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Author Topic: Has anyone seen a finger top with no stem?  (Read 4208 times)

Jeremy McCreary

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Re: Has anyone seen a finger top with no stem?
« Reply #15 on: August 30, 2016, 07:08:22 PM »

I started to read that post and my brain short circuited and eyes glassed over very quickly lol. Ok, here is one thing you state...

Sorry, I tend to have that effect on people, as anyone here can tell you. After decades of thinking in those terms, it's hard to stop. You should see the glaze-over times I get when I talk to my kids about science. Actually, you can't see them anymore, because the kids quickly learned to spot lectures coming. Now they glaze over before the words are even out of my mouth.  ;^}

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

Way over my head as I don't even know what acceleration of gravity means, I thought gravity was a constant. If a scale is not reporting weight but mass, then why do we have the word weight if the word denotes the combination of mass and gravity? Science is great when I can understand it, lol.

Wikipedia is a great resource on such things. Strictly speaking, mass is what gravity acts upon, and weight is the magnitude of the resulting force. For various reasons, a typical digital scale reads out the mass in grams or kilograms, not the weight, which is almost 10 times larger and measured in Newtons.

The acceleration of gravity g is just the proportionality factor between the mass and weight of a given object. The value of g varies very slightly with elevation and latitude and longitude, but generally not enough to worry about with tops. By international agreement, the standard sea-level value of g is exactly 9.80665 m/s2, but your local g will vary. Here in Colorado at almost 1,800 m elevation, my local g = 9.779 m/s2, or about 0.1% less. A typical digital scale really measures weight with its load cells, divides the weight by a pre-programmed value of g to get the corresponding mass, and then shows that mass on the display.

Why distinguish between mass and weight? Well, when comparing forces on a top, you'd better use weight for gravity's contribution, and that means multiplying the mass given by your scale by the appropriate value of g. Otherwise, you'll underestimate gravity's contribution by almost a factor of 10. The gravitational torque on a top is proportional to its weight, which in turn is proportional to its mass. Tip friction also depends strongly on weight.

« Last Edit: August 30, 2016, 07:16:13 PM by Jeremy McCreary »
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Atomic

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Re: Has anyone seen a finger top with no stem?
« Reply #16 on: August 30, 2016, 09:26:58 PM »

Ok, here is the video I made about tops staying upright when stopped.

https://www.youtube.com/watch?v=sZ1rKHmtPFA
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Atomic

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Re: Has anyone seen a finger top with no stem?
« Reply #17 on: August 30, 2016, 10:38:53 PM »

Wikipedia is a great resource on such things. Strictly speaking, mass is what gravity acts upon, and weight is the magnitude of the resulting force.
-This I understand.

For various reasons, a typical digital scale reads out the mass in grams or kilograms.

- This I still don't understand because isn't weight the resulting force that we are measuring? For example in space the mass would be the same but not the gravitational pull and therefore isn't weight at any given time subjective to the strength of a gravitational influence? Hmm, I know for example a top will spin in a vacuum longer, but that is due to the lack of friction with air. But if we took a top to place where there is a  vacuum AND no gravity,how is weight measured? Is that where "weightless" comes from? But it still has mass. So for example what is the thrust needed in space to move a certain mass compared to earth if in a vacuum chamber and if the movement was parallel to the gravitational field, not away from it. Ok, maybe this is getting to carried away, lol.

....not the weight, which is almost 10 times larger and measured in Newtons.
***I will go study about Newtons although I understand that is a unit of measurement, I don't full get what is being measured exactly.
.

Why distinguish between mass and weight? Well, when comparing forces on a top, you'd better use weight for gravity's contribution, and that means multiplying the mass given by your scale by the appropriate value of g. Otherwise, you'll underestimate gravity's contribution by almost a factor of 10. The gravitational torque on a top is proportional to its weight, which in turn is proportional to its mass. Tip friction also depends strongly on weight.
[/quote]

- comparing forces on a top, how many are there other than gravity? Do you mean things like density, inertia, stored kinetic energy?   

- gravitational torque? What torque?  I understand the pull of gravity will have an effect on friction of the contact point but I am still having a hard time understanding where is the break even point (if that's what it is) between the advantage of more weight and kinetic energy to the increased friction it will cause on the tip. Also another thing I struggle with is where the weight distribution is best given a tops size so to get the best ratio of spin time at lower revs but still has the most weight possible. If the increased friction isn't such a big issue especially for a small top, then it seems there would be a math formula for a spinning speed that an average person could hope to attain, which would give the best weight/mass distribution of a given size of top. I think that's way too much for me to get into, but basically as far as I know it's true that if a few tops of exactly the same proportions but of greatly differing weight are spun exactly the same then the heavier one will spin longest. Foreverspin has said their tungsten tops spin longest for example.

