It seems to me that a 180 degree curve, for example, would be a U-turn, but that tells you nothing about the sharpness of the curve.
What am I missing?
- Harry
Sharpness of a curve is measured in inches of radius, at least in model railroading.
Prototypes use a different system.
Dave
Lackawanna Route of the Phoebe Snow
I think (and I could be way off base here) that 1:1 RR's measured curves in degrees of arc, rather than radius or diameter. I'll have to dig up my NYC locomotive book that I found to verify this... or maybe someone else will know off hand.
Like Pheobe said, we measure sharpnes in teh size of the curve radius, and I believe that most commercial sectional track is sold in 22.5 degree sections.
-Dan
Builder of Bowser steam! Railimages Site
HarryHotspur wrote:It seems to me that a 180 degree curve, for example, would be a U-turn, but that tells you nothing about the sharpness of the curve.What am I missing?
Real railroad engineers measure a curve based on 100 foot cords. That is a distance of 100 feet is measured out and the difference (divergence) in the rail at that point from the starting point is the degrees of curvature of the track. In the prototype 10 degrees is a sharp curve, and without redoing the math I recall that equates to about 88" radius in HO scale. 10 degrees is the smallest an SD45 can negotiate. The D&RGW had some 18 degree curves on the Monarch Branch and some 24 degree curves on the narrow gauge.
Why 100 foot cords? I am not certain, but I've heard that it is the length of chain that was practical for the early surveyors to use. It just became the standard.
Elmer.
The above is my opinion, from an active and experienced Model Railroader in N scale and HO since 1961.
(Modeling Freelance, Eastern US, HO scale, in 1962, with NCE DCC for locomotive control and a stand alone LocoNet for block detection and signals.) http://waynes-trains.com/ at home, and N scale at the Club.
gandydancer19 wrote:TZ - Thanks for that bit of information. Would you happen to know the distance in feet at that 100 foot mark that makes up the 10 degree curve? I'm sure there is a formula somewhere for that, but math is not one of my strong suites.
We know the angle is 10 degrees. The number we want would be the adjacent side of the triangle. Adjacent is the cos(), so the cosine of 10 degrees is 0.984. Multiply that by 100 for 98.4 feet. Or if you want to know how far from the original center the track has moved it would be the sine of 10 degrees or 17.3 feet.
radius of 1 degree curve based on cord definition (railroad) = 5729.65'
So, a 1 degree curve in HO = 790.29 inches
A 10 degree curve is really a sharp curve on the mainline, about 30 MPH max. It would be 79 inches radius
So you can see even 48" radius curves in HO are really industrial spur type geometry.
Remember in high school when the geometry teacher said to pay attention because some day you were going to need this?
Today is the day.
R = Radius
D = degree of curvature
R= 50/sin (D/2)
For a 10 degree curve, sin (10/2) = sin 5 = .0872
50/.0872 = 573 ft.
Dave H.
Dave H. Painted side goes up. My website : wnbranch.com
Or if you do not want to go through all of that math, look at this NMRA web site(another free service of the NMRA):
http://www.nmra.org/standards/sandrp/s-8.html
The chart in Standard S-8 shows the prototype degree of curvature, the prototype radius in feet, and below that the radius in inches for each of the major scales.
Jim Bernier
Modeling BNSF and Milwaukee Road in SW Wisconsin
Texas Zepher wrote: HarryHotspur wrote:It seems to me that a 180 degree curve, for example, would be a U-turn, but that tells you nothing about the sharpness of the curve.What am I missing?What you are missing is how curves are measured in the real world. There is no nice flat table top on which someone can map out a radius from a given center point.
Well there is in Kansas. Just kidding. Thanks for the info. If I'm following this correctly, a 90 degree curve would be impossible - it would have zero radius. Is that correct? IF so, it seems that anything more than 90 degrees would have to be measured as two separate curves.
I tried to sell my two cents worth, but no one would give me a plug nickel for it.
I don't have a leg to stand on.
Back to post 1. A 180 degree curve would, in fact, be a U-turn, with a radius of 50 feet and a diameter and chord of 100 feet. There were streetcar lines that had curves that sharp.
A 60 degree curve would have a 100 foot radius (the 100 foot chord and two 100 foot radii would form an equilateral triangle.)
A 90 degree curve would have a radius of (100/square root of 2) feet = 70.7 feet.
If the Mantua/Uintah HOn3 2-6-6-2 is capable of taking the same 68 degree curve as its prototype, it should make it around a 12 inch radius. (Has anyone gotten one around a curve that tight?)
The 100 foot chord was, in fact, measured with a standard surveyor's chain. I recall a tale from the early days of rail development where a larcenous surveyor was using a well worn chain to survey land to be ceded to the railroad which employed him. Later surveys found that his measurements were consistently 3% high - in other words, his 'standard' chain was 103 feet long.
Chuck (modeling Central Japan in September, 1964 - with curves stated in metric radius)
Why are we having this discussion?
