Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[sixtoe]

So, RPM being w.

[sixtoe]

ta0. I would love to help with this data collecting. I have a deep interest in bringing hard data to bear upon the anecdotal system of describing play using fluff words that require a deeper knowledge of skill toy play to understand. I have some issues in helping that need to be resolved first. I don't have an apparatus for MOI and CG measurements and I don't have a library of tops to measure. I have depleted my collection to pay for personal needs since I can't work I don't have any extra money. I did however get a great birthday present and it has added to my toolset. I now have a USB arduino(grin). I will look into making a torsion table using a rotational sensor to measure speed. This will allow me to take direct readings of the velocity over time and automatically calculate the MOI. It will also allow accurate zeroing out of the table and any gear needed to suspend the subject being measured so it will make the measurements more accurate. I should be able to do this at low low cost. I think I can make the torsion table out of a used hard drive platter and bearing. I can get a junk hd that will have a large platter at a local second hand store. The rest should be fairly inexpensive and easy to find. I have a breadboard and the arduino is for breadboard prototyping. If I can get a table built the only thing that would stop me would be the end of my spintop supply. My brother, TonySauce on this forum, has some tops that I don't and we can see how people feel about sending their tops to be measured. Once the table is working it should be a quick and easy thing to get a set of measurements.

[ta0]

I will look into making a torsion table . . .

Wow! That would be a major project! Of course, it would be great to have one, but I don't know if it is worth going into that much trouble considering that the torsion balance is pretty simple and the number of tops is not that large. But I am interested in your electronics hacking abilities. I have been interested in putting accelerometers on a top for a long time . . . We can talk about that later.

I am sorry about your top collection. Hopefully you will be able to slowly replenish it.

I re-read a few things about precession and figured out the following (WARNING to others, nerd curves ahead):

1- The reason the transverse moi does not appear in the equation for precession is that almost always the angular momentum due to precession is assumed to be much smaller than the angular momentum around the top axis and it is neglected. Baldufado just over a week ago posted a technical bibliography on Pulp's forum. A short paper (pdf) has the general case. Interestingly, it shows that there can also be a second stable precession speed (very fast) and the direction depends on whether the transverse moi is larger or smaller than the axial moi.

2 -  Another of Baldufado's finds shows that a top floating in space (i.e., no gravity torque) will precess if initially spun in a direction that does not coincide with the symmetry axis. The rate of this inertial precession depends on the ratio of the transverse and axial moi.

3 - If you replace the torque T on the equation I posted by M.g.d, where M is the mass, g the acceleration of gravity, and d the distance between the tip groove and the center of mass, the equation is also valid when the top is at any angle to the vertical.

So, RPM being w.

Actually the units are radians per second, so you have to take into account the 2.pi (or tau?) factor for both p and w. If you write it in terms of turns per second (hertz) you have to divide by (2.pi)² on the right side.

Let's calculate as an example the precession of a Spintastic's Trompo Bearing.
According to pizquie data, a Trompo Bearing has a weight of 76 gr and the height of the center of gravity is 49mm. Let's subtract 8mm of the cone of the tip to get the distance between the groove of the tip and the CG (the lever arm of the torque): 41mm. Therefore, the torque from gravity will be: 0.076 kg x 9.81 m/s² x 0.041 m = 0.03 Newton x m.
The moi from my measurements above (assuming that the Trompo Grande and Trompo Bearing have about the same) is 3.26 x 10^-5 kg m². 
Let's assume it is spinning at a normal speed of 3,600 RPM = 60 1/s.
The equation gives a rate of precession (in revolutions per second) of:
p = 0.03 kg m²/s² / (0.0000326 kg m² x 60 s^-1 x (2 x pi)²) = 0.39 1/s
So a Trompo Bearing spinning at 3600 RPM should take about 2.5 seconds to precess one turn, what seems about right.

