ii-A] 



WAVE ANALYSIS. 



337 



10. Example. Required to analyze a wave of which the fol- 

 lowing 18 orclinates at 10 intervals are known (see Fig. 2) : 



y = 18.0 y\ = 9.0 



y 5 = + 72.0 y e = + 99-0 



y 10 = + I23.O yn = + I l6.O 



y 18 = + 73.0 y w = + 57-5 



= 1.0 



= + 14.0 



= + I2 7-5 

 = + 1 08.0 y* = + 97.5 



= + 128.0 



l4 = + 85.0 



These are arranged according to the scheme of 7. 



SCHEME. 



+ 32.0 + 169.5 235-0 = 33-5 

 = + 58-5 + 131-0 250.5 = 61.0 

 = + 87.0 128 = 41.0 



= 33-5+ 41 =+ 7.5 



d*' = 18.0+ 9.0 = 9.0 



dS = 50.0 + 25.5 3.0 = 27.5 



dz' = 56.5 + 39.0 4.5 = 22.0 

 d" = 9-0 + 22.0 + 13.0 



TABLE. 



SINE COMPONENTS. 



Check : Ai A 3 + A s Ai + + An = 128.01 ; y* = 128.00. 



2 3 



