Waxetin.— Fallacies in the Theory of Circular Motion. 141 
at the surface of the earth, the point B will be 16 feet from A, the point C 
64 feet, and the point D 144 feet. Let it now be required to make the body 
pass through these several points in half the time. What must be the 
acceleration ? | 
At B,s= 16=4/ (})2=4 f and f=128 
» O,8= 64=$/12 =4 f and f=128 
» D, s=144=3 f (3)? =8 fand f=128 
From this we gather that when the body falls through the same spaces in 
half the times the acceleration must be four-fold, for 128 is four times 82. 
Let the body be now drawn by an accelerating force through the points 
B, C, D, in one third of the times it was drawn through those points in the 
first case. What must now be the acceleration ? 
At B,s= 16=4/ (4)2=,, f and f=288 
» OC, s= 64=3f (3)? = f and f=288 
» D,s=144=}f 12 = 4 f and f=288 
We see from this that when the body is drawn through the same spaces in 
One-third the times that the acceleration must’ be increased ninefold. 
Reverting now to the geometrical figure already given (see page 138), if the 
velocity of the revolving body be increased to twice or thrice the velocity it 
had at first, it will have to be drawn from P to B in half or one-third the 
time. But if the body had not been revolving at all, but had been at rest at 
P, the acceleration would have had to be increased fourfold or ninefold. 
Tt is not necessary, therefore, for a body to be revolving in any orbit to 
Satisfy the condition, that if it be required to draw the body through the 
Same space in one-half or one-third the time, the acceleration must be 
increased fourfold or ninefold. That the acceleration should increase 
directly as the square of the velocity, or inversely as the square of the times, . 
it is not necessary, therefore, that the body acted upon by an accelerating 
force should be moving in any orbit. 
Let a circle be drawn, and let a polygon of n sides be inscribed in th 
circle. Produce each of the sides to a distance equal to itself. This 
lengthened side is divided equally by the circumference—the side of the 
Polygon is one-half, and the part produced outside the circle is the other 
half. From the end of the produced side draw a line to meet the angular 
Point of the polygon opposite to it. This line will not coincide with the 
radius—it will not form part of the radius produced through the angular 
Point of the polygon. Leta particle B be moving with any velocity along 
one of the sides of the polygon, and when it comes to the angular point let 
. be struck by another particle H so as to cause it to move along the next 
Side of the polygon. When the particle B comes to the next angular point 
of the Polygon, let it be struck by another particle (of course equal to H), 
#9 a8 to cause it to move along the next side of the polygon. And so on in 
