How much of an incline could l get away with if I want a train to climb 18 inches between levels.
Thinking of building a helix.
Joe Staten Island West
Hi Joe
A 2% grade that is ideally the steepest you want is approximately 1/4 inch per foot. Therefore you would need 72 feet of run to climb to 18 inches of rise. A 4% grade would be 36 feet of run needed.
Radius compounds grades and I've heard many of times the steepest grade you ever want on a helix is 2 1/2 %.
Your question all depends on what locomotive you're going to use, how many locomotives you're using and how many cars you want to pull.
A shay can navigate a steep grade like 6% pulling a few cars. For that climb you would need 24 feet of run.
Hope this somewhat helps but more information is needed on both ends of your question for the answers you need.
TF
Track fiddlerRadius compounds grades
Returning to model railroading after 40 years and taking unconscionable liberties with the SP&S, Northern Pacific and Great Northern roads in the '40s and '50s.
A simple answer is friction, the flanges of all inside wheels rub the side of the rail, therefore increasing the friction.- NO MATH required.
You might benefit from reading my post about using a 2% incline on a curve:
http://cs.trains.com/mrr/f/11/t/287231.aspx
Bottom line: I installed in and love it. The gradual change in elevation makes the entire layout more eye appealing. While I run freight, the principle is somewhat the same about rise/run.
I'm with you CTD
Scientifically there's actually a mathematical equation to determine radius compounding grades, I called them math radical equations I'm not even sure if I just got the exact terminology above right but I think you get the gist.
For these equations you incert the actual radius and the actual grade percent and it calculates what the grade equivalent would be. I'm actually good in math but nobody has given me the formula yet. There's probably some Plug and Play sites on the internet but I haven't ever looked for them.
In layman terms this is how I understand it. Both grades and radius create resistance for your locomotives pulling power.
Grade resistance is created by gravity. Radius resistance is created by friction. So while your locomotive is fighting gravity on a climb, the radius creates friction on the wheel flanges against the rail to drag your locomotives pulling power down even more. The tighter the radius the more added drag.
I have no idea how finding these mathematical answers could benefit a Model Railroader. My solution to these math problems is just build the slightest grade with the largest radius as you possibly can is all you can do anyway. At that point who needs a number?
Hope that answered your question
TF got it right! I have a small HO layout and wanted a mountain trestle at 10” above my small town and that called for 3½% grade. I have a 32” radius in the up section and a 32” radius helix back down. My norm is counter clockwise direction which is the long grade up and helix down.
My Cab Frowards can easily tow a dozen cars up the 3½% long grade. I have a wye and a loop for reversing direction and my Cab forwards can pull a dozen cars up the helix with a bit of wheel slip.I might add my Rivarossi Cab Forwards have 10oz of added weight with almost 6oz of drawbar so the 3½% helix is a challenge.I have E7s with a bit over 9oz drawbar that can tow 11 heavy passenger cars up the helix easily. Mel Modeling the early to mid 1950s SP in HO scale since 1951My Model Railroad http://melvineperry.blogspot.com/ Bakersfield, California I'm beginning to realize that aging is not for wimps.
Track fiddlerRadius resistance is created by friction. So while your locomotive is fighting gravity on a climb, the radius creates friction on the wheel flanges against the rail to drag your locomotives pulling power down even more. The tighter the radius the more added drag. TF
While there is a slight increase in friction, it's essentially negligible (but increases as the radius tightens). The major part of the apparent increase in rolling resistance is caused by the locomotive pulling on the couplers at an angle. Some of the force is essentially trying to pull the cars sideways off the rails. The force the loco is able to impart to the cars in the direction of travel is the total locomotive pulling force times the sine of the angle between the couplers, which will always be less than the loco's total pulling force. The force imparted to the cars in the direction of travel decreases as the curve radius decreases, and the side force increases.
Here's a diagram about stringlining, but it serves to illustrate the point:
The total effective grade is called the "compensated grade."
For model railroading, it's "CG = G + 28 / radius" where g = actual grade and radius is in inches. This is an empirical formula derived by experimentation. John Allen (yes, that John Allen) derived the first equation many years ago. Some time later, the Layout Design Special Interest Group (LDSIG) did additional experiments and refined the formula to the one above.
Mark P.
Website: http://www.thecbandqinwyoming.comVideos: https://www.youtube.com/user/mabrunton
That is some interesting stuff Mark. Now I have the mathematical formula you Illustrated but I still don't know how to plug and play with it
My brother is a computer research scientist in artificial intelligence and works for the government on military projects. I prefer my younger brother's company better because some of the things my older brother says ties my Noodle in Knotts
I gotta tell you I do take an interest in the things you shared though Mark. I find that stuff extremely fascinating
Mel! I got it right you said! Thanks Mel.