[Iacopo]

I have not studied mathematics nor physics, so I beg your pardon if what I have written here is not perfect. I hope there are not important errors.

The aim of the trifilar pendulum here is to measure the radius of gyration, (indicated here with the letter "r") of a spinning top.   The radius of gyration is a kind of average distance of the mass from its axis of rotation, or, the distance from the axis of rotation where all the mass of the top could be concentrated without changing its moment of inertia. 

The radius of gyration of a full cylinder is  always 7.071/10 of its geometrical radius. That of a full sphere 6.324/10, that of a full cone 5.477/10.  The radius of gyration of an imaginary pipe with all its mass points at the same distance from the axis of rotation, is 10/10 of the geometrical radius, the two radii coincide. 
As a sample, the radius of gyration of a cylinder with a diameter of 12 inches, is 12/2 x 0.7071 = 4.2426 inches.
 
But in the case of more complex objects it is more difficult to calculate the radius of gyration. 
We can use a trifilar pendulum to measure it; in fact the duration of the oscillations of the pendulum is proportional to the radius of gyration.

I made my pendulum with a wooden plate, (diameter mm 85, weight 5.6 grams), and some fishing line, (diameter mm 0.30, but for tops lighter than 100 grams I suggest diameter mm 0.15).  Three lines are fixed to the plate in three equidistant points at mm 40.5 from the center of the plate.
The lines are fixed at their opposite ends to another plate, to which the first plate is suspended by the three lines. The distance between the two plates is mm 1540.
The upper plate is very little, and the three lines are fixed here at a distance of only mm 1.8 from the center of the plate.
You can use different measures of course, but this will affect the oscillation time.  If the oscillating movement lasts for a longer time, the reading will be more accurate, because the angle of oscillation is more stable, not decreasing too rapidly.

To use the pendulum first we need to calibrate it.
The suspended plate itself has its radius of gyration, we can start calculating it; the plate is a cylinder with diameter mm 85; 
85/2 x 0.7071 = mm 30.05 , radius of gyration of the plate.
Let's see how much is the oscillation period of the plate.
I choose to make it oscillate for an angle of 45 degrees:
different angles will not affect much the oscillation period, (unless very large or very little angles), but readings will be more accurate if you make it oscillate always from the same angle.
The plate, (as also could be expected for whatever other object with a radius of gyration of mm 30.05 in this pendulum, but I better explain this later) oscillates 10 times in 36.42 seconds, (time, "t")
The ratio  t/r   is:  36.42/30.05= 1.212 (fixed number)
So, to know the radius of gyration of an object, I could simply divide its oscillation period for the fixed number:
36.42 seconds/1.212 = 30.05 (mm, radius of gyration).

But it is more complicated:
If you put an object on the plate and make it oscillate, you will measure an average radius of gyration given by the object together with the plate:  you need to subtract the effect of the plate to have the correct data.
To do so we first need to calculate the moment of inertia of the plate:
     
The formula for the moment of inertia ("I") is:
r x r x m = I , where m is the mass of the plate, 5.6 grams.
30.5 x 30.5 x 5.6 =  5057 (gram-square millimeter, rounded off)
5057 is the moment of inertia of the plate.

Now I put an object (weight 82.6 grams, I want to calculate the radius of gyration of this object) on the pendulum and I clock the oscillation period;  it oscillates 10 times in 32.17 seconds. 

The weight of the oscillating mass (the object together with the plate) is  88.2 grams.

The radius of gyration of the object together with the plate is:  32.17 seconds/1.212 fixed number= 26.54 mm

The moment of inertia of the object together with the plate is:  26.54 mm x 26.54 mm x 88.2 grams = 62,126

Now we know the moments of inertia of the plate alone and of the object together with the plate.
I subtract the first from the second to know the moment of inertia of the object alone:
62,126 ( I tot) - 5057 ( I plate) = 57,069 ( I object)

To know the radius of gyration of the object alone;
57,069 ( I object) / 82.6 (m object) = 690.9 (r x r)
Square root of 690.9 = 26.28 mm , radius of gyration of the object alone.

