Vertical curve length or radius

[jagged ben]

So I'm working on a track plan where we want a track to transition from 2% downgrade to 2% upgrade, and it will be advantageous to do so in the minimum distance possible that doesn't cause problems.   How would you calculate that distance?  The transition is a trough (as opposed to a crest).    I think I'm primarily concerned with uncoupling, I don't want the coupler faces to be slipping past each other vertically.  The track will need to carry long trains, although I suppose the trough will mitigate the drawbar pull pressure on the couplers.

Thoughts?

[trainforfun]

Personnaly I would try a quick mock up .  Two plank at 2% one plus one minus meeting at an angle .
Then you put a thin plank over them secured by Wise grip or C clamp at the junction of the the curve / tangent .
You dont need to try a long train to see what the couplers will do .
The worst case will be when a car with long coupler meet a car with short coupler .
Give us some feedback on this .

[C855B]

... The transition is a trough (as opposed to a crest). ...

In civil engineering terms, a "sag". I was looking at various vertical curve calculators to see how they could be applied to coupler offsets, but the this-math-is-too-deep headache beat me to arriving at a solution. As per Louis' advice, it would take less time to do a mock-up. Also, just like horizontal curves, transitions matter, and the best solution will be parabolic.

[ednadolski]

I've looked at some online VC calculators, but did not find them very useful for model railroading.   The notion of what will "cause problems" has too many parameters, such as the length/wheelbase/overhang of your longest rolling stock, not to mention the condition and quality of your couplers, track curvature thru the VC, and such.  In theory one could compute a mathematical limit but controlling all the variables in practice becomes unworkable, and is hardly needed in a model. Then there is also the matter of constructing something that conforms to the mathematically computed values (in three dimensions no less).  In the end I just picked something that did not seem too abrupt and went with that.... IIRC, 4x-5x the longest car length.

Ed

[jagged ben]

I've looked at some online VC calculators, but did not find them very useful for model railroading.   The notion of what will "cause problems" has too many parameters, such as the length/wheelbase/overhang of your longest rolling stock, not to mention the condition and quality of your couplers, track curvature thru the VC, and such.  In theory one could compute a mathematical limit but controlling all the variables in practice becomes unworkable, and is hardly needed in a model. Then there is also the matter of constructing something that conforms to the mathematically computed values (in three dimensions no less).  In the end I just picked something that did not seem too abrupt and went with that.... IIRC, 4x-5x the longest car length.

Ed

4x-5x the longest car per what?  Per a certain percentage change?  For any flat to a grade?    I'm talking about a transition equivalent to flat-to-4%, would that apply to any of what you've done? 

Thanks for the replies all.  Thinking about the easiest way to do a mockup quickly that can be redone to test different results.  Might pull out some old code 80 and use track nails.   

[nickelplate759]

Keep in mind that body-mounted couplers will likely be more particular than truck-mounted couplers.

George

[John]

[ Guests cannot view attachments ] 2% grade to a french curve then flat for at least 1.5 car lenghts of the longest car, then gradual increase again to 2%

[ednadolski]

4x-5x the longest car per what?  Per a certain percentage change?  For any flat to a grade?

That's from a 2% up to ~1.8% down, partly on a large (34.5" r.) curve to straight transition.   I'm allowing for autoracks so my transition length is ~ 3 feet. (I have not built it or even tried a mockup, but there is also some room to adjust if I really need to.)

Ed

[jagged ben]

(Attachment Link) 2% grade to a french curve then flat for at least 1.5 car lenghts of the longest car, then gradual increase again to 2%

Sounds convincing.  What's the 1.5 car lengths based on?  Also I guess I'm still wondering how to determine the length of the curve

[robert3985]

If it were me, my first thought is that a spiral easement should work vertically as well as horizontally...from an infinite radius (straight) to a radius that is tangent to the angles of decent and ascent, forming the "dip".  The bottom of the "dip" will be tangent to my level benchwork.

I can reproduce this line easily using Cadrail, but an easier, and much more reliable solution would be to take a thin strip of molding (about 1/8" thick, about 2" wide) and nail or screw the bottom of the "dip" to your level benchwork and then lift the ends until they are at the proper angles to be tangent to your 2% ascending and descending grades.  The "dip" can be made longer or shorter by clamping a 1X3 (with a tapered end) at varying distances from the screw holding the bottom of the "dip".  Clamp a piece of flextrack to this arrangement to find out what works most reliably.  The molding strip will automatically form a spiral easement with the effective radii being controlled by the distance you clamp the 1X3 to it from the screw or screws holding the bottom of the "dip".

I don't know if this will work or not, but it sounds like it would, and, again, would be what I'd do if I had your problem.

Hope this is clear enough.  If not, tell me and I'll draw up a quick diagram...

