# 15 degrees?



## Randy Stone (Jan 2, 2008)

While looking through my new Trains magazine today, I seen a picture of 3 UP Diesels pulling freight cars around a curve said to be 15 degrees.

The curve looks really tight, but how does 15 degrees compare to the curves commonly used in G Gauge?


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## Gary Armitstead (Jan 2, 2008)

Randy,

15 degrees would be a little over a 13 foot RADIUS in 1/29.


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## Greg Elmassian (Jan 3, 2008)

Roughly 15' radius, 30' diameter in 1:29... 

http://www.steamlocomotive.com/model/curve.shtml 

Regards, Greg


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## krs (Feb 29, 2008)

If my math is right, a 15 degree curve would equate to a 13 ft radius (26 ft diameter) curve in 1:29 scale.


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## krs (Feb 29, 2008)

Wow! 

We all agree - that has to be a first. 


As another reference point - on the Swiss RhB Meter gauge line, the tightest curve is 45 meters radius - that in 1:22.5 scale would equate to about a 6.5 ft radius or a 13 foot diameter curve. 

Steepest grade on the RhB is 7%


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## Ltotis (Jan 3, 2008)

If you want to see tight curves the tracks into and out of South Station in Boston would be one place. 
LAO


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## Randy Stone (Jan 2, 2008)

Humm, the 15 degree curve sure looks tighter than any 20 ft diameter curve I've ever seen in G.


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## Greg Elmassian (Jan 3, 2008)

13 foot times 29 for the scale gives 377 feet... from the link (assuming his math is right) gives you maybe 17 degrees... but still looking for a good formula.. 

Some railroads had a maximum curvature of 2 degrees... that's a curve almost 6,000 feet in diameter! (206 feet in diameter in 1:29).


Just gives you some perspective on how sharp our curves are. 


Greg


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## Gary Armitstead (Jan 2, 2008)

Posted By Gary Armitstead on 05 Apr 2012 07:07 PM 
Randy,

15 degrees would be a little over a 13 foot RADIUS in 1/29.









My answer came from using Stan Silverman's Handy Converter V16. Actual answer was 13 feet, 2.5 inches AND this is RADIUS. Prototype radius would be 383.06 feet. Close enough for "combat work". Circa 1967, Vietnam.


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## krs (Feb 29, 2008)

Posted By Greg Elmassian on 05 Apr 2012 07:27 PMbut still looking for a good formula.. 
I used this formula:
Radius in feet = 5729 divided by the degrees of curvature.


Comes from a Trains magazine article.

Knut


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## ThinkerT (Jan 2, 2008)

That would make R1 and R2 curves the sort of thing you'd find only on really steep and twisty narrow guage lines in 1:1.


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## Greg Elmassian (Jan 3, 2008)

Using that formula, an R1 curve, 4 foot diameter, 2 foot radius, now multiply by 29 = 58 feet radius , now that makes the degrees 5729/58 = 99 degrees.... 

Hard to believe that anyone would have that tight of a turn except mining tracks and trolley cars... 

Greg


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## rdamurphy (Jan 3, 2008)

Actually, 1 degree of curvature = a radius of 5729.686 feet. 


Or, ... Divide 50 by the sine of half of the degree of curvature.

So, divide that number by the degree of curvature you want, i.e. 10 degrees = radius 572.9686 feet. 



A narrow gauge 20 degree curve, with a radius of 283.4843 feet, would be a 13.957 foot radius curve in 1/20.3. Hmm, 28 foot diameter!



Robert


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## East Broad Top (Dec 29, 2007)

That would make R1 and R2 curves the sort of thing you'd find only on really steep and twisty narrow guage lines in 1:1. 

*Here's a link* to a great on-line resource for curves, both in terms of prototype and how they translate to the various scales in large scale. The tightest curves for a 3' gauge line were on the Uintah, at around 66 degrees. (They were originally tighter, but the locos had troubles on them). Interesting to note that the Uintah used special coupler mounts to account for the tight curves. There _is_ a prototype for everything. 

