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NathanO
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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
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Does this help?: 

http://www.trainweb.org/freemoslo/Modules/Tips-and-Techniques/degrees_of_curve_to_radius.htm



NathanO
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Michael,

No, it does not give the offset information.

Thanks for the link though.

Nathan


Helmut
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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
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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
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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
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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
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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



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