Cheerio!
Bob Gilmore

[nkalanaga]

Not helpful for the calculations, but if you're doing it by trial and error, consider your locomotives as well.  I have a nasty transition at the top of my mining company's standard gauge track:  8% up ending at horizontal points of a turnout.  2-bay hoppers take this fine, with truck or body mounted couplers.

Life-Like SWs and Atlas Shays/S2s don't have any problems.  My ancient Trix FM doesn't have any problems. 

An Arnold S2 can't handle it.  The trucks rotate fine, so sharp curves aren't a problem, but they don't tilt fore/aft at all.  Thus, they go straight through the turnout, regardless of the points, because the lead axle is in midair. 

In your case, too sharp would lead to the center axles lifting, which might work on straight track, but even there it wouldn't help electrically.  On curves it would be as bad as my hump.

[GaryHinshaw]

If you are using continuous plywood roadbed, I think you would be hard pressed to make a vertical curve that would be sharp enough to be a problem.    I'm using 3/4" ply and I haven't come close to worrying about vertical curves.  In any event, here are some formulae if you are so inclined.

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.

But

sin θ = (W/2)/R,

so

D = OW/2R. 

Note that D goes to zero (no sag) as R goes to infinity (no vertical curve) as it should.  Now consider a typical auto rack with a wheel base of (roughly) W=5", a coupler overhang of O=1", the coupler sag will be

D = (1*5)/(2*R) = (2.5/R)"

where R is the vertical curve radius in inches.  Now an MTL N scale coupler has a vertical pulling face of ~0.1", so let's say we want D to be less than half this dimension: D < 0.05", then we would place a lower limit on the vertical curve radius:

R > (2.5/0.05)" = 50".

So keep your vertical curves to larger than 50" and you should be fine.

[GaryHinshaw]

I just realized that I didn't answer your original question.   

Let's say we have a 50" vertical curve that connects a −2% grade to a +2% grade with no easements (to keep the calculation simple).  A 2% grade corresponds to an angle of 1.1 degrees, so your vertical curve needs to subtend an angle of 2.2 degrees (= 0.038 radians) to connect the two grades.  The arc length of a circular arc of radius R that subtends an angle of θ (in radians) is

L = θR

So a 50" vertical curve would subtend 2.2 degrees in a length L = 0.038*50" = 1.9".  This arc length is also - to a very good approximation - the horizontal length of the arc.  So, if you have 6" to play with between your grades, you would only need a vertical curve of ~150" radius to connect the two grades.  This should pose no problems at all for your trains.

[John]

Sounds convincing.  What's the 1.5 car lengths based on?  Also I guess I'm still wondering how to determine the length of the curve

WAG!!! .. but the rule about S curves applies .. I have an Armstrong track book somewhere .. he details this exact situcation ..  let me see if I can find it ..

some other info

https://www.scribd.com/document/246409974/Chapter-6-Railway-Track-Design

found this ..  here http://cs.trains.com/mrr/f/11/p/256330/2868783.aspx

In Armstrong's Track Planning for Realistic Operation, he describes two values, x & L, defining an easement.  He lists values of x and L for 3 radii

   3/8"    12"   18"

   7/16"  16"   24"

    1/2"   18"   30"

normally, a curve would meet a straight track at some point P and R is the distance between P and the center of the curve.

Armstrong suggests increasing the distance between P and the center of the curve by x,  1/2" for a 30" radius curve (i.e. 30.5")

But he also suggests changing the point where the curve ends and becomes straight by extending the curve L/2 beyond point P.    and the curve starts deviating from its normal radius of R at roughly L/2 from point P.

So the easement begins ~L/2 from point P and ends L/2 beyond P with the center of the curve R+x perpedicular to P.

[CNR5529]

I just realized that I didn't answer your original question.   

Let's say we have a 50" vertical curve that connects a −2% grade to a +2% grade with no easements (to keep the calculation simple).  A 2% grade corresponds to an angle of 1.1 degrees, so your vertical curve needs to subtend an angle of 2.2 degrees (= 0.038 radians) to connect the two grades.  The arc length of a circular arc of radius R that subtends an angle of θ (in radians) is

L = θR

So a 50" vertical curve would subtend 2.2 degrees in a length L = 0.038*50" = 1.9".  This arc length is also - to a very good approximation - the horizontal length of the arc.  So, if you have 6" to play with between your grades, you would only need a vertical curve of ~150" radius to connect the two grades.  This should pose no problems at all for your trains.

LOL Gary! Love seeing the geometry work first thing in the morning. Indeed, that vertical curve should pose no problem for all applications. Just to throw in a bit of prototype reference info, most modern passenger equipment specifications require that the vehicle be able to negotiate a 1000 ft vertical curve in both crest and sag, with no spiral transition to tangent track, while coupled up to another reference vehicle. That vertical curve would scale down to a 75" curve, again with no transition. All this to say that the proposed 150" radius curve should pose no problem, especially if easements are included.