(124) 
de, | IN dy, 
tdi a U 
For these eurves on the surface of saturation I have chosen the 
name of “curves of slope”. This series of curves begins and ends 
with the following curves: 1st the curve p= f (y,) for the pair (1, 3), 
and 2°¢ the eurve consisting of ap = f (r,) part for the pair (1, 2) and 
of a corresponding part for the pair (2, 3). If we draw these curves on the 
vapour sheet, we must imagine that we descend instead of ascending. 
For the projection of these curves on the plane of the triangle I 
have chosen the name of “nodal envelope’. The outmost curves 
of this series are: 1s* one of the sides of the triangle, adjacent to 
the right angle, namely that one corresponding to the third component, 
and 2°¢ a line consisting of the other side adjacent to the right 
angle and the hypothenuse of the triangle. 
For the solution of the differential equation of these curves it is 
required that we may express 7, and y, in, and y,. This is 
possible (p. 7) when the second phasis is a rare gas phasis, if we 
namely assume the functions gw, and w,, to be known. For that 
case the equation we have to integrate may be written: 
du, dy, E 
ees ; 
Ties ps es De 
Bles ik fe hl} y, (1—y,) (e e —1)—u,(e ce —1)} 
or É ; 
dit, dy, 
F Pa, 1) ae pe m2 meus Chee En Wa, En 8 
miley) “*- tye Te 7). 4 id-«,-y)e ™-)-«,¢ =e ) j 
or zi (en en Oh of et (en —é' tij scene 
ay U, Ed 
py! le i SS wu! ] 7 Ein | p 
mi (en D |: Ly d(1 ar 1) zi (em ea dy, is d( LY.) i 
| ik 1—w#,—y, Mn La; 
The last equation may also be written as follows: 
u, v, al U, | 
(e° 4% —1) d log ————_ = (¢ “—1) d log ————_ . i 
1—2z,—y, 1—z,—y, . 5 
: 
; : = pw Pr 
For the case that the liquid sheet is a plane, e° *—l ande’ *—l 
el Paes a 
are constant, and equal to — and — and so the equation of 
Pi Pi 
the nodal envelope will be given by: 
Pa Di Ran PL 
( Ly ) (et ae ( Y, ) Pi 
LE 1—a,—y, 
