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Author Topic: Air drag in a spinning top at different air pressures  (Read 3051 times)

Iacopo

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Air drag in a spinning top at different air pressures
« on: September 10, 2016, 10:45:56 AM »

I have measured the RPMs lost in one minute, at the average speed of 520 RPMs, by one of my wooden tops, at various different values of the air pressure.
The top is made of mahogany, the tip is made of HSS steel, conical shape; spinning surface of tungsten carbide.



There is some leakage in my vacuum chamber and 68 millibar is the lowest pressure I was able to test.
For each timing I considered the minute from 30 seconds before to 30 seconds after the instant the top was spinning at 520 RPMs:  for example at 1011 millibar I considered the one minute lapse from 600 RPMs to 440 RPMs: average speed 520 RPMs.  At 574 millibar I considered the one minute lapse from 581 RPMs to 460 RPMs: average speed 520,5 RPMs. And so on.
1011 is the value of the atmospheric pressure when I made the test.

The results are:

Millibar     RPMs lost in one minute   

 1011                160
  574                 121
  320                  93
  123                  70
  114                  63
   91                   60
   68                   51

I am wondering how many RPMs would have been lost in total absence of air;
maybe 40 RPMs...?

Also I am wondering about the importance of air drag in this top compared to tip friction;
I suppose that torque (friction) is proportional to RPMs lost in a given period of time (acceleration);

160 is proportional to torque from total friction on the top, (tip and air).
40 is proportional to torque from tip friction alone.
The difference ( 120 ) would be proportional to torque from air drag.  Is this correct ?

If so, the torque from air drag in this top spinning at 520 RPMs, would be about 3 times stronger than that from tip friction.


   
« Last Edit: September 11, 2016, 08:09:02 AM by Iacopo »
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ta0

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Re: Air drag in a spinning top at different air pressures
« Reply #1 on: September 10, 2016, 11:57:14 AM »


Also I am wondering about the importance of air drag in this top compared to tip friction;
I suppose that torque (friction) is proportional to RPMs lost in a given period of time (acceleration);

160 is proportional to torque from total friction on the top, (tip and air).
30 - 35 is proportional to torque from tip friction alone.
The difference ( 125 - 130) would be proportional to torque from air drag.  Is this correct ?

If so, the torque from air drag in this top spinning at 520 RPMs, would be about 4 times stronger than that from tip friction.

I believe you are correct. To a first approximation, the change of speed is proportional to the torque (if the speed does not change too much and the time during which the torque is applied is constant). By the way, the torque is equal to the friction force times its lever (distance to the axis.)

Great work, Iacopo, as usual!
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Aerobie

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Re: Air drag in a spinning top at different air pressures
« Reply #2 on: September 10, 2016, 02:58:26 PM »

Nice work.  I was surprised to see pressure reduction has such a large effect at 520 RPM.  I thought aero drag was pretty low at 520. 

For comparison, here is speed loss (RPM's per second) of some of my tops around 520 RPM:

Top                Loss around 520 RPM          Loss around 250 RPM

3", 440 gram                0.60                                    0.3
3", 105g                       0.95                                    0.5
2.5", 124g                    0.73                                    0.4
2", 105g                       0.90                                    0.45
1.5", 105g                    0.84                                    ----
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Iacopo

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Re: Air drag in a spinning top at different air pressures
« Reply #3 on: September 10, 2016, 03:59:45 PM »

I was surprised to see pressure reduction has such a large effect at 520 RPM.  I thought aero drag was pretty low at 520. 

The diameter of this wooden top is 125 millimeters, (about 5 inches), so at 520 RPMs it has about the same speed at its circumference as a 2 inches diameter top spinning at 1300 RPMs.
Then lighter tops (in the sense of specific weight) are more sensible to air drag, I chose this wooden top for the test because I knew that the effect of air drag on it is evident.  Metal tops are far less sensible to air drag. 

In which way are calculated the numbers you give for RPM loss ?
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Aerobie

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Re: Air drag in a spinning top at different air pressures
« Reply #4 on: September 10, 2016, 04:27:37 PM »

Hi Iacopo,

Here is the formula you requested:

RPM loss per second = (RPM1-RPM2) / (delta seconds)

My tachometer can't read well near the end of spin if the top is wobbling, but I can extrapolate the final RPM with the help of this formula.

