Yes, 3.5 is close. However, the helix would build if I was going to have one is very sturdy and has a thin cross section.
A club I was in built a helix this way. It worked very well and was very sturdy. Basically, cut 1/8" thick plywood into 45 degree arcs at the appropriate radius. Glue those sections together with staggered joints and you have a very sturdy, vertical kink free 1/4 inch thick roadbed. Support the helix on both sides with vertical 1x2s with notches or threaded rods with nuts. Since the helix is hidden, there really is no need to use cork or other roadbed material. Just lay the track right on the plywood. Assuming the total height of the track is about 3/16th of an inch, the total roadbed plus track height is just under 7/16 inches, meaning the vertical clearance is just over 3 inches at 3 1/16 inches. The NMRA standard clearance above the rail heads is 3 inches.
deadhead 49 wrote:THANKS MUCH FOR THE TIME . I KNEW THAT ALL YOU HAVE TO DO IS ASK AND ONE FINE PERSON WILL ANSWER. STEVE
Regarding what ericboone said about the helix - 3.5 inches is pretty close separation (in HO), when one takes into account the subroadbed and roadbed, plus the thickness of the track itself. I can be done, for sure, but your subroadbed thickness, especially at splices, might get to be problematic. You would be better off planning on 4" (or even 4 1/2") tier separation for access to equipment on the helix. With a 4" separation, you'd need a curve radius of 32", rather than the 28" you'd have with a 3 1/2" separation.
Mark P.
Website: http://www.thecbandqinwyoming.comVideos: https://www.youtube.com/user/mabrunton
Oooh, crap! That's what I get for trying to cram a reply into the last three minutes before I have to go out to dinner. My really bad!
Forget what I said...it's garbage. Worse, it's confusing garbage.
I was doing fine until I got to the reading of the result in step three. I took a left turn when I should have turned right. The result would simply be your length of run, in inches, which would have been the figure quoted in the earlier post. So, you would have divided 13" by 0.02, and you would get a readout figure of 650".
Aaagh! What was I thinking!
Thanks, Larak, for breaking it to me gently.
selector wrote:These guys are perfectly correct, but I'll offer another variation, or word picture, if it'll help. Sometimes different words work better than others, but they won't be more correct than what the other two gentlemen have said,...just different.Step 1 - take the percent grade you have in mind, and change it to a decimal. That reads as "0.02", or simply "decimal zero two". Step 2 - decide on your height difference....you said 13", so we'll work in inches.Step 3 - on your calculator, input 13, then hit the divide sign (looks like /), and then enter the point, followed by 0 and then 2. Press the = sign, and you will get a number like 0.12345...whatever...the thing to remember is that there is a decimal point in the second position. The next three numbers, to the right of the decimal point, are to be read as " ...percent grade". If the numbers read "0.0142337", you read them as "one point four percent". The first two numbers are the actual percentage, and the left-over is the third number, hence "point four".As was stated, grade is merely the distance horizontally over the rise vertically...up or down. If that helped at all, super. If not...nothing lost.
These guys are perfectly correct, but I'll offer another variation, or word picture, if it'll help. Sometimes different words work better than others, but they won't be more correct than what the other two gentlemen have said,...just different.
Step 1 - take the percent grade you have in mind, and change it to a decimal. That reads as "0.02", or simply "decimal zero two".
Step 2 - decide on your height difference....you said 13", so we'll work in inches.
Step 3 - on your calculator, input 13, then hit the divide sign (looks like /), and then enter the point, followed by 0 and then 2. Press the = sign, and you will get a number like 0.12345...whatever...the thing to remember is that there is a decimal point in the second position. The next three numbers, to the right of the decimal point, are to be read as " ...percent grade". If the numbers read "0.0142337", you read them as "one point four percent". The first two numbers are the actual percentage, and the left-over is the third number, hence "point four".
As was stated, grade is merely the distance horizontally over the rise vertically...up or down.
If that helped at all, super. If not...nothing lost.
Wow Selector, that even confused me for a moment. Did you mix your equations by accident? Your INPUTS are % grade and vertical distance. Won't the output be horizontal distance in the same units as the vertical distance? IE 13 / 0.02 = 650
Now 13/650 (Rise / run) = 0.02 (percent grade).
Karl
The mind is like a parachute. It works better when it's open. www.stremy.net
Formulas for future reference:
Rise / Run = % Grade
% Grade * Run = Rise
Rise / % Grade = Run
13 inches / 0.02 = 650 inches
Good luck,-John
Grade is simply rise over run. A 2% grade rises 2 inches every 100 inches of run or requires 50 inches of travel to climb 1 inch. Thus to climb 13 inches at 2%, you need 650 inches or just over 54 feet.
Ideally, in my opinion, you'll have the mainline gradually climb as it goes around the room, flattening out for town sites. (Unlike the real thing, model train cars don't have brakes to keep them from rolling on a grade.)
However, you may need to build a helix. Assuming 3.5 inches of vertical distance between rail heads, you'll need 175 inches of run per loop at a 2% grade. That correlates to a 56 inch diameter helix.