This example machine has a single, forceclosed, cam with a swinging roller follower. The cam operates two mechanisms. The follower is also designed as a BellCrank. The BellCrank drives a:
The Plunger moves horizontally against a friction force. The Workhead 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 motionlaw 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 Workhead is moved upwards. The spring pulls the tools into the danger zone. If there is a jam, the camfollower 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 MotionLaw 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 

BellCrank lever, roller follower and rocker shaft assembly 
Distance form Pivot to ConnRod 
100 
mm 
Distance from Pivot to Follower Centre 
100 
mm 

Inertia about Pivot 
0.231 
kg.m2 

Angular Stroke 
22.62 / 0.395 
º / rad 


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 

Inertia about Pivot 
0.0747 
kg.m2 

Angular Stroke 
11.31 / 0.197 
º / rad 

Efficiency 
95 
% 
Plunger and Link 
Mass 
3.25 
kg 

Slide Friction Force 
3.15 
N 

Linear Stroke 
32 
mm 
Gear Segment 
Pitch Circle Radius 
110 
mm 

Inertia about Pivot 
0.0855 
kg.m^2 

Angular Stroke 
22.62/0.395 
º / rad 

Efficiency 
98 
% 
Workhead, rack and slide Assembly 
Mass 
3.24 
kg 

Slide Friction Force 
41 
N 

Linear Stroke 
43.5 
mm 
Return Spring 
Springpost to Pivot 
150 
mm 

Force at Extended Length 
300 
N 

Force at short working length (preload) 
120 
N 

Linear Load 
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 
ConnectingRod 
= 0.0077 

Plunger Lever 
= 0.0187 

Plunger Assembly 
= 0.0213 

Gear Segment 
no conversion required 
= 0.0855 
Workhead Assembly 
= 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 bellcrank lever in the rise period is: 
= 

The inertia torque referred to the bellcrank, 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 PeriodRatio is : 

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



Noninertia Load 

Plunger Assembly Friction 

Workhead slide Friction 

Springforce maximum 

Maximum nonInertia load: 

Spring force minimum 

Minimum noninertia load (the spring preload): 

PreLoad Ratio: 

Inertia Ratio: 

Load Mix Coefficient 
Use an Excel Spreadsheet to calculate this value. Please ask, if you want example. Cm= 0.693 

Input Torque Coefficient: 
Use an Excel Spreadsheet to calculate this value. Cc= 1.593 

The Peak Output Loading on the bell crank is: 
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 4Pole 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 fullload 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 wormwheel inertia (all running at 60RPM) is 0.25 kg.m2 The motor and wormshaft 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 bellcrank. 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 bellcrank 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. See CamAnalysis and Design Checklist. 