One important aspect of input-torque (see here) is the existence of:

•a period of negative torque

•a rapid change from positive to negative torque, particularly for some cam-laws (motion-laws).

This has a great influence on the design needs of the input-transmission.

If the input-transmission is excessively elastic and/or it has a significant amount of backlash, an Overrun effect takes place.

What is overrun, and how does it happen?

•During the positive torque phase, the payload lags the cam-shaft - the elasticity of the transmission winds-up and its backlash is taken up.

•As the torque reduces, unwinding occurs until the torque is zero. Then the system goes free while the backlash is taken up the 'other way'.

•The transmission winds-down in the opposite direction, the payload system overruns its drive. The output from the Follower leads the input to the cam.

•When the payload is no longer driven by the motor, or even the payload starts to drive the motor, the speed of the motor can rapidly increase.

This increase in the speed of the motor is called OVERRUN.

The amount the speed increases can be considerable, even in well designed machines.

•If the overrun happens very quickly, there is also a significant cam-shaft acceleration.

The importance of the cam-shaft acceleration cannot be overstated.

Actual Output Acceleration:

Actual Output Acceleration = (Geometric Output Acceleration)×(input velocity)2 + (Geometric Output Velocity)×(input acceleration)

1.The second term (after the +) is usually zero, because the input shaft should rotate with a constant angular velocity.

During overrun, however, the input cam-shaft accelerates. Therefore the second term is not zero.

2.The first term (before the +) will increase in proportion to the square of the increase in the angular velocity of the input shaft.

The two effects increase the Actual Output Acceleration. The inertia loading (proportional to the output acceleration), can increase due to the increase in by as much as , in bad cases.

How to reduce Over-Run?

There are two ways to reduce over-run:

1: Good Input Transmission Design

In order of priority:

•Consider a new Motion-Law - one where the input torque does not change rapidly - for example change to a Mod-Sine from a Mod-Trap motion law.

•Eliminate backlash.

•Increase rigidity.

Even so, with high speed, or high inertia, applications, the elasticity of the input transmission can still be a problem.

2: Flywheel

Add a flywheel near to the cam.

If this is not possible:

•Use a high-ratio reduction gear-box, connected near to the cam, with a small flywheel on the high-speed input to the gear-box

The effective inertia of the flywheel is increased by the gear-ratio-squared, when referred to the cam.

•Connect high quality, low backlash, worm-gearing directly to the motor shaft. The motor armature of the motor becomes an effective flywheel.

An important side-effect of worm-gearing is its low reverse efficiency. The normal forward efficiency of worm gear is about 75-85%. The reverse efficiency - that is when the input torque is negative, is very low, typically 10% for gear-ratios of 30 - 50:1, or about 55% for ratios between 15 - 25:1.

This has a beneficial braking effect on the cam-shaft, which reduces the overrun during the negative torque phase.

Kinetic Energy, Flywheels, Speed Fluctuation

If we ignore any input braking effect and friction loading, there is a constant amount of Kinetic-Energy in the system when it is running at full speed. However, some of the Kinetic-Energy is transferred from the input to the output as the output inertia accelerates, and from output to the input when the output inertia decelerates.

The loss of energy from the input causes the input to slow down, and then when the input regains energy, it speeds up again.

The Output Kinetic-Energy varies from zero, when the load is in its dwell-period, to a maximum, when the load is at its maximum speed.

The Input Kinetic-Energy must fluctuate by the same amount.

We can find how much input-inertia is required to limit the speed fluctuation of the input to 10%.  - a 10% speed fluctuation limit is a fairy arbitrary - you may specify any limitation.

Equating these two expressions gives:

and similarly

Example 1 - Input Transmission Design Improvement

The images show an original and improved input transmission arrangement.

The design has a constant speed conveyor and in indexing table driven by the same motor and work gear.

The original design - left - has long shafts and a potentially slack chain drive between the worm gear and the indexing cam.

The improved design - right - has essentially the same mechanism layout, but has a better position for the motor and work gear.

The improved design puts the motor and worm gear near to the cam to give a far better dynamic performance at high speed.

Example 2 - Input Transmission Design Improvement

Similarly, the Indexing Unit is moved near to the Motor and Reducer.

The power being transmitted at any moment at any point in the transmission is its instantaneous load multiplied by the instantaneous velocity:

where:

  - torque at the cam

- angular velocity at the cam

For most applications, the input velocity is virtually constant, and is usually expressed as the number of cam-shaft revolutions per minute.

The peak power is thus:

Input Velocity =

where:

We can now give an expression for the Peak Power of a cam system in terms of the cam-shaft speed and the input torque, with input torque being a function of the output loading, the input and output stroke and the input torque coefficient of the motion-law as described:

This is the Peak Power, in the system used by the cam. To find the size of a motor to drive the mechanism, we must add any power losses in the input transmission (friction, gearing etc).

But we can take advantage of the fact that there is usually a significant amount of kinetic-energy associated with the cam-shaft.

The power fluctuates during the motion period. But the power to drive the cam is very low during a dwell period.

The reserve of energy, derived from the input inertia, supplies the extra power needed to cope with the peak requirement, provided that the input inertia is adequate, as described above.

It is therefore necessary only to have a power source that can provide the 'average power' of the motion cycle. The average power is never more than half of the peak power, and if the dwell period(s) are much longer than the motion period(s), the input energy reserve can be restored during one cycle with a power input much less than half the maximum.

Where the power source is an A.C induction motor, quite apart from the benefit of a fairly high armature inertia (at high speed), there is the advantage that its peak torque is more than twice its normal torque rating.

Thus, it is always safe to select a motor with a:

This would be for one mechanism. Clearly, you must calculate the power needed for all of the mechanisms in the machine.