Vehicle Dynamics and Torque Distribution

Introduction

In the realm of vehicle dynamics, understanding the impact of torque distribution on acceleration and deceleration is crucial, especially for heavy-duty vehicles like mining trucks and construction equipments. This blog delves into the analysis of torque distribution factors in 4-wheel drive vehicles, focusing on the Caterpillar 797F dump truck as a real-world example. We'll use MATLAB to simulate and analyze these dynamics, incorporating various constraints and cases to optimize performance.

Free Body Diagram and Equations

To start with, we model the vehicle dynamics using a free body diagram. The diagram includes forces acting on the vehicle and moments about the rear and front axles. The primary equations governing the system are derived from the equilibrium conditions:

Here are the equations we derived from the free body diagram:

1. Sum of Forces in the x-direction:

   $$\sum F_x=F_{tr}+F_{tf}=M\ddot{x}$$

   This equation represents the sum of tractive forces ($F_{tr}$and $F_{tf}$) in the x-direction, which equals the mass ($M$) of the vehicle multiplied by its acceleration ($\ddot{x}$).

2. Sum of Forces in the z-direction:

   $$\sum F_z=N_r+N_f=mg$$

   Here, the normal forces ($N_r$ and $N_f$) at the rear and front axles respectively, sum up to the weight of the vehicle ($mg$).

3. Sum of Moments about the Rear Axle:

   $$\sum M_y\text{ (about rear axle)}=l_{r}Mg-l{N}_f=hM\ddot{x}+I\omega ({\ddot{\theta }}_r+{\ddot{\theta }}_f)$$

   This equation represents the sum of moments about the rear axle, where $l_r$ and $l$ are distances, $h$ is the height, $\ddot{x}$ is the linear acceleration, $I$ is the moment of inertia, and $\omega$ is the angular velocity.

4. Sum of Moments about the Front Axle:

   $$\sum M_y\text{ (about front axle)}=-l_{f}Mg+l{N}_r=hM\ddot{x}+I\omega ({\ddot{\theta }}_r+{\ddot{\theta }}_f)$$

   This equation represents the sum of moments about the front axle, where $l_f$ and $l$ are distances, $h$ is the height, $\ddot{x}$ is the linear acceleration, $I$ is the moment of inertia, and $\omega$ is the angular velocity.

Solving for Normal Forces

From the above equations, we can solve for the normal forces $N_f$ and $N_r$:

1. Front Normal Force ($N_f$):

   $$N_f=m\left(\frac{l_{r}g-h\ddot{x}}{l}\right)-\frac{I\omega ({\ddot{\theta }}_r+{\ddot{\theta }}_f)}{l}$$

2. Rear Normal Force ($N_r$):

   $$N_r=m\left(\frac{l_{f}g+\mathrm{hx}\stackrel{}{¨}}{l}\right)+\frac{I\omega ({\ddot{\theta }}_r+{\ddot{\theta }}_f)}{l}$$

Tractive Force at the Rear Wheels:

$$T_w(1-k_f)-\frac{R_w}{I\omega }{\ddot{\theta }}_r=F_{t\mathrm{r}}$$

This equation represents the tractive force at the rear wheels ($F_{tr}$), where $T_w$ is the wheel torque, $k_f$ is a distribution factor, $R_w$ is the wheel radius, $I\omega$ is the moment of inertia, and ${\ddot{\theta }}_r$ is the angular acceleration of the rear wheels.

Tractive Force at the Front Wheels:

$$T_w{k}_f-\frac{R_w}{I\omega }{\ddot{\theta }}_f=F_{t\mathrm{f}}$$

This equation represents the tractive force at the front wheels ($F_{tf}$), using the same variables as above but for the front wheels.

Solving for Normal Forces again,

By solving the equations, we can determine the normal forces acting on the front and rear axles. These forces are crucial for understanding how the vehicle interacts with the ground and how torque is distributed:

1. Rear Normal Force ($N_r$):

   $$N_r=\frac{m}{l}(g{l}_f+\ddot{x}h)+\frac{I\omega }{l}({\ddot{\theta }}_r+{\ddot{\theta }}_f)$$

   This equation calculates the normal force at the rear axle, considering the vehicle's mass ($m$), gravitational acceleration ($g$), distances ($l_f$ and $l$), height ($h$), linear acceleration ($\ddot{x}$), and angular accelerations ($\ddot{\theta}_r$ and $\ddot{\theta}_f$).

2. Front Normal Force ($N_f$):

   $$N_f=\frac{m}{l}(g{l}_r-\ddot{x}h)-\frac{I\omega }{l}({\ddot{\theta }}_r+{\ddot{\theta }}_f)$$

   This equation calculates the normal force at the front axle using the same variables as above but with the front axle distances.

Known Values

• $T_w$: Wheel torque
• $k_f$: Torque distribution factor (assumed)
• $M$: Mass of the vehicle
• $I\omega$: Moment of inertia
• $l$: Distance between axles
• $l_r$: Distance from the center of gravity to the rear axle
• $l_f$: Distance from the center of gravity to the front axle
• $h$: Height of the center of gravity
• $R_w$: Radius of the wheel (assuming it's known or can be measured)

Unknown Values

• $\ddot{\theta}_r$: Angular acceleration of the rear wheels
• $\ddot{\theta}_f$: Angular acceleration of the front wheels
• $F_{tr}$: Tractive force at the rear wheels
• $F_{tf}$: Tractive force at the front wheels
• $\ddot{x}$: Linear acceleration of the vehicle
• $N_r$ Normal force at the rear axle
• $N_f$: Normal force at the front axle

Introducing constraints or specific cases will help reduce the number of unknowns, making the system solvable. Let's outline how to implement these constraints and solve for each case.

Case 1: No Slip at All Wheels

Here, the angular acceleration of the wheels is directly related to the linear acceleration, $\ddot{x}$:

$$\ddot{\theta}_r = \frac{\ddot{x}}{R_w}$$ $$\ddot{\theta}_f = \frac{\ddot{x}}{R_w}$$

Case 2: Rear Wheel Slips

For the rear wheel slip condition, we assume $\ddot{\theta}_r \geq \frac{\ddot{x}}{R_w}$. If slipping occurs, the rear wheel traction force is equal to the friction force:

$$\ddot{\theta}_r \geq \frac{\ddot{x}}{R_w}$$ $$F_{tr} = \mu N_r$$

Case 3: Front Wheel Slips

For the front wheel slip condition, we assume $\ddot{\theta}_f \geq \frac{\ddot{x}}{R_w}$. If slipping occurs, the front wheel traction force is equal to the friction force:

$$\ddot{\theta}_f \geq \frac{\ddot{x}}{R_w}$$ $$F_{tf} = \mu N_f$$

Case 4: All Wheels Slip

For all wheels slipping, both front and rear wheels meet the slip condition:

$$\ddot{\theta}_r \geq \frac{\ddot{x}}{R_w}$$ $$\ddot{\theta}_f \geq \frac{\ddot{x}}{R_w}$$ $$F_{tr} = \mu N_r$$ $$F_{tf} = \mu N_f$$