At least you have room on your layout for a helix. I've always wanted to build one of those as I think it would be really fun. We always need a little more room, don't we?
I don't know if it was the same in every school but do you remember when your grade school teacher would put a star on the edge of your desk after you got a good grade or got something right. Then sometimes you looked around the room and some smarty pants kids had the stars going all the way around their desk and starting a second row That always did give me this really inferior feeling
Track fiddler Mel! I got it right you said! Thanks Mel. TF
When I was in design stage of my layout I made a ‘Must Have List’ and the trestle was high on the list. Getting there was easy, getting back down wasn’t. With limited space the helix was the only way back down.I figured that I was limiting my layout to one direction and that was how it was until I got into remotoring my locomotives. I quickly found out after installing the new rare earth magnet motors the added power let me add weight to my locomotives thus adding traction.A stock out of the box Rivarossi Cab Forward has 2.8oz drawbar, after adding 10oz to the shell the drawbar came up to 5.9oz. That is better than a pair of stock Cab Forwards.The new motors with added traction have conquered my 3½% helix.
Mel Modeling the early to mid 1950s SP in HO scale since 1951My Model Railroad http://melvineperry.blogspot.com/ Bakersfield, California I'm beginning to realize that aging is not for wimps.
Effective Slope caused by curves
greg - Philadelphia & Reading / Reading
The calculations for compensation of curves on a grade were posted on either this or my other "home" forum.
The formula is 32 ÷ radius
I have a 45' long grade on my layout that's at a constant 2.8% grade. However, there are several curves within that incline.
The first two each have a radius of 34" and as horseshoe-type grades, extend somewhat beyond a half circle each.The third curve is the entry into an "S-curve, and relatively short, with a 40" radius. It's followed by two somewhat longer reverse curves, both at 48" radii.
That means that if I run a train as long as that entire grade, the effect of the compensation for the curves shows it to be the equivalent of 6.82%.
I have run such a train on that grade (4 locomotives of varying types and in varying positions with the train, plus 71 freight cars, of various types and weights) but could not reach what would be the actual top of the grade, as the partial second level of the layout had not yet been built.
The train occupied about 42' of what would eventually be the finished grade to the partial second level. In day-to-day operations, I would more likely run trains of 20 cars or less, and in almost all case, at least two locomotives...more if necessary.
Wayne
Hi Mark
I Revisited the thread to check out that Compensated Grade formula because math has always interested me.
The guys at my MR Club were talking about that formula one day and I asked them if anyone had it and nobody did. Actually after looking it seems to be pretty simple but I wanted to check with you if I'm thinking correctly here.
CG = G + 28 / Radius
So on my layout my steepest grade is 2% and my smallest radius is 18.
CG = (2 + 28) ÷ 18
CG = 30 ÷ 18 = 1.667
CG = 1.667 + 2 (original grade)
Adjusted grade = 3.667
Is that correct Mark?
Thanks
Track fiddlerCG = (2 + 28) ÷ 18
the true grade is not divided by 28
effective grade = grade + 32/radius = 2 + 1.77 = 3.77
Greg is correct: in my post, the first curve, at a radius of 34", adds .94%, as does the second curve, also at a 34" radius.The third curve is on a 40" radius, adding another .80%, and the final two, both at a 48" radius, contribute .67% each, for a total compensated grade of 6.82% if the entire grade is occupied by the train.
For shorter trains, only those curves which are occupied need to be figured in, and will, of course, vary with the varying radii.
I have an inkling of recollection that that figure of 32 might have originated, at least to my awareness, by some data offered by Al Krug, dealing with real trains.
doctorwaynethe first curve, at a radius of 34", adds .94%, as does the second curve, also at a 34" radius. The third curve is on a 40" radius, adding another .80%, and the final two, both at a 48" radius, contribute .67% each, for a total compensated grade of 6.82% if the entire grade is occupied by the train.
how are you coming up with 6.82?
did you say the linear grade is 2.8%
the section of a train on a straight section of the grade will be 2.8%
the section on the 34" curve will be 3.7% = 2.8 + 32/34
the section on the 40" curve will be 3.6% = 2.8 + 32/40
the secont on the 48" curve will be 3.46% = 2.8 + 32/48
the average grade will be somewhere between 2.8 - 3.7%
Well that certainly is simple math. I thought this stuff would be a lot more difficult than it really is. I just hope my locomotives don't understand this stuff
2) 32÷34 = .94
.94
32÷40 = .8
2) 32÷48 = .67
.67
Orginal Grade 2.8
Total 6.82
I didn't see the one double at first Wayne? I Gotcha
Track fiddlerI just hope my locomotives don't understand this stuff
another thing to bear in mind is that the resistance of a train with good wheels is ~2% (sometimes worse)
a 20 car train, with 4 oz cars weighs 80 oz. and requires ~1.6 oz of tractive effort from the loco.
when that same train goes up a 2% grade, an additional ~1.6oz of tractive effort, ~3.2oz is required from the loco
the tractive effort of a loco is roughly 20-30% of its weight. this means that that 20 car train with ~2% wheel resistance going up a 2% grade must weigh ~16 oz (5 * ~3.2oz)
gregchow are you coming up with 6.82?
