Here are the details for doing the math:  I had a 48 inch radius helix on my
layout.  It was constructed from ¾ inch plywood with 2x4 risers (perhaps
overkill but I could walk on it if needed!).  To make things easy for
myself, I installed the risers center to center at 12 inch spaces and cut
the top ends on a “chop saw” at the proper angle so the plywood helix pieces
sat on the 2x4’s nicely (did the trigonometry but do not remember what that
angle was!).  The center of each 2x4 was ¼ inch taller than the previous.
(.25/12) x 100 = a grade of 2.083%  - not precisely 2% but worked well for
me.

 

Here is the critical math:   The circumference of a 2*48 = 96 inch diameter
circle = 96*Pi = 96*3.1415…..  = approximately 301.584/12 = 25.13 feet.
Since, in my case, the helix rises ¼ every foot of length, the spacing
between adjacent levels was 25.13 * ¼ = 6.283 inches ------minus the
thickness of the plywood.  This easily accommodated the height of any
rolling stock I was using.  I used the helix to accommodate a mushroom
design.  There were two “layers” to my helix so I gained 2 * 6.283 = roughly
12 ½ inches.  I needed more spacing between the levels of the mushroom but I
had enough straight track between the helix and the mushroom to continue the
climb and get to the upper level.

 

For your case, using 46 inch radius but my 2.083% climb rate design, you
would get 92*Pi = 289.018/12 = approximately 24 feet per level.  24*1/4 = 6
inch spacing between levels.  If you used 3 levels in the helix, instead of
only two, you would actually gain 18 inches in height. Or, depending, on
where you need the ends of the helix to be, you could get 15 inches with 2 ½
levels in the helix. – that is, trains leaving the helix would be traveling
at a direction 90 degrees from whence they entered.

 

You could also “back calculate” the radius if the 15 inch rise is important.
That is, for instance, with a 3 layer helix, you need to rise 15/3 = 5
inches (I am not remembering the NASG standard for height clearance – is
that enough?).  Then 5*4*12 = 240 inches of helix track required.   240/Pi =
a 76.4 inch helix diameter = a 38.2 inch radius, which is probably smaller
than what you would like to use.

 

I hope this has helped.

 

Walt Jopke

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