Any sugestions would be deeply apreciated.
Im building an ho layout in the diesel era 1972-1997 in my garage. Im in the homework/track plan stage, I want to go three levels and the only option I have to go between levels will be a helix. I want to run long trains ( coal,intermodal,manifest ) with midtrain helpers. My question is how do I properly figure grade percentages per foot. I would like to keep the grade to a minimal. The area I have set aside for the helix is 5’6"x7’ And how do you land scape a helix to make it not look like a giant wall? Thanks in advance for everyones help.
Hello. You can’t work with “grade percentages per foot” because grades don’t relate so much to linear distance as they do to changes in elevation. I think you meant the rise per unit of linear measure along a horizontal distance…which is how grades are calculated. A grade’s “percentage” is calculated as the units of change in elevation (the rise) divided by the run, or the distance along the bench. Rise over run. If you intend that your grade rise one inch for every 100 inches of travel, whether straight or curved, then your grade would be one one-hundredth (1/100, rise over run), or 0.01 in decimal notation. That translates to “one percent”. If you need to get to a height of 3.5", the minimum recommended clearance in HO, and have the same 100" to make it happen, then you have a grade of 3.5/100, or 0.035. That translates to 3.5%. If you want to get the same clearance, but have only 80", your grade would be 3.5/80 (three point five divided by eighty) which comes to 0.04375, or 4.4% rounded up.
That’s all fine and dandy if that’s as complicated as grades ever got on our layouts. Unfortunately, if you want your trains to enter and exit your grade smoothly, and not lose traction and/or decouple cars, you must build vertical transition curves at either end of your grades. What that means is your grades are entered and exited on a curve that starts flat, slowly rises to meet your grade, and then reverses at the other end. It also means that during your transitions your height requirements for the clearances are being impinged. Since you are effectively clipping each end of the grade, you won’t get quite the height you need. The way around that is to make the real grade between the transitions even steeper!! Gads!
When I figure grades, I use 1/8 in of rise for each foot of travel for a 1% grade. 2/8 (1/4) equals 2%… etc. I know it isn’t exact (1/8 is, I believe 1.04%), but is easy for my old brain to remember.
Gary
hi, the 1/8" rise per lineal foot formula is plenty accurate for figuring your grades. when curves enter the equation just multiply the diameter of the circle by 3 and you will get a pretty close idea of track length. actually the figure is 3.14 but 3 is close enough. for example a full circle (360 degrees) of 36 inch radius works out as follows. 36" radius x 2 = 72" diameter. 3 x 72 = 216". so if you are making a 90 degree curve of 36 " radius, you have 1/4 of 216 or 54" of track run. at 1/8" rise per foot, you get 9/16" of elevation. or just a hair over 1/2". a full circle of the same track would get you a little over 2" of elevation which is obviously not enough in HO scale so go with 1/4 inch per foot or 2% and you will get just over 4" of separation. got it??
some other things to consider about grades; curves add to the friction between the wheels and track so one must consider the compensation involved. i don’t have a formula for figuring that but, the sharper the curve, the greater the compensation required. also, short grades such as you would encounter coming out of a yard or otherwise are not the issue that long grades are since only a small portion of the train is actually on the grade at any given time while the other cars are rolling on level track.
i am now in the process of testing all my cars to determine their rolling characteristics. i mark them by sticking a tiny piece of colored paper to the tack boards on both sides of the cars. a car that will roll down 1/2% on it’s own accord gets a white mark, 1% free rollers get green, 1 1/2% get yellow and 2% free rollers get a red mark. anything stiffer than that goes back to the work bench for truck and wheel work. this system enables me to “figure the tonnage” much like the prototypes do and assign power accordingly.
grizlump
As others have noted, grades are measured in ‘rise in 100 ft’ - With a helix, you add the drag from the curvature to the mix. The prototype usually uses .05% for each degree of curve and that is ‘added’ to the physical grade that is present already, For example:
A 5’6" diameter helix is 33" radius. That works out to about a 25 degree curve in ‘Proto Speak’. 25 times .05 = 1.25% compensation for the curve. This means that even flat, level track; the ‘drag’ on the train going through that 33" radius curve will be equal to climbing a 1.25% grade.
