Engineering Applying Physics to Vehicle Dynamics: Do Wheel Number Matter?

[erobz]

I fixed up the Latex there. Just reply to this message to see what has changed.

"If we know that Torque = I * a, and so, that Force = Torque / r, then the why wouldn't it be correct for the first term to be $\frac {T 2 I \alpha}{r}$

And the second: $\frac { 2 I \alpha}{r}$"

 

[jack action]

2 * ( ( T_engine - ( I * a) ) / r ) - 2 * ( (I * a) / r ) - ( μ * m_vehicle * g ) - ( ( c_d * A_frontal * ρ_air * v^2 ) / 2 ) = m_vehicle * a

I strongly suggest using a simplified version with the equivalent mass ($m_e$) concept where:
$$F_{net} = m_e a$$
$$F_w - F_r - F_d = m_e a$$
$$F_w - C_{rr} mg - \frac{1}{2}\rho C_dAv^2 = m \left(1+\frac{4I}{mR^2}\right) a$$
$$a = \frac{\frac{F_w}{m} - C_{rr} g - \frac{1}{2}\rho\frac{C_dA}{m}v^2}{1+\frac{4I}{mR^2}}$$
Note that I use a coefficient of rolling resistance $C_{rr}$ instead of $\mu$ which usually represents the friction coefficient (not the same thing).

The force $F_w$ (sum from all driven wheels) is either the lesser between the maximum friction force available on all wheels (the normal force acting on each wheel is important here) OR the total power available over the speed of the vehicle ($F_w = \frac{P}{v}$), which is most likely. Using the total vehicle power available is less confusing than trying to identify how the torque is modified and split through the drivetrain.

The term in parenthesis for the equivalent mass represents the combined inertial effect of the vehicle mass $m$ and rotational inertia $I$ of 4 wheels. this term can be modified to include other rotational components from the brake system or drivetrain (more info here).

If we know that Torque = I * a,

No, the torque is $I\alpha$ or $I\frac{a}{R}$.

 

[Saalz]

I strongly suggest using a simplified version with the equivalent mass ($m_e$) concept where:
$$F_{net} = m_e a$$
$$F_w - F_r - F_d = m_e a$$
$$F_w - C_{rr} mg - \frac{1}{2}\rho C_dAv^2 = m \left(1+\frac{4I}{mR^2}\right) a$$
$$a = \frac{\frac{F_w}{m} - C_{rr} g - \frac{1}{2}\rho\frac{C_dA}{m}v^2}{1+\frac{4I}{mR^2}}$$
Note that I use a coefficient of rolling resistance $C_{rr}$ instead of $\mu$ which usually represents the friction coefficient (not the same thing).

The force $F_w$ (sum from all driven wheels) is either the lesser between the maximum friction force available on all wheels (the normal force acting on each wheel is important here) OR the total power available over the speed of the vehicle ($F_w = \frac{P}{v}$), which is most likely. Using the total vehicle power available is less confusing than trying to identify how the torque is modified and split through the drivetrain.

The term in parenthesis for the equivalent mass represents the combined inertial effect of the vehicle mass $m$ and rotational inertia $I$ of 4 wheels. this term can be modified to include other rotational components from the brake system or drivetrain (more info here).No, the torque is $I\alpha$ or $I\frac{a}{R}$.

Hi Jack,

I'm willing to use torque, to be fair. I've got the electric motor modeled. I have the torque curve across the whole RPM range of the electric motor that propels the vehicle and the drivetrain ratio, so I know the total torque available at the driven wheels.

In that case, I'll guess I could use the F_w in your expression as the total wheel torque available at a speed v, divided by the radius R of the wheels ( F_w = Torque / R ), right?