Question by Nineteen-Twenty by Ten-Eighty Pixels 🙂: Is this a good formula to find the circumference of a model railroad track curve?
Before you say anything, this has nothing to do with homework
I think I’ve got a good formula for figuring it out, but I just want to run it by anyone else into H0 scale model railroading.
If it took 3 curves to make a 90 degree turn then each piece would be 30 degrees.
If it took 4 curves to make a 90 degree turn then each piece would be 22.5 degrees.
So count the number of pieces it takes to make a 90 degree curve
Divide 90 by the number of track pieces needed
90 degrees / 3 curves = 30 degrees each curve
90 degrees / 4 curves = 22.5 degrees each curve
To make sure it is correct, you times the number of curves by the amount of degrees of each piece, and if the answer is 90, then it is right.
X = degrees
Y = per track piece
Z = Track pieces
90X / # of Z needed to make 90X = # of XY
# of Z needed to make 90X x # of XY = 90X
90 / 3 = 30
3 x 30 = 90
90 / 4 = 22.5
4 x 22.5 = 90
Would this formula be suitable to figure out the circumference of each track piece as long as each curve track used is the same?
Answer by Wdlane
Here is a NO math answer that will get you the center line for the best model railroad curves you can make without sending your brain into spastic convulsions. Go to a lumber yard (Or Lowes or Home Depot) and look for long square trim or pine as long and perfectly clear as possible. The end result you want is a 3/4″ x 3/4″ piece of wood. You should be able to get up to 16 feet long.
Tack the wood strip on the center line at the beginning of where you need the curve to start. Tack the other end of the strip on the center line of where the curve needs to stop. Draw the line you get. Put down your flex track to that center line. DONE! You get a VERY natural looking varied radius curve complete with easements with almost no math involved. It will look much better then trying to force the use of sectional track.
Add your own answer in the comments!