12,500 : 15,000 :: 1 : d

where 1 is the resisting length of the plate at right angles to tension, and d, the sum of rivet diameters. Thus suppose we have a plate 13.2 inches wide, to be fastened with nine rivets of 0.8 inch diameter; we have

9 × 0.8 = 7.2 = d,

and the above proportion becomes

15,000 : 12,000 :: 7.2 to 6 inches,

which is the length of plate section at right angles to tension. As there are nine rivets, there will be eight spaces between them, and one space at each edge of the plate, half as large as those between; or reducing all to the same size,

8 × 2 = 16, 16 + 2 = 18;

and as the whole plate section after punching is six inches,

6
18 = .33 or ⅓ inch

for the edge space, and two thirds inch between rivets. Proceeding thus, the result compares with the practice of Mr. Fairbairn as follows:—