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[Aerobie]

In addition to height of Cg, the ball diameter affects topple speed.  I'm in constant trade-off between lower friction (smaller ball) and lower topple speed (larger ball).

Alan

[Iacopo]

Hi Alan,

today I received the package with your top and your AeroPress coffee maker.
It was a nice present, thank you !

This is the spinning top:



I had to paint the upper side black and white, otherwise my tachometer couldn't read the RPM reliably.

I tried about 30 hard spins with this top; the highest initial speed I reached by a single twirl was 1132 RPM.
If I insist I think I can make better than this, but not much better.

The weight of the top is 86.1 grams.
I measured/calculated its moment of inertia, which is kg-m² 0.0000901.
So the energy it has at 1132 RPM is 0.63 joule.
0.63 joule is the maximum energy I could put in this top with a single twirl.

This is a comparison with my tops, (the red dots).
This graph is about the maximum energy I can put in each top with a single twirl of the fingers.
The curved line passes through the best tested tops, (the ones able to receive more energy with a single twirl). 
The green dot is your top.
The numbers near the dots are the weight (grams) of the tops.



I found the twirling system of this top quite good;
The knurl is very large, compared to that of my tops, and the grip is excellent.
The fingers can grasp it securely and I found to spin this top easy and comfortable.
So the top has a nice playability.  Also I like the way the top precesses, making large circles on the spinning surface, thanks to the large ball tip.

Anyway spinning hard this top was a bit difficult; often the flywheel touched the mirror surface, in spite of the large clearance between the flywheel and the mirror.
But this isn't only because of a short stem.
I find the large ball tip quite slippery on the glass mirror, it reminds me my teflon tips.
This is a bit different from spiked tips, which tend to stay in place without slipping around.
This effect, added to the unwanted lateral motions of the hand, and the short stem, make more difficult to control the verticality of the top axis during a hard spinning action.

I suspect that the large knurl could contribute to this difficulty, but I am not sure:
the large knurl makes for a shorter and a more abrupt acceleration.
A more narrow knurl makes for a more gradual acceleration, spread on a longer spinning time action, and, maybe, this could help to control the verticality of the top.


I timed a spin of this top on my concave glass mirror;
I started the top at 1187 RPM by multiple twirls.  At high speed the top slowed down rapidly and I thought it couldn't spin for long, but then at low speed was more efficient, also it toppled down at very low speed.
The spin time was 19:02.
RPM measured minute after minute:
1187 - 995 - 849 - 735 - 641 - 564 - 500 - 445 - 399 - 358 - 322 - 291 - 263 - 237 - 215 - 194 - 175 - 158 - 143 - 129.



 

[Aerobie]

As I've written several times, I expected Iacopo's twirling power to exceed mine. 

My best twirl with this top is rather poor at only 690 RPM.  But I've only recorded three twirls with this top. 

With a 105g version of this top I recorded more twirls and my best twirl with that is 77% of Iacopo's energy.

As I've posted earlier, my longest spins are with 2.25" (57mm) tops, so I haven't pursued this larger 3" size. 

I attach a top view of an earlier iteration of this same top so you can see the wood center and bronze rim.  This iteration had a stem with knurled flutes.  I subsequently changed to a plain knurled stem of slightly smaller diameter.  That's the version which Iacopo tested.

Regards,

Alan

[Jeremy McCreary]

The weight of the top is 86.1 grams.
I measured/calculated its moment of inertia, which is kg-m² 0.0000901.
So the energy it has at 1132 RPM is 0.63 joule.
0.63 joule is the maximum energy I could put in this top with a single twirl....
I timed a spin of this top on my concave glass mirror; I started the top at 1187 RPM by multiple twirls.  At high speed the top slowed down rapidly and I thought it couldn't spin for long, but then at low speed was more efficient, also it toppled down at very low speed.... The spin time was 19:02.
RPM measured minute after minute:
1187 - 995 - 849 - 735 - 641 - 564 - 500 - 445 - 399 - 358 - 322 - 291 - 263 - 237 - 215 - 194 - 175 - 158 - 143 - 129.

