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DYNAMIC METEOROLOGY AND HYDROGRAPHY. 



We then construct a table (table F') according to the equation 



(f) a = arc cos <p 



By use of this table we can easily draw the curves for "integer" values of cos a 



between those for integer values of a (fig. 65). 



Table F'. Arcus-cosine table for passing from the field of an angle to the field of its cosine. 



148. Graphical Algebra with Two Variables. Let a and fi be two scalar func- 

 tions, each represented by a chart of equiscalar curves. The problem is to draw the 

 equiscalar curves representing the field of a third scalar cp, which is determined by 

 the relation 



(a) tp = f(a, j8) 



The discontinuous method of solving this problem will be this: we choose a 

 point, take out from the charts the values of a and /3 and calculate by equation (a), 

 the corresponding values of /(a, (J). This is repeated for a sufficient number of 

 points. The values thus found for ip are inscribed upon a sheet of paper, and then 

 the equiscalar curves tp = 1, 2, 3, . . . are drawn by leading of the values thus 

 found. Evidently the work can be facilitated by the construction of an auxiliary 

 table containing the values of <p tabulated with a and [3 as arguments. 



But a corresponding continuous method can also be given. To see it we solve 

 equation (a) with respect to one of the known quantities a or j3, 



(b) p = F(a,<f) or a = F'(P,<p) 

 According to these equations we construct the auxiliary tables G. 



Let us first follow one of the vertical columns in the table and the corresponding 

 curve a = const, on the chart. We shall then see that the curve a = o will be cut by 

 the curve <p = o at the point where /3 has the value /3 OOI by the curve <p = 1 at the 

 point where /3 has the value /3 IO ,by the curve <p = 2 at the point where 13 has the value 

 /S 20 , and so on. The situation of the points ft*,, /3 IO , /3 20 , ... is seen at once, as the 

 intersection of the curve a = o with the curves j3 = o,f3 = i,j3 = 2 . . . shows where 

 on the curve a = o we have the integer values of /3. Interpolating by eye-measure 

 we can mark the points where the curves a = const, are cut by the curves for integer 

 values of <p. These points being marked, we can draw at once the curves <p = const. 



