You are right about armchair layout transitions!  Imaginary trains can
climb any hill!

The answer I learned was:

First  foot at 1% giving 1/8 inch.
Second foot at 2% giving 1/4 inch.
Third  foot at 3% giving 3/8 inch.

There is my transition to 4%.  The three feet transition gives a total
rise of 3/4 of an inch.  As I need to rise a total of 15 inches (and
there is not room for a helix), I get a total run of 34 feet.

The math: Two three foot transitions = 1.5 inches.  15 - 1.5 = 13.5
inches.  13.5 inches rise at 4% gives 28' 2".  This includes two 90
degree turns.  I have underestimated the length of the curves to both
eat the 2" of length and compensate for the curvature.  28 + 6 = 34.

It would be nice to have longer transitions but I am constrained by a
doorway in the NW corner, access to windows in the NE and SE corners
plus the telephone junction box in the NE corner, all in a 12 * 17 room.

This is the kind of thing I do on my 30 minute bus ride each morning and
evening.  (Lots and lots of time here.  None for actual modeling.)

Thank you for all of your help, this armchair plan is now complete.
Thorin

-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf
Of mhrreast
Sent: Tuesday, February 19, 2008 4:32 AM
To: Marty.Thorin
Subject: Re: Transition segments

SOMEBODY has to say it.....
with an armchair layout you don't NEED to worry about transitions!!
<G>
ted larson 


--- In [email protected], "Marty.Thorin" <[EMAIL PROTECTED]> wrote:
>
> Good morning!
>       I am doing some armchair layout construction.  I will need a 4%
grade 
> for this proposal.  I am worried about going from 0% grade to 4% grade

> without a transition.  I looked at the NMRA website and they do not 
> discuss grades (nor transitions).  I did a web search and did not find

> anything about grade transition.  (I did find Woodland Scenic's foam 
> grades and wondered if these have transitions.)
> 
> How long is the transition section from flat to 4% and how much rise 
> does it yield?  2%?  3%?
> Is there a formula?
> 
> Many thanks in advance for this information, Thorin




 
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