8 Ton Heavy-Duty AGV Drive System Design Example
This engineering example shows how an 8 ton AGV drive system can be evaluated before a differential drive module is selected. The purpose is not to assign a module from vehicle weight alone, but to check tractive force, driven-wheel load, acceleration, slope, turning resistance and duty cycle together.
The vehicle data below are illustrative design assumptions. Final selection should always be verified against the actual chassis geometry, centre of gravity, floor condition and operating cycle.
1. Example Vehicle Requirements
| Design Input | Example Value |
|---|---|
| Gross vehicle mass | 8,000 kg |
| Operating environment | Indoor industrial floor |
| Rolling resistance coefficient | 0.02 |
| Target acceleration | 0.12 m/s² |
| Maximum operating grade | 1% |
| Required motion | Forward / reverse + compact pivot turning |
| Drive architecture | Twin-wheel differential drive |
2. Calculate the Straight-Line Tractive Force
Rolling resistance:
Frr = 8000 × 9.81 × 0.02 ≈ 1,570 N
Acceleration force:
Fa = 8000 × 0.12 = 960 N
Grade force at 1%:
Fgrade ≈ 8000 × 9.81 × 0.01 ≈ 785 N
The combined straight-line requirement is therefore approximately:
1,570 + 960 + 785 = 3,315 N
Applying a 20% engineering allowance gives approximately:
3,315 × 1.20 = 3,978 N
3. Why This Number Is Not the Final Answer
A calculated requirement of about 3.98 kN does not automatically mean that any drive module rated above 3.98 kN is suitable. The available tyre-road friction must be sufficient to transmit that force, which depends directly on the load carried by the driven wheels.
Pivot turning also creates lateral tyre scrub that is not represented by the straight-line calculation. On an 8 ton AGV, wheel width, polyurethane hardness, floor friction and the position of the drive module relative to the chassis centre can significantly change turning torque.
The next step is therefore to check driven-wheel load and turning behaviour before confirming the module.
4. Check Driven-Wheel Load Before Using the Traction Rating
Motor torque does not guarantee usable tractive force. The tyres can only transmit force up to the friction available at the driven wheels:
Fmax = μ × Ndrive
where μ is the effective tyre-floor friction coefficient and Ndrive is the normal load carried by the driven wheels.
For example, if the differential drive module carries 25% of an 8,000 kg vehicle mass, the driven-wheel load is approximately 2,000 kg, or 19.6 kN of normal force. To transmit the calculated 3,978 N requirement, the required effective friction coefficient would be approximately:
μ = 3,978 / 19,620 ≈ 0.20
If the driven wheels carry only 20% of the vehicle mass, the required coefficient rises to about 0.25. This is why centre-of-gravity position and support-wheel layout can change the drivetrain result even when gross vehicle weight stays the same.
5. Compare the Requirement With the Drive Module Rating
For the 8,000 kg configuration in JY Robot's heavy-duty differential drive range, the rated tractive force is 3,795 N. In this example, the calculated straight-line design target is about 3,978 N.
The 3,795 N rating is about 480 N above the 3,315 N base straight-line requirement, giving approximately 14.5% margin. However, it is about 183 N below the illustrative 3,978 N design target after applying the 20% engineering allowance.
Before approving the module, the engineer should check actual rolling resistance, maximum simultaneous slope and acceleration, driven-wheel load, tyre-floor friction, pivot-turn resistance and thermal duty. If the real operating condition is more severe than the assumptions above, a higher-capacity drive solution or a revised vehicle layout may be required.
This is also why an “8 ton” product label should not be interpreted as meaning that every 8,000 kg AGV can use the same drive configuration.
6. Pivot Turning Can Become the Governing Condition
The straight-line calculation does not include tyre scrub during low-speed pivoting. On a heavy AGV, wide polyurethane wheels under high normal load can generate substantial lateral resistance when the chassis rotates in place.
Turning performance therefore depends on wheel spacing, tyre width and hardness, floor friction, support-wheel arrangement and the position of the differential drive module relative to the vehicle centre.
If pivot turning is frequent, the turning condition should be validated separately rather than assuming that a straight-line traction calculation is sufficient.
7. Engineering Conclusion for This 8 Ton Example
For the stated assumptions, the base straight-line requirement is about 3.315 kN, which is below the 3.795 kN module rating. The available margin is about 14.5%. If the illustrative 20% engineering allowance is required, however, the resulting 3.978 kN design target is higher than the module rating.
The final decision depends on whether the driven wheels carry enough normal load, whether the floor provides sufficient friction, and whether turning and duty-cycle conditions remain within the module's mechanical and thermal limits.
For the corresponding product family, see the 1–8 ton heavy-duty differential drive module. For the general sizing method, see our heavy-duty differential drive sizing guide.
8. Design Decision Summary
| Engineering Check | Example Result | Decision |
|---|---|---|
| Gross vehicle mass | 8,000 kg | Heavy-duty application |
| Straight-line force before allowance | 3,315 N | Calculated from rolling resistance, acceleration and grade |
| Design target with 20% allowance | 3,978 N | Use for preliminary comparison |
| 8,000 kg configuration rated tractive force | 3,795 N | Meets 3,315 N base requirement; below 3,978 N target with 20% allowance |
| Drive-wheel load at 25% vehicle mass | ≈ 2,000 kg | Requires effective friction coefficient ≈ 0.20 |
| Pivot-turn condition | Not included in straight-line force | Must be validated separately |
Under these assumptions, the 8,000 kg differential drive configuration meets the base straight-line requirement but does not provide the illustrative 20% traction allowance. The required design margin should therefore be confirmed before final approval.
9. When This 8 Ton Drive Selection Would No Longer Be Suitable
The preliminary selection should be reconsidered if any of the following project conditions become more demanding:
- Higher slope: grade resistance rises directly with vehicle weight and slope.
- Higher acceleration: additional acceleration force quickly consumes the remaining traction margin.
- Higher rolling resistance: floor joints, soft wheels or uneven surfaces can increase the real force requirement.
- Lower driven-wheel load: insufficient normal load can cause tyre slip before the motor reaches its rated tractive force.
- Frequent pivot turning: lateral tyre scrub can become more demanding than straight-line travel.
- Long continuous duty: thermal capacity of the motors, reducers and tyres must be checked separately.
If several of these conditions occur at the same time, the safer engineering response is not simply to add a larger motor. The chassis load distribution, wheel configuration, reduction ratio, module capacity and turning architecture should be reviewed together.