Chapter 13.R, Problem 24E

In designing transfer curves to connect sections of straight railroad tracks, it’s important to realize that the acceleration of the train should be continuous so that the reactive force exerted by the train on the track is also continuous. Because of the formulas for the components of acceleration in Section 13.4, this will be the case if the curvature varies continuously.

(a) A logical candidate for a transfer curve to join existing tracks given by $y=1$ for $x\le 0$ and $y=\sqrt{2}-x$ for $x\ge 1/\sqrt{2}$ might be the function $f\left(x\right)=\sqrt{1-x^2},0<x<1/\sqrt{2},$ whose graph is the arc of the circle shown in the figure. It looks reasonable at first glance. Show that the function

$F\left(x\right)=\left\{\begin{array}{l}1\text{ if }x\le 0\\ \sqrt{1-x^2}\text{ if 0}<x<1/\sqrt{2}\\ \sqrt{2}-x\text{ if }x\ge 1/\sqrt{2}\end{array}\right\}$

is continuous and has continuous slope, but does not have continuous curvature. Therefore f is not an appropriate transfer curve.

(b) Find a fifth-degree polynomial to serve as a transfer curve between the following straight line segments: $y=0$ for $x\le 0$ and $y=x$ for $x\ge 1$. Could this be done with a fourth-degree polynomial? Use a graphing calculator or computer to sketch the graph of the “connected” function and check to see that it looks like the one in the figure.