( 762 ) 
Let us now consider in how far we can find the surface (2) for 
a given striction line. 
a Let r — 7 {&) be the given projection of the striction lino. 
This furnishes, regarding (3), the relation between and /,: M 
_ ML _= „ 
1 + (/,)’ + </,')■ 
,, - <P |1 + (/,)» + (/,')»! 
/! /,' 
Thus: The surface 2 = rf x {0) -f- j* - ^ 1 ^ ^ ■ d6, where 
f x is an arbitrary function of 6, has as (r, ^j-prqjection of the striction! 
line the curve r = p (0). 
b. Let now be given that the striction line must be a plane curve 
lying in the plane z = Ax -{-By -f C or z — r (Accs 6+B sin 0) C. 
By substituting this value for 2 in (2) we must get (3). This 
furnishes between f\ and f t the relation 
a - c = /:// 
A _ A cos e - Bsin 0 1 -f (/,)* + (//,’ 
/; = i / +(/ 1 ) a + (/ 1 ? 
A -c // (A - A cos d-B sin 0) 
So the surface 2 = rf x (0) -\-e J ~ A cos6 “******* -{- C, in which 
A is an arbitrary function of 6, has a plane striction line lying in 
£ = Ax -J- By -j- C. 
c. The most general problem here is: what is the surface (2) for 
which r=ffi{0), 2 = ip {0) is striction line? 
To solve this we substitute these values for r and z in (2) and (3) 
and we obtain then two equations with /, and /, as unknown 
quantities. If we eliminate between these two /„ we retain f x as 
only unknown quantity in the equation: 
•r + viftY + /«¥ - <pAfi = 0. 
If A is solved out of this we can find /, out of: 
*== ?/»+/,. 
We can find the solution in explicit form for the special case 
rp = const., for which constant we take 0. We then find: 
9\ 1 +(/.)•)-^A/i f = o f 
from which ensues 
