of Binaural Audition. 195 



cross sections are situated. Let us call the angles included 

 between that plane and the surfaces f x and f 2 respectively $1 

 and <f> 2 . We know also that 90° — w x =(f> u and 90°— iv 2 = cf) 2 ; 

 and hence the areas of the cross sections f l and/^ are respec- 

 tively/! cos </> 1? and/ 2 cos <f> 2 . 



The remainder of the calculation follows the course thus in- 

 dicated. From figure 6 we learn that 



MM" = Wn = aM sin w 2 , 



, . Wn 



and smw 2 =-iiF' 



fll 



Further, since angle A!an=fi, 



M'n=aMsin(/3-«). 

 But 



aM= aW 



cos 7 ? 

 whence 



flM'sinQg— ») 



sin w 2 = a W ' 



cosy 

 and 



sin w 2 — sin (<3 — a) cos 7. 



Here w 2 is the angle included between the direction of the 

 rays of sound and the surface f 2 , since that angle is, as we 

 know, also the angle which that direction includes with the 

 projection of the surface f 2 . 



Now imagine the surface f 2 rotated about the straight line 

 ad to the right until the line ac forms a continuation of ba. 

 The angle (3 which this surface included with the line of sight 

 is thereby changed into the angle (180— /3), while the surface/2 

 has changed into the position of the surface f lt Hence, in 

 order to find the angle u^, which is included between the 

 direction of the rays of sound and the surface / 1? we have 

 merely to replace ft in the preceding formulae by the value 

 (180°-/3). 



We have therefore 



sin Wi= sin [180° — /3— a] cos 7, 

 = sin (et + /3) cos 7. 



Let (/>! and cf> 2 be the angles which are respectively included 

 between the surfaces / x and f 2 and the plane which we have 

 taken normally to the direction of the rays of sound ; then, 



