In the world of computers we call this a binary search. Repeatedly cutting the part to be searched in half after eliminating one of the existing halves as not having what’s being searched for.
With trains, if you’re at the head end, there’s no need to start halfway back. Walk back 5-10 cars, close an anglecock, apply brakes. If no kicker - then walk ahead a few more cars and try again. Repeat ad nauseum. It’d be silly to walk to the 50th car of a 100 car train only to find out the kicker was the 4th car.
GFood observation, zug. Reminds me of Bill Cosby’s early comedy bit about him and his younger brother Russell playing the game of “20 Questions”: Cos goes at it methodically and logically , uses up all 20 questions, doesn’t get it; but Russell guesses randomly and gets it on the 3rd one . . . [:-^]
This is an illustration of some deep mathematical principles in the nature of statistics, probability, game theory, “linear programmming”, “real math”, etc., but I’m too fried right now to remember and assign the correct name. Suffice it to say that in this exercise (literally !), the “starting position” does matter, and the choices for the next trail or test are not equal, nor is the decision process "free’ - it involves time, and the effort of someone walking, and then the time for each ‘cycle’ of closing the next anglecock, pumping up the air, and applying the brakes, etc. - kind of like friction in physics, in that the intellectual models assume there is no friction, whereas there of course is in the real world. All of that is to say that given a single car with a certain characteristic somewhere randomly and evenly distributed in a train of x cars, there is likely a rational method to work out the fastest and easiest way to isolate and identify that particular car, which is a hybrid of the above 2 methods. Someday I might figure out what that is . . . [swg]
- Paul North.
If you had instant access to each car at random then the “Binary Search” method would be best, but since access to the cars is linear then a linear search is best.
But the more I think about it… it also depends on how long each test takes. If it takes longer to do the test than to walk half the length of the train then the Binary Search would be faster. At some point between how long it takes to walk some distance and how long it takes to perform a test one or the other method might be faster.
…or just tell the engineer to not touch the air if he/she can help it.
Yeah - you (Semper Vaporo) understand the analytical dilemma/ challenge with this: What are the relative risks/ benefits of adopting a particular method and testing a specific car - in terms of the time already expended to do that, and the most probable amount of time to finish the isolation process ? SUuperficially, this problem seems similar to finding a specific card in a randomly shuffled deck, and could be solved by writing an equation in the nature of:
Total time = time walking + time testing = (No. of cars walked x average time to walk 1 car-length) + (No. of cars tested x average time to test 1 car), or = NWTW + NTTT.
Then, substitute the probability equations for the NW and NT terms, and differentiate the resulting equation with respect to the no. of cars walked (NW), and solve for the optimum/ minimum Total Time.
Note that the number of cars walked has no fixed relationship with the number of cars tested (except that they both have to be less than or equal to the total number of cars in the entire train length), and further, the time to walk 1 car-length and the time to test 1 car are also unrelated in any way. So there might be multiple valid answers, such as if the equation is also solved for minimum Total Time with respect to the number of cars tested (NT).
Intuitively, I suspect the answer will be related to the inverse of the ratio of the average time to walk 1 car-length to the average time to test 1 car. So, if it takes 20 seconds to walk 1 car-length (60 ft. at 3 ft./ sec., about 2 miles per hour), but 1* minute to perform the average brake pump-up and release, that would be a ratio of 1/3. So, the optimum or ‘balance point’ (in calculus terms, slope of function = 0) might be around where equal amounts of time are spent on each task - walking and te
Ugh… too much reminder of calculus.
Indeed - if you’re using the binary (or half-split) method on a circuit, you’re not usually having to cover 5,000 feet overall in the process…
This is elementary line of road railroading generally at 3 AM with the temperature at 10 degrees and a 30 MPH ‘breeze’. Brakemen and Car Knockers are not using calculus to solve the situation…they don’t have the education or time for it.
[quote user=“Paul_D_North_Jr”]
Yeah - you (Semper Vaporo) understand the analytical dilemma/ challenge with this: What are the relative risks/ benefits of adopting a particular method and testing a specific car - in terms of the time already expended to do that, and the most probable amount of time to finish the isolation process ? SUuperficially, this problem seems similar to finding a specific card in a randomly shuffled deck, and could be solved by writing an equation in the nature of:
Total time = time walking + time testing = (No. of cars walked x average time to walk 1 car-length) + (No. of cars tested x average time to test 1 car), or = NWTW + NTTT.
Then, substitute the probability equations for the NW and NT terms, and differentiate the resulting equation with respect to the no. of cars walked (NW), and solve for the optimum/ minimum Total Time.
Note that the number of cars walked has no fixed relationship with the number of cars tested (except that they both have to be less than or equal to the total number of cars in the entire train length), and further, the time to walk 1 car-length and the time to test 1 car are also unrelated in any way. So there might be multiple valid answers, such as if the equation is also solved for minimum Total Time with respect to the number of cars tested (NT).
Intuitively, I suspect the answer will be related to the inverse of the ratio of the average time to walk 1 car-length to the average time to test 1 car. So, if it takes 20 se
Another secnerio: If there weren’t a kicker, and the train could still move, the headbrakie would get on the ground and let the train roll by then hit the cushions in the caboose if nothing found and ride there to the next destination where he could or would have to return to the locomotive.
Has anyone ever tried to find a dynamiter/kicker? I’ve never heard of it being done. There’s all kinds of variables too. Some dynamiters go every time you touch the air, some don’t. Some only go during slow speeds, others speed doesn’t matter. You could check every car and find nothing, then the first time you use the air again, it dynamites.
Besides, sometimes the DPU is the one throwing the train into emergency.
Old Wive’s Tales (or maybe Old Hoghead’s Tales) type remedies. None guaranteed to work.
Go to the first car behind the engine(s) and cut out the air brakes. Some say this only works when the car is a long car, like a 89 ft flat or autorack.
When setting air, first place the brake valve in minimum reduction for a second then back to release. Repeat once more. Then make your brake application.
Again, the experts will say those, and any other methods, won’t prevent a dynamiter. If they seem to work, it’s just coincidence. Well, I’ve used the second method a time or two. It doesn’t always work, but I’ve had a few trains where every time I tried it the train didn’t dynamite and the times I didn’t the train went into emergency. I actually haven’t had a train with a dynamiter in it for a while and have probably jinxed myself now.
Jeff