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I am trying to write a program that simulates the physics of a locomotive (diesel, not steam). I already have a method to get the tractive effort of the locomotive. If anyone must know, the one I am using in this example is the EMD F7-A.

I am trying to model wheel slip and all the forces that play into it. I already know I can use the adhesion factor (30% I believe in this case) to find the wheel adhesion. For an F7, it is 69,000lbs. Since the wheels can grip no more than 69,000lbs, how do I account for all the forces that may add up to 69,000lbs? I know that if the rolling resistance of the whole train adds up to over 69,000lbs the wheels slip, but what about other things?

The thing I am having the most trouble with is locking up the brakes. It's like a car, but on a larger scale. If I lock up the brakes on the train, chances are the locomotive is going skid. But what force is this? I am guessing it has something to do with kinetic energy, but since that's in Joules, how do I get the force?

Any and all help is appreciated. And feel free to correct me if I got any concepts wrong, I am kind of new at this.

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  • $\begingroup$ Small additional thing to consider: if your locomotive is pulling a load, you need to consider torque (point where you pull the cars is higher than the point where the rails apply force to the wheels) that affects how much of the weight is on each of the wheels. In the limit you would lift the front wheels and all the normal force would be on the rear wheels. $\endgroup$
    – Floris
    Commented Oct 27, 2015 at 18:32

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When you hit the brakes, you're essentially changing the coefficient of friction for your vehicle. Specifically, instead of a rolling coefficient of friction, you're going to use the kinetic or sliding coefficient of friction. The coefficient change should help you account for the sliding and slipping. That ought to be enough for some Newtonian modeling.

Alternatively, you could model this using energy and work. The braking force is doing work on the train, decreasing its kinetic energy over some distance. Work, in case you don't remember, is $W=\vec{F}\cdot\vec{d}$. Work has units of Joules, and it represents how a force increases or decreases the energy of a system. If your force changes with time (or distance), you can integrate to get the correct values. I hope that gives you some ideas.

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  • $\begingroup$ So could I model this with a cube on a table with a given coefficient of friction, and the net force on the train is the net force on the cube? Then if the cube slides on the table, the trains wheels are sliding by the same amount relative to the rails. Am I understanding this correctly? $\endgroup$
    – K4KFH
    Commented Oct 27, 2015 at 14:00
  • $\begingroup$ @K4KFH That sounds fundamentally correct. You can also use the energy model I proposed, but it's up to you. $\endgroup$
    – PipperChip
    Commented Oct 27, 2015 at 15:36
  • $\begingroup$ apparently I misread that, upon reading it again that actually sounds like a better idea. Since kinetic energy is measured in Joules, how would I convert that to force to calculate acceleration? $\endgroup$
    – K4KFH
    Commented Oct 27, 2015 at 17:51
  • $\begingroup$ @K4KFH use the principle of work, or force applied to an object over a distance gives you energy (work). W=Fd I'll add that to the answer. $\endgroup$
    – PipperChip
    Commented Oct 27, 2015 at 18:19
  • $\begingroup$ How would I handle this if I don't have a defined distance? (I don't know how long the train's going to keep going) $\endgroup$
    – K4KFH
    Commented Oct 27, 2015 at 22:13
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The quarter car model, that is the equations of motion modeling each wheel's dynamics in the simplest possible sense are given as $$\tau = I_w \dot{\omega}_w - F_w R_w$$, where $\tau$ is the accelerating, braking or net torque on the wheel (acted through the driveshaft), $I_w$ is the moment of inertia of the wheel, $\omega_w$ is the wheel speed, $F_w = \mu F^z_w$ is the friction force that the ground acts on the wheel contact surface, with $\mu$ being the (kinetic when skidding) friction coefficient and $F^z_w$ being the normal force acted on the wheel by the car pressing down on it. If there is no slipping or skidding, $v_g = \omega_w R_w$, where $v_g$ is the ground speed of the car's center of mass. Therefore, the wheel slip is defined as $$\lambda = \frac{\omega_w R_w - v_g}{v_g}$$. Imagine that for a given constant negative torque $\tau$ (as in braking), the wheel encounters ice, thus decreasing $\mu$ by a large value. Consequently, the value of $\omega$ will drop in the short term (over the next few seconds) and cause the wheel to slip ($\lambda < 0$). This is the technical description of a skid. ABS and traction control are systems which contain such undesired phenomenon. A good idea to do traction control is to use classical or robust nonlinear controllers since they have guarantees or stability margins. The literature you are looking for includes this comprehensive book: https://www.elsevier.com/books/tire-and-vehicle-dynamics/pacejka/978-0-08-097016-5

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