# Degrees in curves?



## NTCGRR (Jan 2, 2008)

OK you smarties.

I was reading last night ,, yes I do read captions. 
any way they said a train was heading into a 10 degree curve.

I know radious and dia. I even know how to figure degrees when cutting rafter tales.
So in our small world of 10 ft radius curves. (20' dia)
WHAT WOULD A 10 degree curve be???

thanks


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## Gary Armitstead (Jan 2, 2008)

10 degrees equals 19 feet, 9 and 3/8 inches radius, according to Stan's Converter ( in 1/29).


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## Semper Vaporo (Jan 2, 2008)

Posted By NTCGRR on 03/10/2009 5:06 PM
OK you smarties.

I was reading last night ,, yes I do read captions. 
any way they said a train was heading into a 10 degree curve.

I know radious and dia. I even know how to figure degrees when cutting rafter tales.
So in our small world of 10 ft radius curves. (20' dia)
WHAT WOULD A 10 degree curve be???

thanks



In the real world, meaning 1:1 scale, measuing the radius of a curve would entail stretching a line several hundred feet long which would be difficult given the weight of the line, and trees, gullys, mountains, houses and other objects getting in the way. 
So, they measured out some distance along the track and noted the angle formed at that distance from a tangent at the starting point. That angle then is the degree of curvature. Early on, each railroad (or track contractor) used whatever distance was convenient for them, such as 50 feet, 100 feet, 25 yards, or 10 chains, etc. Thus one railroads 10 deg curve might be 8 deg or 15 deg for some other railroad.

I think that today, the distance is 100 feet.


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## Gary Armitstead (Jan 2, 2008)

Semper,

Again according to Stan's Handy Converter, 10 degrees is about 573.69 feet in full size.


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## Semper Vaporo (Jan 2, 2008)

http://mysite.du.edu/~jcalvert/railway/degcurv.htm


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## chrisb (Jan 3, 2008)

Its a way of expressing the sharpness of a curve. See http://www.tpub.com/content/engineering/14071/css/14071_243.htm
R = 5729.58/degree of curve.
Haven't used that in years. Now you enter the radius in a data collector and it figures the curve for you.
I think most RR lines used spiral curve to transition into the circular curve. 
For 1:20.3 scale
Radius in inches=(5729.58/D)(12)/20.3


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## Semper Vaporo (Jan 2, 2008)

Posted By gary Armitstead on 03/10/2009 5:40 PM
Semper,

Again according to Stan's Handy Converter, 10 degrees is about 573.59 feet in full size.



Posted By gary Armitstead on 03/10/2009 5:40 PM
Semper,

Again according to Stan's Handy Converter, 10 degrees is about 573.59 feet in full size.



Take a straight track and starting at some point begin a uniform curve with a radius of 573.59 ft. A straight line to a point on that curve that is 100 ft from the starting point will form a 10 degree angle to the straight track.


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## NTCGRR (Jan 2, 2008)

Woow, thanks 
I had one eye pop out while tring to read that first one. Then I got a head ach. 

I always thought a 20' rad. would be great on a modern double track main.


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## GG (Jan 1, 2009)

OK, the ultimate answer: 

"10 degrees is just that". So if understood then so be it. 

Pull out a protractor.... measure 10 degrees offset. Move protractor 1 degree forward along the line and once again measure 10 degrees forward offset. 


Call me when you get through the curve...










Second option..... get the garden hose out and ball park it.... 

gg


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## Semper Vaporo (Jan 2, 2008)

Posted By GG on 03/10/2009 7:02 PM
OK, the ultimate answer: 

"10 degrees is just that". So if understood then so be it. 

Pull out a protractor.... measure 10 degrees offset. Move protractor 1 degree forward along the line and once again measure 10 degrees forward offset. 


Call me when you get through the curve...










Second option..... get the garden hose out and ball park it.... 

gg




Sorry, GG, using the protractor would depend on the diameter of the protractor... besides, how "LONG" is a degree (that one you said to move the protractor forward)?

But I agree... get out the garden hose and ballpark it.

Then again, the first time I layed out mine, I used graph paper scaled to 1/4-inch to the Foot and drew it with my highschool drafting class compass. Then I gave up on that and drew it using my computer and Microsoft's PAINT program (the freebie with Winders) scaled to 1 pixel per inch, but I had to set the display to 1040x680 to get it to fit all at one time on the screen.


