How is a curve measured in degrees?

[dstaley]

Guys, this is not a triangle.  It's a curve.  The 10 degree angle isn't achieved until the end of the segment of track, so the calculations based on a triangle are not accurate.

 The math is actually not significantly much more complicated, we just need to use the right equations.  The equation for a circle is (x-h)^2 + (y-k)^2 = r^2, where:

 h and k are the x- and y- coordinates of the circle center and r is the radius.

 You can algebraically solve for the arc length using the known coordinates, and can then calculate the angle, or work backwards using just the known arc length because you know the angle and the arc length- just solve the perimeter equation for radius.

Given 10 degree (= 1/36 of a circle) and the length of the arc is 100', then the perimeter of the circle is 3600 feet.  The diameter of a circle is (pi) * diameter.  Divide 3600 feet by pi and you get about 1145 feet, so the radius is half of that or 572'.

 In HO scale that comes out to 6.586 feet (79.03 inches).

[jsotto]

There is also a convenient shortcut trackmen use in the field that can be useful for model railroad measurement.  Stretch a 62-foot string as a chord between two points on the curve.  Measure the offset at the midpoint of the string back over to the rail.  Each inch of offset equates to 1 degree of curvature.  So if the offset is 6 inches that is a 6 degree curve.

On a model railroad, a solid straight edge 62 scale feet long would be more convenient.  Then measure the offset of the midpoint in scale inches and you have the translation into degrees.

It's not that you would build a model railroad that way, just that it can be interesting to some to know how it compares to your prototype.

My EMD SD40-2 Operators Manual specifies that a single unit with single shoe or clasp brakes could negotiate a 193' radius or 30 degree curve. Two units coupled could negotiate a 22 degree curve but a unit coupled to a standard 50' boxcar could only handle 16 degrees (or like a model railroad, the wider swing of the pilot would pull the boxcar off the track sideways).

My EMD F7 Enginemen's Operating Manual simpy says that the minimum radius is 274 feet.

My recollection reading somewhere is that the GE C-truck was rated at 21 or 22 degrees, and a DRGW K-36 was rated at 24 degrees.

Jeff Otto

Missabe Northern Ry. 

[rickster.ca]

As an old exsurveyor, a chain was a unit of measure used in surveying, is still found on old plans, and is 66' and yes old chains for  did stretch. As for his curves, just use inches of radious, or are we just complicating things,

[markpierce]

Guys, this is not a triangle.  It's a curve.  The 10 degree angle isn't achieved until the end of the segment of track, so the calculations based on a triangle are not accurate.

 The math is actually not significantly much more complicated, we just need to use the right equations.  The equation for a circle is (x-h)^2 + (y-k)^2 = r^2, where:

 h and k are the x- and y- coordinates of the circle center and r is the radius.

 You can algebraically solve for the arc length using the known coordinates, and can then calculate the angle, or work backwards using just the known arc length because you know the angle and the arc length- just solve the perimeter equation for radius.

Given 10 degree (= 1/36 of a circle) and the length of the arc is 100', then the perimeter of the circle is 3600 feet.  The diameter of a circle is (pi) * diameter.  Divide 3600 feet by pi and you get about 1145 feet, so the radius is half of that or 572'.

 In HO scale that comes out to 6.586 feet (79.03 inches).

For measuring simple curves, one can measure the degree of the curve using the length of the chord (triangle method) or by the arc.  Using the chord method, the curve is the angle of the center for a given 100-foot chord (a 100-foot straight line crossing the curve at two points).  Using the arc method, the curve is the angle at the center for an arc of 100 feet.    Practical methods of measuring the length of an arc is not practical before construction, so the chord method was what was used by railroads.  (Source: Clement C. Williams's [Professor of Civil Engineering, University of Illinois] book The Design of Railway Location, copyright 1917, 1924.) Therefore, a  an X-degree curve using the chord method has a different radius than an X-degree curve using the arc method.

