Determining motor/control options to act on a spring

[TPayne]

I'm building a power hammer for a buddy's forge and am working through the design phase. We are going to use a 1200lb leaf spring to actuate the upper anvil. Pretty much got the mechanical design down, but having some issues with the selection of motor and control.

The leaf spring we selected has a Spring Rate of ~300lb/in (force required to compress 1"). We would like the anvil to move 4-6" (travel of the spring is 5"--but this design pivots at CL through some pillow bearings). I feel that this means we need 1000lbf to mechanically act on the leaf spring but THROUGH a v-drive belt pulley system. I found these at automationdirect.com:

https://www.automationdirect.com/ad.../ac_motors/conveyor-z-farm_duty/mtf2-003-1b18

https://www.automationdirect.com/ad...quency_drives_(vfd)/general_purpose/gs21-23p0

We will have 230V single phase power, that's why the phase converter for a 3ph AC induction motor. I really feel like farm-duty is the way to go for the motor, but am not confident that a 3HP motor will provide enough power...especially because (with the motor speed being 1800RPM) I will be doing a speed reduction (6:1?) via the v-belt pulleys to get the RPM down. Any further speed control will be via the VFD with the goal to get all the way to <60RPM strikes at the anvil.

Before throwing $1000 at a motor and controller I wanted to see if anyone with more physics/mechanical/electrical engineering experience could confirm that this motor/controller setup I linked meets our intended needs (Listed:
-speed control down to 40RPM via VFD
-single phase 220V input
-motor HP sufficient for force required to actually strike the anvils with force

 

[jrmichler]

The calculation procedure is as follows:

1) Assume stroke is 5 inches, peak RPM is 300, and peak force on the connecting rod is 1000 lbs.

2) The maximum torque at the drive pulley is 1000 lbs X 2.5 inch radius = 2500 inch-lbs. This includes the worst case assumption that peak force will be with the crankshaft at 90 degrees to the connecting rod. If you can positively guarantee that the peak force will be at a different crank position, then you need to calculate the worst case torque.

3) Horsepower at the drive pulley = 2500 X 300 / 63,025 = 12 hp. The 63,025 is the conversion factor to calculate horsepower given RPM and torque in in-lbs.

4) You need to check with the VFD manufacturer (it might be in their installation manual) to find the minimum percent RPM at which the motor/drive combination will develop full torque.

Keep in mind that TEFC and ODP motors generate heat in proportion to the actual torque, but the cooling capacity is proportional to RPM. Extended running at low RPM and high torque will overheat the motor. It's not a problem if the duty cycle is low enough - just shut it down if the motor gets too hot.