TARGET STRENGTH OF A SPHERE 



349 



Now assume that this sound energy is reflected uni- 

 formly in all directions. At a distance r from the 

 center of the sphere, it will be spread uniformly over 

 the surface of a spiiere of radius r or over the surface 

 area Att^. Since the intensity h of the reflected sound 

 equals the total energy irA-U reflected by the target 

 sphere per unit time, divided by the area 4irr^ over 

 which it is distributed, then at a distance r from the 

 sphere 



^' ~ 4,rr^^" - 4/"- 



But, from equation (1) 



Ir = k- 



(7) 



(8) 



where r is the distance from the target to the point 

 where the echo is measured. Therefore by substitu- 

 tion 



A' 



and 



k = 



r = 10 log A; = 20 log 



(f). 



(9) 



(10) 



where T is the target strength and A the radius of the 

 sphere. With the yard chosen as the unit of length, 

 A becomes the radius of the sphere in yards, and from 

 equation (10) it is evident that the target strength is 

 the echo level of the target in decibels above the echo 

 level from a sphere 2 yd in radius. Target strengths 

 for spheres of various radii are shown in Figure 2. 



19.2.2 



Rigorous Derivation 



This derivation explicitly assumed that sound is 

 reflected from a sphere uniformly in all directions. To 

 justify this assumption, consider the same sphere of 

 radius A in Figure 3. Two adjacent rays x and y 

 separated by a distance dz are traveling parallel to 

 00' and strike the sphere at X and Y respectively, 

 making angles of 6 and 6 + dd with the sphere radii 

 drawn to the points X and Y. From Figure 4, 



dz = XY cosd = A cos ddd. 



(11) 



Now rotate rays x and y about 00'. These rays 

 will describe circular cylinders and dz will generate 

 an area ds between them, where 



ds = 2irXWdz = 2iT {A sin 6) {A cos BdO) (12) 

 or ds = 2irA'^sm.dcosede. (13) 



The total energy dJ striking the sphere at angles 

 between 6 and 6 -\- d6 will be the product of the in- 



w= » -f 222X 



f— z— ( 



ECHO 



»i=x + 2z = 3x = 32 



H 



i 



X 



i 



PULSE 



I— iH 



ECHO 



-AxW- c h i 



w = X + 2zs2z 



Figure 4. Effect of pulse length on target strength, 

 echo length, and echo structure. 



tensity la of the incident sound and the cross-sec- 

 tional area ds, or 



dJ = lads = 2TrA^h sin 8 cos Bde. (14) 



Now consider the reflected rays x' and y' making 

 angles of 2d and 2d + 2de with 00' in Figure 3. At a 

 distance r from the center of the sphere, x' and y' will 

 be separated by a distance dZ. At a distance much 

 larger than the radius of the sphere, dZ becomes much 

 greater than XY; and x' and y' may be replaced by r. 



dZ = r 2(dd) = 2rde. (15) 



Again rotate the rays about 00', and dZ will generate 

 the area dS between them, where 



dS = 2TrPQdZ = 2w(r sin 261) {2rde) (16) 



