436 ANALYSIS OF CURVES. 



6, cos + 6 3 cos 3 + 6 5 cos 5 = 26*5 

 .'. &, + & 3 + b s = 26-5 



i.e., b, + 20-7 + 5-78 = 26'5 and 6, = -02 

 This may be assumed = 0, since readings are only taken 

 to 1st place of decimals. Hence 6, = 0. 



Sine Term. 



When 6 = 90 3 = 270 5 B = 450 



Sin 6 = 1 Sin 3 = - 1 Sin 5 = I 

 a t sin 6 + a 3 sin 30 + a., sin 50 = 41'4 

 i.e., a l a 3 + a 5 = 41'4 

 and a, - 23-0 - 1-58 - 41-4 

 .-. a, = 66-0 



We have now obtained the equation to the curve in 

 the following form : 



/ (0) = 66 sin + 23-0 sin 3 + 20'7 cos 3 - 1-58 



sin 5 + 5-78 cos 5 



This may be simplified as previously described. 

 Thus : 



3 sin 3 + b 3 cos 3 0=vV+V sin { 3 + tan j ( -^ 

 = 31 sin (3 0+42) 



sin 5 0+& 5 cos 5 = Va 5 - + & 5 2 sin j 5 



= 6 sin (50+105), 

 and our equation thus becomes 



/(0) = 66 sin 0+ 31 sin (3 0+42)+6 sin (5 0+105), 

 which agrees with the expression obtained by the method 

 of summation (page 4i7). 



Exactly the same operation as just described for the 

 determination of the third and fifth harmonics was 

 carried out for the seventh, ninth, and eleventh harmonics, 

 by dividing the curve into 7, 9, and 11 parts. In each case 

 the values of the coefficients was found to be zero, or 

 practically zero, showing that these harmonics were not 

 present. 



A single check reading has been taken above to 

 confirm the result of the calculation of each harmonic. 

 It would be preferable in most cases to take sufficient 

 points to enable a curve to be plotted from them. If the 

 curve is found to be a sine curve, the readings are shown 

 to be trustworthy. If thijj is not so, a mean curve should 

 be plotted through the points obtained, 