Later, 

K


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## Ironton (Jan 2, 2008)

The simplest way to figure radius is with trigonometry, not a problem with the built in calculator on your computer. If you look a the link Kevin gives you can see two right triangles. On both of them you know the angle at the center of the circle (is 1/2 of the stated angle) and the opposite side (which is 50 feet). No fancy calculations to get either of these figures. The radius is the hypotenuse. So let's see, we know angle and opposite and are looking for hypotenuse, oh yeah that is tangent. So tangent of 1/2 the angle is 50 divided by the radius. Or radius is 50 divided by tangent of 1/2 the angle. That is the formula I use. 

So from the calculator, 50 / tan(7.5) is 379.7877 feet actual.This is 13.096 feet in 1:29, or 18.71 feet in 1:20.3. 

The tightest 66 degree curve works out to be 76.9932 feet actual. Or in 1:29 is 2.65 feet. In 1:20.3 is 3.79 feet. 

Hope this helps.


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## East Broad Top (Dec 29, 2007)

Rich, if I'm understanding your formula correctly, it's not 100% accurate. It appears that you're using right-triangle geometry with an "opposite" side of 50'. The problem is the "opposite" side isn't 50'. The 50' comes from half of the 100' arc be measured along the arc of the curve, not the opposite end of a right triangle that subtends the full 100' arc as drawn in the drawing. That distance is going to be smaller than 50'; proportional to the degree of the curve. The tighter the radius, the larger the difference in this number from the measurement along the arc. Thus--as your numbers reflect--the variance at 15 degrees is fairly minimal (379' vs 383'), but at the tight end of 66 degrees, they're much farther apart (77' vs 92'). 

*Another link* to a chart showing curves in terms of degree from 1 to 90, and their corresponding radii in feet (and HO scale, which for 1:29, you'd just multiply by 3). 

Later, 

K


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## rdamurphy (Jan 3, 2008)

Simplest way, but not the correct way: 

R = 50/sin(A/2) to find the radius from the curvature where R = Radius and A = degrees of curvature.. 

A = 2 * arcsin(50/2) to find the degrees of curvature from the radius. 

Robert


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## Greg Elmassian (Jan 3, 2008)

Just to put perspective in again, in 1:29, a 10 foot diameter curve is a 40 degree curve, 4 to 5 times sharper than most mainlines ... no wonder 100 car trains are tough... 

Greg


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## rdamurphy (Jan 3, 2008)

You're correct, Greg, and it's not only curves. Turnouts also suffer from "selective compression." A #15 turnout is considered a 30mph restricted turnout. To get a high speed turnout, you'd have to go to a #20 closed frog turnout to get to a 45 mph limit. 

Want a 79mph speed limit? Just use a #24 wye (equilateral) turnout. With a #48 frog. 


But... Just for fun:


http://books.google.com/books/reade...put=reader


Enjoy!

Robert


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## bnsfconductor (Jan 3, 2008)

Posted By rdamurphy on 06 Apr 2012 09:56 AM 


You're correct, Greg, and it's not only curves. Turnouts also suffer from "selective compression." A #15 turnout is considered a 30mph restricted turnout. To get a high speed turnout, you'd have to go to a #20 closed frog turnout to get to a 45 mph limit. 

Want a 79mph speed limit? Just use a #24 wye (equilateral) turnout. With a #48 frog. 


But... Just for fun:


http://books.google.com/books/reade...put=reader


Enjoy!

Robert 

And people wonder why I'm building a #9 turnout to scale... Going through a high speed turnout at 50mph is kind of fun! Although it sure jerks you around quite a bit. I 'think' the only person I know of that is modeling prototypical curves is Dirk (user name SD90WLMT)
http://www.mylargescale.com/Communi.../121905/afv/topic/afpgj/3/Default.aspx#242809
Craig


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## Greg Elmassian (Jan 3, 2008)

Yeah, I also model in Z scale, but we try not to use anything less than a #6 on the mainline... we are making #10 for most mainline turnouts... 

Using the small scale to have larger curves and turnouts. At shows we usually run 50-60 cars minimum and 100 is not uncommon. The 50-60 car trains will run all day unattended. 

Greg


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## SD90WLMT (Feb 16, 2010)

Thanks Craig, !! 