So, now we know how to measure and calculate the radius of gyration and the moment of inertia of a spinning top or whatever other object using a trifilar pendulum.

But still the readings of the oscillation periods are not very accurate, we have a problem. 
In fact fishing line is a bit elastic, and lengthens differently depending on the weight of the object put on the pendulum.  When the lines are a bit longer, the pendulum oscillates a bit more slowly, and the readings will be misleading.
It is possible to calibrate the pendulum for different weights, I did so, using the various aluminum cylinders you have seen in the video.

Example:
An aluminum cylinder weighing 494 grams, diameter mm 69.5, oscillates in the pendulum 10 times in 33.55 seconds;  I want to calculate the fixed number for this weight:

The radius of gyration of the cylinder is:
69.5/2 x 0.7071 = mm 24.57 (r)

Its moment of inertia is:
24.57 (r) x 24.57 (r) x 494 (m) =  298,220  ( I )

Moment of inertia of cylinder and plate together:
298,220  ( I cylinder) + 5,057 ( I plate) = 303,277 ( I tot)

Mass of cylinder and plate together:
494 (m cylinder) + 5.6 (m plate) = 499.6 (m tot)

Radius of gyration of cylinder and plate together:
303,277 ( I tot) / 499.6 (m tot) = 607.04 (r x r)
Square root of 607.04 = mm 24.64, (r tot)
 
Fixed number:
33.55 (t) / 24.64 (r tot) = 1.361 (fixed number for 499.6 grams).

Other fixed numbers I obtained with different cylinders:
5.6  grams:    1.212
100 grams:    1.293
200 grams:    1.333
300 grams:    1.349
400 grams:    1.357
500 grams:    1.361

Still there is some lack of precision in clocking manually the oscillation time, with errors until about 1 %, but making the average of more timings more accuracy is achieved.

With this pendulum I can now measure and calculate the radius of gyration and the moment of inertia of my spinning tops:

Example:
My top Nr. 20 weighs 269 grams and oscillates 10 times in  31.19 seconds.
   
Total oscillating mass:  269 (m top) + 5.6 (m plate) = 274.6 grams.

Fixed number for 274.6 grams:  1.344

Radius of gyration of top and plate together:
31.19 (t) / 1.344 (fixed number) =  23.21 mm

Moment of inertia of top and plate together:
23.21 (r tot) x 23.21 (r tot) x 274.6 (m tot) = 147,928 ( I tot)

Moment of inertia of the top alone:
147,928 ( I tot) - 5,057 ( I plate) =  142,871 ( I top)

Radius of gyration of the top:
142,871 ( I top) / 269 (m top) = 531.12 (r x r)
Square root of 531.12 = 23.05 mm ( r top)

That's all.
(see how to use the value to calculate: How much energy you can put into your spinning top ?

[ta0]

I have not studied mathematics nor physics, so I beg your pardon if what I have written here is not perfect. I hope there are not important errors.

I can tell you that you could have been a very good engineer or experimental scientist!

Your 3-string pendulum is more versatile than the single string torsional pendulum I used as you can just place the top on the plate (well, centered and aligned correctly) while I had to find a way to attach the string at the correct position.

You said that you used a very small plate at the top of the balance. Any reason for that? I looked up trifilar pendulums on google and most of what I found have parallel wires. But I cannot see a problem and the equations should remain basically the same. Just that in yours the restoring force will be smaller for the same rotation angle as the plate will rise less. By the way, although the trifilar pendulum rotates, it behaves more like a regular pendulum than a torsional pendulum.

Amazing work!     

