KINEALY A SPIRAL ON A TORUS. 5 



a' . ra' V 



From page 2 n z=z — , which gives tan ^ = ± jr- = ,^ ■> 



where V is the velocity of generating point in space, due to its 

 rotation about its axis, and V is its velocity, at the same instant, 

 due to its rotation about the axis of the generating curve. 



na is always equal to d. Let n' be the number of turns the 

 spiral makes about the torus while a increases from to 2 ^ , 



ft 2nn' 

 When a = 2 rr , =z2 -«/ ; but n =. = ~ — = n'. 



a ■" ^ 



The helix is a special case of a spiral on a torus, where the 

 radius of the circular axis is equal to infinity. To obtain the 

 equation of the helix, transfer the origin of coordinates from to 

 (?' in Fig. II. Equations (4), (5) and (6) become 



X := [7? -{- r cos na^ sin a _ _ - _ - (13) 



jy ■=. [7? -)- r cos na~\ cos a — 7? - - - - (14) 



z :=. r sin na ........ (i^) 



Now if in these equations Ji is made infinite, n will equal 

 infinity and a will equal zero ; but na = always, and Ra 

 S , sin a = a. 



Making these changes in (13), (14), and (15), the equation ol 

 the helix is obtained as 



X =z R ?,\\i a z= Ra ^ S (16) 



y =z r cos d ..-.._-- (17) 



z = r sin 6 - • - - (i8) 



Sin (16) means simply the distance out from Oy , that the 

 point has moved ; and, as S =1 R a = Rat = Vt , where a = 

 angular velocity of the plane, and V the linear velocity of the 

 generating point, which in this case is constant, it is seen that the 

 helix is not formed by a combination of two rotations, but by a 

 rotation and translation, as it should be. 



In the general equation (i), (2), and (3), let, now, y(x;ya) rr: 

 i?, a constant, and f{yzna) ^ r' — cRa = r. 



This would give us the equations of a spiral on a horn-shaped 

 surface. 