Mark

[dehusman]

Guys, this is not a triangle.  It's a curve.  The 10 degree angle isn't achieved until the end of the segment of track, so the calculations based on a triangle are not accurate.

I beg to differ.  Every prototype railroad for the last hundered years has used the methods I described to measure curves.  the method that Jeff Otto outlined is essentially the same thing i described, its just that by using a 62 ft chord, the offset is proportional to the degree of curvature in inches.  Same math.

 The math is actually not significantly much more complicated, we just need to use the right equations.  The equation for a circle is (x-h)^2 + (y-k)^2 = r^2, where:

 h and k are the x- and y- coordinates of the circle center and r is the radius.

And that is the benefit of the methods that "use triangles".  All you need are two measurements which are easily obtained with simple measuring tools, the chord length and the offset.

That's why real railroaders have used the chord method for decades.  You can find tables of offsets in most engineering manuals.

Dave H.

[cregil]

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.

I wish I could figure out what you mean, here.  I can not get .0657 to turn into 3.76 degrees to save my live-- not using inverse function, not using tangent, not by holding my eyes just right. 

But (!)

I do get the formula

Radius = 1/2 chord length / SIN (1/2 angle in degrees)

4.19" section (of known radius of 24") is a 10 degree section (one of 36 to make a circle) and to test:

Radius = 2.095/SIN(5 degrees) = 24 

That works!  Cool!  Thank you

Crews 

[Alan Robinson]

When building a railroad through mountainous country, or any rough country for that matter, it would clearly be impossible for the surveyor to swing a trammel bar in order to produce a curve of a known sharpness. This is why the method of measuring curves using the 100 foot cord was used. Laying out a curve consisted of moving down the rough roadbed in 100 foot intervals and swinging the transit at each interval through a certain angle. Then the track layers would essentially "follow the stakes" when placing the ties and rail. Since rail isn't exactly real flexible, simply placing the rail along the stake line created very smooth curves of constant radius, which was the object of the exercise.

Steam locomotives were not so forgiving of rough track and sharp curves as a four axle diesel with two swiveling trucks. Nevertheless, all Baldwin standard gauge locomotives had to be able to traverse a curve of sixteen degrees in order to be able to manage turnouts and other special trackwork. This would often be done at quite slow speeds to avoid derailments. Some locomotives, of course, could handle much sharper curves. For example, the Denver and Salt Lake Railway, which I model, had curves of twenty degrees on its "temporary" line over the main range west of Denver. This line crossed the continental divide at an elevation of 11,660 feet and was used from 1904 until the Moffat tunnel was built many years later. The line was sharp enough that consolidations, ten wheelers and unique 2-6-6-0 mallets were used here. Some low wheeled mikado locomotives were too stiff for this line even though they had only 55" drivers and quite short wheelbases.

The grades on this line were four percent so the speeds were of necessity very slow in order to be able to operate safely. A speed of twenty miles an hour was pretty fast for this line.

The 100 foot cord method was used for curves up to about twenty degrees or so and perhaps even sharper in some situations. When the radius began to approach three or four hundred feet, the radius was expressed in feet just as we use when we measure radii in inches on our models. This kind of track would be common on streetcar and interurban lines and on some specialized railroads. Certainly the tracks used by the B&O "Little Joe" 0-4-0T switchers was in the order of 100 feet radius or so. On curves this sharp cars might have to be pulled around a curve with a chain as the couplers wouldn't swing far enough, or else poled around the curve if the car was being pushed.

This topic of degree of curvature is not only of academic interest. I'm thinking of modeling in P87, and in this type of modeling using scaled prototype curvature allows the model to be built to more or less exact proportions including all the running gear. If the real locomotive could squeeze around a 24 degree curve, a correctly built model can certainly be made to do so. The fact that we keep bending 2-10-2 freight hogs around 18" radius curves just goes to show how far from prototype construction methods we must deviate in order to achieve such operation. Is it any wonder that some of our locomotives look a little "funny" as they negotiate such curves? We do well to have in the back of our mind just how much of a compromise our tight model curves forces on our models.

