turnout angle and crossing degrees question

Okay, imagine I am on the left-hand track of a two-track main. I want to have a turnout to the right, a crossing over - but not connecting to as with a turnout - the other parallel track, and then continuing on of course.

I know that if I use a #4 right turnout and a 12.5 degree crossing, I can get all the angles lined up right.

However, I use min #6 on my mainline… Does anyone know what degrees of crossing that would require? And maybe where to get one (handlaying not being my bag)…?

Thanks,

Kris

Im pressuming then, you want a diamond crossing to go across the right hand mainline.

The frog of the diamond will be the same angle as the frog on your right hand switch. The Walthers Shinohara #6 has a frog angle of 9.30 degrees but unfortunately they dont make a #6 diamond. Not that ive seen anyway! However, Peco do (in code 83 and 100 rail) make a #6 switch aswel as a #6 diamond. This would give you your required angles without any alignment problems.

opposite/adjacent = 1/6 = 0.1666, Arctangent(0.1666) = 0.1655 radians or 9.46 degrees.

So a 10 degree crossing could probably be fudged in.

P.S. I get 14 degrees for a true number 4 turnout. 12.5 degrees is for a #4.5 turnout like the Atlas custom line.

Nailed it, TZ. It’s a custom-line I was talking about. The #6 is too, if that alters anything…

Now I just have to go find a 9.5-degree code 83 diamond…

Errr, i just told you above[:D]

Yes, Gary. Thanks very much!

I meant literally find it. The Walthers/Shinohara diamond sounds great… Finding someone with it in stock, now that’s the trick… I am finding a Peco #6 diamond crossing, however, that might serve well.

Kris

Ah, sorry bud, i misunderstood[:)]

No sweat, I appreciate the help! [8D]

According to my Atlas reference a custom line #6 is a ten (10) degree departure, which means it is slightly tighter than a strict #6. I don’t have any information on their super-track #6.