Simple torque transfer question

[wakadarc]

So I am specing out a motor that will be connected to a flywheel with a belt. Is the following formula correct?

T_Flywheel = (R_out / R_in ) x (T_in)

 

[Mech_Engineer]

If you're ignoring efficiency losses, that's about right.

 

[wakadarc]

Im trying to calculate the torque needed for the flywheel. Attached is a picture of the flywheel (large circle) and the eccentric shaft (medium circle). The small circle is just the projectory of the eccentric shaft rotating around the center of the flywheel. If the load (platform and box on top) is 150kg, what torque would I need? Worst case would be the dead point once the cam reaches the bottom.

 

[billy_joule]

Im trying to calculate the torque needed for the flywheel. Attached is a picture of the flywheel (large circle) and the eccentric shaft (medium circle). The small circle is just the projectory of the eccentric shaft rotating around the center of the flywheel. If the load (platform and box on top) is 150kg, what torque would I need? Worst case would be the dead point once the cam reaches the bottom.

That is not the worst case...ie that is not the point where the applied torque is greatest for constant ω.
You need to draw some free body diagrams and analyse forces.

I see you've made a few threads on this design, It would probably be easier to assist if you kept it all in one thread.

 

[wakadarc]

Where would the worst case then? When the eccentric shaft is coming up? (right side of the circular path?) Or probably when it is about to reach the top? I simplified the FBD with the diagram as the cam is pretty much the taking the load at the top pin.

Keep in mind the platform and box are restricted so that they can only move up and down (only y axis)

 

[billy_joule]

https://en.wikipedia.org/wiki/Free_body_diagram

A FBD will show the torque required at the top and bottom of the platforms motion is zero (ignoring bearing friction etc)
There will be points where, depending on ω, the torque required could be positive or negative (ie the motor will need to apply braking).

Start with a FBD of just the platform, what forces are required for the simple harmonic motion?

 

[wakadarc]

Well there will be a downward force on the platform (gravity). Same with the box. The cam itself will project a force onto the center line of the eccentric shaft regardless of position it is in on the circular path it takes.

Omega will ultimately depend on the box being to able to lift off the platform 15mm...so omega will definitely vary from box to box.

It is not clear in my drawing here of the connection between the cam and the platform but the platform will be moving strictly up so the cam will tilt on its way down and on its way up.

 

[billy_joule]

Well there will be a downward force on the platform (gravity). Same with the box. The cam itself will project a force onto the center line of the eccentric shaft regardless of position it is in on the circular path it takes.

What you are looking for is how the linear force varies with time to produce the sinusoidal motion of the platform. This will lead to the torque required for motor selection and the forces for shaft design.

Omega will ultimately depend on the box being to able to lift off the platform 15mm...so omega will definitely vary from box to box.

Have you done any math to prove this? Recheck it, the mass of the box should have canceled out...

 

[wakadarc]

Wouldn't only the normal force of the shaft cancel out the mg force of the cam/box/platform load when it reaches the bottom of the circular path it takes? See attached. Thanks again

 

[billy_joule]

Wouldn't only the normal force of the shaft cancel out the mg force of the cam/box/platform load when it reaches the bottom of the circular path it takes? See attached. Thanks again

I'm not sure exactly what you are responding to here.

Omega will ultimately depend on the box being to able to lift off the platform 15mm...so omega will definitely vary from box to box.

Put two items of different weight in you palm and lift your hand up and down like your platform. Increase the speed until they rise off your palm.
They both lift off at the same time. That is, omega will not vary from box to box, however, the torque required will. The math will confirm this.

If you actually plan on building this thing you'll really need a mechanical engineer to do a proper force analysis. This is a slider crank mechanism, which are covered ad naseum in ME. If you are interested in learning how to do it yourself I'd recommend Engineering Mechanics: Dynamics by Meriam & Kraige and also Shigleys Mechanical Engineering Design.

 

[wakadarc]

Thanks billy_joule. I have seen vibration simulation/testing machines and a lot of them seem to use 1 HP motors. For my purposes, I just need a motor that can handle the worst case scenario load...I just need to determine a max torque so I can appropriately apply a safety factor when purchasing the right motor. I will look into the those books. Thanks again.

 

[wakadarc]

here is an update on the mechanical design