I use a spreadsheet and enter RPM and time of RPM reading about once of minute.  The spreadsheet is programmed to solve this formula and the exponential formula which I posted weeks ago.

Alan

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Iacopo

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Re: Air drag in a spinning top at different air pressures
« Reply #5 on: September 10, 2016, 05:22:14 PM »

By the way, the torque is equal to the friction force times its lever (distance to the axis.)

Interesting.  I have found this formula for torque:

Torque (newton meters) = moment of inertia (kg m^2) x angular acceleration (radians/sec^2)

The moment of inertia of the wooden top of this test is  0.000255 kg m^2.
The RPMs lost around 520 RPMs in one minute are 160.
160 RPMs are 16.75 radians/sec , so the average angular acceleration during that minute was 16.75 radians/sec^2.

The torque for air and tip friction on this top at 520 RPMs should be:

0.000255 x 16.75 = 0.0043 newton meters (rounded off)

Since one newton is 102 grams, 0.0043 newtons are 0.44 grams (rounded off).
So the total drag on this top at 520 RPMs could be thought as a force of 0.44 grams at one meter from the axis.

We have seen that about 3/4 of it, is air drag, and about 1/4 tip friction.

3/4 of 0.44 grams are 0.33 grams;
this torque acts at various distances from the axis, so it can't be calculated for a single distance from the axis, but, just to have an idea of the intensity of the torque, the radius of this top is 62 mm, and, at this distance from the axis the torque for air drag is 5.6 grams.

1/4 of 0.44 grams are 0.11 grams;
The diameter of the contact point seems to be one tenth of millimeter and not larger than so.
At 1/20 mm from the axis, 0.11 grams become 2200 grams, (more than two kg !).
I wonder if this data is correct...  the top weighs 113 grams... is it possible to have more than two kilograms of friction force on its tip ?   Certainly the tip is relatively sharp and the weight of the top is concentrated on that tiny contact point with a pressure of about 10 kg/square millimeter (113 grams on 1/10 millimeter diameter contact point)..
in fact it is no surprise if there is wear of the contact points.  I don't know what to think about this.
 
« Last Edit: September 11, 2016, 08:13:24 AM by Iacopo »
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ta0

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Re: Air drag in a spinning top at different air pressures
« Reply #6 on: September 10, 2016, 06:26:24 PM »

There is an error on your calculation of the average acceleration (deceleration).
If the total change in velocity was 16.75 radians/sec in 60 seconds, you get the acceleration by dividing one by the other: 0.28 radians/sec2.
So your final result will be 60 times smaller.

PS: I don't think Jeremy will like your use of grams force.  :P
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Iacopo

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Re: Air drag in a spinning top at different air pressures
« Reply #7 on: September 11, 2016, 03:53:19 AM »

There is an error on your calculation of the average acceleration (deceleration).
If the total change in velocity was 16.75 radians/sec in 60 seconds, you get the acceleration by dividing one by the other: 0.28 radians/sec2.
So your final result will be 60 times smaller.

You are right...
I had to divide two times by 60:
1 radians/sec2 are 60 radians/sec.min which are 3600 radians/min2

So total torque at 520 RPMs should be

0.000255 x 0.28 = 0.000071 newton meters

If 3/4 of it is due to air drag, (0.000053 newton meters), the drag force at 62 mm from the axis would be 0.00085 newton, (0.09 grams).

And if 1/4 of it is due to tip friction, (0.000018 newton meters), this would be equivalent to a friction force on the tip of 0.36 newton (36.7 grams) at 1/20 mm from the axis.  Which is a much more credible result in fact.

I am not used to these calculations, I easily get confused...
 
« Last Edit: September 11, 2016, 08:18:06 AM by Iacopo »
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Iacopo

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Re: Air drag in a spinning top at different air pressures
« Reply #8 on: September 11, 2016, 05:30:11 AM »

Top                Loss around 520 RPM          Loss around 250 RPM

3", 440 gram                0.60                                    0.3
3", 105g                       0.95                                    0.5
2.5", 124g                    0.73                                    0.4
2", 105g                       0.90                                    0.45
1.5", 105g                    0.84                                    ----

I add some of my data:
Loss calculated as RPMs lost in one second.

