jclayton
May 13th, 2003, 06:12 AM
I read in a very old book which I have (ca: 1909) that rail weight was (is?) determined by the following formula:
E/D/2250 lbs (long ton) X 10 = rail weight per yard
where:
E is the weight of the locomotive in pounds,
D is the total number of "drivers" on the rails,
2250 is the number of pounds per long ton (a constant),
10 is a constant.
All of the examples in the book were of steam locomotives near the turn of the 20th century and the rail weights were given in a table whose maximum was 100 pounds per yard ... which makes sense since locomotives of that era were under 200K pounds.
Question: I know that the UP Challenger and Big Boy locomotives nearly tipped the 1M pound range. With that formula in mind, does that mean that rail weight in some areas was between 270 and 300 pounds per yard? In addition, do you know if that same formula is applied to today's modern motive power units? If not, how?
... Jim
E/D/2250 lbs (long ton) X 10 = rail weight per yard
where:
E is the weight of the locomotive in pounds,
D is the total number of "drivers" on the rails,
2250 is the number of pounds per long ton (a constant),
10 is a constant.
All of the examples in the book were of steam locomotives near the turn of the 20th century and the rail weights were given in a table whose maximum was 100 pounds per yard ... which makes sense since locomotives of that era were under 200K pounds.
Question: I know that the UP Challenger and Big Boy locomotives nearly tipped the 1M pound range. With that formula in mind, does that mean that rail weight in some areas was between 270 and 300 pounds per yard? In addition, do you know if that same formula is applied to today's modern motive power units? If not, how?
... Jim