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NathanO Registered

I need some help in finding the proper formula to calculate a radius. I know two items, the degree of curve, and the offset. Lets say I need a 15 degree curve that will create a 3/4" offset. Nathan 

Michael M Registered

Does this help?: http://www.trainweb.org/freemoslo/Modules/TipsandTechniques/degrees_of_curve_to_radius.htm 

NathanO Registered

Michael, No, it does not give the offset information. Thanks for the link though. Nathan 

Helmut Registered

The degree of a curve is the degree of curvature and sets the curve radius. So whatever the offset is, it will not affect the curve's radius. The link given in post #2 gives the wrong formula for degree determination in railroad practice, as the chord length of 100' is defined as 1 station, and not the arc. In the following I refer to the terminology outlined here. The deflection angle I of a curve is affected by the offset x, as I is the angular distance between PC and PI. So the thing one can ask for is: 1. "What degree of curvature do I get when I set an offset x and a deflection angle of 15°?" 2. "What deflection angle I do I get when I set an offset x and a curvature of 15°?" The answer to 1. Is straightforward as the degree of curvature is 15° The answer to 2. Can be derived from the formulas given in the link, and renders sin(I/2)=SQRT(x/2R), where R is taken from the curvature table. 

NathanO Registered

What I am trying to do, as I stated, is find the radius of a curve, that with a 15 Degree section, will give me a 3/4" offset. The Offset is not the length of the section on the Y axis, the offset is the X axis. Nathan 

NathanO Registered

I finally just decided to use my CAD program, that I use for laying out track plans and layouts, for the modular layout group I am part of. I was able to get to within .004" from the ideal 3/4" offset with a 22" R at 15 degrees. The length of the Y axis was within .006" of 5.7" I will now start to work on the other offsets I need: 1/2", 1", 1.5", 2" and 4 " Yes, it will take a while. Nathan 

Helmut Registered

What else than 21.95", to be exact, would you have expected? You can calculate that Radius within 2 minutes, when you just look at the geometrical dependencies. Look at the drawing given in the link: x/L = L/2R = sin(D/2) So x=L*sin(D/2) =2R*sin²(D/2) or R= x/(2*sin²(D/2)) Take your values: 3/4" = 19mm, D=15°, sin(D/2) = 0.13, sin²(D/2)=0.017 Insert in formula for R: R = 19/(2*0.017) = 558.8mm = 22" That's all. Addendum: To calculate y, just remember that x/y = tan(D/2) and so y= x/tan(D/2) In our case, y= 0.75/0,132 = 5.7 As easy as pie. 

NathanO Registered

Helmut, Thanks for the info. I am now using it, along with my CAD program, to see what looks like it will do the job. Nathan 