Top    Diameter   weight   flywheel       Tip         Loss around 520 RPM   Loss around 250 RPM
 Nr.        mm          gr      material    position

  5         125         113    mahogany   normal               2.70                              1.05
  6          78          329        tin          normal               0.75                              0.60
  8          45          110     copper       normal               1.05                                -
  9          99          847        tin          normal               0.35                              0.25
 12         52          172      brass       recessed              0.65                              0.40
 15         51          186       lead        recessed              0.50                              0.30
« Last Edit: September 11, 2016, 08:34:22 AM by Iacopo »
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Jeremy McCreary

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Re: Air drag in a spinning top at different air pressures
« Reply #9 on: September 11, 2016, 07:24:39 PM »

PS: I don't think Jeremy will like your use of grams force.  :P

I can handle brief exposures to non-SI units, but I feel safer when my asthma inhaler's handy.  ;^}
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Playing with the physical world through LEGO

Jeremy McCreary

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Re: Air drag in a spinning top at different air pressures
« Reply #10 on: September 15, 2016, 12:56:06 AM »

For spin decay experiments like the ones we're discussing here, it's hard to beat the accuracy and reliability of the video method documented by ta0 below. (His post is well worth quoting in its entirety.) The method is easily adapted to angular deceleration measurements.

This is my setup for taking the measurements of RPM decay using a laser tachometer (with no digital output) and a camera recording the measurement.


Not only the data is recorded against the video time base, but also the instant of the boomerang to establish time zero. The original post of the setup is here. Actually, Jeff started it here.

This was the main result (thread):


Measurements on a Gates top seemed to show that a little tape can affect the results due to air friction (I am still surprised about this):


Or on a logarithmic plot to show them as lines (the slopes give the decay constants):


Also took measurements for a Takeshi BK mod:


The decay constant varies just a bit between the initial high rotation regime and lower speeds, but seems to be a useful characteristic to compare bearing tops. For example, the half-life of the Gates is more than double that of the BK.

Happily, the video method also works the other way around -- i.e., with the top spinning on a pedestal/mirror/table and the laser tachometer in your hand. Either way, you'll be recording the real-time tach display with a tripod-mounted video camera. I just use my smartphone and a cheap remote control from Amazon.

The beauty of the video method is that it captures a high-resolution stream of (time,speed) points from the top's actual spin decay curve (SDC), and it does so with very little effort. You can then plot the entire SDC, as ta0 did above for several different tops, or process just a few carefully chosen points, as you would do to assess the angular deceleration at a particular time or speed of interest. To extract the desired (time,speed) points from the video, all you need is pencil and paper (or a spreadsheet) and a video player that shows the frame time stamps to the nearest 0.1 sec or less. YouTube won't do that, but Microsoft Movie Maker (free) will.

The video method is especially well-suited to angular deceleration measurements. Recall the torque equation

Q = I3 a = I3 dw/ dt,

where Q is the net torque about the top's spin axis, I3 is the axial moment of inertia in kg m^2, a is the resulting angular acceleration in rad/s^2, w is the spin rate in rad/s, and t is time in s. The validity of the far right hand side of this equation rests on the assumption that dt (the time interval) and dw (the speed change over dt) are both vanishingly small.

We violate that assumption least when dt and dw are the smallest possible fractions of the spin time and release speed, respectively. Short of fancy lab equipment, the video method is probably the best way to achieve that.
« Last Edit: September 15, 2016, 01:53:36 AM by Jeremy McCreary »
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Iacopo

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Re: Air drag in a spinning top at different air pressures
« Reply #11 on: September 15, 2016, 01:04:07 PM »

For spin decay experiments like the ones we're discussing here, it's hard to beat the accuracy and reliability of the video method documented by ta0

Thank you Jeremy !  You (and Ta0) have inspired me.  I will remember this.

I want to set up a method for evaluating how much friction I have from the air and from the tip in my various tops;
these data would be useful to develop my design.  In fact when one my top spins longer, or shorter, than another one, I am not always sure what is the cause, if it is the tip, the aerodynamics or what else: 
I need to collect data for understanding.

My biggest problem with accuracy is that tip friction is not stable, it changes continuously while spinning (probably because there is always a bit of wear).  In single spin times there can be up to 10 % of difference from the average spin time, and some times even more.  For example this is a sequence of spin times of one of my tops, (Nr. 12), started always from 1250 RPMs;  30'10"  26'07"  25'05"  31'14"  30'47"  30'02"  31'54"  33'46"  29'45".

