msb@lsuc.UUCP (Mark Brader) (01/30/86)
Various people write: > [in the UK] the train normally goes forward on the left-hand track. A rare exception to this is a short segment of the London Underground, part of the Northern Line City Branch, somewhere around London Bridge. When this was being built as the City and South London, they figured they'd most likely use cable traction rather than the very young electric technology, and right-hand running in that area gave them a better cable alignment. (But they did go with electricity, the first electric subway, opened in 1890.) > What is surprising is that this scheme is also followed in France, > on the SNCF (Government operated long-distance rail lines) system, > despite the fact that motorists drive on the right-hand side of the > road over there. I query this. I've certainly read of right-hand running on at least some SNCF lines -- I remember it being noted as a minor problem with the Channel Tunnel, for one thing. And in fact, I've traveled on the right on SNCF myself at least once. What I do know is that they have a lot of reversibly-signaled lines. Can we have an authoritative answer on this from somebody? Incidentally, the Paris Metro trains run on the right, but the Lyon Metro trains run on the left. > In the good 'ole US of A, the Chicago and North Western runs trains > on the left track in double-track territory. If you're a commuter > in Chicago, your station is therefore on the "wrong" side of the tracks. > Running left hand - C & NW, because of they were originally financed by > English investors in the 19th century. I've read that this is a myth. The explanation I read is as follows. The line was originally built as single track, with fairly substantial stations all or mostly on the north side of the track. When the C&NW decided there was enough traffic to double-track, they naturally laid down the second track on the south side. Now, most people boarding the trains along the way would be going TO Chicago, and most people getting off along the way would be coming FROM Chicago. People getting off don't need waiting rooms and ticket agents. So the north track, where the stations were, was made the track TO Chicago, so the trains had to run on the left. > Finally, speed calculations - just time the train between mileposts. > I won't insult a group of computer people by supplying a formula. Well, I will. Speed in mph = (3600 x (number of miles)) / (time in seconds) If timing to the nearest second, try to use 2-mile intervals above about 75 mph, otherwise you'll lose accuracy. For 1-mile intervals we have the pairs 60-60, 55.5-65 approx, 48-72, 45-80, 42.5-85 approx, 40-90, 38-95 approx, 36-100, 33-109 approx, 30-120, 29-124 approx. In each case if one number is the time in seconds the other is the speed in mph; memorize the pairs. If you're going above 125 mph the mileposts are in km anyway. Mark Brader
jis1@mtgzz.UUCP (j.mukerji) (01/31/86)
> Speed in mph = (3600 x (number of miles)) / (time in seconds) > > If timing to the nearest second, try to use 2-mile intervals above about > 75 mph, otherwise you'll lose accuracy. For 1-mile intervals we have the > pairs 60-60, 55.5-65 approx, 48-72, 45-80, 42.5-85 approx, 40-90, 38-95 approx, > 36-100, 33-109 approx, 30-120, 29-124 approx. In each case if one number > is the time in seconds the other is the speed in mph; memorize the pairs. > If you're going above 125 mph the mileposts are in km anyway. > In Km post territory, the speed in mph is obtained by the following modified version of the equation posted by Mark: Speed in mph = (2250 x (number of Km)) / (time in seconds) The one that I happen to use more often, since I am more used to Kms than miles is for calculating the speed in Kmph in milepost territory, and that is Speed in Kmph = (5760 x (number of miles)) / (time in seconds) Jishnu Mukerji