In this topic, we review:

• | the harmful effects of Overrun. |

• | the Power we need to drive a cam mechanical system. |

One important aspect of input-torque (see here) is the existence of:
This has a great influence on the design needs of the input transmission. If the transmission is excessively elastic and/or the transmission has a significant amount of backlash, an overrun effect takes place.
This is OVERRUN. The amount and speed of the overrun can be considerable, even in fairly well designed machines.
The importance of the cam-shaft acceleration cannot be overstated. When we look at the equation for 'real output acceleration'... Real Output Acceleration = (Geometric Output Acceleration)×(input velocity)2 + (Geometric Output Velocity)×(input acceleration) The second term (after the +) is usually zero, because the input (rotating shaft) is intended to be at constant velocity, and thus, the input acceleration is zero. During overrun, however, there is high input acceleration. But that is not all, the first term in the expression, increases in proportion to the square of the increase in 'input velocity'. The two effects can increase the output acceleration, and therefore the inertia loading, by as much as tenfold (x10) in bad cases. We can consider two ways to reduce over-run : with and without a flywheel. Reduce Overrun: Good Input Transmission Design: In order of priority:
Even so, with high speed, or high inertia, applications, the elasticity of the input transmission can still be a problem. Reduce Overrun: Flywheels
If this is not practicable:
The effective inertia of the flywheel is increased by the gear-ratio 'squared', when referred to the cam.
An important 'side-effect' of worm-gearing is its low reverse efficiency. The normal forward efficiency of work 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 to 50 to 1, or about 55% for ratios between 15 and 25:1. This has a beneficial braking effect on the cam to reduce the overrun during the negative torque phase. Kinetic Energy, Flywheels, Speed Fluctuation Ignoring 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 energy is transferred from input to output as the output inertia is accelerated, and transferred back again when it is decelerated. 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, determined by the motion-law. 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%. [10% Speed Fluctuation Limitation is fairy arbitrary industry standard - you may specify any limitation]. Equating these two expressions gives: and similarly Example 1 Input Transmission Design Improvement |
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Original Input Transmission |
Improved Input Transmission |
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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. |
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Example 2 Input Transmission Design Improvement |
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Similarly, the Indexing Unit is moved to closer 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: For most applications, the input velocity is virtually constant, and is usually expressed as the number of camshaft revolutions per minute. The peak power is thus: where:
We can now give an expression for the Peak Power of a cam system in terms of the camshaft 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: or 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 camshaft. The power fluctuates during the motion period, and 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. |