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## Gary Armitstead (Jan 2, 2008)

C.T.,

I'm going to try something wih my CAD program. I hope I understand your premise. I'll draw a radius of 573.69 thru two pointa on a 100 ft. straight line-thru the endpoints of the 100 ft. line. Correct? I'll get back with the answer.


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## GG (Jan 1, 2009)

Posted By Semper Vaporo on 03/10/2009 7:12 PM
Posted By GG on 03/10/2009 7:02 PM
OK, the ultimate answer: 

"10 degrees is just that". So if understood then so be it. 

Pull out a protractor.... measure 10 degrees offset. Move protractor 1 degree forward along the line and once again measure 10 degrees forward offset. 


Call me when you get through the curve...










Second option..... get the garden hose out and ball park it.... 

gg




Sorry, GG, using the protractor would depend on the diameter of the protractor... besides, how "LONG" is a degree (that one you said to move the protractor forward)?

But I agree... get out the garden hose and ballpark it.

Then again, the first time I layed out mine, I used graph paper scaled to 1/4-inch to the Foot and drew it with my highschool drafting class compass. Then I gave up on that and drew it using my computer and Microsoft's PAINT program (the freebie with Winders) scaled to 1 pixel per inch, but I had to set the display to 1040x680 to get it to fit all at one time on the screen.











Semper, you kill me with your wit... 

OK, the diameter is a function of the square root of the radius multiplied by the abismus of the cosyn of the radius. As such the length of the length of the degree is relative to the perspective of the protractor. 



Get my point? Think switches and misguided ones....










Clearly demonstrated is the lack of abismus of the cosyn of the radius... 


gg


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## GG (Jan 1, 2009)

Posted By Semper Vaporo on 03/10/2009 7:12 PM
Posted By GG on 03/10/2009 7:02 PM
OK, the ultimate answer: 

"10 degrees is just that". So if understood then so be it. 

Pull out a protractor.... measure 10 degrees offset. Move protractor 1 degree forward along the line and once again measure 10 degrees forward offset. 


Call me when you get through the curve...










Second option..... get the garden hose out and ball park it.... 

gg




Sorry, GG, using the protractor would depend on the diameter of the protractor... besides, how "LONG" is a degree (that one you said to move the protractor forward)?

But I agree... get out the garden hose and ballpark it.

Then again, the first time I layed out mine, I used graph paper scaled to 1/4-inch to the Foot and drew it with my highschool drafting class compass. Then I gave up on that and drew it using my computer and Microsoft's PAINT program (the freebie with Winders) scaled to 1 pixel per inch, but I had to set the display to 1040x680 to get it to fit all at one time on the screen.











Semper, you kill me with your wit... 

OK, the diameter is a function of the square root of the radius multiplied by the abismus of the cosyn of the radius. As such the length of the length of the degree is relative to the perspective of the protractor. 



Get my point? Think switches and misguided ones....










Clearly demonstrated is the lack of abismus of the cosyn of the radius... 


gg


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## Gary Armitstead (Jan 2, 2008)

C.T.

I just layed out a radius of 573.69 with endpoints going thru a line with a length of 100. The angle checks at 9.99992 degrees. Pretty close to what Stan gets on his converter. I did the layout using MasterCAM software.


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## NTCGRR (Jan 2, 2008)

woow, now I'm seeing double????? 

Its 8:30 pm and my cardboard pizza (dinner) is ready, thats how busy I am.


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## Gary Armitstead (Jan 2, 2008)

Marty,

I believe that C.T. was saying that the 100 ft. chord is always used as the basic for the angle. My layout checks pretty close to the Converter. I'm not about to do the math. To late in the day for that. AND I'm too old! I have a headache now!









Actually you can lay it out in feet, inches or miles. The angle will always be the same.


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## GG (Jan 1, 2009)

Semper... foot note here. 

" How LONG" is also a Chinese person... 


My argument is so strong here. the garden hose concept wins.









gg


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## stanman (Jan 4, 2008)

Pretty close?


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## NTCGRR (Jan 2, 2008)

Funny how a thread goes on with out me.. hope this one does not get locked also. 
I get the idea. I think????


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## Treeman (Jan 6, 2008)

Here is a way to look at this in model RRing. A LGB R1 curve turns 30 degrees in one section. Twelve of them makes a full circle 360 degrees.


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## Gary Armitstead (Jan 2, 2008)

Marty,

Silly me! I thought you were actually asking a legitimate question. No wonfer people are hestitant in answering. Yawn. I think I'll go in the yard and play with my trains.


