Henry Ernest Dudeney/Modern Puzzles/222 - A Mechanical Paradox/Solution

by : $222$

 * A Mechanical Paradox
 * A remarkable mechanical paradox, invented by about the year $1751$, ought to be known by everyone, but, unfortunately, it is not.


 * It was contrived by him as a challenge to a sceptical watchmaker during a metaphysical controversy.


 * "Suppose," said, "I make one wheel as thick as three others and cut teeth in them all,
 * and then put the three wheels all loose upon one axis and set the thick wheel to turn them,
 * so that its teeth may take into those of the three thin ones.
 * Now, if I turn the thick wheel round, how must it turn the others?"
 * The watchmaker replied that it was obvious that all three must be turned the contrary way.
 * Then produced his simple machine, which anyone can make in a few hours,
 * showing that, turning the thick wheel which way you would,
 * one of the thin wheels revolved in the same way, the second the contrary way, and the third remained stationary.
 * Although the watchmaker took the machine away for careful examination, he failed to detect the cause of the strange paradox.

Solution

 * The machine shown in our diagram consist of two pieces of thin wood, $B$ and $C$, made into a frame by being joined at the corners.
 * This frame, by means of the handle $n$, may be turned round an axle $a$, which pierces the frame and is fixed in a stationary board or table, $A$,
 * and carries within the frame an immovable wheel.


 * Dudeney-Modern-Puzzles-222-solution.png


 * This first wheel, $D$, when the frame revolves, turns a second and thick wheel, $E$,
 * which, like the remaining three wheels, $F$, $G$ and $H$, moves freely on its axis.
 * The thin wheels, $F$, $G$ and $H$, are driven by the wheel $E$ in such a manner that when the frame revolves
 * $H$ turns the same way as $E$ does,
 * $G$ turns the contrary way,
 * and $F$ remains stationary.
 * The secret lies in the fact that though the wheels may be all of the same diameter,
 * and $D$, $E$ and $F$ may ($D$ and $F$ must) have an equal number of teeth,
 * yet $G$ must have at least one tooth fewer, and $H$ at least one tooth more, than $D$.