physics question

[Massive Lee]

Circumference is 2 x pi x radius. Unlike surface measurement, it increases linearly.

Therefore, the circumference at 3" is half of the circumference at 6". Linear speed follows circumference progression.

[Massive Lee]

Circumference is 2 x pi x radius. Unlike surface measurement, it increases linearly.

Therefore, the circumference at 3" is half of the circumference at 6". Linear speed follows circumference progression.

[Conki]

http://wimp.com/metricsystem/

[TwoJ's]

ok so no matter what size pulley I use, it will always be spinning at the driving rpm anywhere on the pulley. Its the linear speed that varies. Thus a 6" pulley is spinning slower than the 3" driving pulley because its coupled with a belt moving literally linearly.

Keep in mind that although the rotational speed is the same, the belt will rotate twice as much as it would with the 3" pulley. So whatever the other end of the belt is on would be stepped up.

[Wh33lhop]

Keep in mind that although the rotational speed is the same, the belt will rotate twice as much as it would with the 3" pulley. So whatever the other end of the belt is on would be stepped up.

Depends on which pulley you are changing out.

[TwoJ's]

All I meant was a switch from a 3" to a 6" pulley will move the belt twice as fast if it's a driving pulley (or double the load if it's driven).

[b*saint]

So let me ask this then. Lets say Im driving a 2:1 ratio system. Will a 6" driving 12" driven be the same load on the driving motor as a 3" driving 6" driven system? Will it take more effort to drive physically bigger gear train with the same ratios?

[Farbin Kaiber]

I'm glad to see so many people already posted what I wanted to.

[TwoJ's]

Yes -- well for the most part. There will be a bit more rotational inertia to overcome because of the increased mass of the pulleys, but the drag from the load will be the same as long as you change the pulley size proportionally.

[Wh33lhop]

In the sense that you mean, yes. However, when you get down to it the parasitic drag from the inertia of the larger diameter pulleys, the larger belt required and the higher belt speeds of the 6"/12" system will be greater. Think of it like a really heavy flywheel.

Edit: god damn it, that's what happens when you make breakfast in the middle of writing a response.

[b*saint]

Alright I believe you but Im just wondering why people with bigger tires (36" off road vs the stock size tires) on trucks, break axles and gears with bigger tires easier? More leverage on the outside of the tire to stop the shaft from turning?

[Wh33lhop]

You are not talking about a pulley system with axles/tires. The axles remain the same diameter, while you are increasing the rolling diameter requiring more torque acting on the tires for the same force on the road (accelerating/braking). So the axles remain at the same torsional strength while you are asking for more torque through them.

[b*saint]

So it would be the same idea applied to the whole 3" and 6" and 6" and 12". We would be increasing the rolling diameter of the pulleys thus requiring more shaft torque to do the same amount of work.

[TwoJ's]

It is the increased moment of inertia. It puts a much greater load on the axles. Also, in rotational situations, increased inertia exponentially increases the torque required to start motion. You can see this with power factors on electric motors... the power required to start a high torque motor is huge.

The equation for the moment of inertia (resistance to motion) of a disc (wheel/tire combo) is .5 x mass x radius^2. Not only does the mass increase it, but the radius increases the MOI exponentially. So if you double the radius and mass, you effectively increase the moment of inertia by a factor of 8.

[TwoJ's]

So it would be the same idea applied to the whole 3" and 6" and 6" and 12". We would be increasing the rolling diameter of the pulleys thus requiring more shaft torque to do the same amount of work.

The difference in inertia of two small pulleys is not nearly as great as large wheels and tires. An electric motor that was spec'd for the 3"/6" setup is *likely* to be just fine with the 6"/12". You can calculate the difference in moment of inertia and find the added load.

Nothing is changing aside from the added mass and diameter. you are halving the force in the belt by doubling the radius, but it doesn't change the amount of torque exerted on the larger pulley's shaft.