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## aceinspp (Jan 2, 2008)

In the real world of 1to 1 a cord of 62 ft is used to determine the degree of curve. Placing the string on the high side of the curve on the gage side 5/8 of an inch below the top of rail and stretching one would walk back to the center of the cord being 31 ft. Taking a tape measure and then measure the distance from gage side of rail to center of string. What ever the measurement is tells you the degree of curve. For 1/29 use the equations to convert . Simple and you will be surprised at the results. Later RJD


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## NTCGRR (Jan 2, 2008)

Gary 
I really did want to know, just thought it was a simple A +B=C, not all these different things.


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## Gary Armitstead (Jan 2, 2008)

Marty,

That's why I never worried about laying it out using degrees. It was always too much trig for me! But with Stan's Converter and my CAD software, life is MUCH easier.









I just wish that large scale track manufacturers would designate their track by RADIUS, tather than R1, R3, etc. or by diameter.


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## GG (Jan 1, 2009)

Posted By NTCGRR on 03/11/2009 7:55 PM
Gary 
I really did want to know, just thought it was a simple A +B=C, not all these different things.






Marty... 

You got your answer here. 


Cudo's to you and you opened up the conversation and thread for others to develop and grow on the original question... As you can see by the thread, there are many ways to approach the issue.. and many people will take different approaches. 


Shutting down the thread would be counter productive... your initial question has allowed others to learn... ( and a few to have fun... like me







) 


Thank you for bringing this up. 


gg


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## Semper Vaporo (Jan 2, 2008)

Posted By NTCGRR on 03/11/2009 7:55 PM
Gary 
I really did want to know, just thought it was a simple A +B=C, not all these different things.


It is not that they are all different things, but rather different ways to find what you want. Like 2*3 is a different way of saying 2+2+2 or 3+3 or counting pairs (2,4,6) or counting triplets (3,6)... all give the same answer, given the same data, but one may be an easier way to arrive at the answer than another depending on the circumstances.

aceinspp's use of 62 ft is a quick way to eliminate the need for a calculator or a Surveyor's transit. The use of 62 is a "magic number" in this instance because if you work through all the calculations (which are really beyond me to do easily right now... but I did do it once just to prove to myself that the person that I respected who had told me of the "trick" had not just pulled the wool over my eyes and had risked losing my respect) uh... back to my thought... if you work the formulae when you take that measurement in inches, the number you get (in inches) can just be called the degree of curvature (in degrees) i.e.: you get the value you want, using only a 62-ft long line and a tape measure.

Attach your line to the outside rail at some point, move to the other end of the 62-ft long line and attach it to the outside rail (both places 5/8ths of an inch down from the top of the rail, then walk back to the center of that line (at 31-ft) and measure the distance from the center of the line to the rail at 5/8ths of an inch down from the top of the rail and the number of inches between the line and the rail is the number of degrees of curvature of the curve.

Great! Right? What if your curve is not linear? What if your curve is shorter than 62 ft? Well, if it is non linear then you can only arrive at some average value and if it is not at least 62-ft long then... well, hmmm... get a Surveyor's transit?

The concept of degree of curvature is not all that hard, it is just outside of the normal way the lay person thinks about curves. We tend to think of radius or diameter and usually we can see the whole circle in one glance... a round table top, a round garden plot or a circle inscribed within our limited outdoor yard space.


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## GG (Jan 1, 2009)

Posted By Semper Vaporo on 03/11/2009 8:58 PM
Posted By NTCGRR on 03/11/2009 7:55 PM
Gary 
I really did want to know, just thought it was a simple A +B=C, not all these different things.


It is not that they are all different things, but rather different ways to find what you want. Like 2*3 is a different way of saying 2+2+2 or 3+3 or counting pairs (2,4,6) or counting triplets (3,6)... all give the same answer, given the same data, but one may be an easier way to arrive at the answer than another depending on the circumstances.

aceinspp's use of 62 ft is a quick way to eliminate the need for a calculator or a Surveyor's transit. The use of 62 is a "magic number" in this instance because if you work through all the calculations (which are really beyond me to do easily right now... but I did do it once just to prove to myself that the person that I respected who had told me of the "trick" had not just pulled the wool over my eyes and had risked losing my respect) uh... back to my thought... if you work the formulae when you take that measurement in inches, the number you get (in inches) can just be called the degree of curvature (in degrees) i.e.: you get the value you want, using only a 62-ft long line and a tape measure.

