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 Post subject: Radius of arch-"N" scale
PostPosted: Sat Aug 01, 2015 10:40 am 
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Joined: Thu Jul 30, 2015 1:15 pm
Posts: 1
How many degrees of arch does each "N" 24' radius cover. On my Ntrak modules, I use a 24" radius to move the 3 lines from their normal position to a position some 12-16 inches toward the front of the module (the module is oversized). This is, of course, sweeping the rails into to a 24 in radius followed by a reversing radius to straighten out the run across the front of a 2 module set up. All this is reversed on the second module to bring the tracks back specified position. I am trying to figure how many pieces to order.
flexiflyer


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PostPosted: Sat Aug 08, 2015 7:15 am 
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Joined: Thu Feb 23, 2006 4:51 pm
Posts: 84
Hello flexiflyer,

Each SweepStick is approximately 10" long and there are 2 SweepSticks in each package for a combined length of 20". To determine how many would be required to create a 90-degree arc with an 24" diameter, for example, use the following formula:

C over d equals Pi where C = Circumference, d = diameter and Pi = 3.141592

therefore,

C = (24" x 3.141592)/4 = 18.850"

and,

18.850" = 1 package

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- Terry -
Customer Service - Fast Tracks HobbyWorks Inc.
service@fast-tracks.net


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PostPosted: Tue Aug 18, 2015 11:26 am 
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Joined: Sat Aug 12, 2006 7:04 pm
Posts: 45
Wanna try that again Terry. With a diameter of 24" the Circumference is determined by C = Pi * D, C = 3.14 * 24 = 75.39. with each Sweepstick being 10" long you would need 8 total to complete the full circle or 4 packages. That is unless there is some new form of math I am unfamiliar with.


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PostPosted: Fri Aug 21, 2015 7:54 am 
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Joined: Thu Feb 23, 2006 4:51 pm
Posts: 84
To All,

Sorry, my mistake. The formula should be as follows:

C = Circumference, d = diameter and Pi = 3.141592

therefore with a diameter of 48",

C = (48" x 3.141592) = 150.796" and a 90-degree arc would be 150.796"/4 = 37.699"

and,

37.699" = 2 packages (rounded up)

_________________
- Terry -
Customer Service - Fast Tracks HobbyWorks Inc.
service@fast-tracks.net


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