_860 
Now the curvature of (27) is given by: 
ke (28) 
(a? + b?)*/2 ke 
that of curve (26) by: 
2abd—ate—b'e— da't + b*r —2abs) 
et Geli: [ 
As (28) and (29) have the same denominator we, in order to 
compare the curvatures of both curves, only want the numerators. 
For the sake of brevity we write: 
2 abd — ae bee. ee 
(29) 
and 
2 abd — ate — b'e — À(a't + br — 2 abs) =Q— AS KS 
If, by means of the known values of a and 4 we calculate the 
value of S we find: 
S = (rt — s?) (a’r + 2 aBs + Bt) 
hence, S is always positive. 
In order to find the direction of the curvature we calculate the 
coordinates & and % of the centre of the curved circle and ascertain 
at which side of the tangent this centre is situated. Therefore, we 
call the origin of the coordinate system the point which in this case 
represents the vapour, 0. We now find the following: the liquidum line 
is curved in the point / towards O when Q< QO; it is curved in 
FP away trom QatiQ >. 0. 
A consideration of Q shows that this can be positive as well as 
negative; hence, the liquidum line can be curved in /’, away from 
O as well as towards U. 
In order to find the saturation line under its own "apour pressure 
we will consider two cases. 
Owing to the small value of V—v, A will generally have 
a large positive value. In Fig. 1, wherein for the moment we 
disregard the curve d’F¢, the liquidum line is represented by dFe; 
the point QO is supposed to be somewhere to the left of this curve 
dFe so that this is curved towards’ OQ; Q is consequently negative. 
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