BTW BarstowRick. My math is so simple a caveman can do it ! 32 inch wide layout. Leaving 'derail buffer zones' on the edges....using Unitrack...15" radius fits perfect. 13 3/4" for inside parallel track with correct spacing and... BAM...layout done...

'Elementary my dear Watson' Already explained, but the link contains all the steps. https://mysite.du.edu/~jcalvert/railway/degcurv.htm So the degree of curvature, the angle, of a scaled curve is the same in any scale. Not so useful info, but true.

Not sure we are all on the same page. Degree of curvature is the sharpness of the curve, not how far it curves. Higher the degree, sharper the curve. A five degree curve could end up turning 180 if it goes far enough. This is grossly sharp "for illustration purposes only", pretty sure it is correct. Out in the real world the center of the arc could be inside a mountain or off over a swamp - not terribly practical for laying out a railroad, right? So the shortcut way is to bend that rail until the number of inches from the center of the 100' chord to the rail is equal to the degrees of curvature specified by the surveyors: The specs would have said how far to continue the curve.

I would think the surveyors would have staked out the centerline of the track and the roadbed crew would just follow paying attention to both the alignment as well as the elevation. Once the roadbed was completed then the track laying crew would follow.

That's interesting. Never heard that one before. On the one hand, it doesn't work. Consider a 180° curve. That's a U-turn in a hundred feet. According to that, the rail would only deviate 180"--fifteen feet--from the chord, when a circle with a one hundred foot diameter would have a radius of fifty feet, not fifteen. On the other hand, no railroad has anything anywhere near 180° curvature. Within a narrow range, that system might land you "in the ballpark".

The fallacy in your position is that you are equating the fifteen feet to the radius. It is not the radius but rather the distance from the mid point of the chord to the mid point of the arc. Also where did you come up with the hundred feet measurement? A 180 degree turn can have any radius depending on its degree of curvature. A 1 degree curve has a radius of 5729.651 feet while a 9 degree curve has a radius of 637.275 feet. This holds true as long as the curvature remains uniform throughout the curve. .

A one eighty is a semicircle. A chord across a semicircle is the diameter. A line from the mid point of that chord to the mid point of that arc is the radius. Read the thread. Railroad curves are measured in change of direction in degrees across a hundred foot chord. Just saying the curve turns a train ninety degrees doesn't tell us anything about how sharp it is. Does it do this in forty acres? Does it do this on a postage stamp? I'm not talking about 180° turns, I'm talking about 180° curves. Railroads never build them, because they have no equipment that can negotiate them. But they do exist. We've both driven cars through them. And they have a radius of fifty feet. The fallacy is believing a linear method like that one will always solve what is clearly a parabolic equation. Yes, it can work well enough to get the job done in a narrow range of choices, though.

Now you know why I went into Electrical Engineering and not Civil Engineering! But one of the reasons the kept it at a fixed chord length was to minimize the curvature of the surface. They were laying track on a three dimensional surface. While this is a very interesting subject And there is some applicability to N Scale layouts, but in the end it is the space constraints that usually determine every thing when building a model. And George you are correct, a long train rounding a 48" radius is just amazing to watch! But so is a #12 crossover at full speed. So there are ways to sneak in some coolness even on smaller layouts where you are not so constrained by dimensions. Watching a passenger train slip through a #12 crossover without hardly a quiver is a really cool sight.

Yep, and that is why I am(was) a chemist. Differential equations and boundary value problems were a piece of cake (for me), but this practical 1:1 Civil Eng makes me need full info to get.

Okay, I misunderstood you. I thought you were referring to the fifteen feet as the radius. However, Your example is an impossibility. The lowest radius that the formula applies to is 70.711 feet. That is for a 90 degree curvature. So your semicircle of a diameter of 100 feet would have a radius of only 50 feet. With anything less than a full 180 degree semicircle the 100 foot chord would never touch two points on the arc. No matter how many degrees the arc curves the longest 100 foot chord covers only 90 degrees and you need a radius of over 70 feet to have the ends of the chord touching two points on the arc.

Well, I finally can't resist jumping into the math. The way acptulsa figured the radius from the "degree of curvature" is not exactly the way the surveyors laid out track. The surveyors had a chain that was 100' long, and they stretched it out in a direction that was the specified number of degrees to the left or right of the direction the track was headed at the starting point. They did not lay the chain along the curved path of the track. So, the 100' measurement is not the fraction of the circle's circumference, but rather what is called a "chord", which is a straight line across a circle from one point on its circumference to another. Without getting into the geometry of the derivation, the formula for the radius that is associated with a 100' chord that changes direction by an angle expressed in degrees is: 100 feet / [ 2 x sin(angle/2)] when the sine function is based on degrees (not radians or grads). Of course, those feet could be scale feet or prototype feet, because it is all the same geometry. In N scale, 100 scale feet = 7.5 real inches. The difference between this and what acptulsa posted is tiny for large radius curves, but, by the time we get to tight model curves, it is noticeable. For example, a 44 degree curve in N scale has a 9.77" radius by acptulsa's method, and a 10" radius by the chord method. Not really any big deal. To put some of this into perspective for how a model looks going around a curve, that tiny real 0-4-0 "Docksider" was rated for 50 degrees, and up to 82 degrees "slow", which equate to radii of 9.6" and 5.72" in N scale. So, it does look right going through sharp 90 degree turns on our models of city streets in warehouse districts. An 0-6-0 switcher might be rated for 30 degrees, which is 14.5" radius in N scale. But, something like an 0-8-0 or 2-8-0 is rated for 12 degrees at-speed and up to 17 degrees (slow), which is 36" at-speed and 25" slow in N scale. Going to large locomotives, even big articulated engines, maximum curvature ratings did not change that much from that 2-8-0. So, to look prototypical, mainlines should be at least 36" radius and yards at least 24" radius in N scale unless you use 0-6-0 switchers, which can look right down to 14.5" radius. [I hope mtmtrainman brought a big bag of popcorn to this post.]