Unless you are building your layout outdoors or in a warehouse, you don't have room for prototypical curve radii.
Because sometimes things like this are nice to know.
AND, if you quit learning, you die!
dehusman wrote:R = RadiusD = degree of curvatureR= 50/sin (D/2)For a 10 degree curve, sin (10/2) = sin 5 = .087250/.0872 = 573 ft. Dave H.
I did not take trig but seem to be wishing I had in designing my model railroad. Here is what I have learned through research:
What I am missing is if I know the length of a tangent of an arc (the linear length of a section of curved track) and I know it’s amount of divergence from that tangent, what formula is there for determining the radius of that arc?
Anyone?
Thanks,
Crews
Phoebe Vet wrote:Why are we having this discussion?Unless you are building your layout outdoors or in a warehouse, you don't have room for prototypical curve radii.
I can't figure out why this discussion is not more common. I cannot imagine how y'all are designing model railroads to standards without this math.
So, to give one of many answers:
Most of my learning of trig has to do with determining how a manufactured turnout fits in a curve so that I can know how to reverse the angle of the divergent track for proper parallel track spacing and without turning inside my minimum radius.
cregil wrote: Slope of Angle = TAN(angle in degrees)Frog Number = (1/2)1/TAN(angle in degrees)=COTAN(angle in degrees)/2Divergence of curve from Tangent = radius-SQROOT(radius squared-length squared)Length of Arc = 2*pi*r/Portion of Circumference
I don't underatnd what you are measuring in the above (I know what a tangent is, I haven't ever heard of a "slope" of an angle).
You are also mixing angles. The angle of a frog has nothing to do with the degree of curvature.
What I am missing is if I know the length of a tangent of an arc (the linear length of a section of curved track)
The tangent of an arc is a line that touches the arc on the "outside". The chord is the straight line between the two end points of the arc, its on the inside of the curve. The length of the arc is the length of the curved line.
and I know it’s amount of divergence from that tangent, what formula is there for determining the radius of that arc?
Rather than fumbling around trying to figure out what you are trying to measure, here's how you can figure out the radius.
Mark the center line of track at two places on the arc.
Measure the distance between those places. That's the chord of the arc.
Bisect the chord and mark the center of the track to find the center of the arc.
Measure the offset from the center of the chord to the center of the arc.
Angle = A
Chord length = C
Offset = F
Radius = R
Tan A= F/(1/2 C) (to find the angle)
R= (1/2 C)/sin A
To make this a little more confusing, railroad engineers use degree in two ways refering to horizontal curves.
The first is what we have been talking about in which degree is just another way to specify the radius of the curve. The primary reason for it's use is because in the old days horizontal curves were layed out by using a tape to measure the cord and a transit to meassure a deflection angle. When an engineer refers to a "10 degree curve" thats' what they are talking about.
The other way degree is used is to describe how straight a segment of the route is. All the central angles of the horizontal curves are added up. Depending on how long the segment is you can get a route with thousands of "degrees of curvature". There was no direct connection to how sharp the curves are. Generally the less degrees of curvature the better as curved rails wear faster and wheels wear faster on curved track. Railroads would generally try to reduce the degrees of curvature as part of long term improvement plans for their main line.
selector wrote: Any deviation from the shortest distance between two points (line/tangent) means more costs in preparation, laying, and then maintaining a greater distance of roadbed and track, not to mention the increased operating costs for those longer distances.
Actually few railroads followed the "shortest distance between two points" because it was ironically, in most cases, MORE expensive than a longer route. The railroads followed the FLATEST distance between two points. It was cheaper to haul a car 50 miles further on a flat route than to haul it up a hill on a short route. That's why so many railroads wind across the countryside, they follow the terrain, minimizing grade, while roads and highways are often much straighter go up and down the hills rather than around them.
Mostly true but if you read railroad engineering planning studies they often refer to the total degrees of curvature on a route before and after a proposed improvement or line change. The railroads did care about reducing the total degrees of curvature but as you stated it had to be balanced against other factors (costs, both construction and operating).
tomikawaTT wrote:Back to post 1. A 180 degree curve would, in fact, be a U-turn, with a radius of 50 feet and a diameter and chord of 100 feet. There were streetcar lines that had curves that sharp.A 60 degree curve would have a 100 foot radius (the 100 foot chord and two 100 foot radii would form an equilateral triangle.)A 90 degree curve would have a radius of (100/square root of 2) feet = 70.7 feet.If the Mantua/Uintah HOn3 2-6-6-2 is capable of taking the same 68 degree curve as its prototype, it should make it around a 12 inch radius. (Has anyone gotten one around a curve that tight?)The 100 foot chord was, in fact, measured with a standard surveyor's chain. I recall a tale from the early days of rail development where a larcenous surveyor was using a well worn chain to survey land to be ceded to the railroad which employed him. Later surveys found that his measurements were consistently 3% high - in other words, his 'standard' chain was 103 feet long.Chuck (modeling Central Japan in September, 1964 - with curves stated in metric radius)
Thank you Tomikawa and others. Very interesting to me.
