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The drive wheels may slip when starting with a full load.
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Insufficient wheel load can reduce traction and affect driving performance.
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The motor may not provide enough torque during acceleration.
1. Understanding the Load of an Under-Ride AGV
Mtotal = mAGV + mload
Where:
- mAGV: AGV body mass (kg)
- mload: Mass of the rack and cargo carried by the AGV (kg)
- Mtotal: Total vehicle mass (kg)
2. Load Distribution on Under-Ride AGV Drive Wheels
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The mounting positions of the drive wheels and support wheels.
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The center of gravity of the AGV and its cargo.
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The design of the lifting mechanism.
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The drive wheel suspension and spring preload.
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Changes in load under empty, full-load, and uneven-load conditions.
2.1 Actual Load on the Drive Wheels
Nd = λ × Mtotal × g
Where:
- Nd: Total normal load on all drive wheels (N)
- λ: Proportion of the total vehicle weight supported by the drive wheels (e.g., 0.5 for 50%)
- Mtotal: Total vehicle mass (kg)
- g: Gravitational acceleration (9.81 m/s²)
Example:
An AGV has a total mass of 800 kg, and its two drive wheels support 50% of the vehicle weight.
Nd = 0.5 × 800 × 9.81
= 3,924 N
2.2 Drive Wheel Pressure

3. AGV Running Resistance and Traction Force Calculation
3.1 Rolling Resistance
Fr = Crr × Mtotal × g × cos(θ)
Where:
- Fr: Rolling resistance (N)
- Crr: Rolling resistance coefficient
- Mtotal: Total vehicle mass (kg)
- g: Gravitational acceleration (9.81 m/s²)
- θ: Slope angle
On a flat floor (θ = 0°), the formula becomes:
Fr = Crr × Mtotal × g
3.2 Slope Resistance
Fg = M × g × sin(θ)
3.3 Acceleration Force
Fa = M × a
Where:
- Fa: Acceleration force (N)
- M: Total vehicle mass (kg)
- a: Target acceleration (m/s²)
Example:
An AGV with a total mass of 800 kg and a target acceleration of 0.3 m/s² requires:
Fa = 800 × 0.3 = 240 N
3.4 Total Traction Force
Ftotal = Fr + Fg + Fa
The Full Formula:
Ftotal = M × a + Crr × M × g × cos(θ) + M × g × sin(θ)
4. AGV Drive Wheel Traction and Output Torque
4.1 Traction Force Per Drive Wheel
Fwheel = Ftotal / n
Where:
- Fwheel: Traction force required from each drive wheel (N)
- Ftotal: Total vehicle traction force (N)
- n: Number of drive wheels
4.2 Drive Wheel Output Torque
Twheel = (Ftotal × r) / n
Where:
- Twheel: Output torque of one drive wheel (N·m)
- r: Effective rolling radius of the drive wheel (m)
- n: Number of drive wheels
4.3 Matching the Motor and Gearbox
5. Drive Wheel Adhesion and Slip Prevention
Fadhesion = μ × Nd
Where:
- Fadhesion: Maximum available traction force (N)
- μ: Ground adhesion coefficient
- Nd: Total normal load on the drive wheels (N)
To prevent wheel slip, the following conditions must be met:
Ftotal ≤ Fadhesion
6. Engineering Example: Drive Wheel Calculation for an 800 kg Under-Ride AGV
Fr = 0.02 × 800 × 9.81 × cos(1.72°) ≈ 156.89 N
The slope resistance is:
Fg = 800 × 9.81 × sin(1.72°) ≈ 235.33 N
The acceleration force is:
Fa = 800 × 0.3 = 240 N
Therefore, the total traction force required is:
Ftotal = 156.89 + 235.33 + 240 = 632.22 N
Since the AGV has two drive wheels, each wheel needs to provide approximately 316.11 N of traction force.
With a wheel radius of 0.08 m, the required output torque per drive wheel is:
Twheel = 316.11 × 0.08 ≈ 25.29 N·m
Based on a 20:1 gear ratio and 90% transmission efficiency, the required motor shaft torque is:
Tmotor = 25.29 / (20 × 0.9) ≈ 1.405 N·m
Next, check whether the drive wheels have enough grip. If the two drive wheels support 50% of the total vehicle weight, their combined normal load is approximately 3,924 N.
With an assumed ground adhesion coefficient of 0.25, the maximum available traction force is:
Fadhesion = 0.25 × 3,924 = 981 N

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