[BroadwayLion]

The LION makes simple curves. 24" radius is my minimum curve at the edge of the table, so any curve I lay on the table will be more than that. (snicker snicker!) The LION then lays out the tangent tracks along the table edge, and curves it around the curve. He flexes it a few times until it is resting naturally, and then spikes it down.

The LION has nice smooth trackwork, but it is a good thing that this LION models subway trains with 50' cars. Some of the curves are sharper than his design thought for.

BTW: the real IRT subway cars can make some pretty darn sharp turns underground. The South Ferry Station *is* a loop, just like in HO, and is actually, just as tight.

[Texas Zepher]

Tan A = .163/2.48 = .0657 ... That gives you an angle A of 3.76 degrees.

I wish I could figure out what you mean, here.  I can not get .0657 to turn into 3.76 degrees to save my live-

Good thing your life doesn't depend on it.  It doesn't turn into it - it IS it.
  0.0657 radians = 3.76 degrees.

I can't think of how many times I've made that same umm ummm "oversight" .

 

[cregil]

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.

I wish I could figure out what you mean, here.  I can not get .0657 to turn into 3.76 degrees to save my live-- not using inverse function, not using tangent, not by holding my eyes just right. 

 

I see!  The INV with the TAN function on the MS Windows calculator is not a 1/x inverse, it is a programmed function to reverse the calculation.

My TI hand held calculator has the same function (TAN^-1).  From that, I get the 3.758930464.

I do wish that I had a formula as that will allow me to understand the mathematical relationships whereas a button on a keypad looks simply like magic.

So, how was this done before electronic calculators?  Slide rule?  And before a slide rule?

Crews

 

[pkeppers]

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.

I wish I could figure out what you mean, here.  I can not get .0657 to turn into 3.76 degrees to save my live-- not using inverse function, not using tangent, not by holding my eyes just right. 

 

I see!  The INV with the TAN function on the MS Windows calculator is not a 1/x inverse, it is a programmed function to reverse the calculation.

My TI hand held calculator has the same function (TAN^-1).  From that, I get the 3.758930464.

I do wish that I had a formula as that will allow me to understand the mathematical relationships whereas a button on a keypad looks simply like magic.

So, how was this done before electronic calculators?  Slide rule?  And before a slide rule?

Crews

 

 

 

Inv tan on a calculator is what is called the arctan function.

When you hit tan the number the calculator gives is the ratio of the appropriate sides of a triangle with the angle you plugged in.

When you hit inv tan you are plugging in the ratio and the calculator is telling you what angle produces that ratio.

Prior to scientific calculators trig tables were used. Here is an online one

http://www.industrialpress.com/en/tabid/63/default.aspx

 

[Texas Zepher]

So, how was this done before electronic calculators?  Slide rule?  And before a slide rule?

You mean you don't have a book of trig and logarithm tables sitting on your most used books shelf?     Seriously, at one time someone had to sit down and manually do the long division so-to-speak for each entry put into the tables.

[fwright]

So, how was this done before electronic calculators?  Slide rule?  And before a slide rule?

Crews

Being older than dirt, in a British middle school equivalent, I was taught that by memorizing just 4 entries in the log tables (log 2, 3, 5, 7), you could do most any operation.  So we memorized them to 4 figures.  And memorizing a few critical square roots (2, 3) and pi gave you the trig tables for 30, 45, and 60 degree angles.  Finally, knowing that tangent and sine functions for less than 10 degrees are pretty close to each other, and directly proportional to the angle took care of the real small angles.

In my junior year of high school I learned slide rules, both circular and straight (I prefer circular).

