rate. 4 Shaw: Microtechnical Methods 255 



This measurement of the periphery was reduced to terms of 

 average intercellular distance by multiplying it by the number of 

 intercellular distances in the selected row. From this measure- 

 ment of the periphery^27rr he calculated r and then 4irr 2 . He 

 thus obtained the area of the spheroidal surface in terms of the 

 average area occupied by a single cell assuming the latter area 

 to be equal to the square of the intercellular distance. 



Janet 6 has recently applied to the above method the assump- 

 tion, more in accord with fact, that each cell occupies a hexagonal 

 area. Taking e the average intercellular distance, and d the 

 mean diameter of the sphere, he gives the formula 7 for the 

 total number, N, of cells as : 



N = 3.627 (<*) 2 . 



He also gives the formula for the calculation of the number of 

 cells, N, from the number, n, of cells counted in the great circle 

 which forms the visible contour of the median optical section. 

 Based on the assumption that each cell occupies a hexagonal area 

 of the spherical surface, the formula 8 is : 



N = 0.367 n\ 



8 Janet, C, Le Volvox. Ducourtieux et Gout, Limoges (1912), 28. 



T This formula may be derived from those for the area of the surface 

 of a sphere in which A is the area, r the radius, and d the diameter: 



A = 4T 2 = ttcP. 



Since the area of a hexagon having a diameter of unity is equal to 

 the sine of 60°, which is 0.86603, the number, N, of hexagons of unity 

 diameter in the spherical surface is: 



A ird 2 r 



N = = = d 2 = 3.627 d 2 . 



0.86603 0.86603 0.86603 



The coefficient in this formula is, then, ir divided by the sine of 60° ; and 

 d over e is the diameter of the sphere in terms of the average diameter of 

 the area occupied by a single cell. 



8 This formula may be derived from that for obtaining the area of the 

 surface of a sphere from the circumference of a great circle. A being that 

 area, and c the circumference: 



~fc) 



c 2 1 



A = T ( ) = 7T - = - = - C 2 . 



Taking account of the fact that the area of a hexagon having a diameter 

 of unity is 0.86603, the formula for the number, N, is : 



l ° 2 l 



A IT IT 



N = = = c 2 = 0.367 c*. 



0.86603 0.86603 0.86603 



The coefficient in this formula is, then, the reciprocal of w divided by 

 the sine of 60°; and n is the circumference of a great circle in terms of 

 the average diameter of the area occupied by a single cell. 



