- #1
supernaught
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Greetings, lurked for a long time, but I need some help here so I joined up. Please have some patience with me if I miss any important information.
I am working on a belted pulley drive system where the driveN pulley will move relative to the driveR while in the stopped state.
An outside idler pulley is used to take up the slack as the distance between R&N changes.
I did not design this system, but I am tasked with making it work.
The idler force is managed using a proportional valved pneumatic cylinder, controlled by a PLC which also knows the distance between the R & N pulleys.
I need to know the force to apply to the cylinder in order to maintain correct tension on the belt to prevent skipping and excessive wear.
Here are the conditions
The belt is 1750mm long, 8mm pitch, synchronous(ribbed), 30mm wide
The pitch diameter of the R&N pulleys are the same ~180mm
The idler pulley(smooth) diameter is ~160mm
The distance between the R & N pulleys is variable between 485mm and 560mm
The next part that makes this more difficult (at least in my head) is that the idler does not apply force perpendicular to the line between the pulleys. The DriveN pulley and the idler move on a parallel axis, offset by ~185mm(from pulley centres).
To illustrate this, picture an XY plane with the driveR fixed at (X0 Y0). The driveN pulley can travel between (X-400 Y400) and (X-286 Y400). The idler then travels along a rail at Y215, parallel to the N pulley just to keep tension.
Power
The power and speed transmitted from driveR is variable. There is a 2 speed gearbox(4:1 & 1:1) where speed is limited (4:1 is 0-1500rpm), 1:1 is 1500-6000rpm. This is a speed controlled motor, so maximum power available is 45KW peak through a very wide rpm range.
So all I really need to know, is what is the ideal proportional curve/slope of the idler force given the geometry.
Feel free to ask for more information, I am interested in sorting this out. Also if there is a calculator somewhere that could help me figure this out(as well as forces acting on the pulley shafts at maximum power output in 4:1 gear), it would be greatly appreciated.
Many thanks,
Rob
I am working on a belted pulley drive system where the driveN pulley will move relative to the driveR while in the stopped state.
An outside idler pulley is used to take up the slack as the distance between R&N changes.
I did not design this system, but I am tasked with making it work.
The idler force is managed using a proportional valved pneumatic cylinder, controlled by a PLC which also knows the distance between the R & N pulleys.
I need to know the force to apply to the cylinder in order to maintain correct tension on the belt to prevent skipping and excessive wear.
Here are the conditions
The belt is 1750mm long, 8mm pitch, synchronous(ribbed), 30mm wide
The pitch diameter of the R&N pulleys are the same ~180mm
The idler pulley(smooth) diameter is ~160mm
The distance between the R & N pulleys is variable between 485mm and 560mm
The next part that makes this more difficult (at least in my head) is that the idler does not apply force perpendicular to the line between the pulleys. The DriveN pulley and the idler move on a parallel axis, offset by ~185mm(from pulley centres).
To illustrate this, picture an XY plane with the driveR fixed at (X0 Y0). The driveN pulley can travel between (X-400 Y400) and (X-286 Y400). The idler then travels along a rail at Y215, parallel to the N pulley just to keep tension.
Power
The power and speed transmitted from driveR is variable. There is a 2 speed gearbox(4:1 & 1:1) where speed is limited (4:1 is 0-1500rpm), 1:1 is 1500-6000rpm. This is a speed controlled motor, so maximum power available is 45KW peak through a very wide rpm range.
So all I really need to know, is what is the ideal proportional curve/slope of the idler force given the geometry.
Feel free to ask for more information, I am interested in sorting this out. Also if there is a calculator somewhere that could help me figure this out(as well as forces acting on the pulley shafts at maximum power output in 4:1 gear), it would be greatly appreciated.
Many thanks,
Rob