Well, it's easy to get degrees from the frog number. But the problem is then figuring which side of the angle you're actually on. Either Atlas rounds their numbers or they don't EXACTLY fit to make a double crossover - small gaps are usually not a problem, but I don't think things fit completely square if you attach all 4 turnouts tightly to the crossing. One of the old Custom Line plan books I have has a section in the back that shows how to make all sorts of combinations using the stock parts.
Otherwise, it's right, you need the frog angle of the crossing to be double that of the turnouts to form a double crossover, a a #3 crossing for #6 turnouts, or a #4 crossing for # 8 turnouts. If I had an easy way to sketch it out...
So on one main, you have 2x #8, one left, one right. Tracks fromt he diverging routes cross in the middle. That forms a triangle. The sum of angles in a triangle is 180. A #8 is 7.125 degrees. SO what's left is 180-7.125-7.125, or 165.75.
With me so far? Extend one of those lines through to the other side, like the crossover really is. The sum of two angles on the same line is also 180 - you know, turnign around and going back exactly the way you just came is pulling a 180. So the smaller angle, the one that is the angle of the crossing - the onees facing the direction of tracvel along the mains, is 180-165.75, or 14.25.
Draw two parallel lines and an X between them, then you cna see what I'm talking about.
The frog number is 1/TAN(angle) Don't make me to into expansion of expressions for trig functions, but if the angle of the frog is ARCTAN(1/frog) then the frog number is 1/TAN(angle). So that's 1/tan(14.25) which might as well be 4. Thus - indeed you do need a #4 crossing.
Math is fun. As long as you stop at integrals.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.