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On Determining When Trains Pass Each Other

4th draft

July 31, 2006

Charles J. Higgins, Ph.D.

As an amateur train/rail photographer, I like pictures/videos of trainspassing each other. However, it can be a guess as to when (and thus where) trains

will pass each other in unfamiliar locales. At this juncture, the classic mathproblem (and often a joke setup) comes to mind, but lets see what the answer

really is. As a first approximation, it seemed that averaging the departure and

arrival times of both trains in each direction seemed to be effective; but upon

trying the timings of trains with different speeds it became apparent that theapproximation was just that and no longer very useful. Do note that trains on a

grade do have differing speeds and that downhill trains often go slower for brake

control reasons. Of course, here given the lack of more information no

consideration will be made for interval changes in acceleration. For a more exactcomputation, consider train A going from X to Y with departure and arrival times

of T1 and T2 with a duration of DA = T2 -T1 and train B going from Y to X withrespective times of T3 and T4 with a duration of DB = T4 -T3, or:

T1 A===================> T2X f >< 1-f YT4

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The application of equation 5) is more precise than the mere

approximation obtained by averaging the four timings of the two trains. An

example would be that of a fast train A which leaves X at 1pm and arrives at Y at4pm and a slow train B which leaves Y at 3pm and arrives at X at 9pm. With

equation 5), the time the trains would pass each other is (4x9 3x1)/(4-1+9-3)

which is 3.666 or 3:40pm. The average method would have given 4pm which isclearly wrong.

However, when the timing durations are equal (DA = DB) it may not be

apparent as to why the averaging the four timings is effective. One could use acomputational shortcut to use twice the duration D for the trains for those

stretches where the travel times in both directions are equal (say in flat straight

segments) which then simplifies equation 5) to:

6) t = (T4T2 -T1T3)/2D.

In this case where the durations are equal, an example would be: train A

goes north and leaves X at 8:10 and arrives at Y at 8:50; train B goes south andleaves Y at 8:20 and arrives at X at 9:00. The computation (now in minutes past 8

oclock) for when the trains pass each other is:

(60x50 10x20)/(2x40) = 35 or 8:35.

When trains have identical speeds (as in this second example), theapproximation of averaging the four timings will be equivalent, or:

(10+50+20+60)/4 = 35; when otherwise, equation 5) is more accurate. To prove

that the average,

7) A = (T1+T2+T3+T4)/4,

is equivalent to equation 5) when the durations (DA = T2 -T1, DB = T4 -T3) are

equal, I examined whether a non zero difference exists between these two

equations 5) and 7), or

8) = (T4T2 -T1T3)/(T2 -T1+T4 -T3) - (T1+T2+T3+T4)/4 and that

9) T4 = T3 + T2 T1 when the durations are equal.

When 9) is substituted into equation 8), it becomes

10) = (T2T3+T22-T1T2 -T1T3)/2(T2 -T1) - (T2+T3)/2.

After manipulation to provide a common denominator and expansion of terms,equation 10) resolves to:

11) = 0/(T2 -T1). Q.E.D.

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Since T2 > T1 then is always zero and for equal durations as a special case, the

average of all four train timings is equal to (or per the proof, is not different from)

the general case which would be described as the product of the arrival timesminus the product of the departure times divided by the sum of the durations.