Maybe I am focusing to much on the weight issue and not enough on other factors that can have significant impact on a top, such as air resistance. I watched one video where the guy only highly polished the top and increased the spin time dramatically.

Ok, my brain hurts, hope all that doesn't sound too stupid. Thanks for your replies and I will go read about Newtons to kill off any brain cells that still think they have any intelligence :)
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ta0

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Re: Has anyone seen a finger top with no stem?
« Reply #18 on: August 31, 2016, 11:26:14 AM »

Quote
....not the weight, which is almost 10 times larger and measured in Newtons.
***I will go study about Newtons although I understand that is a unit of measurement, I don't full get what is being measured exactly.

A scale that uses a spring or a load cell measures weight which is a force.
From this point of view your bathroom scale works everywhere: when it says that you weigh 65 kg on the Earth and 24.5 kg on Mars, both are correct: you are talking about kilogram force (kgf or kilopond) which is a different unit than kg mass. It just happens that something with a mass of 1 kg weighs 1 kgf on the Earth (or 9.8 Newtons, which is the correct unit for force in the same system of units as the kg mass).
If the table you brought from Earth says it withstands 50 pounds, you can use the scale you brought from Earth to check if you can stand on it  :P
« Last Edit: August 31, 2016, 11:34:29 AM by ta0 »
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Jeremy McCreary

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Re: Has anyone seen a finger top with no stem?
« Reply #19 on: August 31, 2016, 06:56:04 PM »

Maybe I am focusing to much on the weight issue and not enough on other factors that can have significant impact on a top, such as air resistance.

You have to worry a lot about weight in a throwing top, but it's generally a minor consideration in a high-performance pocket-sized finger top meant to spin in place. In the latter case, it's generally much more productive to fret over center of mass height, mass distribution about the spin axis, aerodynamics, and ground clearance. Tip friction is the only performance factor that depends on weight per se without regard to how the weight is distributed through the top, and you can generally limit tip friction by choosing a very hard, low-friction tip material, maximizing tip smoothness, and minimizing tip radius of curvature. Bear in mind, though, that the overall performance hit due to aerodynamic drag is generally much bigger than that due to tip friction -- even in very smooth tops.

This I still don't understand because isn't weight the resulting force that we are measuring? For example in space the mass would be the same but not the gravitational pull and therefore isn't weight at any given time subjective to the strength of a gravitational influence?

Yes, and the acceleration of gravity g measures the "strength" of the gravitational field (influence) at a given point in space.
 
But if we took a top to place where there is a  vacuum AND no gravity, how is weight measured?  Is that where "weightless" comes from? But it still has mass.

The vacuum is irrelevant here. Where there's no gravity, the local g = 0. And since the magnitude of the weight W = m g, the local weight W = 0, too -- regardless of the mass m involved.

....not the weight, which is almost 10 times larger and measured in Newtons.
***I will go study about Newtons although I understand that is a unit of measurement, I don't full get what is being measured exactly.

Weight is by definition the force of gravity. Forces are properly measured in Newtons (N) in the International System of Units -- "SI" for short. (Don't get me started on Imperial units.) The SI unit of mass is the kilogram (kg). The kilogram-force (kgf) unit mentioned by ta0 is not an SI unit, and the SI folks discourage its use. Hence, I prefer to interpret digital scale readouts in the way I outlined earlier. For scales near the Earth's surface, my interpretation is for all intents and purposes equivalent to ta0's.

- comparing forces on a top, how many are there other than gravity? Do you mean things like density, inertia, stored kinetic energy?

Aerodynamic drag and tip friction are ultimately due to viscous, inertial, frictional, and elastic forces acting on various parts of the top. If the top is thrown, or is twirled in a breeze or on a moving surface, additional forces enter the picture. And if a top has moving parts, like the centrifugal tops below, or has inadequately secured parts or too much flex, then centrifugal forces also come into play.

https://www.youtube.com/watch?v=nTNyuuhw04s

All of these forces can affect top behavior in visible ways.

