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PostPosted: Thu Oct 20, 2005 3:12 am 
Sorry but I am not used to this system of#.

How do you relate it to a radius in centimeters ???

Thanks


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 Post subject: radius in ARA turnouts
PostPosted: Thu Oct 20, 2005 4:01 am 
Most straight turnouts using ARA specs have a radius so large that there is no point in worrying about it in N or HO scale. However number 6 or 4 turnouts might be a concern for some equipment, anything larger has a radius so large that its hardly seen in model railroading.
Turnout generally have a set of straight points leading to a gentle curve and then to a straight frog. the angle of the frog indicates the spread say for a number 10 it would go like this, take a measurement, say 1/2 inch and use it as a constant, for a number 10 turnout you would x ten the 1/2 inch then 90 degrees to that a 1/2 inch measurement in inserted and the ends aligned. thus completing the triangle.
Is there a specific reason your concerned about the radius or is just for alignment purposes? If its for equipment concerns, number 10 turnouts was used by the Union Pacific terminal where they had their biggest locomotives, (articulated etc etc)

Rob


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 Post subject:
PostPosted: Thu Oct 20, 2005 4:25 am 
Sorry but in Europe we use the angle but given in tangeant figure or the radius of curve to know wich type of equipement can use a turnout.
For example most of turnout have a radius of at least 400 m.


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 Post subject: translating #10....
PostPosted: Wed Jan 16, 2008 11:46 am 
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Joined: Sat Jan 12, 2008 7:43 am
Posts: 1
Location: Germany
That's not correct:

The #10 is written as 1:10 in Europe.... Just visit a german Site: http://www.gleisbau-welt.de/site/weichen/weichengeometrie.htm

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PostPosted: Sun Nov 23, 2008 3:25 pm 
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Joined: Fri Nov 21, 2008 1:32 pm
Posts: 4
If you want to know the radius that the diverging track is effectively following, download the templates of the turnouts. They include that information. They are free and can be found on this website. For example, this is what a few of them indicate:

#5: 14" (35.56 cm)
#6: 23" (58.42 cm)
#7: 27" (68.58 cm)


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PostPosted: Sun Nov 23, 2008 6:10 pm 
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Joined: Sat Aug 13, 2005 10:48 am
Posts: 365
Location: East Texas - USA
True, an equivalent radius will be smaller than the closure radius.

The minimum radius of the closure rail in a normal US AREA turnout will be the following formula:
n= frog number
g= gauge in inches
R= radius of closure rail center line
R= 2*g*(square(n))

Thus for number 6:
N=6 (6*6=36)
R=(2*56.5*36)
R= 4,068” (4068 /160 = 25.43” radius of closure in N) - about 645,92mm

Thus for number 10:
N=10 (10*10=100)
R=(2*56.5*100)
R= 11,300” (11,300 /160 = 70.63” radius of closure in N) - about 1794,0mm

You only need to divide the prototype by your scale factor and any scale will give the proper minimum radius.

-ed-

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COSLAR RR - http://www.coslar.us/
NMRA Standards and Conformance Department
PROTO & FINE Scale Coordinator
I estimate I have about 5 pounds of coupler springs somewhere in the vicinity of my workbench.


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