152 The Farmstead 



If the square is laid on 12 times at 9 and 6 

 inches, it will give the length of the rafter, for 

 12 times 9 is 108, half the width of the build- 

 ing, and 12 times 6 is 72, the height of the 

 peak above the plates. If the square is laid on 

 18x12 inches, the proportion is preserved, and 

 hence the angles ; the square would only have 

 to be laid on six times. 



Consider a building 20 feet broad and 6 inches 

 above one -third pitch. The half of 20 feet 

 equals 10 feet, or 120 inches. Seven feet 2 

 inches (86 inches) is the height of the peak 

 above the plate. It is quickly seen that this 

 problem, like the other, can be solved in more 

 than one way. If the long end of the square is 

 laid on at 20 inches and the short end at 14% 

 inches, and this is repeated six times, both the 

 bevels and the length will be secured (Fig. 64), 

 for 6 multiplied by 20 equals 120 inches, half 

 the width of the building, and 6 multiplied by 

 14% equals 86 inches, the height of the peak. 

 Or the long end of the square might be laid on 

 at 24 and the short end at 15i five times, but 

 squares are not marked in fifths of inches, hence 

 the previous method would be best.* The same 

 results would be reached by laying the square 

 on at 15 and 10% inches ; eight steps would 



* Since the square is laid on, see Figs. 61, 62, in the same manner as for 

 cutting a stair ; each one of these spaces is called a " step." 