- gravitational torque? What torque?  I understand the pull of gravity will have an effect on friction of the contact point but I am still having a hard time understanding where is the break even point (if that's what it is) between the advantage of more weight and kinetic energy to the increased friction it will cause on the tip. Also another thing I struggle with is where the weight distribution is best given a tops size so to get the best ratio of spin time at lower revs but still has the most weight possible. If the increased friction isn't such a big issue especially for a small top, then it seems there would be a math formula for a spinning speed that an average person could hope to attain, which would give the best weight/mass distribution of a given size of top. I think that's way too much for me to get into, but basically as far as I know it's true that if a few tops of exactly the same proportions but of greatly differing weight are spun exactly the same then the heavier one will spin longest. Foreverspin has said their tungsten tops spin longest for example.

We'll have to take this up another day, but consider an ideal top with its center of mass (CM) and tip exactly on the spin axis. If that axis deviates even a hair from the vertical defined by local gravity, the CM will no longer be directly over the point where the tip contacts the ground. Since gravity acts vertically through the CM, it will then exert a torque on the top about its tip. This gravitational torque is what ultimately brings the top down.
« Last Edit: August 31, 2016, 07:19:23 PM by Jeremy McCreary »
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Atomic

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Re: Has anyone seen a finger top with no stem?
« Reply #20 on: September 01, 2016, 03:34:17 PM »

Thanks Ta0 and Jeremy for the detailed answers. The only thing I don't get is the 0 weight in 0 gravity as it must take some sort of force to move something.

What I have learnt is that one can make whatever one wants and that's ok, but for best spin times a top with a low Cg and most of the weight towards the outside diameter will spin longest, all other things being equal. I admit though, I forget why the weight to the outside works better. Does it make a larger area over all for the centre of gravity?
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Jeremy McCreary

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Re: Has anyone seen a finger top with no stem?
« Reply #21 on: September 02, 2016, 03:17:20 PM »

Thanks Ta0 and Jeremy for the detailed answers. The only thing I don't get is the 0 weight in 0 gravity as it must take some sort of force to move something.

You're welcome. That's where mass comes in, and why mass is so important in throwing tops. Mass measures a body's inertia, or more properly, its resistance to acceleration (or deceleration) by a force acting in a straight line. So double the mass, and you'll need twice the force to achieve the same straight-line acceleration.

This is the gist of Newton's 2nd Law, which I encourage you to read about on Wikipedia. The law applies to all bodies with mass, including weightless ones. After all, weight is just one force among many possible forces.

What I have learnt is that one can make whatever one wants and that's ok, but for best spin times a top with a low Cg and most of the weight towards the outside diameter will spin longest, all other things being equal. I admit though, I forget why the weight to the outside works better. Does it make a larger area over all for the centre of gravity?

Pushing a top's mass "to the outside" at design time greatly increases a measure of mass distribution known as the axial moment of inertia (AMI). AMI is to torque as mass is to force, in that AMI measures a body's resistance to angular acceleration due to a torque about the spin axis. So double the AMI by any means, and you'll need twice the torque to achieve the same angular acceleration. Angular acceleration is just the change in rotation speed per unit time. During a twirl from rest, the average angular acceleration is the speed gain divided by the time it took to get there. Or put another way, the faster you get to 1,000 RPM from rest, the greater the angular acceleration up to that point in the twirl, and the greater the average torque you had to apply to achieve it.

During spin-up, you apply an accelerating torque to the top about the spin axis. After release, the top experiences a braking torque about the spin axis due the combined action of aerodynamic drag and tip friction. The greater the AMI, the more muscle (torque) needed to hit a given release speed in a given amount of time during spin-up. But that works to your advantage after release, as the greater AMI then does a better job of resisting the braking torque's effect on top speed during spin-down.

But wait! There's more! In a well-balanced top, a greater AMI also reduces the minimum speed needed for a stable, wobble-free spin. Hence, as long as you can still twirl the top to high speed, increasing the AMI at constant mass will prolong spin time in 2 different ways.

AMI is proportional to top mass, but 2 tops with the same mass but different mass distributions relative to the spin axis can have drastically different AMIs. I strongly encourage you to read up on moment of inertia on Wikipedia, as it's generally far more important than mass in finger top design.
« Last Edit: September 02, 2016, 10:42:25 PM by Jeremy McCreary »
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Atomic

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Re: Has anyone seen a finger top with no stem?
« Reply #22 on: September 02, 2016, 04:12:53 PM »

Ok, read Newtons second law and when the brain cools down will go for the AMI.

Actually the 2nd law is pretty easy to generally understand in concept in that all the factors are in a way equal in resultant action, but I need to think about a few of the fine details so will look at a few more explanations on this.

The AMI topic, I'll get back on that one, need batteries...
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