The discussion on TT length prompted me to lok closely through a booklet on steam locos I have, now it says on each of the locos spec sheets what type of curve they were designed for (eg 18 degree, 30’ curve for a J3a Hudson).
I am led to assume that this means a the loco was designed to fit through curved sections that made up roughly 18 degree arcs. What would that 30’ be used to describe?
I have no idea. Perhaps you should try a post on the prototype section of this forum. I would like to know as well.
N:
Thirty “minutes”. Sixty minutes make up one degree, and sixty seconds make up one minute.
Degree of curvature is a curve-sharpness dimension. It measures the degree turned in 1 chain (100 feet) of travel.
http://www.steamlocomotive.com/model/curve.shtml
It’s easier to do this than to get out the furlong compass (generally sold at places that distribute Avogadro tweezers and unobtainium).
As that page shows, the eighteen-degree curve mentioned is about a 44" HO radius. Physics being unscalable, we can get away with a lot more. In fact, the real J3 would have to go pretty slowly on a curve that sharp.
Perhaps the " 30’ " is meant to describe the minutes portion of the arc.
18° 30’ 00" would be 18 degrees, 30 minutes, 00 seconds.
-John
John, my boy, you hit the nail right on the head.
Eighteen and a half degrees equates to a (prototypical) three hundred and nine foot radius curve - and even with superelevation that is likely to be rule-booked a slow order curve. That computes to an HO Scale radius of forty two and eleven sixteens of an inch.
Thanks guys [:)]. Never could quite grasp measurements when they started using minutes and seconds… especially since the symbols are the same as those used for feet or inches.
The NMRA has a nice chart the shows what radius matches the ‘degree of curvature’ for the prototype and for each scale:
http://www.nmra.org/standards/s-8.html
Jim