I'm using O 31 tubular on my layout and am planning a small expansion. I want to create an S curve (opposing curved tracks) that will require the track to land in a specific spot. Aside from trial and error, I wanted to know if there is a formula to calculate what is necessary to situate the new track approximately 4" off the centerline of the existing track. Using full curved O 31 will create more that a 4" distance between tracks. I'm thinking I'll need half curves and a small cut section of straight to acheive the 4" differential.
Any thoughts on how to calculate.
One thing to keep in mind is that an S curve could cause problems with train derailments. The scale modellers will tell you to avoid it. If you press ahead with this, You need to sit down and draw this out to scale using a track planning template. Then you need to measure the linear distance covered in inches and then divide it into 4 inches which is the rise you want to acheive. That will be your gradient. You can obtain a gradient measuring instrument from the Micromark catalog and use this to keep the gradient correct as you build this track segment. It will sit on the track and you would build the risers to match. I hope this helps. Good luck.
George
The distance d between the tracks equals 2 * r * (1 - cos(a)), where r is the O31 radius of sqrt(2) * 10 inches and a is the angle of one of the curves. Solving for a gives a = acos(1 - d / 2 / r) = 30.823 degrees.
An easy way to measure a curved section is the chord between the ends of the center rail. This is c = 2 * r * sin(a / 2) = 2 * r * sqrt((1 - cos(a)) / 2). Substituting the first equation into this one gives c = 2 * r * sqrt(d / r / 4) = 7.521 inches.
If you want to use half curves, the tangent between them will be 2.612 inches long.
(I didn't notice anything about grades in your question.)
Bob Nelson
Bob:
I need to digest the numbers .... math not being my forte but I'm sure your formulas work. Right ... no grades to consider. Thanks
Jim
lionelsoni The distance d between the tracks equals 2 * r * (1 - cos(a)), where r is the O31 radius of sqrt(2) * 10 inches and a is the angle of one of the curves. Solving for a gives a = acos(1 - d / 2 / r) = 30.823 degrees. An easy way to measure a curved section is the chord between the ends of the center rail. This is c = 2 * r * sin(a / 2) = 2 * r * sqrt((1 - cos(a)) / 2). Substituting the first equation into this one gives c = 2 * r * sqrt(d / r / 4) = 7.521 inches. If you want to use half curves, the tangent between them will be 2.612 inches long. (I didn't notice anything about grades in your question.)
Overkill mean anything ? Bob you help a lot of people ( and thank you for that ) but bring it at a level most can understand, please.
Come on we all know you are smart, but, Please speak English or anything that the lower / working class people can understand. I am sure some may get it but
A lot of - stuff on posts today, I need to spit,,,, again.
Joined 1-21-2011 TCA 13-68614
Kev, From The North Bluff Above Marseilles IL.
SRGuy try this put the extra curve in position that is curving in then take another curve and place it in the direction to make the s curve then overlay till you get what you want look for the spot where they match perfectly as possible and mark it on both tracks thats where and the angle you need to cut, use a ruler over the track to give you a straight line to cut. You will probably need like a dremal to get a straight cut.
KEV hows that for laymens terms?
Life's hard, even harder if your stupid John Wayne
http://rtssite.shutterfly.com/
If he wants the numerical answers, they're there. If, as I suspect, he (or anyone else) is interested in where they come from, he's got an outline of the analytic geometry involved, without any spoilers. (It's not a trivial problem with a simple intuitive answer.) I'm happy to answer any specific questions anyone may have.
Thanks, RT, Even I can get that.
That is what I do, no math, lay it out how I want it and cut some track to make it work. Not a problem and no rockets to the moon. Same way I did it at ten and it still works. Why they call them TOY trains, I think.
Bob,
Thank you for working out the answer in addition to supplying the formula srguy asked for. I think I can use a protractor to layout 30.8 degrees for cutting back two 45 degree sections of O-31 curve. If I can do it, I think most can do it. [opinion]
May I ask you about O-31 radius being the sqrt(2) * 10 inches equals 14.142 inches? I measure 28.75 inches center of center rail to center of center rail (14.375 inch radius.) I could not get an accurate measurement because;
1) the joints will not all close tight and
2) the circle is not perfectly round (see 1).
How did you get sqrt(2)*10?
..........Wayne..........
You guys are funny. After brief contemplation on the formula, I decided that approach is too "heady" for me. I remeasured and the actual distance is 5" and I'm sure I'll be able to lay it out on the workbench and make it work. Bob, I do appreciate you input as well as yours Kev. Thanks
Wayne, it comes from the design of the turnouts. They of course have a 45-degree curved section superimposed on a straight section. The distance traveled in the direction of the straight section is exactly the same between the straight and curved sections. So, if you put a reverse curved section after the curved section of the turnout (as Jim is doing, but without the constraint of a certain distance between the tracks) the track joints will line up exactly between the two tracks. This makes it possible to create a passing siding without cutting track. (Unfortunately, the tracks are a whopping 8.284 inches apart, which is clearly why Jim is interested in a closer spacing.)
That distance in the direction of the straight section is r * sin(a), where r is the radius and a is 45 degrees and, for O31, equals 10 inches exactly. so r * sin(45 deg) = 10. The sine of 45 degrees is 1 / sqrt(2), so r = sqrt(2) * 10 inches.
Note that, if you add the tie length to the diameter, 2 * sqrt(2) * 10 + 2.25, you get 30.534 inches, almost as close to 30 as to 31. In fact, the track is sometimes called O30.
The same relationship applies to O27 track, but there the radius was made to be a round number, causing the straight section's length to be weird. Twenty-seven inches, less 2 inches for the tie, then divided by 2, gives a radius of 12.5 inches. Divide that by sqrt(2) and you get the length of the straight section, 8.839 inches. There has been a lot of variation in actual O27 track lengths among manufacturers, probably from not understanding the geometry behind that number.
(Anyone trying to cut curved track using a protractor is in for a lot of frustration. The chord method is much easier.)
Jim, there were some mistakes in the numbers I gave you for 4-inch track spacing. The angle a should have been 30.843 degrees; the chord c was correct, 7.521 inches, and the straight section between the half curves should have been 1.999 inches.
If you're interested, the corresponding numbers for 5-inch track spacing are 34.591 degrees, 8.409 inches, and 3.082 inches.
The formula for the chord can be simplified to c = sqrt(d * r).
Bob, keep up the good work. There are some of us that like to use your analysis, and even though I could do it, it is easier to copy from you.
With that said, if I had this problem, I would probably lay it out with uncut pieces of track, and then cut the track to fit. Tubular track is very forgiving.
Thank you for the information. Of course I had to check it out--CHECK! Now I am totally awed by your knowledge.
Very nicely done, Wayne.
I'm with 'servoguy'...rather than pull out the old sliderule it's just easier to copy off Bob's paper. And, it's also fun to see his analyses and presentations. Thanks for all your contributions, Bob!
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