Attach your line to the outside rail at some point, move to the other end of the 62-ft long line and attach it to the outside rail (both places 5/8ths of an inch down from the top of the rail, then walk back to the center of that line (at 31-ft) and measure the distance from the center of the line to the rail at 5/8ths of an inch down from the top of the rail and the number of inches between the line and the rail is the number of degrees of curvature of the curve.

Great! Right? What if your curve is not linear? What if your curve is shorter than 62 ft? Well, if it is non linear then you can only arrive at some average value and if it is not at least 62-ft long then... well, hmmm... get a Surveyor's transit?

The concept of degree of curvature is not all that hard, it is just outside of the normal way the lay person thinks about curves. We tend to think of radius or diameter and usually we can see the whole circle in one glance... a round table top, a round garden plot or a circle inscribed within our limited outdoor yard space.














OK Semper, here you go again with your wit and knowledge. 


How many times do I need to lecture you !



Good and great... Now please reproduce your film c/w a switch that allows a live steam loco to ..... not fall on the ground.










You are alway enjoyable and yes like me.... never off the hook. If in doubt speak with my family. 



gg 



EDIT and as an after thought.... 

GARDEN HOSE CUTS THE MATH OUT. 



I REST MY CASE AND I AM GOING TO BED...


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## Semper Vaporo (Jan 2, 2008)

Posted By GG on 03/11/2009 9:04 PM
OK Semper, here you go again with your wit and knowledge. 


How many times do I need to lecture you !



Good and great... Now please reproduce your film c/w a switch that allows a live steam loco to ..... not fall on the ground.










You are alway enjoyable and yes like me.... never off the hook. If in doubt speak with my family. 



gg 



EDIT and as an after thought.... 

GARDEN HOSE CUTS THE MATH OUT. 



I REST MY CASE AND I AM GOING TO BED...











"BED"? "Lecture"? Same thing for me!

"Off the hook"??? Don't you mean, "Slipped the leash"?


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## Hagen (Jan 10, 2008)

Posted By Semper Vaporo on 03/11/2009 8:58 PM
Posted By NTCGRR on 03/11/2009 7:55 PM
Gary 
I really did want to know, just thought it was a simple A +B=C, not all these different things.


It is not that they are all different things, but rather different ways to find what you want. Like 2*3 is a different way of saying 2+2+2 or 3+3 or counting pairs (2,4,6) or counting triplets (3,6)... all give the same answer, given the same data, but one may be an easier way to arrive at the answer than another depending on the circumstances.

aceinspp's use of 62 ft is a quick way to eliminate the need for a calculator or a Surveyor's transit. The use of 62 is a "magic number" in this instance because if you work through all the calculations (which are really beyond me to do easily right now... but I did do it once just to prove to myself that the person that I respected who had told me of the "trick" had not just pulled the wool over my eyes and had risked losing my respect) uh... back to my thought... if you work the formulae when you take that measurement in inches, the number you get (in inches) can just be called the degree of curvature (in degrees) i.e.: you get the value you want, using only a 62-ft long line and a tape measure.

Attach your line to the outside rail at some point, move to the other end of the 62-ft long line and attach it to the outside rail (both places 5/8ths of an inch down from the top of the rail, then walk back to the center of that line (at 31-ft) and measure the distance from the center of the line to the rail at 5/8ths of an inch down from the top of the rail and the number of inches between the line and the rail is the number of degrees of curvature of the curve.

Great! Right? What if your curve is not linear? What if your curve is shorter than 62 ft? Well, if it is non linear then you can only arrive at some average value and if it is not at least 62-ft long then... well, hmmm... get a Surveyor's transit?




Could you scale that 'trick' down?
the 62 feet then becomes 74.4cm or 744mm (I always did like centimeters and millimeters, smaller measures " align="absmiddle" border="0" />) 

Then do the same as you mentioned, 'walk' halfway back and at that point 37.2cm or 372mm, measure the number of millimeters to the track, the degree curvature is the same as the number of millimeters?

Hmmmm
In my theory then 1mm represents one inch

one feet is twelve inches (62feet = 744 inches) 

Therefore my line is 744mm long 

Those numbers would work better in our scales if you can do it like that? that would work down to 2' radius


not sure about this at all, just throwing out a theory that could work in our scales?


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## Semper Vaporo (Jan 2, 2008)

Posted By Hagen on 03/12/2009 1:34 AM
Posted By Semper Vaporo on 03/11/2009 8:58 PM
Posted By NTCGRR on 03/11/2009 7:55 PM
Gary 
I really did want to know, just thought it was a simple A +B=C, not all these different things.