Then I went to college to gain more knowledge.  I bought my own copy of the CRC tables, which has all the formulae, tables, and stuff I would ever need plus some.  Incoming freshman were required to take a short course in how to use a slide rule.  In my junior year, HP came out with their scientific calculators (HP-35 and HP-45) discounted to the amazing prices of $220 and $300.  Overnight, slide rules became obsolete, and the use of round numbers for homework and tests ended.  Some profs really objected to calculators and computers because folks were putting answers down with 8 digit precision.  One of my graduate profs started marking wrong any problem where your answer had more precision than the entering data did.

Now, when I tutor senior high school and junior college math students, I usually take my CRC tables with me.  I'm still often faster than they are with their graphing calculators.  And I do prefer my HP-45 over all other calculators I've ever used, and still use it regularly.  The Reverse Polish Notation, 4 register stack, 9 memory registers, and easy polar to rectangular and back conversions make it a joy to use, and are about all my aging memory can handle.

remembering the old ways

Fred W 

[cregil]

Thanks for the trig and algebraic formulas on this thread. The 62’ chain trick is really neat!
These were the missing pieces in my ability to seek precision and foresee compromises in my layout planning.
I am pleased to learn that other model railroaders find this both interesting and useful. It took me since January, working by myself, to get to learn and understand enough of trigonometry to reach this point, and while I loved the process- not all will.
This leads me to want to provide a compilation of these and the other calculations I have discovered to be useful to the ldsig site.
Thanks, also, to Harry for starting this thread just as I was wrestling with this last bit.
Crews

[R. T. POTEET]

Hotspur, my boy, you are mixing apples and oranges here.

Let me give you an example; if you have a circle and it traverses 100 feet in every degree then your circle is going to be divided into 360 segments and the circumference will be 36,000 feet (360 ÷ 1 X 100). Divide by pi (3.141592654) and you get a diameter of 11,459.16 feet and a radius of 5,729.58 feet which equates to an HO-Scale radius of 789.51 inches.

Lets change the figures a little bit and create a circle that traverses 100 feet in every 15 degrees; your circle is divided into 24 segments and the circumference will be 2,400 feet (360 ÷ 15 X 100). Dividing by pi gives us a diameter of 763.94 feet and a radius of 381.97 feet which equates to an HO-Scale radius of 52.63 inches.

Here's where apples and oranges get mixed up: both circles are 360 degrees around - that's what a circle is - but I think that you will agree that that 15 degree circle is going to produce a far tighter/sharper curve than the 1 degree one. If you were in the parking lot at your local Wal-M at 2 in the morning you could spin a U-turn at a heck of a lot faster speed than you can cut a U-ey in the street in front of your house.

You can do the same math on the dynamics of a 4-8-8-4 Big Boy which was engineered for 20 degree curves: 360 ÷ 20 X 100 ÷ pi ÷ 2 X .137795 = 39.48 inches which is a very broad curve in HO-Scale yet we engineer our model locomotives to negotiate 32.90 degree - that's 24 inches -  curves, nearly 64.5% tighter than the prototype's capability.  

[NeO6874]

All this math has my head spinning... and I'm majoring in computers (guess that is a failing of using calculators for everything....)

 

Let me see if I understand the math right R.T.  Using a NYC K-11 Pacific as the locomotive of choice, it was engineered for an 18° 30' curve - which is 18.5° if I remember the conversion from minutes to degrees (I think it is 60' per 1°). 

That means the K-11 would be able to negotiate a 286.5 foot (radius, rounded up to nearest tenth) curve (1:1 scale) or that same 39.48" curve (rounded up to nearest hundredth) in HO scale, correct? 

[pkeppers]

Let me give you an example; if you have a circle and it traverses 100 feet in every degree then your circle is going to be divided into 360 segments and the circumference will be 36,000 feet (360 ÷ 1 X 100). Divide by pi (3.141592654) and you get a diameter of 11,459.16 feet and a radius of 5,729.58 feet which equates to an HO-Scale radius of 789.51 inches.