I could register data of more spins both in air and in the vacuum, and calculate the average spin time, but it takes much time.

One method I have tried recently is to time short periods of spinning alternately in air and in the vacuum, (for example from 1000 to 900 RPMs), in this way the top doesn't have much time for changing its efficiency, and comparison between these spin times is more reliable, (but still not perfect).
I think I will try shorter still periods of spinning, and making a video for more accurate data.



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Jeremy McCreary

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Re: Air drag in a spinning top at different air pressures
« Reply #12 on: September 15, 2016, 05:53:23 PM »

I want to set up a method for evaluating how much friction I have from the air and from the tip in my various tops;
these data would be useful to develop my design.  In fact when one my top spins longer, or shorter, than another one, I am not always sure what is the cause, if it is the tip, the aerodynamics or what else: 
I need to collect data for understanding.

My sentiments exactly! It's no different with LEGO tops: So many practical and interesting mechanical questions -- and so few of them easy to answer. We should collaborate on some of our more pressing engineering issues -- including this one.

My biggest problem with accuracy is that tip friction is not stable, it changes continuously while spinning (probably because there is always a bit of wear).  In single spin times there can be up to 10 % of difference from the average spin time, and some times even more.  For example this is a sequence of spin times of one of my tops, (Nr. 12), started always from 1250 RPMs;  30'10"  26'07"  25'05"  31'14"  30'47"  30'02"  31'54"  33'46"  29'45".

I also see a surprising amount of variability in my tops, though more in some than others. Tip processes surely have a lot to do with it, but I think it's more complicated than that. For starters, we have no direct evidence bearing on either (i) the importance of aerodynamic drag near topple speed or (ii) the aerodynamic instabilities that might affect spin-down. Structural soft spots and resonances enter my picture much more than yours.

My sources all agree that friction and rolling resistance differ in important ways but change little with speed. If so, the braking torque Qtip due to tip processes will still be near its maximum at the topple, while the braking torque Qaer due to aerodynamic drag will be near its minimum. I'd love to know what Qtip / Qaer and Qtotal = Qtip + Qtip really look like in the "low-speed" regime leading to topple, and what controls them. I know you would, too. We'll learn a lot just by looking at good quality video spin decay curves, and spot analyses will help us get at Qtotal. Making headway on Qtip / Qaer will be much harder -- in part, because Qaer may still be a player at your lowest pressures. We need a fast, accurate way to estimate moments of inertia -- especially AMI.

I could register data of more spins both in air and in the vacuum, and calculate the average spin time, but it takes much time.

Agree, spin times need to be averaged over at least several runs. None of this is easy without data logging equipment and software. Luckily, precision test tops and test rigs are often pretty easy to build in LEGO. A good rig can take some of the sting out of data collection.

One method I have tried recently is to time short periods of spinning alternately in air and in the vacuum, (for example from 1000 to 900 RPMs), in this way the top doesn't have much time for changing its efficiency, and comparison between these spin times is more reliable, (but still not perfect).
I think I will try shorter still periods of spinning, and making a video for more accurate data.

I think you're going to like smaller time and speed increments. Please let me know how the video part works for you.
« Last Edit: September 15, 2016, 05:56:50 PM by Jeremy McCreary »
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Iacopo

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Re: Air drag in a spinning top at different air pressures
« Reply #13 on: September 16, 2016, 12:45:12 PM »

Jeremy,

I have ordered a new little vacuum chamber from China; my actual one has a leakage problem, with the new one I will reach at least 38 millibar, (not perfect but better than 68), but maybe less if I will find a leakage also around the vacuum pump (which I think there is).  I want a little chamber because my actual one is large and it takes time to empty it all, it's ok for moulds but not so much for tops.
When I will have the new chamber I will start the tests.  I will publish the results in this site.

I am not sure, but I think that the drag from air stays relatively unvaried from spin to spin, and that it is the tip friction that changes, mainly.  The wooden top I tested here has less stable spin times at the lowest pressures of the air.  I suppose tops with a larger ratio Qaer/AMI should have generally more stable spin times.
But we will see this better in the future.
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