This data describes a spin decay curve (SDC) with an excellent exponential fit (R² = 0.9942):

N(t) = N₀ exp(-t / T),

where N is the speed in RPM, t is the time after launch in sec, N₀ = 1,037 RPM is the fitted launch speed, and T = 533 sec is the characteristic decay time. The last means that for each and every 533 sec interval during the exponential decay described by this equation, the ending speed will be 36.8% of the starting speed.

Prior to 300 sec or so, the actual SDC lies slightly above and is somewhat steeper than the fitted exponential decay curve. After ~550 sec, however, the fit is almost perfect, with no sign of a purely frictional linear tail.

Hence, the total braking torque acting on this top still had a significant speed-dependent component down to the lowest recorded speed (129 RPM at 1,140 sec). If you used no lubricant, that component was most likely aerodynamic.

[Aerobie]

Nice observation Jeremy.

I mentioned recently that many of my tops decay about 10-12% of their RPM per minute and that I was surprised how constant it is over the entire spin history.

Alan

[Iacopo]

As I've written several times, I expected Iacopo's twirling power to exceed mine. 

I am an artisan, a manual worker, and also for this reason I have robust hands..
But I am not extraordinary.  A person who has one of my tops can spin it a bit stronger than me.

the total braking torque acting on this top still had a significant speed-dependent component down to the lowest recorded speed (129 RPM at 1,140 sec). If you used no lubricant, that component was most likely aerodynamic.

I used a very thin layer of oil.  I believe that in this top air drag is quite more important that tip friction.

[Jeremy McCreary]

the total braking torque acting on this top still had a significant speed-dependent component down to the lowest recorded speed (129 RPM at 1,140 sec). If you used no lubricant, that component was most likely aerodynamic.

I used a very thin layer of oil.  I believe that in this top air drag is quite more important that tip friction.

By all accounts, dry tip friction doesn't depend on speed. If that were the only braking torque acting on the top at the end of the spin, the SDC would have a straight line for a tail. But this top's tail is curved with an ever-decreasing slope. So the braking torque must still be decreasing as the speed continues to decay. We know that aerodynamic drag behaves this way. So can lubricated friction. I'm guessing that the aerodynamic torque would outweigh the other lubricated frictional torque with just a thin film of oil, but I don't know that for a fact.

Repeating the experiment with no oil might shed some light.

[Aerobie]

Jeremy,

You should join us lathe turners.  Turning parts on a lathe is one of the most enjoyable pursuits around.  You can buy a small lathe for surprisingly little cost.  I'll bet that if you take the plunge, you'll thank me for this nudge.

Alan

[Iacopo]

By all accounts, dry tip friction doesn't depend on speed. If that were the only braking torque acting on the top at the end of the spin, the SDC would have a straight line for a tail. But this top's tail is curved with an ever-decreasing slope. So the braking torque must still be decreasing as the speed continues to decay. We know that aerodynamic drag behaves this way. So can lubricated friction. I'm guessing that the aerodynamic torque would outweigh the other with just a thin film of oil, but I don't know that for a fact.

Interesting observation.  My preliminary tests in the vacuum chamber seem to confirm what you say.
I haven't tried without oil, but even with the oil the tip friction doesn't change a lot during the spin.
Air drag instead changes radically during the spin, it is very high at the start, and decreases more rapidly at the beginning of the spin.  I confirm that in the whole air drag is quite greater than tip friction.
When ready I will post all the measurements of air drag and tip friction in various tops in a new thread.

[Jeremy McCreary]

You should join us lathe turners.  Turning parts on a lathe is one of the most enjoyable pursuits around.  You can buy a small lathe for surprisingly little cost.  I'll bet that if you take the plunge, you'll thank me for this nudge.

Thanks for the invitation, Alan! LEGO tops and turned tops are in many ways complementary. I'd love to work on your side of the design space, too, but my wife would have to park her car on the street to make room for a lathe at home.

