Hi -- While it is not as accurate, there is a way to almost figure out these things in your head. For example, a 12 degree curve indicates that there would be 30 (360 divided by 12) 100 foot segments or 3000 feet circumference (close, not perfect) in a complete circle. Divide that by pi (3.14159) or for rough and dirty, 3 to get the diameter – roughly 1000 feet. Therefore, the radius is roughly 500 feet (less because of the rounding of pi). 500 divided by 64 will give you the REAL feet in S, or again rounding considerably, a bit less than 8 feet or a bit less than 96 real inches. In actuality, the real answer is a shade under 89 inches, or an error of about 10%. Still, this method will give you an approximation for planning purposes – if you are into that sort of thing. FYI, a 96” radius in S is close to a 10 degree curve, considered to be a minimum radius for standard gauge mainlines.
I think there is also a table of these values in one of the NMRA data sheets. Have fun! Bill Winans --------------------------- Jim is correct, the degrees of curvature applies to a 100' chord. If you work through the math, for S scale it works out to: R = 75/(8*sin (A/2)) Where R = radius in inches for S scale; A = degrees of curvature for the prototype Since I know more about narrow gauge: a typical minimum design curvature for an early narrow gauge railroad was 24 degrees. The sharpest curve on the D&RGW narrow gauge was a 25 deg 30" curve between Marshall Pass and Gunnison. This corresponds to a 42" radius in S. However, the Uintah, also in Colorado, had a 66 deg curve over Baxter Pass, and this was after it was straightened when they purchased their articulated locomotives. Previously they had used Shays for the portion of the mainline over the pass. This corresponds to 17" in S, which most modelers would consider very sharp. Of course, different equipment was operated on these two lines. Dave Heine Easton, PA
