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PostPosted: Wed Aug 27, 2014 11:24 am 
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I'm building a slightly complex set of turnouts and so I printed some templates and taped them down to poster board in attempt to plan the whole thing and have something to work against. I very carefully lined them up and taped them down, but then I did some check measurements and found out I was way, way off.

A generic answer would be great, but a specific one would also be fine. I am building N Scale FastTracks #12 turnouts, in a simple arrangement to connect two tracks, like so:

Image

My question is, if the track centerlines are 1.5", how long should the crossing segment be, ideally via some easy to measure point on the turnout? Frog point to frog point would be great. End of quick sticks to end of quick sticks would work.

Basically after I hand laid that out the differences in my "looks perfectly straight to me" by eye layout were 0.75" across several of these! I can make them all the same, but if I am going to that trouble I should make them right so the segment between tracks is properly straight.


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PostPosted: Thu Aug 28, 2014 1:01 pm 
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I took a stab at this, can someone check my math?

http://www.ufp.org/~bicknell/nscale/TurnoutGeometry.pdf

Basically it looks like:

d= linear distance frog to frog I want to know
a= angle of divergence (this requires it to be the same on both turnouts)
c= center line to center line distance
g= gauge of track

The result is:

d = (c - g - (g / cos(a))) / sin(a)

Putting the values for a FastTracks N Scale #12 into the formula, it becomes:

(1.5 - .36 - (.36 / cos(4.76 degrees))) / sin(4.76 degrees)

And dumping that into Wolfram Alpha gives a result of 9.385 inches.

Giving that a eyeball with my ruler, it looks good.

Note that this assumes that from the frog on the rail is straight, which I understand is the case with FastTracks geometry but may not be the case for all switch types that exist in the world.

Does that look right?


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PostPosted: Thu Aug 28, 2014 3:10 pm 
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A valiant effort, but wrong. I printed out a template with these dimensions as well as a end-of-quicksticks line I calculated, lined them up, put down the ruler, and was clearly off.

Image

I'm wondering if I missed a width-of-rail issue, but the theoretical point of the turnout is a point, so I'm not quite sure where that would be...


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PostPosted: Wed Oct 08, 2014 2:03 pm 
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Location: Phoenix (metro area) Arizona
I drew two N Scale FastTracks #12 switches fixed at 1.5 inches center to center in Anyrail and came up with a straight connecting rail of 6.43 inches. Admittedly this is not a concise method for doing this calculation but it works in practical applications.

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PostPosted: Wed Jun 03, 2015 2:12 pm 
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I'd really like to know the proper math on this, and would be happy to turn it into an Excel and give it to our hosts so their customers could do the math in any scale for any turnout size.


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PostPosted: Wed Jun 10, 2015 11:01 am 
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As far as I know the only "point" you can use is the centrelines at the height of the frog. These centrelines crosses each other at the angle deriving from the point number. The crossing point is called "the mathematical point", but this is somewhere between the pointblades and the frog.

From the tip of the pointblades to a point short of the frog is a curve. From shortly before the frog till shortly after the frog is, in principal, a straight line.

I could not quickly find a quick and dirty method for translating a physical point dimension to a schematic approach. If I come across a method in the older books I have I will let you know


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PostPosted: Wed Sep 30, 2015 9:04 pm 
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I think this page has the right formulas, but I have to wrap my head around it...


http://mysite.du.edu/~jcalvert/railway/turnout.htm


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PostPosted: Fri Oct 02, 2015 8:29 am 
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The frog is not at a ratio height from the tangent . . . the frog is an isosceles triangle measured asymmetrically. This will complicate the math of location. (The frog location will be further from the points, lengthening the turnout lead.)

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PostPosted: Thu Oct 15, 2015 5:07 pm 
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I printed some #12 turnout templates and when laid out at 1.5" centerlines, I get about 13" length from frog point to frog point.

That is crossing rails to get point to point, but does give you a reference point.

Not sure my method is correct, but everyone is coming up a little different. :shock:

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