102 DYNAMIC METEOROLOGY AND HYDROGRAPHY. 



of (p corresponding to the line-element ds measured off as ordinates. If the length 

 of this element is given, we can read off on the axis of abscissae the corresponding 



value of the ratio <p = -j- . Instead of measuring the length (fa between the hyperbola 



4 and the axis of abscissse, we can also measure it between the two symmetric hyper- 

 bolae da = +2 and da = 2. This will as a rule be preferable. 



For practical use we engrave the diagram on the under side of a transparent 

 sheet of celluloid, and cut a narrow slit in this sheet along the axis of abscissae. The 

 slit should just be broad enough to make it possible to make marks with a sharp 

 pencil on the paper below the sheet. The sheet is placed so that the line-element ds 

 is parallel to the ordinates of the sheet. In the case da = 4 it will have one end-point 

 on each of the two hyperbolae da = +2 and da = 2. The reading on the axis of the 



abscissa gives the value <p = -j- , which the derivative has in the central point of the 



line-element ds. This point can be marked through the slit. It will be clear how 

 different hyperbolae should be used according to the occurring values of the differ- 

 ential da. The procedure is illustrated by the upper line of fig. 82, where the points 

 for integer values of the function a are marked on the upper side of the line, while 

 the values determined for the derivative are noted on the under side. 



VL-8 a fS SO 2i 28 28 24 20 >6 12 



1 , ' I ' I ' 1 ' 1 ' ' 1 ' 1 ' I ' I ' 



<p- 56 5.U >/.2 3.2 2.1 -AS -2.8 ->I.O -5.1 



I I I I 1 1 1 1 1 ' I 1 



<p- S 'i 3 2 / -/ -Z -3 -'l -5 



Fig. 82. Linear differentiation or integration. 



When the line-elements ds are short, a small error in the placement of the points 

 where the given function has integer values will cause great errors in the values 

 of the derivative. It will then be an excellent method of reducing these errors to 

 measure two or more elements simultaneously. Thus if the points for all integer 

 values a = 1, 2, 3, . . . are given, we measure the corresponding elements two by 

 two between the hyperbolae + 1 and i ; or we can measure them four by four 

 between the hyperbolae +2 and 2, and so on. 



As it is seen, the direct use of the sheet gives the derivative at points where it 

 has all sorts of fractional values. But it will be easy afterwards by interpolation 

 to find the points where the derivative has certain integer values. In the example 

 of fig. 82 these points are marked on the lower line. 



We can now treat the inverse problem, the linear integration. Then let the 

 function <p(s) be given. The problem is to determine any function a(s), which is 

 in the relation to the given function <p which is defined by equation (b) or (c) . Evi- 

 dently this can be done by the same divided sheet. For the sheet at once gives those 

 lengths ds to which integer increases da will correspond. The process of integration 

 must begin at a certain initial point s = s and we presume that at this the required 

 function has a given value a = a . 



Now let the value of the given function in the region of this initial point be 

 <p = <p . In order to find the point where a has the value a +4 we set off from the 