My main line curves run about 4.4 degrees for the minimum ( 45 ft radius ), up to 3.27 degrees on the larger end ( 60 ft. radius ). These are for the mainline only however.. Most other track-age is of course smaller in size. The Present day line is scaled to 1/29 scale. All mainline switches are no. 14's. 

My Narrow Gauge line is also running 17 foot radius curves, and scaled for 1/20.3 stuff - so everything will fit mostly!! It runs no. 8 & 10 switches. 

Been working out on the N/G today, getting ready for a tunnel, and close to lay'n rail this spring!! woooHoo! At last.... 

Dirk 
DMS Ry.


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## Greg Elmassian (Jan 3, 2008)

Dirk, I apologize if you listed this before, but do you have a web site or photo archive? 

Would love to see way more of your work! 

Greg


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## bnsfconductor (Jan 3, 2008)

Posted By Greg Elmassian on 06 Apr 2012 02:33 PM 
Dirk, I apologize if you listed this before, but do you have a web site or photo archive? 

Would love to see way more of your work! 

Greg 
Greg,
I don't think he does. I've been emailing him back and forth for a while, and Dirk's got some ambitious plans for his layout! We need to convince him to set up an online photo sharing account!


Craig


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## SD90WLMT (Feb 16, 2010)

hmmmmm, Greg, apologize, ... for what? :~} 

You guys will have to come to AZ to twist my arms off!!!!!!!!!!!!!!!!!!!!!! 

Have not got around to pick'n a cloud for My pix yet! Web site sounds rather time consuming ... 

But He does keep working on the ambitious layout here, That's what I hear anyway! 

Dirk 
DMS Ry.


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## bnsfconductor (Jan 3, 2008)

If you want to share the wealth, Dirk send a couple to me and I'll post them for you... If not I won't say anything.


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## NTCGRR (Jan 2, 2008)

Thanks Kevin for the chart info I printed it out. 
very helpful.


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## Ironton (Jan 2, 2008)

Kevin, 

I was told the chord was 100 feet, probably measured with a 100 foot chain. That is where the formula I use comes from. It is not my formula, it is what I got from various sources. Maybe I am wrong, but that is what my research came up with some years back. 

Sorry to disagree.


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## krs (Feb 29, 2008)

I'm surprised to see all sorts of slightly different results to this question both in this thread and also on the internet. 

This is a simple basic mathematical problem and there is one and only one correct answer. 

http://www.handymath.com/cgi-bin/arc18.cgi 

Knut


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## Greg Elmassian (Jan 3, 2008)

The link does not work, but I figured out which calculator you used, the "complete circular arc calculator".... unfortunately the formula is not exposed, and it is simple trig, but it does not do the thing we want, enter the "degrees of curvature" to get the radius... What you have is entering the "degrees of arc", completely different. 

Greg


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## krs (Feb 29, 2008)

I hate that when I copy the displayed URL and it doesn't work. 

This one should be the right one. 
http://www.handymath.com/cgi-bin/arc18.cgi?submit=Entry 

And Greg, I don't understand why you say this is totally different of what we are looking for. 

This calculator lets you calculate every possible parameter of a segment of a circle. 
When I put in the appropriate values, I get 5729.65067 (to five decimal places) of the "magic" value which I had posted as 5729 based on a Trains magazine article and someone else posted here as 5729.651 

All ties together quite nicely I thought - where am I wrong? 

Knut


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## Greg Elmassian (Jan 3, 2008)

show me how to enter a 10 degree curve and get the radius in feet.... that one input and that one output... that's the calculator we are looking for... that's what was asked, and that was in the chart Kevin referenced. 

You cannot with that calculator, because it clearly states you need to enter 2 values.... also, again, there is NO input for degrees of curvature, it is degrees of arc... completely different... read it again... 

Greg


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## krs (Feb 29, 2008)

Posted By Greg Elmassian on 07 Apr 2012 10:25 AM 
show me how to enter a 10 degree curve and get the radius in feet.... that one input and that one output... that's the calculator we are looking for... that's what was asked, and that was in the chart Kevin referenced. 

You cannot with that calculator, because it clearly states you need to enter 2 values.... also, again, there is NO input for degrees of curvature, it is degrees of arc... completely different... read it again... 

Greg 


Ah - I see what you mean.

