
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


NathanO
Registered

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


Helmut
Registered
Joined:  Sun Feb 17th, 2013 
Location:  Friedberg, Germany 
Posts:  1180 
Status: 
Offline


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.
____________________ Regards, H.


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
Joined:  Sun Feb 17th, 2013 
Location:  Friedberg, Germany 
Posts:  1180 
Status: 
Offline


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.
____________________ Regards, H.


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


Current time is 10:22 pm  