1. Because from time to time I have read about sharp curves on prototype railroads stated in degrees and wondered how, in fact, they would compare to curves on a model layout.
2. Because I recently read an article which said the Uintah Railway ran Malleys on 66 degree curves and wondered what the HO equivalent would be. (Apparently it's about 14-15 inches, small enough for most layouts.) I guess the standard follow up is, "Yeah, but what would they look like?" and in this case, the answer is, "Just like the prototype."
3. Because knowledge is never wasted, and
4. Because numerous people were kind enough to answer my question.
Sorry:
No insult was intended. I truly wondered what the purpose of the discussion is and why it is not in the prototype section. I did not mean to infer that those engaging in it were doing anything wrong.
Phoebe Vet wrote: I truly wondered what the purpose of the discussion is and why it is not in the prototype section.
Math is the same whether its 1:1 or 1:87 or 1:160. 8-)
dehusman wrote: cregil wrote: Slope of Angle = TAN(angle in degrees)Frog Number = (1/2)1/TAN(angle in degrees)=COTAN(angle in degrees)/2Divergence of curve from Tangent = radius-SQROOT(radius squared-length squared)Length of Arc = 2*pi*r/Portion of CircumferenceI don't underatnd what you are measuring in the above (I know what a tangent is, I haven't ever heard of a "slope" of an angle).You are also mixing angles. The angle of a frog has nothing to do with the degree of curvature....Dave H.
If you are using graph paper to construct a turnout template, it is much easier to plot a slope than it is to use a protractor and try to duplicate a quarter of a degree. The tangent function of an angle is the slope: rise over run.
The angle of a frog gives you angle of divergence, and if you are then wanting to reverse the angle with a reciprocal curve to bring the track parallel to the straight leg of the turnout (such as the last leg of a yard ladder) a curve of the same degree will do that.
Of course, I can freehand that reciprocal curve, but I want to insure that the radius is at or above my minimum, and balance that with my intended track spacing.
Likewise, if I have a dogbone-type reverse loop planned, I want to be able to calculate the smallest amount of real estate necessary, and that bringing that loop back into the divergent leg of the turnout where the loop begins and ends, requires an “S” turn which is dictated by my minimums. Mathematically treating the divergent leg of a turnout as a curve allows me to determine where that turnout is placed, and where the “s” breaks to reverse on itself.
The Atlas Custom turnouts for N=scale are interesting to do this math upon:
Thanks for the formula—I am looking forward to my break time at work for “fun with railroad math!”
Phoebe Vet wrote:Sorry:No insult was intended. I truly wondered what the purpose of the discussion is and why it is not in the prototype section. I did not mean to infer that those engaging in it were doing anything wrong.
No problem.
Anyone notice there's an error in the "Railway Track and Maintenance" book? It says to start at the Point of Tangency (Point B), then line back 100' to Point A, then put the chain at A and move the other end in to point D. Actually the chain will need to be at B, and the other end swung in to point D. (This all makes sense if you look at the diagram!)
Remember: In South Carolina, North is southeast of Due West... HIOAg /Bill
dehusman wrote:Mark the center line of track at two places on the arc.Measure the distance between those places. That's the chord of the arc.Bisect the chord and mark the center of the track to find the center of the arc. Measure the offset from the center of the chord to the center of the arc.Angle = AChord length = COffset = FRadius = RTan A= F/(1/2 C) (to find the angle)R= (1/2 C)/sin ADave H.
Okay, here is what I have tried.
I have a 5" section of 19" radius track, 24 of which makes a loop (2pi19=120” circumference, and 120”/5”= 24), so I know the angle is 15 degrees. (360 degrees/24 sections = 15 degrees). Is that right?
According to my divergence formula (previous post) the divergence should be .6696972202. I cannot measure to that mathematical precision, but that seems to fit.
I lay that out on graph paper, mark end points, mark center of curve, draw chord, measure distance between chord and midpoint of curve—It is about .5/32” = .156”
Tan A = .156/(1/2 * 5”) = .0624
That is not working for me.
Moving on…
Take the 15 degree angle
R = ½ * 5”/sin(15 degrees))
R= 2.5/.258819045
R=9.659 (nope)
BUT!
If I use C/sin(angle)
I get 5/.25881904= 19.31” A 19’ Radius. Yes.
So, what have I done wrong?
cregil wrote:So, what have I done wrong?
Oops. I made an error in the formula for the Radius (last formula). Correct one is:
R= (1/2 C)/sin 2A
You have to double the angle.
In your case:
For a 15 degree section of 19" radius track, the chord is 4.96" (half chord is 2.48") and the offset is .163", so you were close.
Tan A = .163/2.48 = .0657
If you do this using the "scientific" view of the MS Windows calculator, check the "Inv" box and click the "tan" button.
That gives you an angle A of 3.76 degrees. Double A = 7.5 degrees.
Radius = 2.48/ sin 7.5 degrees
Radius = 19.00002 inches.