Why yes...yes I did . https://media.tenor.co/images/b83f96fdba42cfe9c3bb505417756b40/raw Just remember........... Pi r round...cake r squared

OOOOPS!!! In my long explanation, I did not quite say what I meant about how the surveyors actually placed the 100' chain. If the curve was specified to be a certain number of degrees in 100', then the chain is laid at half that angle from the starting point. That results in the track being at the specified curvature angle at the end of the chain. If you draw it out on a piece of paper, you will see that the angle of the track relative to its starting point at the end of the chain will be double the angle of the chain from the track at the starting point. Basically, that is the reason that the formula needs to take the sine function for one-half the angle of curvature but then double the result to compare it to 100 feet. I would have changed the previous post, but apparently my edit period expired while I slept on it.

Similar to my gaff, what NtheBasement wrote also seems to have a 'not exactly what I meant to say" issue: Although this seems to say that the distance between the chord line and the track centerline will equal (in its number of inches) the same number as the number of degrees of curvature, that is not true. What is true is that the distance from the chord line to the track centerline can be specified in a table for set distances along the chord line, with different entries for each degree of curvature, so that a crew can do offset measurements from the chord line to find the track centerline when actually laying a smooth curve of track. For example, the radius for 10 degrees of curvature in 100 feet comes out to 573.7 feet. The distance from the center of the chord line to the track centerline will be 573.7 feet times [1-cos(5 degrees)] = 2.18 feet. That is 26.2 inches, not 10 inches. I'm not trying to be picky, just trying to make sure nobody tries to lay out model track using the idea that they can just use the intended number of degrees of curvature to be the number of scale inches to deflect 100 scale feet of track.

Well this has been a fun discussion and I thank all who have joined in. The reality though is that I doubt if anyone is employing a survey crew to lay out their model RR curves. For us, using a trammel and calculating in inches works. We may get fancy and include an easement at each end of the curve but that's about it. Hopefully we have added to the knowledge of how the real railroads do it and when we come across something in the literature about a curve of 9 degrees being a sharp main line curve we have something to relate it to. Maybe next we can discuss how to determine the amount of super elevation on a curve. Do you think?

I am not sure about Super Elevation, I got that track set from Kato and had nothing but problems with it. Not event their locomotives and stack cars would run on it! Maybe they over super elevated it?

The secret to superelevating track is to make the transition from level track gradual and smooth. Too much superelevation is defined as when cars parked on the curve fall over. All other problems are all about a too-abrupt transition.

Super elevation on a real railroad serves a real purpose in the dynamics of the train's motion and how it is felt by people inside the train (and I guess animals in stock cars, too). It is offsetting some of the centrifugal force with gravity by tilting the floor towards the inside of the curve. That also helps keep the train from derailing to the outside of the curve when at high speed. But, on a model railroad, the only purpose for super elevation is to make the model look like the prototype. So, we can just model what our prototype railroad did for the amount of super elevation on curves of various radii with various speed limits, and that will look right on our layouts. The reason that there is no beneficial effect of super elevation on model performance is that we are not scaling down the force of gravity when we scale down everything else. Think about it this way: In the full-size world, gravity is trying to accelerate everything at 32.2 real feet per second each second, including our model trains, so they are seeing what amounts to a scale gravitational force that is (for N scale) 160 times "g" in scale feet per second squared. But, we are scaling down the mass that gravity is acting on, too, so the model does weigh less than the prototype. However, the centrifugal force is the mass times the velocity squared divided by the radius of the turn. So, the effect of reducing the mass from the prototype to the model really has no net effect on the relationship between the gravitational force and the centrifugal force. What does matter is that the velocity is reduced by the scale factor and that is squared, and the radius is reduced by the scale factor, but it is in the denominator, so, the net result is that the centrifugal force per unit of mass is reduced by the scale factor, while the gravitational force per unit mass is not. Therefore, the amount of tilt needed to get the same apparent "down" direction felt by a model passenger on a model going through a scale curve at a scale speed is reduced by the scale factor compared to the amount of tilt required to get the same downward direction on the prototype. So, to get the same force balance, the amount of tilt created by the super elevation would need to be 160 time smaller than the prototype for N scale. So, say there is 6" of super elevation on a prototype curve. Scaling down the geometry makes this look right at 6" / 160 = 0.0375", but to make it feel right to an N scale passenger, we need to apply the scale factor again and get only 0.00023" of super elevation! Not going to see that. In reality, super elevation that looks right on model curves is probably worse than useless with regard to keeping the train on the tracks, because it tends to make the train want to fall towards the inside of the curve. That just adds to the effect of string-lining to derail the cars. I have wondered if it would be useful to put in counter super elevation (tilting the track to the outside of the curve) to help resist string-lining on a model helix. Obviously, it can't be too much elevation, or it would make cars tend to fall off the track to the outside of the curve when trains are going slow, and there is probably a bigger effect that wheel flanges are trying to catch rail joints and climb over the outside rails,and those would be lower. So, the idea may not be practical. It might help, or it might hurt with respect to keeping the wheels on the rails. I don't think I can do a theoretical calculation to test the theory, so I may try some experiments - - unless somebody here tries it and tells me the results before I have the setup and free time to try it, myself.

Another benefit to super elevation is that it effectively reduces the track curvature by introducing a vertical element to what would be a purely horizontal curve. It also reduces flange wear.