It is not that they are all different things, but rather different ways to find what you want. Like 2*3 is a different way of saying 2+2+2 or 3+3 or counting pairs (2,4,6) or counting triplets (3,6)... all give the same answer, given the same data, but one may be an easier way to arrive at the answer than another depending on the circumstances.

aceinspp's use of 62 ft is a quick way to eliminate the need for a calculator or a Surveyor's transit. The use of 62 is a "magic number" in this instance because if you work through all the calculations (which are really beyond me to do easily right now... but I did do it once just to prove to myself that the person that I respected who had told me of the "trick" had not just pulled the wool over my eyes and had risked losing my respect) uh... back to my thought... if you work the formulae when you take that measurement in inches, the number you get (in inches) can just be called the degree of curvature (in degrees) i.e.: you get the value you want, using only a 62-ft long line and a tape measure.

Attach your line to the outside rail at some point, move to the other end of the 62-ft long line and attach it to the outside rail (both places 5/8ths of an inch down from the top of the rail, then walk back to the center of that line (at 31-ft) and measure the distance from the center of the line to the rail at 5/8ths of an inch down from the top of the rail and the number of inches between the line and the rail is the number of degrees of curvature of the curve.

Great! Right? What if your curve is not linear? What if your curve is shorter than 62 ft? Well, if it is non linear then you can only arrive at some average value and if it is not at least 62-ft long then... well, hmmm... get a Surveyor's transit?




Could you scale that 'trick' down?
the 62 feet then becomes 74.4cm or 744mm (I always did like centimeters and millimeters, smaller measures







" align="absmiddle" border="0" />) 

Then do the same as you mentioned, 'walk' halfway back and at that point 37.2cm or 372mm, measure the number of millimeters to the track, the degree curvature is the same as the number of millimeters?

Hmmmm
In my theory then 1mm represents one inch

one feet is twelve inches (62feet = 744 inches) 

Therefore my line is 744mm long 

Those numbers would work better in our scales if you can do it like that? that would work down to 2' radius


not sure about this at all, just throwing out a theory that could work in our scales? 







I have NOT worked this out completely yet... just thinking out loud here... (kind'a hoping someone with more math smarts will jump in!)... 
I think your numbers will work if your scale is 1:25.4

Not sure if scale affects this or not.

When the Tavist-D wears off and I can think more clearly and I am not having to catch my nose (it is running all over!) I may try to do the math again and see how scale affects it.

P.S. Maybe it is too late now, but... Don't touch your computer monitor while you are reading this... I'd hate for you to get this cold!


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## Hagen (Jan 10, 2008)

I was rather hoping it would work regardless of scale, as it's basicly only taking into account the relationship between two (three) lengths and an arch.

Just as a triangle with sides 3*4*5 always will have a 90deg angle, no matter if it's inches, mm or miles 

(I didn't touch the monitor, honest)


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## NTCGRR (Jan 2, 2008)

See 
back when I was starting to build with styrene I had to learn what thousands was, now I can look at a sheet and know. 
After getting into G I had to learn Dia to go with the flow. 
Now if I read 10degree curve in real life I can say thats a ???1,000 ft curve?? or 5,000 ft curve. 
Just like the basics of looking at a F unit, FA, PA, E etc even tho I don't know the details of each different . I have the basics. 
Just like saying garden railroader simple means ,loves to eat, has choo choo shirts , wares speedos and knee pads..


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## chrisb (Jan 3, 2008)

Seeing that metric measurements have been mentioned, this reminds me of an arguement we had at work years ago concernit a metric construction project. Definition degree of curve is not used in the metric system.
Thats because its based on 100 foot chord length. 100 ft is a little over 30 meters. 3.281 feet per meter. Instead the radius is refered to. 

The old timers could layout curves all kid of ways using a steel tape and a 20 second transit. Now you don't have to set up on the center line. Its all by coordinates with 2 guys instead of 4.

The old timers were good but not perfect, they fudged once in a while.


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## GG (Jan 1, 2009)

Guess what... yes 

Old timers didn' t have advanced math, GPS and associated concepts however.... the Trans Canada Railroad was built across the Rockies using transits and basic common sense. 

Call this the "daa" approach. Worked well. 


Now... back to my basic garden hose concept.... 


To small an idea for the Rockies however perfect for the back yard. 


gg


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## chrisb (Jan 3, 2008)

I would say that prior to the use of scientific calculators , total stations, data collectors and gps, surveyors were better with advanced math. They used trig tables to use figure angles, used all kinds of trig functions that are now obsolete. I suspect the line and grade for the trans c. RR's was layed out this way. This type of layout works well with HO, not practical in the garden. Some people count rivets, 
some like to calulate curves. I glad that people still talk about it.


