G Scale Model Train Forum banner
1 - 2 of 2 Posts

· Premium Member
Joined
·
5,578 Posts
I have Llagas Creek code 215 Nickle-silver track. I used the plain slip-on rail joiners. When I assembled the track, I put a penny or dime between the rail ends as a gauge to allow for expansion.

When the track heats up and the rails get longer due to expansion of the metal, the small space dissappears, as was intended.

Unfortunately, when the track then cools down and the rails contract, the one rail joiner with the least resistance to slipping is the only place that opens up. Thus all the places where I have joiners, only one creates a gap and it is now the sum of all those dime/penny thickness gaps I stated with.

To be done right (at least as far as controlling expansion joints) each rail needs to be solidly affixed to some sort of underlayment ONLY at the very center of the length of the rail. Thus expansion and contraction would only be allowed from the center to the end and the whole length of rail could not be pulled toward one end. The expansion joints would have to be designed to handle the absolute maximum and minimum length the rail might reach. Of course, the underlayment may have some expansion/contraction coeffiecient to contend with too.

I am now of the opinion that there should be no expansion joints except in very rare and special cases. The whole track should just float completely such that loops or circles, in total, get bigger or smaller in diameter, which is what happens on my track now anyway. Unfortunately, after mine expands, upon contracting again, something will catch and keep the circle from reducing in diameter and one joint somewhere around the loop will come apart.
 

· Premium Member
Joined
·
5,578 Posts
Posted By armorsmith on 06/02/2008 6:11 PM
If I may, let me present a little science/math demonstration. In Ken's RGS Southern above, the track that buckled needed to be gapped as follows with the following known information.
Known:
Coefficient of expansion of aluminum rail = .0000123 inches per inch of length per degree of temperature rise.
Length of track needed to compensate for expansion
total temperature range
So.....
If Ken's striaght line track is 10'-0" long (120 inches), and say the total temperature variation is from say 20 degrees for a low and 100 degrees for a high for a total temperature variation of 80 degrees. Hence, 80 degrees times 120 inches times .0000123 equals 0.11808 inches expansion.
These numbers do not consider rail mounting (tight/loose/fixed) or roadbed (floating in ballast/attached to stringers/etc). And yes, the prototype railroads have the same expansion/contraction issues. I am familiar with Ken's section of the country, I was raised in NE Jersey. I spent many days and nights railfanning Pensy, and later Conrail.
Hope this is enlightening to some.
Bob Cope

Just some fun calculations based on the numbers...
Assume the following:
You have two 10-ft sections of nice straight track.
It is firmly attached at each end and cannot go anywhere AT THE ENDS.
The rail jointer in the middle is just a flimsy slip joint.
The rail itself is extremely stiff and inflexible such that if it is lifted in the middle it will become two perfectly straight ramps to the peak formed by lifting it.
It was laid when it was at the extreme cold temperature and no room for expansion was allowed.
It is now at the extreme high temperature of your example.
Thus, each section is now 0.11808 inches longer.
The only place for the expansion to go is to lift the middle up such that the rail is now the hypotenuse of a right triangle formed by the track bed and the elevation in the middle of the track.
How high is that middle of the track, now?
Pythagorean Theorem applies... A^2 + B^2 = C^2
"A" is the length of the track base (original length of the rail).
"C" is the length of the rail after expanding.
"B" is the height the middle of the track has raised.
Arrange the formula to solve fr "B": C^2 - A^2 = B^2
A= 120^2 = 14400
C = 120.11808^2 = 14428.35314
14428.35314-14400 = 28.35314285
The square-root of 28.35314285 = 5.324766933 INCHES!
Divide the rise by the run and you have a 4.437305777 percent grade from one end to the middle.
If the rail is not real real stiff, then it will sag and that will mean the length of the base will be less and thus the middle will be even higher.
Practical example...
My first track was that silly plastic stuff of 1-ft sections. I had it on four 10-ft 1x6 boards laying on the ground. I was afraid it would blow off and break the tabs that hold it together, so I nailed it down on the extreme ends of the track.
I don't know the temperature when I did this nor do I know the coefficient of expansion for the plastic.
I do know that a few days later is was quite warm and when I came home from work and drove into the driveway I noticed the track had assumed the shape of a bell curve with the middle at close to THREE FEET above the boards.
Like an idiot I walked over and touched the track and it fell over and broke several of the plastic tabs that hold the track together. /DesktopModules/NTForums/themes/mls/emoticons/crying.gif" border=0>
 
1 - 2 of 2 Posts
This is an older thread, you may not receive a response, and could be reviving an old thread. Please consider creating a new thread.
Top