I know there is someone here that can give me an answer. If 2 units in 100 is a 2% grade, (according to books) How can I have a train do an under over in 48"? Is that still a 2% grade? [%-)]And on a straight run, how many feet do I need to go from 0" to 6" high?. This is to round for my square head to figure out [banghead]. Thank you.
In order to have an over & under, figuire 8, in a 4ft by 8 ft area you will have a grade of 6% or better. Not good with grade percentages, also you might need magnatraction to keep the engine on the track going around curves.
I had a scout set from the 1960’s with a # 235 scout engine that had magnatraction and it came with a figuire 8 track & trestle set. Came with an engine & tender with four other cars, a missle launching car, a satelite car and an explosives boxcar with a brown caboose, a # 6017.
Lee F.
Thanks Lee, you’ve clarified it a bit for me. Now all I need to know is how long a run of straight track I need to make a 2% grade to clear a 6" structure. How do I calculate this?[
] The book said I could make an UP curve in a 48" circle. I don’t understand where they get the figures from.
BTW that’s a nice set you have.
Larry
To get a 6-inch grade separation with a 2-percent grade, you need 6/.02 inches, or 300 inches, or 25 feet. That’s why toy-train grades are often steeper than 2 percent. With 3 percent, you need 16 feet 8 inches. With 4 percent, which is about as steep as you would want to go, you need 12 feet 6 inches.
Don’t forget that you won’t have all of those 6 inches for the train. That has to include the track and its support too, a minimum of 1/2 to 1 inch probably. This probably rules out double-stacks and raised pantographs.
What counts is the distance along the track. If you’re talking about a 48-inch-diameter circle, that’s 48 inches multiplied by pi, or about 151 inches. That’s long enough for a 6-inch rise with a 4-percent grade: 6/(48*pi) = .040 = 4 percent.
If you’re talking about a 3/4-circle loop and a crossing at right angles, that’s 161 inches. The grade is 6/(48pi3/4 + 48) = .037 = 3.7 percent.
Perhaps your book was talking about a 48-inch radius (96-inch diameter). In that case, the required grade would indeed be 2 percent.
Thanks Bob,
I needed to know the formula. You have helped me greatly. Thank you for the expert explaination and examples. [tup]
Larry
Are you using locomotives with traction tires? Magnetraction only? What type of rails? How many motors doing the hauling? And how much weight are you going to pull: die cast gondolas, plastic cars with plastic trucks/couplers, 21" Kline passengers? Smooth transitions from the flats to the grades is always key in preventing uncouplings. What radius turns going up the grades? Got enough transformer power?
Depends, depends, depends …
Back in the 50’s my 1957 CP F3 (dual horizontals, magnetraction) was able to haul 5 Aluminums L. Passengers using the D side of a 1956 ZW. up Lionel’s iconic graduated trestle set, using classic Lionel 031 tubular track. As they say: " that was then … today things are different! [:D]
No Traction tires, Magnetraction only, Lionel 0 tubular track, 1 motor possibly 2 (depending on loco used), maybe 8-12cars, med weight freight cars, no turns going up or down, top & bottom of curves are 054 to make 1/4 circle, plenty of transformer power.
Guess I’m old fashioned, not sure I like command control.
That’s why I sit next to lionelsoni…[(-D]
If you aren’t going to run any excess height equipment, then why go up 6 inches? Why not 5? Start your incline/decline as close to the underpass as possible, and keep it steadily increasing/decreasing towards the apogy, using the entire run.
I need 6" to go over a structure at the top of the run. I just needed to know how long the straight run had to be, in order for me to clear the structure.
Thanks