Your calculations are correct but, as I mentioned, if the train is as long (or longer) than the entire grade, all curves need to be included as the train traverses them.The number will increase as the train proceeds, until it occupies the total grade, and then, if that's the limit of the train's length, will begin to decrease as each portion exits their respective curves.
At the time of the initial tests, the train did occupy the entire grade, but couldn't go further, as the upper portion of the layout was not yet built.
I also did another test where the same train was stretched over a number of up-and-down grades, and multiple curves in various directions.
It was nerve-wracking to watch the accordion-like slack run-in and run-out in various parts of the train, many occurring at the same time, but there were, surprisingly, no derailments.
Track fiddler 2) 32÷34 = .94 .94 32÷40 = .8 2) 32÷48 = .67 .67 Org. Grade 2.8 Total 6.82
Org. Grade 2.8
the average grade is NOT the sum of the grades under different sections of the train.
This response is only tangentally (get it?) related to the topic, but since there will be no math involved I'll chime in . . .
Anyone here old enough to have watched "Death Valley Days" on television? Back in the late '50s and early 60s? One of the staples of that series was a 20-mule team pulling a wagonload of borax out of the desert. To make hand soap, or something.
Well . . . when the teamster wants to turn right, he lashes the lead mules to pull right, but he also lashes the middle mules to pull left. The hindmost mules just follow the rest. If you ain't a leader, the scenery never changes.
My grandfather was a Cracker, and he told me that.
Robert
LINK to SNSR Blog
doctorwayneif the train is as long (or longer) than the entire grade, all curves need to be included as the train traverses them.
of course, but only in determining the max grade
if you have a straight length 50' of track that rises 1" 1', that is a 2% grade
if you have the same lenght of track where only half, 25' is a 32" radius curve, the grade is 2.5% = 2 + (32/32)/2
if you have the same length of track with a continuous 32" radius curve (helix) the grade is 3% = 2 + 32/32.
you wouldn't consider that case with a continuous 50' curve as 2 sections of 32" radius curve and therefore 4% = 2 + 2 * (32/32)? what if you considered the continuous curve as 4 adjacent 32" curves, would it be 6% = 2 + 4 * (32/32)?
I knew there would be math.
gregcif you have a straight length 50' of track that rises 1", that is a 2% grade
I don't think so, Greg. The grade percentage for those figures is 0.1667%, although you would be correct if that 50' were 50".
your right ... 50" or 1' rise
but what about the curve part of the comment
2.39 If my calculations are correct Greg.
For the 50' 2% grade we're only half is a 32" radius.
i don't see where you get 0.39 from (you do mean 32", not 32')
the effective grade of curvature is 32/R.
so the effective grade of a 32" radius is 1%
if it's only over half the grade then the average is 0.5% = (32/32") / 2 and the total grade would be 2.5%.
Nope, I don't know what I did the first time. I just got back from the drawing board. Your answer is correct and mine was not.
gregcyou wouldn't consider that case with a continuous 50' curve as 2 sections of 32" radius curve and therefore 4% = 2 + 2 * (32/32)? what if you considered the continuous curve as 4 adjacent 32" curves, would it be 6% = 2 + 4 * (32/32)?
No, because it's still only one curve, while the first two curves on my layout are separated by straight track and curve in opposite directions, and likewise with the third curve, which is separated from the second by a short length of straight track and curves in the opposite direction.
The final two curves are separated from the third curve by a length of straight track over a bridge, again curving in the opposite direction of the previous one, and then segueing directly into the final curve, with no straight track, again in the opposite direction of the previous curve, a true "S" bend.
The direction of each curve counters that of the previous one, and all but the last two are separated by various lengths of straight track. A locomotive running light would be in only one curve at a time, the radius of that curve only affecting the compensation of the grade in that area. Where it goes through the two final curves, which are not separated by straight track, I can't say if the transition from a left curve directly into a right curve both add to the compensated grade or if the second one cancels out the previous one...by the time that happens, the loco will already be in the final curve, and onto level track again.My original tests never even took curves into consideration, as I was unaware, at the time, of the formula to determine the effect of curves on grades, although I guessed that it must have had some effect on performance.
In some instances, the loco on a curve can pull better than on straight track, especially if the engine is on the curve, but the trailing train, at the same time, is mostly on straight track...less friction on the straight, more on the curve.