39.5"

[ericsp]

I was looking through an old issue of Trains (05/97) and found information about real railroads and curves. I thought I would do some calculations and post the results.

The sharpest curve a four axle locomotive can run on when coupled to a car is 20 degrees, which is 39.5" in HO scale and 21.5" in N scale.

Mainline curves are usually one to two degrees. This is 395" to 790" in HO scale and 215" to 430" in N scale.

[R. T. POTEET]

I was looking through an old issue of Trains (05/97) and found information about real railroads and curves. I thought I would do some calculations and post the results.

The sharpest curve a four axle locomotive can run on when coupled to a car is 20 degrees, which is 39.5" in HO scale and 21.5" in N scale.

Mainline curves are usually one to two degrees. This is 395" to 790" in HO scale and 215" to 430" in N scale.

I don't know about the one to two degree curves but your cited figure for a four axle locomotive coupled to a car sounds about right - of course I'm not sure if those parameters would be correct if that car is an eighty-nine foot autorack.  I try to adhere to the 3X specifications layed down by the NMRA and that works out to an N Scale figure of a twenty inch radius curve for that eighty-nine foot car; that calibrates to a 21.5º curve. Where possible I try to maintain a thirteen inch minimum radius on industrial trackage - 33º - but have had to go down to eleven and a half inches upon occasion - 37.37º. The NMRA standard for that radius curve is for fifty foot cars.

Some years ago one of the hobby mags published a photo of a KCS passenger car being eased around a one hundred and seventy nine foot radius curve in Kansas City.  In N Scale that boils out to 13.43 inches - 32º. The caption cited that this curve was an operational nightmare because. if I remember correctly, it took the better part of an hour to turn an eighty-five foot passenger car on this curve.

[johncolley]

It is worth remembering that railroad surveyors used a 100' chain and measured track centerline curves by the offset of their transit from tangent (straight). A one degree curve has a radius of 5729 feet. the radius of any curve can be found by dividing that radius by the degrees of offset in 100 ft. A ten degree curve has a radius of 730 feet and scales out to 88" radius in HO. So you can see why our attempting to use anything under 36" radius makes our equipment look toylike. I have a 72" radius curve with easements and superelevation that doesn't look too bad, but would like to see some really big curves. There is a modular group in the east with a minimum radius of 88" and it looks great! I also prefer large turnouts and note that CVT turnouts #9 has a longer point to frog distance than Wally's #10, hence a longer smoother curve even though there is a bit more angle of divergence.                                                  jc5729 JOhn Colley, Port Townsend, WA

[Tilden]

I think I'll start looking for old warehouse space.....

[reklein]

JohnColley says," The radius of any curve can be found by dividing that radius by the degrees of offset in 100ft." I understand the offset thing, but how can the radius be found by dividing it by the offset. We don't know the radius. Clarify please?

[Gandy Dancer]

Oh darn I edited over the top of my old message instead of posting at the end of the thread ... How stupid. 

I always thought the 100 feet was a chord, but can't make the math work to get the numbers others are throwing around.

OK, I think I got it.   Had to draw a picture and resort to a Geometric conversion and then Trigonometry.  The radius ends up being the hypotenuse of a triangle with the apothem as the adjacent side. The values for the apothem and sagitta didn't even come into play.  At least it is within numerical tolerance due to computer rounding errors.  The sine of an angle is equal to the opposite over the hypotenuse.  In this case the opposite happens to be 1/2 the cord.  The angle is 1/2 the degree of curvature.

The radius can be derived from the degree of a curve as follows:

Radius = (chord length/2) / sin(degree of curvature/2)
In railroading the chord length is always 100 feet.

so a 10 degree curve is:

R = (100 feet/2) / (sin(10 degrees/2) =  50 feet / sin(5) = 50 feet / 0.087 = 574 feet
R = 574 feet * 12 inches/foot = 6884 inches
R = 6884 inches / 87 HO scale = 79" radius in HO scale.

I don't know how the prior poster got a value of 88".

 

[reklein]

I looked in a surveyors manual this morn. And the instructions are pretty complicated to lay out a horizontal curve. It increased my respect immensly for those old time engineers. The problem is complicated even further by having to go around things you can't see through. Theres a lotta trig involved .Its a lot easier to layout a curve on a model railroad with a yardstick or a string.

[selector]

...Its a lot easier to layout a curve on a model railroad with a yardstick or a string.

.....yeah.....uhh....sure it is.  Agree completely. 

Yessirree.

