Example: Let L (transect length) = 2,034 nm ^ 

 Bight area = 25,000 nm 

 % surveyed =7.8 



N = 350 



X = Bight-wide population 



•. 7.8 : 350 :: 100 : x 

 7.8x = 35,000 

 X = 4,487 



This simplistic ratio yields population estimates that are in 

 significant agreement with figures obtained using either 

 formula #2 or #3. Its usefulness may also be extended to 

 smaller geographic zones where the assumed range of a species 

 may be less than the entire study area. 



Formula #5 . Gates I; 



N - 1 = density 

 2 L • 7 



where N = number of animals observed; 



where L = linear distance flown in nm, multipled by 2 if both sides of 

 the aircraft are manned. 



where x= the mean of right angle sightings distances, obtained from 

 records kept at the right angle distance to each sighting from transect 

 1 i ne. 



Example: Let N = 350 



L = 2,034 nm 



x= 0.25 nm 



350-1 = 0.34 animal s/nm 



2 • 2,034 • 0.25 



The above d = 0.34, when extrapolated to the area of the SCB 

 (25,000 nm ), yields an estimated abundance of 8,500 animals. 

 These figures for d and Bight-wide abundance are approximately 

 double those obtained in formulae #2, #3, or #4. 



Formula #5, utilizing the mean right-angle sighting distance, appears 

 to artifically inflate the animal density number by drawing all 

 sightings into a very narrow corridor, in this case 0.25 nm wide. 



General remarks on data analysis 



Throughout the remainder of this report and within the individual 

 species accounts, animal density and area-wide population estimates are 



i-73 rZH, 





