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I am programming a vehicle system for a company, but I cannot find the right formulas to be able to have access to the resistance pulse of a wheel attached to an axle during acceleration / deceleration. All I know is that this value decreases with acceleration and increases with deceleration, but I can't get my finger on it. I am able to find the angular momentum, the rolling coefficient, the friction coefficient and the rolling resistance, but I cannot find a formula allowing me to use these in order to be able to find this value.

The vehicle travels at a speed of $2{km/h}$ with wheels of a radius of $0.59{m}$

The torque sent by the engine with a revolution of 900 rpm during an acceleration ($200{N m}$) sent to the transmission: $ T = 200{N m}$

The transmission multiplies the torque received from the engine by a transmission ratio (8.03) : $ T = (200{N\cdot m}) (8.03) $

The differential again multiplies the torque received from the transmission (3.36): $ T = (200{N\cdot m}) ((8.03)(3.36)) $

I do not take into account in this version the loss due to friction.

In this case, the Fz is the force (in newton) of the suspension which is connected to the vehicle of $14014{N}$.

$ Fz = 14014{N} $

$ Vx{(m/s)} = (2{km/h} (3.6)) $

$ r = 0.59{m} $

$ rpm = 8.9917 $

$\Omega{(rad/s)} = r * \omega = {0.59m * (rpm({8.9917})/9.5492)}$

I'm trying to find the Fx with the formulas mentioned above as well as the context.

$ Fx = ?$

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  • $\begingroup$ A diagram would be helpful to understand the conditions where an impulse is applied. $\endgroup$
    – John Alexiou
    Commented Feb 4, 2020 at 18:01
  • $\begingroup$ The reaction impulse at the wheel axle is zero during smooth acceleration. Impulses imply a step in velocity and infinite acceleration. So something must hit the vehicle to produce a reaction impulse at the wheel. $\endgroup$
    – John Alexiou
    Commented Feb 4, 2020 at 18:03
  • $\begingroup$ @ja72 It is in the context that an engine of a vehicle that sends a torque to each wheel for acceleration / deceleration and I will need to know the right formula with all the others that I mentioned above in order to have the necessary impetus to rotate the wheel according to its angular speed and the torque it receives. $\endgroup$
    – Jonathan Dion
    Commented Feb 4, 2020 at 18:34
  • $\begingroup$ I mentioned above in order to have the necessary impulse** to rotate the wheel according to its angular speed and the torque it receives. $\endgroup$
    – Jonathan Dion
    Commented Feb 4, 2020 at 19:11
  • $\begingroup$ Please add to the post you current work (the formulas you are looking at, and a diagram of the model with loads/torques, displacements and angles). $\endgroup$
    – John Alexiou
    Commented Feb 5, 2020 at 13:58

1 Answer 1

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To answer this problem, first do a free body diagram that includes all forces and motions acting.

fbd

Now assume that the axle bears some fraction of the total mass of the vehicle, $m_{\rm axle}$.

Also assume no slipping, which means $v - \Omega R =0$ or it terms of accelerations

$$ \dot{\Omega} = \frac{\dot{v}}{R} $$

where $R$ is the radius of the tyre, and $\dot{\square}$ designates the time derivative.

Now form the three equations of motion with care taken to consider what is considered a positive or a negative sense.

$$ \begin{aligned} F_x & = m_{\rm axle}\, \dot{v} \\ F_y - m_{\rm axle}\, g & = 0 \\ - \tau + R F_x & = I (-\dot{\Omega}) \end{aligned}$$

where $I$ is the mass moment of inertia of the tyre and wheel.

The solution of the above equations is

$$ \begin{aligned} \dot{v} & = \frac{R\, \tau}{I+m_{\rm axle} R^2} \\ F_x & = m_{\rm axle}\, \dot{v} \\ F_y & = m_{\rm axle}\, g \\ \end{aligned} $$

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  • $\begingroup$ Do all of them take into account rolling resistance? $\endgroup$
    – Jonathan Dion
    Commented Feb 6, 2020 at 18:11
  • $\begingroup$ No rolling resistance is not present in the equations. You can add it, by reducing the torque applied $\tau$ by the equivalent amount. $\endgroup$
    – John Alexiou
    Commented Feb 6, 2020 at 22:41

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