[johncolley]

REKlein et al: My bad for not explaining. The problem with surveying a curve is that there is no place to put the center pin and use a string for a compass like we modellers can. Think of a bow with the curve as the track centerline, and the string as a chord. The surveyor's chain is pulled tight ( and there are actually tables to compensate for shrinkage or expansion due to temperature differences from standard! but this is not needed here) So consider a 10 degree curve as mentioned earlier: the first stake is driven with the transit lined up with the tangent (straight track) behind it then reversed to look ahead, the surveyor swings the transit 10 degrees and another stake is driven in line with the first. The chain is moved out another 100 feet and the transit is moved to the new point just staked, aligned back to the start point, reversed and the process is repeated as often as needed, What you end up with is a series of straight lines 100 feet long, joined together each at 10 degrees from the ones adjacent to it. An arc connecting the joints makes the track centerline. If you actually wanted to lay out your curve on plywood, using 100 scale feet and a protractor, this way, you could for a better understanding of the process. Like anything else in this hobby there is nothing like practicing a few times, eh?                                                 jc5729 John Colley, Port Townsend, WA

[johncolley]

Gandy Dancer, sometimes the simple method is the easiest. 5729 feet divided by 10 (degrees)rounds out to about 573 feet radius. so x 12" per ' and divide by 87 (our scale in HO) does indeed give about 79 and 1/32" Thank you, now I will have to see what 88" radius equals. So 88 times 87 divided by 12 = 638' gives 8.979 degrees or almost a 9 degree curve. Good to know and share, eh? This too is happy railroading! jc5729 John Colley, Port Townsend, WA

[reklein]

 

John,thanks for the comeback. I've done a little surveying back in the day,but under the guidance of a civil engineer.He said do this and this aand you'll get this.Then he said"now go do it again and be within a tenth in a mile". I understand now how one lays out the segments of the curve . How do the real guys get the arc between segments or point on line or whatever that turning point is called.

[ericsp]

The calculations are based on the assumption that the 100 feet is the arc length. I do not know if this is correct, but that is what Trains had.

[Gandy Dancer]

the simple method is the easiest. 5729 feet divided by 10 (degrees) rounds out...

I don't understand why that works?  Usually circular functions aren't linear.

[ericsp]

the simple method is the easiest. 5729 feet divided by 10 (degrees) rounds out...

I don't understand why that works?  Usually circular functions aren't linear.

That comes from assuming the 100 feet is the arc length, as opposed to the chord. The equation used is arc length = radius * angle (in radians). If you divide both sides by the angle you get radius = arc length/angle. For a 100 foot arc length radius = (100 feet)*180/(PI*angle), this is approximately 5729.578/angle (in degrees). 1 degree = PI/180 radians

[johncolley]

The surveyor's chain is always stretched tight and is the chord = 100 feet. There are simple trig formulas to give the height of the arc from the chord so an intermediate stake can be driven at the center point.

For those inclined to pursue it: remember that you must use half of the total angle so if looking for a 10 degree curve you must use calc's for 5 degrees to get the height of the center point of the arc.

It helps visualize it to draw a bow and arrow, with the arrow at the center of the bow. X= half of the chord, y= the distance from the chord to the center if the arc were continued to make a circle.

x/y=tan <, x= r. sin <, y= r. cos <,

r. - y = h (height of arc at center)

It would be a lot easier to understand if I could draw it here.                                       jc5729 John Colley, Port Townsend, WA

[Gandy Dancer]

the simple method is the easiest. 5729 feet divided by 10 (degrees) rounds out...

I don't understand why that works?

That comes from assuming the 100 feet is the arc length, as opposed to the chord.

Ah! So the sharper the curve the more inaccurate the estimate calculation.

[Anonymous]

Well, 4 axle units are different leinghts; consider a 44 tonner and a GP60!

[RDG1519]

Gandy Dancer, you gotta love this forum, I can't remember the last time I used "apothem" and "sagitta". It goes to show the depth you can go in this hobby.

Poteet brings up a good point. As to extremely tight radius on industrial trackage it is useful to remember that many times the locomotive was never intended to traverse the curve, instead the switching crews would use other frieght cars as "handles". I have calculated from drawings 12 and 13 inch radius curves (HO scale) from turnouts into buildings. Many times the locomotives particularly steam locomotives were not allowed to enter the buildings for obvious reasons. Add to this what I am told the "ALCO" salesmen would say" an S-1 can go any where a 40 foot box car can go". Chris

[johncolley]

Wow, this discussion is getting interesting and is bringing back a lot of old memories. When I was a major food company's Project engineer we had one job relating to the addition of a spur into one of our warehouses, and I happen to still have the letter and specs from Indiana Harbor Belt Line for industrial spurs. At the time ('80's-early '90's) they were still using SW1200's or SW1500's with slugs. Their minimum allowable radius for on site industrial spurs was 12 1/2 degrees which as I recall came out to be 458'. So in HO that would be about 62 1/4" radius. Thanks for refreshing some good times, jc5729John Colley, Port Townsend, WA

[reklein]

To belabor a subject,I got to thinking "didn't my cousin give me an old book she found on RR engineering?" I looked around and found 'Railroad Curves & Earthwork' By C.Frank Allen copywright 1931 by McGraw Hill. Expalins the whole thing. However I don't have the math to do the problems,but interesting nonetheless.  The book even has how to layout RR spirals and how to compute cuts and fills. VeryInteresting. It would be fun to get a transit and go out with a coupla guys just to see whats involved.