Vertical curve length or radius

[jagged ben]

Gary thanks for the math.  I imagine I could have doped that out over the course of a couple days, but it's much easier when someone else does it.      I have to say I'm shocked by the small size of the answer, but I can't poke any holes in your math.  (Well, actually if you do the math with straight radians instead of rounding off degrees then you end up with 2" instead of 1.9".  But big whoop.  )   

However, I think the answer corroborates what others have said about how locomotive construction is probably more of a constraint than couplers.   I surveyed a few of my locos.  The 'winner' so far is an Athearn SD75 whose pilot only sits about a hundredth above the rail.  The math on that came out to a 290" vertical curve radius required to keep the plow from bottoming out.  (W=3.875, O=.75, D=0.01).    That comes out to 11.6 inches of curve length, which is more like the answer I was expecting to begin with. 

And not only that, but this is a club environment so I've got to consider all sorts of locomotives that I don't own.    It would be nice to hear from folks who own Big boys and Challengers what the driver trucks wheelbase is, distance from wheelbase plow, and most importantly height above rails of the plow.

The space constraint really argues for not allowing the subroadbed material to determine the length of the curve.  I might go with 1/2" ply or something even thinner to get a minimum length curve.  Going into this discussion I was hoping the answer wouldn't be more than 18", or as little as a foot if possible.  Looks like sticking with something like 18" would be wise.   

[jagged ben]

John, thanks for your post as well.  That's similar to what I've seen in XtrackCAD although I had no idea it (probably?) came from Armstrong.  Interesting to know.

One thing that baffles me a little is that the offset is larger for a larger radius.  I would think as the radius gets larger the easement is less of a concern.  I mean, on a 290" radius (see previous post) do I really need an easement?  I think I will rely on a natural easement created by the material.

[nkalanaga]

For those who don't like radians, or trigonometry, the formula L = θR can be rephrased as:

(length of arc) = (degrees/360)*(2*pi*radius)

It's the same thing, as (2*pi*radius) is the circumference of the total circle, and (degrees/360) is the portion of the curve you need.  For even simpler math, (2*pi) = 6.283, so if you write that, and the angle for various percent grades, on a note somewhere, you can do all the calculations with nothing but a basic calculator.  To simplify it even further, write the (angle/360), and you won't have to divide that either. 

At that point, multiply the 6.283 by the (angle/360), and you have a constant for any vertical curve between that combination of grades, but since there's an almost infinite combination of grades, it probably isn't worth the trouble.

In a case like this one, where there's a non-0 grade on both sides, add the two grade angles to get the total angle, which will allow for any combination.  If you're combining two grades in the same direction, such as 3% down to 2% down, subtract, as the total change is only 1%.

I learned that trick in high school electronics, just before electronic calculators became common. 

[jagged ben]

For those who don't like radians, or trigonometry, ...

Hey I got no problem with any of that...   I just don't where you guys find the theta symbol on your keyboards.   

[GaryHinshaw]

The 'winner' so far is an Athearn SD75 whose pilot only sits about a hundredth above the rail.  The math on that came out to a 290" vertical curve radius required to keep the plow from bottoming out.  (W=3.875, O=.75, D=0.01).    That comes out to 11.6 inches of curve length, which is more like the answer I was expecting to begin with. 

I think you may have an extra factor of 2 somewhere.  For D<0.01" I get:

R > OW/2D = (0.75*3.875)/(2*0.01) = 145"

which requires a bit under 6" of run.  But more is better.

[ednadolski]

Let R be the radius of the vertical curve, W be the wheelbase of your car or loco, and O be the overhang of the coupler from the truck centre, as in this diagram:



The main quantity of interest is the length D (in red): the distance the coupler sags below its nominal height above the rails.  This is given by

D = O sin θ.

I'm probably overlooking something elementary, but I am not seeing where you get this from the drawing.

Ed

[jagged ben]

I think you may have an extra factor of 2 somewhere.  For D<0.01" I get:

R > OW/2D = (0.75*3.875)/(2*0.01) = 145"

which requires a bit under 6" of run.  But more is better.

Oops, you're right.  And that's good.  This is why it's better when someone else does it.

[John]

Just lay the track and see if it works .. all this trig and geometry is making my head hurt 

[jagged ben]

I'm probably overlooking something elementary, but I am not seeing where you get this from the drawing.

Ed

Yeah I noticed that Gary is over simplifying a bit. But for our purposes it doesn't matter. Visualize a line perpendicular to R  where it meets O and the circle.  The angle between that line and O is Theta. I'm not going to write up a proof but it would involve properties of triangles.    As to whether D should be perpendicular to O or to this new line, it might matter in another application.  But at one or two degrees the difference in choice of hypotenuse is 0.1%, which is well within the margin of error of what we're trying to calculate.

[nkalanaga]

In Windows XP, Theta can be found in Programs/Accessories/System Tools/Character Map.  Copy and paste it from there to your document or web page.

Θ is the Arial upper case version, θ is the lower case.  What your screen will show depends on the font your browser uses, how Railwire stores text, and whether your font even has Greek letters.  Not all do.

[GaryHinshaw]

I'm probably overlooking something elementary, but I am not seeing where you get this from the drawing.

As jb notes, my expression for D neglects the curvature of the track over the distance of the coupler overhang, O, so it neglects terms of order (O/W), which can be as large as 20% for, e.g., an auto rack.   At the risk of alienating 98% of our members with some more algebra, we can get an exact expression for D using the Pythagorean theorem.  Let L be the (vertical) distance from the centre of the circle to the mid-point of the car, then the inner triangle sides satisfy

L² + (W/2)² = R²

while the outer triangle sides satisfy

L² + (W/2 + O)² = (R+D)².

Subtract the first equation from the second to eliminate L², then, after some algebra, we get

2RD + D² = OW + O².

Now, the second term on the left-hand-side is very small compared to the first: D/2R ~ 0.001, so we neglect it (and make a ~0.1% error in the process).   The second term on the right-hand-side isn't quite as negligible: O/W ~ 0.1-0.2, depending on the car, so let's keep it.  We then have

D = OW/2R + O²/2R

This differs from my earlier answer by a factor (1+O/W) = 1.2 for the auto rack example, and more like 1.1 for a more typical overhang.

And now back to trains.   

[tom mann]

Something tells me that if you understand @GaryHinshaw 's math here, you would do well in one of his classes.

[trainforfun]

Just lay the track and see if it works .. all this trig and geometry is making my head hurt 

I agree , but also impressed by some Railwire people with math !!!

[ednadolski]

Thanks Gary!   I used to be better at this sort of thing but that was a really long time ago <sheepish>

Ed

[nkalanaga]

I suspect that the civil engineers have a book full of tables for all of this, and they just look up the appropriate numbers!