Is the diverging track from a #6 turnout approximately at a 22" radius? Im specifically looking at an Atlas #6in HO.
No. I would guess closer to 33". But the bottom line is that a standard North American turnout does not have a radius. The proper way to do it is to create two separate radius. One from the end with the points and the other from the end of the diverging track. The two centers of the radii would be offest by the degree curvature of the turnout (9.46 degrees for a real #6 and 10 degrees for an Atlas) and the length of that piece of track.
Edit - OK I can't stand it.
An Atlas #6 (Mark II style not the superswitch) has 10 degrees of departure angle (according to them not my measurement). This means the frog is a 5.6 or 0.1767 radians. Call this theta.
The cord of theta can be measured by the square root of (1 - COS(theta) squared) + SIN(theta) squared or 0.1796
The cord of an Atlas measures 9 7/8".
The formula is: radius cord(theta) = 9 7/8
so reconfiguring to get Radius = 9 7/8 / cord (theta).
then substituting values from above, Radius = 54.9"
Then to check it out practically,
I got a protractor and laid two straight edges at 10 degrees.
I took an Atlas #6 turnout and moved it perpendicular along one edge until it fit. It fit at 54".