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## Ironton (Jan 2, 2008)

O.K. Here is the explanation of degrees of curvature on a 1:1 scal railroad: Degrees of curvature 

The 100 foot was probably chosen because that was the length of a surveyor's chain, IIRC. That made it an easy distance to base things on.

The degree of curvature could be easily found by finding the heading of the tangent track leading inot the curve, and then finding the heading of the 100 foot chord. The degree of curvature is twice the difference of these two headings. I am too lazy to do the geometric proof of this however.


Hope this helps.


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## SteveC (Jan 2, 2008)

Rich

Depends on just which 'chain' you're referring to...

Land Surveyor's Chain (i.e. Gunter's Chain) = 66 ft.

Engineer's Chain = either 50 ft. or 100 ft. length

Vara Chain (i.e. metric) = either 10 M or 20 M length


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## Mike Reilley (Jan 2, 2008)

Posted By GG on 03/12/2009 6:56 PM
Guess what... yes 

Old timers didn' t have advanced math, GPS and associated concepts however.... the Trans Canada Railroad was built across the Rockies using transits and basic common sense. 

Call this the "daa" approach. Worked well........ 



gg




There was no advanced math involved anywhere in the development of railroads....only trigonometry, which is hardly advanced math. 

The hard problems were measuring distances or elevations accurately. To do that required a LOT of surveyed points...10s of thousands of em. Each surveyed point nominally required a minimum of three independent cross fixes from other already surveyed points which had been established from three independent cross fixes from other already surveyed points, each of which had been established from other already surveyed points by three independent cross fixes from other already surveyed points, each of which had been established from other already surveyed points by three independent cross fixes from other already surveyed points, each of which had been established from other already surveyed points by three independent cross fixes from other already surveyed points..............you get the idea.

The point is, the math is way easier than the effort it took to get surveying parties into the areas and all these measurements taken. And remember, each cross fix laid was to a three dimensional point in space...always at a different elevation...so you not only mapped the terrain in X and Y....but also in Z (elevation). It's easy to understand how you lay a curve using a chain and transit...what's hard to understand is how these folks figured out WHERE to lay the curve given all the gullies, glutches, canyons, slopes, hills, etc to get up and over a mountain. I find it pretty amazing. Finding the pass though the mountain range is hard enough...figuring out how to get the track up that hill...or down...is mind boggling.


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## Steve Stockham (Jan 2, 2008)

Hmmm.........I only have two words for this: "_Flex Track!!_"


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## Al McEvoy (Jan 3, 2008)

This is a great discussion. One key reason that the 1:1 engineers used trig/geometry to calculate curvature was (and still is) because they simply cannot measure the large radii required for mainline alignments in so many situations. Such as going up around a mountain - the center of radius of a particular curve may well be a point deep inside the mountain, or it may be a point out in mid air several hundred or several thousand feet off the valley floor. You can easily think of many situations where the curve could not be measured using a center of radius. Not to mention the tediousness of trying to measure varying radii when constructing the spiral lead-in to the tightest section of the curve. We face literally none of those hurdles when building our model railroads hence the success in using the various methods as described above. I echo the sentiments expressing wonderment and admiration of those men who built the railroads (many of which are still in use) over a century ago using tools more crude than what a gradeschooler has access to today.


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## GG (Jan 1, 2009)

Posted By Steve Stockham on 03/13/2009 11:09 AM
Hmmm.........I only have two words for this: "_Flex Track!!_" 







Totally agree Steve: Flex Track AND Garden Hoses. Works every time. 

gg


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## Steve Stockham (Jan 2, 2008)

Okay, _five_ words!


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## GG (Jan 1, 2009)

Posted By Steve Stockham on 03/13/2009 5:04 PM
Okay, _five_ words!













LOL


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## chrisb (Jan 3, 2008)

When the surveyor sets the transit over the point of curvature, sights his back sight, flops the scope, he dials in a deflection angle. The chord length measured out most likely will not be a hundred feet.
If anything it would be the chord length to get to the next 50 foot station along the curve. 100 foot stations are too far apart to build to. The error between arc and chord is understood. I realize this example does not include the spiral but its the same idea.
Curves can be layed out by tangent offset instead of deflection. Tangent offset works good on curved bridges that have straight beams. Just measure off the center of the middle beam.


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