[Jeremy McCreary]

Interesting observation.  My preliminary tests in the vacuum chamber seem to confirm what you say.
I haven't tried without oil, but even with the oil the tip friction doesn't change a lot during the spin.
Air drag instead changes radically during the spin, it is very high at the start, and decreases more rapidly at the beginning of the spin.  I confirm that in the whole air drag is quite greater than tip friction.
When ready I will post all the measurements of air drag and tip friction in various tops in a new thread.

Really looking forward to seeing your data, Iacopo! I think we'll learn a lot from it.

[Jeremy McCreary]

...my tops decay about 10-12% of their RPM per minute... I was surprised how constant it is over the entire spin history.

You can use this observation to estimate the characteristic decay time T as defined a few posts ago, under the reasonable assumption of a near-exponential decay...

T = (t₂ - t₁) / ln(N₁ / N₂),

where t is the time in sec after launch, ln() is the natural log function, N is the speed in RPM, and the subscript "1" points to the earlier time-speed measurement. In your case, t₂ - t₁ = 60 sec, and N₁ / N₂ ~ 1 / (1 - 0.11) =  1.12, where I took the middle value of 10-12%. The estimated decay time is then

T = 60 sec / ln(1.12) = 515 sec,

which isn't far from the value of 533 sec obtained by fitting an exponential curve to Iacopo's spin-down data on the top you sent him.

Decay times like this make it easy to compare tops with near-exponential spin decay curves, as many of our finger and throwing tops seem to have. To my knowledge, ta0 was the first to propose this. The beauty of the method is that you can use any 2 time-speed points from the exponential-looking part of the curve and get similar results.

[Iacopo]

T = 60 sec / ln(1.12) = 515 sec

How can I calculate the natural log function ln() ?
A special calculator is needed ?

[Jeremy McCreary]

How can I calculate the natural log function ln() ? A special calculator is needed ?

It's on many calculators, including the one that comes with Windows and most of the calculator apps available for Android phones. Same is probably true for Macs and iPads and iPhones, and it's defnitely on all handheld scientific calculators -- even the cheap ones for school kids.

The ln() function is also available in Google Sheets, the free online Google Docs spreadsheet app -- which you might want to check out for your data processing anyway. It's a powerful, well-documented Excel knock-off, but getting started is easy.

[Iacopo]

Really looking forward to seeing your data, Iacopo! I think we'll learn a lot from it.

These are the first coming data.  The top is my Nr. 26b, which is similar to that sent me from Alan, but a bit smaller and more thick, and with a carbide spiked tip, spinning on a carbide base, with a thin film of oil.  Weight 103.8 grams. 
Spin time from 1250 to 150 RPM,  28'07" with air and 116'49" in the vacuum.



Everything is new, the pump, and the chamber.  The ultimate pressure indicated in the plate of the pump is 0.3 Pascal, (0.003 millibar).  It would be excellent but I don't know if this is true.   
The vacuometers I have are far too rough, I can have 20 millibar of error with them.
So I am thinking if there is a way for knowing the real ultimate pressure, without spending a lot of money for an accurate vacuum gauge.

I tried making an absolute pressure gauge with a transparent tube closed at one hand, filled of water, with the same layout of the Torricelli barometer;  10 mm of difference in height of the levels of water is about 1 millibar of absolute pressure.  But water boils in the vacuum so it doesn't work.  I tried motor oil instead of water but motor oil boils too.  Then I found ethilene glycol doesn't boil so I tried it and it works much better in fact.
It seems that the ultimate pressure is really low, a fraction of millibar.
But I am not totally sure, because at times there is some bubbles formation even in the ethilene glycol; the bubbles form always in the same spots of the plastic surface in contact with the liquid, so it is not the ethilene glycol by itself, but some interation of the liquid with the plastic surface in these spots; maybe there is some pollution there, maybe a fingerprint, I don't know.  I tried to clean but it is difficult to eliminate completely these bubbles formations.

Someone has ideas ?
It is not important to know if there is a tiny difference like 0.005 millibar instead of 0.003, but it would be good to know for sure at least that there aren't a lot of millibars in the ultimate pressure, (10, 20 or more), because this would make inaccurate the measurements.