Designing a conveyor system involves more than just bolting a motor to a frame. If you undersize the motor, it won't start under load due to breakaway torque . If you oversize it, you waste thousands on electricity and oversized VFDs. In this guide, we will walk through the engineering math required to size a conveyor motor and gearbox correctly, specifically focusing on the critical "Dynamic Tension" resulting from inertia. Table of Contents 1. The Physics: Effective Pull (Te) 2. Calculating Motor Power (Worked Example) 3. The Inertia Problem: VFD vs DOL 4. Gearbox Ratio Selection 5. Frequently Asked Questions Advertisement 1. The Physics: Effective Pull (Te) The first step in any sizing calculation is determining the Effective Pull ( T e ) . This is the sum of all forces resisting the motion of the belt. The Basic Formula: T e = F friction + F gravity + F material...
Figure 1: Deriving the "secret sauce" of smooth motion control requires calculus.
In [Polynomial Cam Function (Introduction) - Part 1], we discussed the fundamental law of cam design: continuity of acceleration.
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In this post, we are going to derive the exact equations for the Fifth-Degree Polynomial Cam Function. This is the mathematical foundation used in Motion Control Algorithms for high-end servo drives to ensure smooth, jerk-free movement.
The General Equations
We start with the general polynomial equation. To make the math handleable, we normalize the input angle as a ratio (x = θ / β), where x goes from 0 to 1.
s = C0 + C1x + C2x2 + C3x3 + C4x4 + C5x5
Where:
s = Displacement (mm)
x = Ratio of cam angle (θ / β)
β = Total angle in sector (rad)
To find Velocity (v) and Acceleration (a), we differentiate with respect to the angle.
(Note: The chain rule applies. v is in mm/rad. To convert to mm/s, multiply by angular velocity ω).
Velocity Equation (v = ds/dθ):
v = (1/β) · [ C1 + 2C2x + 3C3x2 + 4C4x3 + 5C5x4 ]
Acceleration Equation (a = dv/dθ):
a = (1/β2) · [ 2C2 + 6C3x + 12C4x2 + 20C5x3 ]
Applying Boundary Conditions
To solve for the 6 constants (C0 to C5), we need 6 known conditions. This is where Kinematics Analysis comes into play. We define what happens at the start (x=0) and end (x=1) of the motion.
Figure 2: Visualizing the 6 boundary conditions. Note the ability to define non-zero start/end velocities (v0, v1).
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Start Conditions (x=0)
We start at zero displacement with zero acceleration. Crucially, we allow a non-zero velocity (v0).
- s = 0
- v = v0
- a = 0
End Conditions (x=1)
We end at the total lift height (hm) with a target end velocity (v1) and zero acceleration.
- s = hm
- v = v1
- a = 0
Solving for Constants
Step 1: Apply Start Conditions (x=0)
By substituting x=0 into the general equations, terms with x vanish immediately. This quickly solves the first three constants:
- From Displacement: C0 = 0
- From Acceleration: C2 = 0
- From Velocity: v0 = C1 / β → C1 = βv0
Step 2: Apply End Conditions (x=1)
Now we have a system of 3 linear equations with 3 unknowns (C3, C4, C5). Solving this simultaneous system (via substitution or matrix methods handled by engineering software) yields the final coefficients:
The Solved Coefficients
C1 = βv0
C3 = 10h - β(6v0 + 4v1)
C4 = -15h + β(8v0 + 7v1)
C5 = 6h - β(3v0 + 3v1)
The Final Polynomial Equation
Therefore, the 5th-degree polynomial cam function that satisfies all boundary conditions is:
Figure 3: The final mathematical model for smooth cam motion control.
s = C1x + C3x3 + C4x4 + C5x5
With this equation derived, we can now plug in any start/end velocity we want. This is critical for designing "Flying Shears" or "Rotary Knives" where the tool must match the speed of the product line perfectly.
In the next post, we will analyze the characteristics of this curve and graph the results.
Continue to Part 3:
Polynomial Cam Function (Analysis & Graphs) - Part 3
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