamic sense is quadratic in speed, the spin decay will be hyperbolic rather than exponential. The 2 decay curves look somewhat alike when plotted on regular axes, but the hyperbolic decay is steeper at first and shallower in the tail, where both can look rather linear.

I'm sure this sounds like nit-picking, but subtle differences in the shape of a decay curve can reflect substantial differences in the underlying dynamics.

[Jeremy McCreary]

The very low spin behavior may have to do with the difference between static and dynamic friction (compounded by the fact that there is oil there). The surfaces are probably "sticking". That is a very complicated behavior but not of much interest for our tops. The higher speed data is what I would have expected and it is very clean.

Iacopo didn't mention any lube in this last test, but if viscous enough, oil could easily introduce a speed-dependent dissipation at the tip.

I'd love to hear your take on the exponential decay at high speed in the vacuum data.

[Iacopo]

Geremy, you have made a great job with the data ! 
There was a thin layer of oil, (motor oil), in the spinning surface.
The residual pressure in the vacuum chamber was 3.5 millibar, so there was still some air, but very little.
At low speed there is a bit of wobbling while at medium and high speed this top spins perfectly vertical.
 

[ta0]

I'd love to hear your take on the exponential decay at high speed in the vacuum data.

I edited my reply from yesterday night. You are right, the exponential decay in vacuum contradicts the assumption of power loss proportional to spin rate (or constant friction torque) that would result in a linear velocity decrease. I need to think more about it.

[Jeremy McCreary]

Geremy, you have made a great job with the data ! 
There was a thin layer of oil, (motor oil), in the spinning surface.
The residual pressure in the vacuum chamber was 3.5 millibar, so there was still some air, but very little.
 

Iacopo: My pleasure, and thanks for the additional data.

WRT residual drag in the vacuum chamber: At 3.5 mb, the residual air density inside the vacuum chamber was ~0.35% of its standard value at sea level. Since aerodynamic drag (here, a braking torque) is sure to be proportional to air density regardless of its dependence on other factors, you effectively reduced the braking torque due to drag by the same factor -- i.e., 0.35%.

And that makes me doubt that the residual drag in your chamber had much to do with the high speed exponential decay in the vacuum data. A thin layer of motor oil on the substrate also seems unlikely as a major player here. My gut tells me that something else is going on.

You've mentioned tip wear in past tests, and ta0 wondered about the possibility of stick-slip behavior at the tip. Any thoughts along those lines in this particular case?

And just curious, how are you lucky enough to have access to a vacuum chamber? My wife won't let me have one.

========  Correction  ========

Sorry, all, the strike-though above is incorrect, as I forgot about some implicit ways in which braking torque depends on air density. The ultimate dependence is at most a proportionality to the square root of density and could be less in tops with large body height to radius ratios.

If so, Iacopo's vacuum chamber cut the aerodynamic braking torque on his top to as little as ~6% of its value at atmospheric pressure. To first order, this same reduction would apply at all speeds.

[Iacopo]

At 3.5 mb, the residual air density inside the vacuum chamber was ~0.35% of its standard value at sea level. Since aerodynamic drag (here, a braking torque) is sure to be proportional to air density regardless of its dependence on other factors, you effectively reduced the braking torque due to drag by the same factor -- i.e., 0.35%.

Interesting, I didn't know, thanks..
About wear, yes, there is wear, very little but there is, in the HSS tip.
Instead, in the tungsten carbide base, apparently there is no wear at all.
One fact to consider could be that, generally, spin times are not stable, and that the same top, starting from the same RPM, doesn't spin always for the same time: for example, the following is a sequence of spin times obtained by my top Nr. 14 starting always from 1250 RPM:
29'29"  34'37"  37'36"  35'54"  36'48"  35'43"  37'41"  33'24"  30'01"  32'33"  32'36"  35'46"  31'06"  35'45"  34'09"  35'55"  35'20"  33'19"  32'54"
There was always a thin layer of motor oil.  To me it's a mistery this variability of spin times.  Since I suppose that the drag from the air doesn't change (or it changes very little, considering temperature) from one spin to another one, it would seem that it is the tip friction that varies from one spin to another, (and at times also during the same spin I believe). I suspect that the shape of the surface of the contact point of the tip, wearing out while spinning, changes continuously, and depending on the shape, which I bet is never perfectly smooth and even at microscopical level, also its friction varies.
Carbide is not a homogeneous material, (it's carbide powder in a matrix of.. cobalt, maybe.. I can see the matrix at the microscope), this also could play a role.. ?
So, since I have clocked only two spins, (with and without air), I don't know how reliable are the results, I should clock various spins to have more reliable data, (which I will do, but first the video about moment of inertia).
I have a vacuum chamber for job reasons. I make moulds of statues, and I use the vacuum chamber to avoid bubbles in the silicone for the moulds.
     

[Jeremy McCreary]

========  Correction  ========

...aerodynamic drag (here, a braking torque) is sure to be proportional to air density

Sorry, just posted a correction to the strike-through above in its original post at the link above.

[Jack]

========  Correction  ========

...aerodynamic drag (here, a braking torque) is sure to be proportional to air density

Sorry, just posted a correction to the strike-through above in its original post at the link above.

sorry to say anything but im afraid you spelled "toque" wrong 

[Jeremy McCreary]

sorry to say anything but im afraid you spelled "toque" wrong 

Here in Colorado, "toque" means something different.

[Jack]

sorry to say anything but im afraid you spelled "toque" wrong 

Here in Colorado, "toque" means something different.

[Jeremy McCreary]

.... One fact to consider could be that, generally, spin times are not stable, and that the same top, starting from the same RPM, doesn't spin always for the same time: for example, the following is a sequence of spin times obtained by my top Nr. 14 starting always from 1250 RPM:
29'29"  34'37"  37'36"  35'54"  36'48"  35'43"  37'41"  33'24"  30'01"  32'33"  32'36"  35'46"  31'06"  35'45"  34'09"  35'55"  35'20"  33'19"  32'54"
There was always a thin layer of motor oil.  To me it's a mistery this variability of spin times....

Iacopo: I plotted up your spin time series for top No. 14 after converting to seconds.



I'm inclined to pass the apparent periodicity off as an artifact of the data set. If correct, the spin times are pretty random within the 1,769 - 2,261 sec range.

The spin time variability is a mystery to me, too -- especially with such a high-AMI top and with spins this long. I'd guess that drag fluctuations due to aerodynamic instabilities would be averaged out over spin times this long.

Agree that the contact patch is a good place to start looking for answers.