 

 

Not that this really matters for the purpose of converting prototype degree of curve to model but railroads all (to the best of my knowlege) used cord definition so the radius of a one degree railroad curve would be 5729.65'

 

Just want to make sure someone who is following the math doesnt get confused.

[Texas Zepher]

All this math has my head spinning... and I'm majoring in computers (guess that is a failing of using calculators for everything....)

??? Majoring as in college?   A BS in computer science usually requires a minor in mathematics.  Calc I, II, III, and either differential equations or linear algebra. If your advistor hasn't mentioned this you might want to speak to them about it.   What we are dealing with on this thread is simple geometry & trig.  All very essential to computer science especially if one gets into graphical programming for image creation, recognintion, and processing.  

Oops going off topic here.   This is just another way that model railroading is a great source for educating young people.

[NeO6874]

All this math has my head spinning... and I'm majoring in computers (guess that is a failing of using calculators for everything....)

??? Majoring as in college?   A BS in computer science usually requires a minor in mathematics.  Calc I, II, III, and either differential equations or linear algebra. If your advistor hasn't mentioned this you might want to speak to them about it.   What we are dealing with on this thread is simple geometry & trig.  All very essential to computer science especially if one gets into graphical programming for image creation, recognintion, and processing.  

Oops going off topic here.   This is just another way that model railroading is a great source for educating young people.

Maybe I just need to stop frequenting the forums after a particularly exhausting bout of programming... makes so much more sense now.  OTOH, my major is more focused on business application rather than the nitty gritty of straight programming (ie web design, database integration, general programming, etc without a real focus on the theory or the advanced maths).

 

I still think calculators are the bane of everything, 2 years ago (when I was still taking math courses) I don't think I would have been running headlong into a brick wall thinking about this... 

[dehusman]

Maybe I just need to stop frequenting the forums after a particularly exhausting bout of programming... makes so much more sense now.  OTOH, my major is more focused on business application rather than the nitty gritty of straight programming (ie web design, database integration, general programming, etc without a real focus on the theory or the advanced maths).

In the railroad business this IS a "business application".

Dave H.

[R. T. POTEET]

Let me give you an example; if you have a circle and it traverses 100 feet in every degree then your circle is going to be divided into 360 segments and the circumference will be 36,000 feet (360 ÷ 1 X 100). Divide by pi (3.141592654) and you get a diameter of 11,459.16 feet and a radius of 5,729.58 feet which equates to an HO-Scale radius of 789.51 inches.

 

Not that this really matters for the purpose of converting prototype degree of curve to model but railroads all (to the best of my knowlege) used cord definition so the radius of a one degree railroad curve would be 5729.65'

 

Just want to make sure someone who is following the math doesnt get confused.

I didn't really do the math to figure precisely what the arc would be for a 100 foot chord on one degree curve but it may well measure out to 5729.65 feet. 

[R. T. POTEET]

All this math has my head spinning... and I'm majoring in computers (guess that is a failing of using calculators for everything....)

 

Let me see if I understand the math right R.T.  Using a NYC K-11 Pacific as the locomotive of choice, it was engineered for an 18° 30' curve - which is 18.5° if I remember the conversion from minutes to degrees (I think it is 60' per 1°). 

That means the K-11 would be able to negotiate a 286.5 foot (radius, rounded up to nearest tenth) curve (1:1 scale) or that same 39.48" curve (rounded up to nearest hundredth) in HO scale, correct? 

360 ÷ 18.5 X 100 ÷ 3.141592654 - 50% X .137795 = 42.68 inches.

360 ÷ 18.5 X 100 = 1,945.95 feet; divide that by 3.141592654 and subtract 50% renders a prototype radius of 309.71 inches which equates to a 42.68 inch HO-Scale radius curve.

A round number for pi has always been 3.1416; HO-Scale is 3.5mm to a foot and a millimeter is .03937 inches which renders an HO-Scale decimal measurement of  .137795 inches but your measurement is not going to come out more than a couple thousandths off from my calculations.  

