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10

You need to consider what happens if a hose bursts, and your spring-applied brake locks the wheels on one or more cars - how do you get that train back to the depot? There could be a manual brake release lever, but this would have to be on the brake cylinder, and there are many circumstances where crawling under the train to access this lever would be ...

3

Because the system has worked for a long time and railroads don't want to spend money on improving a technology is time tested. Most railroads run locomotives that is over 25-50 years old. (Most north american railroads still have a fleet of EMD GP-40's that were manufactured in the late 60's to 70's). Most railroads cut corners to save a dime so I don't see ...

10

This answer addresses the following sentence: So I'd expect to see the brakes applied by a spring that is capable of stopping the wheels of a fully loaded car, and a single-acting cylinder that works against that spring to release the brakes. IMHO, one very basic reason against using the spring is the fatigue and the degradation of the elastic properties ...

4

With an "spring apply / air release" system the spring pressure may not be enough to apply adequate braking force. Using compressed air to apply the brake allows much higher braking forces. The reservoir system when discharged allows rough shunting of the cars without the trouble and delay of having to couple the cars completely. Hump shunting ...

0

If you know second-order systems you can always add a time delay to obtain the third order as all real systems have them. Often a Pade approximation is used for a time delay. For the Laplace transform {e^-st} the first order Pade approximation for time delay, t, is (1-t/(2s))/(1+t/(2s)). Time delays always cause degradation of system performance due to delay ...

0

Observation: Bandwidth is defined as the frequency range $[\omega_1 \ \omega_2]$ over which the control of the system is "effective". Usually, $\omega_1 = 0$ and then, by definition, $\omega_2 = \omega_B$ is the bandwidth. Definition: The (closed-loop) bandwidth, $\omega_b$, is the frequency where the norm of the sensitivity function, $|S(j\omega)|$...

1

If the flexibility and mass of the cables are small compared to the masses, the early stage could be just winding the cables around the hub. Basically, we need to calculate the equation of a system of 4 whips lashing. The mass per unit length and bending moment stiffness in relation to the 4 end masses will plug into a differential equation with a large ...

0

EUREKA!! I FOUND IT!! (And, I broke all the old links 😕) Here is the continuous-time system http://airvigilante194.sdf.org/Scripts/deadbeatJeff01.slx Vector field Critical point Continuous time matrices And, here is the script to computer the gain matrix http://airvigilante194.sdf.org/Scripts/deadbeatJeff.m and here is the discrete-time system ...

1

SMALL SIGNAL Let's assume a typical loop. Controller and plant in forward path, unity feedback. Let's add a pre-filter. Let C = controller transfer function; P = plant transfer function; F = input filter transfer function (side note: it's possible to put part of the controller in the reverse path. Those configurations could be rearranged to small-signal-...

2

Perhaps this is an X/Y problem. But here are some suggestions. 1 Design an over damped PID closed loop system. An over damped system will have poorer rise time. Assumptions required for this to work are The target pressure is a constant; i.e. the command to the system is a step input. The PID (particularly the D) can be designed to result in a stable over ...

0

You can minimize these peaks by using a Kalman Filter. Kalman Filter is used to estimate the next step (k+1) in each step at (k). Also, it is used for cases like this, to reduce sensor noise. There are plenty of books out there to learn how to implement a Kalman Filter.

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