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/20² + 75² = 78 " [TeX: $\root{20^2 + 75^2} = 78$]
\/

The weight upon each set of braces is that due to one panel, or 3000
x 15 = 45000 lbs., half of this, or 22500 lbs., is the weight for one
truss only--and, as there is a brace under each of the 4 chord sticks,
we divide by 4, and have 5625 lbs. per stick of the brace;--now,
correcting for inclination, we shall have

20 : 25 :: 5625 : 7031 lbs.
20 : 37 :: 5625 : 10406 lbs.
20 : 49 :: 5625 : 13781 lbs.
20 : 64 :: 5625 : 18000 lbs.
20 : 78 :: 5625 : 21937 lbs.

The weights fouud show the compressional strains on the several
braces;--and, were the pieces to be proportioned for compression
only, their Scantling would be quite small--but on account
of their elasticity, they require larger dimensions.

These braces should not be fastened to the verticals,--but
should be confined both laterally and vertically, where they pass
them. The length of beam, for which we have to guard agains
flexure, is the length between verticals in any panel.

In panel No. 1, it will be 25 feet,
" " 2, " " 18 "
" " 3, " " 17 "