﻿ General Design Information > Cam Mechanisms > Dynamics of Cam Mechanical Systems > Example 3: Power Calculation for Cam Mechanism

# Example 3: Power Calculation for Cam Mechanism

## Power Calculation

 This example machine has a single, force-closed, cam with a swinging roller follower. The cam operates two mechanisms. The follower is also designed as a Bell-Crank. The Bell-Crank drives a: •Plunger, operating horizontally, with a linkage transmission •Work-head, operating vertically, with a rack and gear segment transmissionThe Plunger moves horizontally against a friction force. The Work-head and its rack and slide are lifted. The roller follower is held in contact with the cam by a spring. The spring is strong enough to provide the operating force for both mechanisms. The motion-law is a Cycloidal motion. Design Note: The Cam drives the two tools 'positively' away from the 'danger zone' - the Plunger is moved to the left, and the Work-head is moved upwards. The spring pulls the tools into the danger zone. If there is a jam, the cam-follower cannot 'return' move with the cam. This is likely to prevent further damage to the mechanisms. The cam is driven at 60RPM, by a motor and worm gear with a 24:1 ratio. The motor is an AC Squirrel Cage motor running at 1440RPM. The Challenge: Find the peak loading in the rise period of the cam - the most heavily loaded period - which occupies 72º of camshaft rotation. The Lift segment period = (60RPM/60)*72º/360) = 0.2sec. The mechanism does not distort the Cycloidal Motion-Law very much. The rigidity of the output transmission is estimated to be approximately 12,000Nm/rad. Find also the cam torque and the required size of the electric motor to provide sufficient power and input inertia.

Component or Assembly

Parameter

Value

Units

Bell-Crank lever, roller follower and rocker shaft assembly

Distance form Pivot to Conn-Rod

100

mm

Distance from Pivot to Follower center

100

mm

0.231

kg.m2

Angular Stroke

22.62 / 0.395

Efficiency

90

%

Connecting Rod

Mass including end joints

0.75

kg

Linear Stroke

40

mm

Plunger Lever

Pivot to Conn Rod

200

mm

Pivot to Plunger

160

mm

0.0747

kg.m2

Angular Stroke

11.31 / 0.197

Efficiency

95

%

Mass

3.25

kg

Slide Friction Force

3.15

N

Linear Stroke

32

mm

Gear Segment

110

mm

0.0855

kg.m^2

Angular Stroke

22.62/0.395

Efficiency

98

%

Mass

3.24

kg

Slide Friction Force

41

N

Linear Stroke

43.5

mm

Return Spring

Spring-post to Pivot

150

mm

Force at Extended Length

300

N

Force at short working length (pre-load)

120

N

60

mm

Convert all loadings to values referred to the bell crank lever, to simulate a simple oscillating mechanism.

The instantaneous velocity of a component is the cam law velocity factor at that point in the motion multiplied by the 'Stroke of the component ÷ by the motion period (time)'.

Since the period and velocity factor are the same for all components, the velocity of a component is proportional to its stroke, and velocity ratios are the same as the stroke radius.

Equivalent Inertia Calculations

Conversion

kg.m^2

Bell Crank

no conversion required

= 0.0231

Connecting-Rod

= 0.0077

Plunger Lever

= 0.0187

Plunger Assembly

= 0.0213

Gear Segment

no conversion required

= 0.0855

= 0.2183

= 0.3746

Equivalent Torque Calculations

In the Rise Period, , for Cycloidal Motion is 6.283,  the output stroke is B = 0.395rad, and the period is 0.2seconds.

Therefore, the maximum acceleration of the bell-crank lever in the rise period is:

=

The inertia torque referred to the bell-crank, is:

Natural Frequency

The natural frequency of the system can be estimated from the inertia and the rigidity to allow for dynamic response vibration :

The Period-Ratio is :

The Torsion-Factor Ct is found with the Torsion Factor Empirical Equation. :

Plunger Assembly Friction

Spring-force maximum

Spring force minimum

Inertia Ratio:

Cm= 0.693

Input Torque Coefficient:

Use an Excel Spreadsheet to calculate this value.

Cc= 1.593

The peak input torque (on the cam) allowing 90% cam and follower efficiency is:

The most economical drive unit is a worm gear with a 4-Pole Squirrel Cage Motor. It runs at 1440RPM at full speed (50Hz Supply).

This must be a reduction gear ratio of1440/60 = 24:1 Gear Ratio.

The 24:1 worm gear unit has a typical forward efficiency of about 75% and reverse efficiency of 30%.

Peak Power :

The motor with sufficient full-load power to cover this peak power must allow for the reduction gear efficiency of 75%.

A standard 0.4kW motor will be adequate. However, if we assume there is enough inertia at the input shaft, then we can choose a motor that has only the average power. We must consider the efficiency of 75%.  The maximum power is reduced.

Reduced Maximum Power :

We can choose a standard 0.25 kW motor.

We should check the input inertia.

The estimated camshaft, cam, coupling and worm-wheel inertia (all running at 60RPM) is 0.25 kg.m2

The motor and worm-shaft armature inertia is estimated at 0.0012 kg.m2.

The equivalent input inertia referred to the cam is:

The minimum inertia required to be sure that the speed fluctuation is not more 10% is found from this equation:

In this case the input inertia is adequate. It is interesting to note, however, that without the high speed motor inertia it is probable that the speed fluctuation would be unacceptable. In this example, the calculations are for rotary motion on the bell-crank. To design the Cam and Follower we must estimate the contact force.

The peak force on the follower in the direction of motion (that is, the useful force) is the peak torque divided by the length of the bell-crank arm:

This must be divided by the Cosine of the Cam Pressure Angle to get the Contact Force, which can then be checked against load capacity of the Cam.