Is constructing a helix track a simple matter of creating your desired radius, (22", 30" etc), then just increasing the grade at each level at 1 inch per 7 or 8 feet as it ascends upward?
i think the first thing you must do is determine how much head room you need between levels. then knowing the difference in elevation between the top of the rails, you can calculate the radius needed to achieve that as well as the grade it imparts. all three must be included in the calculation. give us some hard figures and someone can work it out for you.
grizlump
on second thought let me try to help you figure this out. i have never built a helix but here goes.
say you want 4" between levels and your roadbed is 1/2’ thick making a total of 4.5" overall difference between rail tops. using 36" radius curves, that means you must climb 4.5" in 226 inches. (the circumference of a 72 inch diameter circle) remember pi? 36x2x3.14. 4.5 divided by 226=.0199 and change or roughly a 2% grade.
obviously, the tighter the radius, the steeper the grade for the same amount of headroom.
perhaps someone on here can tell us of his experiences and just how much clearance you need between levels for track maintenance and construction and (God forbid) re-railing. might vary with the size of your hands.
grizlump
Three glasses of wine tells me for a 2% climb you will need about a 31.5 inch radius on your helix. I stand to be corrected. Hic! [:)]
Brent
Constructing a helix is simple, but there are a few cautions:
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Establish the grade BEFORE the track enters the helix if you are building with a tight overhead clearance.
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Carry the grade OUT of the helix before transitioning to level track.
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While the HEIGHT increases, the GRADE within the helix is, or should be, constant.
With a constant grade, the track will rise X in one turn. If X = 4 inches*, a 22 inch radius will have a 2.9% grade and a 30 inch radius will have a much easier 2.12% grade. X has to be the total overhead clearance, plus the thickness of subgrade, roadbed and track (rails plus ties) - aka the railhead-to-railhead height.
You can’t think of increasing height in terms of a linear distance (“1 inch per 7 or 8 feet.”) You have to think in terms of 1 inch per 90 degrees of curvature. To translate that to track length, it’s Pi x R / 2, where R is the radius and Pi is that wonderful Greek letter that means 3.1415… If you try to set an arbitrary even-percent grade you end up with some odd vertical dimension per turn, much more difficult to design and build than a simple 1 inch per 90 degrees or 4 inches per turn.
- 4 inches with a track structure < 1 inch thick will leave NMRA standard clearance in HO. HOWEVER, this might not be enough if you run double stacks and auto racks. The absolute thinnest track structure I have built is about 1/4 inch thick - Atlas flex laid inside a steel stud, anchored with latex caulk and reinforced with twists of #22 wire through the track nail holes. At 4" per turn, this leaves 3.75 inch railhead to overhead clearance - enough for most purposes, but still probably too little for a loaded Schnabel car.
Chuck (Modeling Central Jap