Customer Project: Backloading AGV with a Total Weight of 2 Tons, Maximum Speed of 2 km/h, Single-Wheel Drive Solution
Based on the specific requirements provided by the customer, our experienced team quickly selected our standard AGV steering drive wheel, model PLT167P102SF30-29-22HS, through precise calculations. Below is the calculation parameter table and steering drive wheel parameter diagram:
After selecting the AGV steering drive wheel, the customer raised a new concern-specifically asking for reference data on the braking ability of the AGV in emergency situations to ensure safety. While we typically calculate the traction force of the drive wheel, braking force calculations, especially for direct braking during AGV operation, are less common.
Customer satisfaction is our service philosophy, and we took this opportunity to analyze the emergency braking performance of the AGV. Before conducting the calculations and evaluations, the following premises must be clarified:
Avoid Wheel Skidding: Skidding means braking failure.
Consider External Factors: These calculations serve as directional references due to various influencing variables. Detailed site-specific data enables more accurate evaluations.
Ensure System Stability and Reliability: Proper measures must be taken during unconventional actions like emergency braking to ensure smooth operation.
Using our PLT167P102SF30-29-22HS AGV steering drive wheel, we calculated the braking ability for the customer's AGV as follows:
Step 1: Calculate the AGV's Inertial Force
The total weight of the AGV is 2 tons. Using the formula F = ma, where a = (Vt - V0) / t, we assume the AGV decelerates to a stop within 1 second. The braking acceleration is a = Vt, and the inertial force is:
F (Inertial Force) = ma = mVt = 2000 × 2 / 3.6 ≈ 1111 N
This means the braking force must exceed 1111 N to bring the AGV to a stop at its maximum speed.
Step 2: Calculate the Steering Drive Wheel's Braking Force
The brake's rated torque is 6 Nm, the gearbox ratio is 29, and the wheel diameter is 167 mm. The braking force at the wheel edge can be calculated as:
F (Braking Force) = 6 × 29 / (0.167 × 2) ≈ 2083 N
This indicates the braking force provided by the AGV steering drive wheel is 2083 N, sufficient to overcome the AGV's inertial force.
Step 3: Prevent Wheel Skidding
For the steering drive wheel to exert the calculated braking force, skidding must be avoided. Skidding is influenced by the normal force on the wheel and the coefficient of friction with the ground. Assuming a normal force of 500 kg and a static friction coefficient of 0.4, the maximum allowable thrust is:
F (Thrust) = 500 × 10 × 0.4 = 2000 N ≈ 2083 N
Step 4: Calculate Minimum Normal Force
The minimum normal force required to meet the customer's braking requirements is:
N (Normal Force) > 1111 / (10 × 0.4) ≈ 278 kg
Based on the methods outlined in "AGV Suspension Analysis①: Spring Stiffness Coefficient and Preload Calculation", the spring stiffness coefficient and preload must ensure this minimum normal force is maintained in all scenarios.
Summary
This article provides an overview of the braking force of our PLT167P102SF30-29-22HS AGV steering drive wheel and discusses suspension design requirements for emergency braking.
Additionally, since the motor remains in high-speed rotation during braking, we recommend disengaging the motor drive current before applying the brake. Commands such as "free stop," "quick stop," or "emergency stop" in the motor driver should be triggered to prevent potential driver faults or damage.
In conclusion, specific solutions should always be tailored to the situation at hand. Readers are advised not to oversimplify or misinterpret this article's content as absolute reference material.
