Quote from: James in FL on August 26, 2014, 11:35:02 PM
Welcome back to trigonometry 101.
OK James. After working through the trig, I may have been able to answer my own question. If the figure 8 is enclosed in a rectangle that runs between the limiting track centers, the width of that rectangle is just twice the track radius, or 22.5 inches. The length of the rectangle is two plus twice the square root of two times the radius, which gives about 53.3198 inches. Do you agree?
Of course, to get the length and width to the edge of the roadbed, I would have to add the total width of the roadbed to these two results. I think that is an extra 1.5 inches that must be added to the length and the width.
Is that about right?