[Kirk]

Very nicely done.  I was not aware of this type of instrument.  I like the idea of the smaller top plate reducing the restoring torque.  Of course you could make the strings longer to do the same thing but then the unit would be quite large and more subject to air currents.  Does the small plate on top make it easier to see if the test specimen is centered? You could also use the same instrument to measure longitudinal moment of inertia. (with some way to support the specimen)
By the way look for "dyneema fishing line". Compared to nylon it is thinner, more flexible and much less stretch. I found some on eBay for less than $6 landed cost. --edit-- also on Amazon without the China shipping time.

[Jeremy McCreary]

I can tell you that you could have been a very good engineer or experimental scientist!
...
Amazing work!     

My sentiments exactly, Iacopo!

[Jeremy McCreary]

Excellent work as usual, Iacopo! You've inspired me to abandon my bifilar LEGO pendulum for a trifilar -- much easier to work with.

For anyone interested in a refresher on how mass and moment of inertia control our tops...

Mass, axial moment of inertia (AMI), and axial radius of gyration (ARG) are related inertial properties that affect top behavior and performance in distinct but practical ways. Another related property that I find useful in maximizing sleep times is the specific AMI, or AMI per unit mass.

Mass: Mass m is the easiest of these properties to understand and measure. Its most direct influence on top behavior is through the top's weight

W = m g,

where g is the acceleration of gravity. Any scale with a readout in grams or kilograms is reporting mass, not weight.

Two important torques grow with weight and therefore mass: (i) The gravitational torque about the tip responsible for precession, wobbling, and the top's eventual fall. (ii) The braking torque about the spin axis due to friction and rolling resistance at the tip. AMI, specific AMI, and ARG have no direct influence here.

Axial moment of inertia: The Wikipedia page on moment of inertia is excellent. Tops have at least 2 distinct moments of inertia, each tied to a specific reference axis. The axial moment of inertia (AMI) refers to the spin axis, which is usually the top's axis of rotational symmetry. The AMI is often written as I₃, with the subscript "3" denoting the spin axis.

Iacopo's trifilar pendulum is set up to measure AMI. One has to go to such lengths because 2 tops of identical mass can have radically different AMIs. Why? Because what really determines AMI is how closely the top holds its mass to the spin axis: For a given total mass, the farther away, the greater the AMI.

And why worry about AMI? Because it ultimately measures a top's resistance to angular acceleration or deceleration by a net torque Q about the spin axis. From the rotational form of Newton's Second Law:

dw/dt = Q / I₃,

where dw/dt is the time rate of change of the angular speed w measured in rad/sec.

EDIT: That last line is a bit confusing WRT units. The angular speed w is in rad/sec, and the angular acceleration dw/dt is in rad/sec^2.

If Q_d is the driving torque applied about the spin axis by the fingers or a spin-up tool, and Q_b is the total braking torque from all sources of dissipation, then the angular acceleration during spin-up is

dw/dt = (Q_d - Q_b) / I₃,

Clearly, the greater I₃, the slower the spin-up, and the slower the spin-down when Q_d is removed. In general, Q_d and Q_b are functions of the instantaneous angular speed w(t).

NB: Where a top's mass resides along the spin axis is irrelevant for AMI, specific AMI, and ARG but has everything to do with the transverse moment of inertia (TMI), which in a top context refers to an axis perpendicular to the spin axis and through the tip, not the center of mass. The TMI/AMI ratio strongly influences the rate and amplitude of wobbling.

Axial radius of gyration: Every moment of inertia I has an associated radius of gyration K about the same reference axis given by

K == sqrt(I / m)

The axial radius of gyration (ARG) is the one associated with the AMI. To get a physical feel for it, consider a thin ring of mass m and radius R spinning about its circular symmetry axis. ("Thin" here means that the radial thickness of the ring is just a tiny fraction of its radius.) This special mass distribution has the simplest possible formula for AMI:

I₃ = m R²,

which gives

K₃ = R

Hence, the thin ring's ARG is just its radius. And the ARG of a top is just the radius of the thin ring having the same AMI and mass as the top. The axial length of the ring is relevant to its TMI, but not to its AMI, specific AMI, or ARG. In effect, the ARG describes the mass distribution of a top without regard to the total mass involved.