Well, mathematically, to calculate any of the values related to the arc of a circle, one always needs two values as a starting point.

That's a simple fact - nothing one can do to change this.

In the specific case we are talking about, the way the American Railroads calculated the radius of a 1 degree (or any degree curve) was based on a chord of 100 feet, ie the 100 foot chain mentioned earlier.

So one does have two values one can plug into that on-line calculator, the fixed value of 100 feet for the chord (which is called width in that on-line calculator) and of course the angle which is the angle subtended by the arc (called degrees of curvature by the railroad folks)


So inputting that into the calculator, it comes up with a value of 573.68566 feet for the radius of a 10 degree curve.

Same result as shown in this table that was posted in this thread earlier.
http://www.trainweb.org/freemoslo/Modules/Tips-and-Techniques/degrees_of_curve_to_radius.htm 
Knut


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## Greg Elmassian (Jan 3, 2008)

That simplifies it a bit, just would be nice to have a simple function, but good enough, just keep 100 in the "width of arc"... 

And I see that the degrees of curvature is indeed the degrees of arc, so they were really the same thing all along... sorry! 

Thanks for finding that calculator, nice to have it, rather than print up a sheet. So hard to find a calculator with sinusoidal functions any more. 

Greg


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## krs (Feb 29, 2008)

Greg - You were actually looking for the formula to calculate this..... 

The formula is 

Radius = 100 / (2 * sin ( å/2)) 

where å is the degree of curvature. 

One value to input, one output. 

If I use that to calculate the radius I get 573.685662208349278 with the scientific calculator on my computer. 

Knut


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## JackM (Jul 29, 2008)

When a high school kid whines that math is dull, just show him this thread. 

JackM 

I, too, have trouble with my sinusoidal functions. Miserable during pollen season.


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## Gary Armitstead (Jan 2, 2008)

Posted By JackM on 07 Apr 2012 11:44 AM 
When a high school kid whines that math is dull, just show him this thread. 

JackM 

I, too, have trouble with my sinusoidal functions. Miserable during pollen season. 

My wife is a recently retired high school math teacher (43 years) and she is having a great laugh at this thread right now!


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## Randy Stone (Jan 2, 2008)

Humm, she's laughing and I'm getting a headache. 

I'm sure glad I got my answer before you guys started throwing around all those calculations.


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## Greg Elmassian (Jan 3, 2008)

The calculator Knut found makes it easy, if you put 100 in the right place... you can also work backwards to see what your curves scale out in the prototype. 

Greg


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## rdamurphy (Jan 3, 2008)

Cool! Can we move on to transition curves now? 

http://www.oregon.gov/ODOT/HWY/GEOM...andouts/2008/RailRoadTaperAlignments.pdf?ga=t 

Robert


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## krs (Feb 29, 2008)

Posted By rdamurphy on 08 Apr 2012 12:16 AM 
Cool! Can we move on to transition curves now? 

http://www.oregon.gov/ODOT/HWY/GEOM...andouts/2008/RailRoadTaperAlignments.pdf?ga=t 

Robert 


Neat - what you found. 

On slide 9 they show the same equation I derived mathematically and posted earlier: 

The chord definition of the degree of curvature is used for both the taper and the center curve according to the following formula: R=50/(Sin D/2) 
␣ Where R = radius and D = degree of curvature 
␣ Since 30-foot chords are used for the tapers and 50-foot chords are used for the central curve, the radius is calculated according to the following formula: R=25/(Sin D/4) 
␣ The central angle for each 30-foot chord of the taper can be calculated by the following formula: 
∆=2[Sin-1 (0.6 Sin D/4)] 

In case someone missed it, the constant is always the length of the chord divided by 2. 
The last formula didn't copy correctly, Sin-1 should be the arcsin (don't know how to do a superscript on this website) 

Knut


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## bnsfconductor (Jan 3, 2008)

Posted By rdamurphy on 08 Apr 2012 12:16 AM 
Cool! Can we move on to transition curves now? 

http://www.oregon.gov/ODOT/HWY/GEOM...andouts/2008/RailRoadTaperAlignments.pdf?ga=t 

Robert 
That's a whole new can of worms...


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