Wakelin. — Fallacies in the Theory of Circular Motion. 141 



at the surface of the earth, the point B will be 16 feet fi'oni 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 ? 



AtB,s= 16 = i/(i)2 = |/and/=128 



„ C, s= 64=i/12 =^/and/=128 



„ D, s = 144 = J/(f)2 = |/and/=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 32. 

 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)2=^3g/and/=288 



„ C,s= 64=J/(f)2=^/and/=288 



„ 0,5 = 144 = 1/ 12 =J/and/=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. 

 Eeverting 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 fom'fold or ninefold. 

 It 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 ui 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 diversely 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 cncle be drawn, and let a polygon of n sides be inscribed in the 

 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 hne will not coincide with the 

 radius — it will not form part of the radius produced through the angular 

 point of the polygon. Let a particle B be moving with any velocity along 

 one of the sides of the polygon, and when it comes to the angular point let 

 it 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), 

 so as to cause it to move along the next side of the polygon. And so on in 