[NeO6874]

Ah hah!

I see where my mistake came in, I was looking at the K-11 (18.5° minimum curve) but plugging in the numbers for the big boy in your example... and not double checking what I was doing.  

OK, I think I have it now... just a bit slow on the pickup

[cregil]

Let me give you an example; if you have a circle and it traverses 100 feet in every degree then your circle is going to be divided into 360 segments and the circumference will be 36,000 feet (360 ÷ 1 X 100). Divide by pi (3.141592654) and you get a diameter of 11,459.16 feet and a radius of 5,729.58 feet which equates to an HO-Scale radius of 789.51 inches.

 

Not that this really matters for the purpose of converting prototype degree of curve to model but railroads all (to the best of my knowlege) used cord definition so the radius of a one degree railroad curve would be 5729.65'

 

Just want to make sure someone who is following the math doesnt get confused.

I didn't really do the math to figure precisely what the arc would be for a 100 foot chord on one degree curve but it may well measure out to 5729.65 feet. 

 

I am still having a ball with this stuff!

Radius of 1 degree curve (100’ chord).

Angle from tangent point to second point of chord = .5 degree

• R = ½ (chord)/SIN(Angle of chord from tangent)
• R = ½ (100’)/SIN(.5)
• R – 50’/.008726535

R = 5729.65’

By the way, you only need 429.72 inches (about 36 feet) in N-Scale!

Crews

 

[wireflight]

From the far, far reaches of the wild, wild west I am: rtpoteet

Question: What's the difference between Political Correctness and Mindless? Answer: Thirteen Letters!

Only dead fish go with the flow! - - - - - Sarah Palin

============================

Political Correctness and Mindless:

Politica ~ Mindless [8 letters each]

That leaves the final l from Political, plus all the letters in Correctness.

Counting in groups:

lCo + rre + ctn + ess = 12 letters, by my count: the space between the words isn't a letter (in English).

Of course, if you intended that as a joke indicating what's wrong with our nation (from the proletariat and the bourgeoisie to the bureaucracy), then you did pretty well.

[Trace Fork]

All.

The math to calculate curves is not particularly difficult. Spiral curves (what model railroaders incorrectly call easements) are somewhat more complicated. Just keep in mind that if any two elements of a curve are known, all other elements can be calculated.

The radius of a 1 degree chord definition curve = 5729.65'

The radius of a 1 degree arc definition curve = 5729.58' 

To obtain the radius of a 5 degree curve, simply divide the above radii by 5, a 15 degree curve by 15 etc. Then all other elements can be found. (including stationing).

During my career, I have put many design curves (albeit arc definition curves) onto the ground by "walking the curve" which required the survey instrument to be moved along the center of the arc at 100' intervals (or intervals divisible by 100), and calculations performed to set the next centerline point(s). I have also staked curves with the survey instrument located at the P.C., P.T., and P.I. of design curves, and calculations performed to determine points on the center of the arc at varius intervals. As I said, these calculations are not difficult, but a basic understanding of the math involved must be learned. Your scientific calculator, or PC is not going to teach this.

These methods can also be used with a protractor (not a high end farming impliment) and rule directly on the layout surface, without need to locate the radius point of the curve. This is in fact how I lay out curves on my layout.

As a note of clarification, the standard surveyor's chain was 66' long and had of 100 links 7.42" long. Many railroad engineering forces had chains 100' in length, with 100 links 12" long. These chains were VERY heavy, and cumbersome to use. They were suseptable to expansion and contraction due to temperature variation, thus correction calculations were frequently required. They were intended to measure horizontal distances, thus had to be used "level" to acheive their highest degree of accuracy. This could prove very difficult in steep terrain, and "breaking chain" was often required. These two factors may go a long way in explaining why a certain chain was 3% long as noted in one post above.

Jim J. 

[TBat55]

http://www.1728.com/circsect.htm

circle calculator