THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

15.17 Theoretical Target Strength of a Sphere 439in sea water, bending by temperature gradients (with consequent focusing andspreading out). Therefore, the actual transmission loss does not equal 20 log ralone, but includes other effects of attenuation. The function 2TL represents amore general situation than 40 log r in Equation (15.31), which is then recast as10 log I r = 10 log K + 10 log F − 2TL. (15.32)As mentioned earlier, because of the great variance of oceanographic conditions,the quantity 2TL can be established only through careful measurements. Settingtarget strength TSand the echoand source levelTS = 10 log K,E = 10 log I r ,SL = 10 log F,we obtain the first sonar equation of Table 15.4, which constitutes the fundamentaldefinition of signal strength:TS = E − SL + 2TL. (15.33)As it entails only directly measurable quantities Equation (15.33) is particularlyuseful in the computation of TS from data measured at sea.A sphere presents a perfectly symmetric target. The echoes it returns to a soundsource are completely independent of its own orientation, and it is for this reasonthat spheres make convenient experimental targets in echo-ranging measurements.In a simple derivation, we consider a plane wave of intensity I o striking a sphereof radius a and cross sectional area πa 2 . The total sound energy intercepted bythe sphere is thus πa 2 I 0 in the ideal case of perfect reflection. Now let us assumeuniform reflectivity in all directions. At a distance r from the sphere’s center theacoustic energy will be spread uniformly over the surface of a sphere of radius r orover the surface area 4πr 2 . Because the intensity I r of the reflected sound equalsthe total energy πa 2 I 0 reflected by the target sphere per unit time divided by 4πr 2over which it is distributed, then at a distance r from the sphere’s centerI r = πa2 I 04πr 2= a24r 2 I 0.But from Equation (15.28), I r = KI 0 /r 2 , where r represents the distance from thetarget to the measurement point,and soK = a24TS = 10 log K = 20 log(a/2).