# Gas station without pumps

## 2017 June 20

### Fidget spinners

Filed under: Uncategorized — gasstationwithoutpumps @ 17:34
Tags: , , ,

I recently bought two fidget spinners from Elecrow:

The 5-spoked wheel spinner weighs 32.88±0.03g, and the 3-spoke brass spinner weighs 61.14±0.02g.

The heavier 3-bladed spinner cost $8.90 and is milled out of brass (though the site claims “pure copper”, the material looks like brass and is slightly magnetic, so I’m sure it is brass).The lighter 5-spoke spinner cost$6.90.

The lighter spinner is easier to get to high speed, spins longer, has more gyroscopic effect, and has a dimple for balancing it on a pencil point, so makes the better fidget spinner in many ways.

I was curious whether I could characterize the fidget spinners electronically. I have a photointerrupter (an aligned LED and photodetector) from Sparkfun with a 1cm gap that the spinners just fit in.

Here is the 3-spoke spinner mounted in the Panavise Jr, with the photointerrupter counting 6 ticks per revolution.

Here is the 5-spoke spinner with the photointerrupter counting 5 ticks per revolution.

I set up PteroDAQ to record a timestamp on every rising edge of the photodetector, which counts 5 uniformly spaced ticks per revolution for the 5-spoke wheel, but 6 ticks (in 3 pair of closely spaced ones) for the 3-bladed spinner. I can then plot the angular position of the spinner as a function of time in gnuplot:

plot '3-spoke-spin-down-ticks.txt' u 1:(\$0/6.)


I tried fitting the spin-down using constant deceleration (a quadratic), using deceleration proportional to velocity (exponential decay), and using a model that has both terms: $v_{0}\tau(1-e^{-t/\tau})+a t^2 /2$.  I expressed position as number of turns (that being simpler to interpret than radians), and so the initial velocity $v_{0}$  is in turns/sec, acceleration $a$ is in turns per second per second, and the decay time $\tau$ is in seconds.  I got terrible fits with the constant deceleration, decent fits until the spinning got slow with the exponential decay, and quite a good fit with the combined model:

The longer spin time for the wheel is partly due to higher initial velocity, but the time constant for the decay is also much longer for the wheel, indicating better bearings.

I’m not quite sure how to interpret the slightly higher contact friction term for the 5-spoke wheel.

1. Without really thinking, I would attribute the longer time constant to a smaller ratio of bearing drag (a torque linear in the angular velocity) to the larger? moment of inertia for the wheel. Interesting to think about how that could be smaller while the larger acceleration parameter suggests that the ratio of the constant friction torque to the moment of inertia is larger for that same bearing. I know zip about the engineering or physics of bearings.

Comment by CCPhysicist — 2017 June 21 @ 08:06

• The time constant does indeed come from the assumption that the drag is proportional to velocity. If $\dot v = -k v$, then $v = v_{0} e^{-k t}$ and $\tau = 1/k$. Using just the velocity-dependent drag model fits the data very well until the last few seconds, when both spinners slow down quicker than the exponential model suggests (the exponential model would never have them come to a complete stop). I think that the slight eccentricity of the 3-spoke spinner may make a difference here, as I stopped the recording as soon as the spinner reversed direction (switching to acting like a pendulum), so the difference in the dry friction term may be an artifact of the measurement—the zero velocity came from including some gravitational forces rather than strictly frictional ones.

Comment by gasstationwithoutpumps — 2017 June 21 @ 11:11

• Sorry for the delay, but the fact that it reverses direction means the constant torque is due to some component of gravity (object is not symmetric) rather than constant friction, which means that linear drag is explaining most of what is going on.

Thanks for posting that familiar differential equation. Could help others who find this posting understand that that problem is also exactly solvable with calculus.

Comment by CCPhysicist — 2017 June 25 @ 11:13

• I’ve been thinking about redoing the measurements with the spinners horizontal (to reduce the gravitational effects due to eccentricity). I think that the bearing drag will also change, as the loading on the bearings will be different, with the weight of the spinner pressing the ball bearings against the sides of the races instead of against the inner faces of the races. (The 3-blade spinner seems to me to have higher drag in this orientation, but I’ve not measured it yet.)

Comment by gasstationwithoutpumps — 2017 June 25 @ 12:10

2. You have wayyyy too much time on your hands.

Comment by gflint — 2017 June 21 @ 08:06

• I had to do something fun after a 5-day grading marathon. The grades were turned in on time (by noon on Tuesday), and I needed a break.

Comment by gasstationwithoutpumps — 2017 June 21 @ 10:54

3. I would think the air resistance for the two spinners should be quite different: the 3-spoke spinner is constantly pushing some air out of the way (and probably creating some vortices and other interesting effects) whereas the wheel really only has tiny viscous boundary-layer resistance.

Comment by secretseasons — 2017 June 21 @ 10:26

• The spokes of the wheel are also encountering some air drag, but I don’t think that air drag is the main difference between the spinners. The 3-spoke spinner is very slightly off center (when stopped it always ends up with the same blade pointing down) and it has a much lower quality bearing (you can feel some vibration that is not just the eccentricity of the spinner if you hold the spinner).

Also, unless you hold the 3-spoke spinner exactly vertically (as I did in the vise for these tests), the axle support rubs slightly against the sides of the spinner, slowing it down more drastically.

The biggest differences, I think, are in bearing quality. That might be testable, by replacing the bearing in the 3-spoke spinner with a higher quality one—everything on that spinner unscrews, so bearing replacement is possible. (The 5-spoke wheel seems to be glued, rather than screwed, together, so replacing its bearings would be tough.)

Comment by gasstationwithoutpumps — 2017 June 21 @ 11:02

4. […] Fidget spinners, I wrote about measuring and modeling the acceleration of two fidget spinners, 5-spoke spinner that […]

Pingback by Fidget spinners revisited | Gas station without pumps — 2017 June 25 @ 17:55

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