Now lets look at the ‘physical’ grade for tha 33" radius helix. This works out to 207" for one complete loop of the helix. With a 4" seperation between helix levels. This works out to just under 2%. So, adding 2% plus 1.25% gives you a ‘compensated’ grade of 3.25% on that helix. Hopefully my math did not fail me this early in the morning. Compensation values can differ from engineering dept to engineering dept of each railroad, and the coefficient of friction is different on HO n/s rail than the prototype’s steel rail/wheels. So tha actual ‘drag’ from the compensation can vary(I am just trying to get ‘in the ball park’ here). Our club layout has a 33"/36" double track helix(5 turns) that climbs 5" on every turn of the helix. This works out to 3.75% with the ‘compensation’ - a pretty stiff grade. We use 3-4 engines to reliably move a 25-30 car train up that ‘hill’. Many of the club members did not believe my ‘numbers’ at first. After running trains for a few months, they either add enough ‘power’ to the train or stay off of the helix! This is also why my eyes ‘roll’ when folks mention that they want to use a 18" radius helix on their track plan ideas. One of our club members likes passenger trains. He was running is 12 car ‘Empire Builder’ with 4 powered Stewart F7’s on
Thanks again to everyone for all the input, It certainly helped me on my plans.
You have hit on the question that could very well give people an aneurism, or at least a bad case of spinning head syndrome. If you are not interested in learning the whole trig thing, then keeping it simple is the key. Grades are based on 100 whatevers in length (inches, feet, miles, mm etc). The important thing is keeping all the measurements in the same units. You will need to know two of three things to figure the third. The length, the height, or the grade. It doesn’t matter which 2, but you need to know 2 of them. For the following equations, the grade has been divided by 100. It just means moving a decimal point over two places. A 1% grade = .01, 2%=.02, 1.573%=.01573 etc.
To find the;
GRADE, take the height and divide by the length. Take the answer and multiply by 100 for the grade
eg: 1" / 100" = .01. .01 x 100 = 1%grade
LENGTH, take the height and divide by the grade.
eg: 1" / .01 (1%grade)= 100"
HEIGHT, take the grade and multiply by the length.
eg: .01 (1%grade) x 100"= 1"
These formulas are accurate, but do not take in for easements. You don’t want to go from flat (0% grade) to a 3.5% grade without doing it gradually or the locos will end up hanging up on it or just running into it.
For the di-hard mathmeticians, yes, I know the length and height and ev
Hi,
Jim Bernier, great explanation. Thank you very much always wondered about the drag created when moving a train through a curve.
Frank
I have a question: what is the equation to find a grade on a full sized RR (is it 1’ rise over 100’= 1%, instead of 1" over 100")?
it doesn’t matter. 1% is 1%. that is 1 unit of rise per 100 units of distance. as long as you use the same definition of unit, it matters not whether it is feet, inches, miles, money, marbles or chalk.
the equation is; divide the rise by the distance of run. example 1 inch rise in 50 inches of run. divide 1 by 50 and you get .02 which translates to 2% 6 inches of rise in 400 inches of run-divide 6 by 400 and you get .015 or 1.5%
real railroads often expressed grades in feet per mile instead of percentages. in that case you divide the number of feet of rise by 5280 (the number of feet in a mile) to get the %.
grizlump
Get yourself one of those foot-long electronic levels which gives the angle in percent. They are inexpensive and are a great aid for installing your track subgrade for accurate and smooth grades. For instance when using cookie-cutter plywood subroadbed, “play” with the risers using C-clamps to hold them temporarily until your grades are “right” and then fasten them permanently. You will probably find the maximum grade will be greater than anticipated because of the need for transitions between grade changes. Don’t forget that track curves create more drag, raising the effective grade on/near the curves. Testing trains for slippage while playing with the risers can help make grade adjustments so that effective grade can be constant.
Mark
If the helix is not going to take up the full 66 inch width, you could wrap a track around the outside at one or more levels.
as far as making it not look like a giant wall, i would consider using foam (pink or blue not the white stuff as it “pills”) and make a mountain out of it. just add a couple of scenes, say a deer hunting or farm scene or the like along with some rocks made of Woodland Scenics castings and you will be good to go
Hi,
sounds like you have quite the project that you are planning. I put three levels into a room that is 9 feet by 11 feet. To connect the levels a 30 inch radius (outer) helix that was installed. It has been in operation since early summer and my experience is that the bigger the radius the better. A 25 car freight train of hoppers and box cars can safely make it up and down. After that it gets dicey as per how reliable the train will go up and down the helix. It is a matter of cars being pulled off of the tracks and the couplers-McHenry’s-holding out. From my experience trains reaching the 35 car length is the upper limit and one has to be very careful of the operation. Unless of course a pusher or mid train helper is added.
Also my three main levels -lowest yard, middle northern Ontario, upper farm country- are only seperated by 12 inches, subtract from that 2 inches for the support frames. This does not leave a lot of space for buildings, hence had to move a mine to the upper level. If the seperation were 18 inches it would leave a lot more choices for scenery as well. Realistic trees are quite a bit taller then the train. Hope that this helps in the design stage.
Frank