Specific AMI: Even simpler than the ARG is the "specific AMI" or AMI per unit mass given by

J₃ == I₃ / m = K₃²

Importantly, the thin ring described above has the highest specific AMI of any mass distribution, period! So if you need to wring the highest possible AMI out of a given mass, put as much of the mass as possible into a thin ring of the largest possible radius, and connect the ring to the tip and stem with the lightest possible suspension system, like so...




Specific AMI and specific TMI loom large when trying to reduce the minimum spin rate w_min for stable sleep:

w_min² = 4 (I₁ - I₃) m g H / I₃²,

where I₁ is the TMI as described above, and H is the distance from the tip to the top's center of mass. All other things being equal, the lower w_min, the longer the top will sleep before precession and wobbling set in. Makers of high-end metal pocket tops like those from BILLETSPIN work very hard to minimize w_min, and the last formula shows just how to go about it.

But this formula carries a lot of excess baggage. For starters, mass cancels out completely, leaving

w_min² = 4 (J₁ - J₃) g H / J₃²,

where J₁ is the specific TMI.

In other words, w_min depends not on the absolute AMI and TMI, but on their counterparts per unit mass. In a top made of just one material, this means that w_min depends only on the size and shape. The density of the material has nothing to do with it!

It's possible to isolate the modest dependence of w_min on absolute size by using the top's maximum radius R from the spin axis as a scaling factor. Let

G_i == J_i / R² = (K_i / R)² = I_i / m R²

I don't know of a name for G₃, but it's the AMI relative to that of a thin ring of the same mass and radius spinning about its circular symmetry axis. For a solid circular disk or cylinder spinning about its symmetry axis, G₃ = 1/2, and for hollow and solid spheres, G₃ = 2/3 and 2/5, respectively. Engineers often refer to a quantity like this as a "geometric factor", as it depends on the system's geometry (i.e., mass distribution) but not on its total mass or absolute size.

If we also define the relative CM height

h == H / R,

then for many top shapes of practical interest (and not just for solids of revolution),

w_min² = 4 (G₁ - G₃) g h / R G₃²

The radius in the denominator here is the only remaining dependence on absolute top size! All other things being equal, the larger the maximum radius, the lower the w_min, and the longer the sleep time. There's no getting around this.

In summary, mass, AMI, ARG, and specific AMI tell us different but related things about the same top. Conceptually if not quantitatively, all are potentially useful to a top builder interested in playing the trade-offs entailed in designing for both appearance and performance to best advantage.

[ta0]

Jeremy: sometimes it is just better to refer to Wikipedia 
What was your reference for the w_min?

[Jeremy McCreary]

Jeremy: sometimes it is just better to refer to Wikipedia 
What was your reference for the w_min?

You have a point, of course, but Wikipedia doesn't really lay out how mass and AMI and its cousins differ in their effects on readily observed top behaviors -- at least not in one place. It took me a couple of years of pretty intensive research to put all that together.

The first expression for w_min came from J. Ginsberg, Advanced Engineering Dynamics, easily found free online. The subsequent expressions are mine, easily derived.

[Iacopo]

you used a very small plate at the top of the balance. Any reason for that?

Very nicely done.  I was not aware of this type of instrument.  I like the idea of the smaller top plate reducing the restoring torque.  Of course you could make the strings longer to do the same thing but then the unit would be quite large and more subject to air currents.  Does the small plate on top make it easier to see if the test specimen is centered? You could also use the same instrument to measure longitudinal moment of inertia. (with some way to support the specimen)
By the way look for "dyneema fishing line". Compared to nylon it is thinner, more flexible and much less stretch. I found some on eBay for less than $6 landed cost. --edit-- also on Amazon without the China shipping time.

I wanted the pendulum to oscillate slowly, I think it is at least a tiny bit more accurate in this way.
The problem isn't the duration of the oscillations anyway, (I could clock 15 or more of them, if it oscillates more rapidly). The problem is the angle of oscillation, which is maintained for a longer time if the pendulum oscillates slowly. Especially with light objects the pendulum slows down more rapidly, the angle of oscillation decreases during the clocking, and the reading becomes a bit less accurate.
Since the pendulum is already as tall as a person and it becomes not much practical to have it much more tall, I made the upper plate little.

The little upper plate causes the lines to be slightly tilted towards the inside, but the grade is little.
They help centering only large objects like the largest and tallest cylinder in the video.
Anyway the only reason for the little plate is the one explained above, not centering.

Kirk, thank you for the tip: I didn't know dyneema fishing line.

[Iacopo]

dw/dt = Q / I₃,
where dw/dt is the time rate of change of the angular speed w measured in rad/sec.

Geremy, if you are willing, could you make a sample with numbers for this formula ?
I would like to understand it better.

About scales, I think that the ones with spring mechanism measure weight, and those with counterweights measure mass. The readings of these scales on the moon, for example, would be very different.

[cecil]

I'm glad I'm stupid. I have more time to make BIG TOPS.

[Jeremy McCreary]

dw/dt = Q / I₃,
where dw/dt is the time rate of change of the angular speed w measured in rad/sec.

Geremy, if you are willing, could you make a sample with numbers for this formula ?
I would like to understand it better.

About scales, I think that the ones with spring mechanism measure weight, and those with counterweights measure mass. The readings of these scales on the moon, for example, would be very different.

Happy to provide an example, Iacopo. First, let's rewrite the equation like so:

dw = (Q / I₃) dt,

where Q is the net torque, I₃ is the moment of inertia, dw is a change in angular speed, and dt is the time interval over which that change occurred. The equation assumes only that Q, I₃, and dw refer to the same axis.

Now consider a flywheel with I₃ = 10 kg m^2 spinning freely at constant angular speed w₀ = 7 rad/s. The constant speed tells us that under these conditions, the net torque at the flywheel shaft is zero. If we then add a constant torque Q = +5 N m to the flywheel shaft for a period of time dt = 4 s, the flywheel's angular speed will increase by an amount

dw = (+5 N m / 10 kg m^2) (4 s) = +2 rad/s,

thus boosting the flywheel's speed to

w = w₀ + dw = 7 + 2 = 9 rad/s

If the torque had been negative relative to the wheel's original speed, the speed would have been lowered by 2 rad/s instead. Does that help?

Yes, exactly what you're actually getting from a scale isn't always clear. Apparently, my digital scale measures gravitational force (weight) with some load cells. It then divides the weight by an assumed acceleration of gravity g to get the corresponding mass. Then it reports the mass in grams on the display.

The value of g programmed in by the manufacturer is presumbably representative of their market somehow. A lunar model would use a much smaller g. I could use one calibrated for Denver, as our local g is ~0.3% below the standard sea level value.

[ta0]

The problem isn't the duration of the oscillations anyway, (I could clock 15 or more of them, if it oscillates more rapidly). The problem is the angle of oscillation, which is maintained for a longer time if the pendulum oscillates slowly. Especially with light objects the pendulum slows down more rapidly, the angle of oscillation decreases during the clocking, and the reading becomes a bit less accurate.

I wouldn't be concerned with the angle of oscillation changing. To a first (or second?) order the period of oscillation should remain constant regardless of the angle, as long as you start with a relatively small angle (that is what makes pendulums so useful). But I guess that is something you could easily test.

The first expression for w_min came from J. Ginsberg, Advanced Engineering Dynamics, easily found free online. The subsequent expressions are mine, easily derived.

Thanks, Jeremy. I downloaded the whole text book and printed the spintop section. It looks like a good reference.