To obtain the number of board feet in any piece of timber, the length, in inches, should be multiplied by the end area, in square inches, and the result divided by 144. For example, the number of feet B. M. in a floor joist 20 feet long, 3 inches thick, and 10 inches deep is 240 inches (=20 feet × 12) multiplied by 30 square inches (the end area) divided by 144, or 50. The following rule is used by most contractors and lumber dealers: _Multiply the length in feet by the thickness and width in inches, and divide the product by 12._ Thus, a scantling 26 feet long, 2 inches thick, and 6 inches wide contains 26 × 2 × 6 ---------- = 26 feet B. M. 12 This rule, expressed in a slightly different manner, is more convenient for mental computation: _Divide the product of the width and thickness in inches by 12, and multiply the quotient by the length in feet._ Thus, a 2" × 10" plank, 18 feet long, contains 2 × 10 ------- = 30 feet B. M. 12 × 18 =44. Prices of Lumber.=—Owing to the continual variation in the prices and grades of lumber, it is impossible to give prices here that will not vary from day to day. The architect before starting to estimate should first be sure that he has the latest lumber prices obtainable. These prices can always be secured from the local lumber dealer. =45. Studs.=—To calculate the number of =studs=—set on 16-inch centers—the following rule may be used: _From the length of the partition, in feet, deduct one-fourth, and to this result add 1. Count the number of returns, or corners, on the plan, where double studding is required, and add 2 studs for each such return._ (The reason for adding 1 is to include the stud at the end, which would otherwise be omitted.) The sills, plates, and double studs must be measured separately. [Illustration: FIG. 4] For example, the number of studs required for partitions only, shown on the plan, Fig. 4, is computed in the following manner. 30 ft. 6 in. 10 ft. 6 in. 9 ft. 6 in. 5 ft. 0 in. 4 ft. 6 in. ------------ 60 ft. 0 in. Deducting one-quarter from 60 feet, the remainder is 45 feet; adding 1 stud, the result is 46 feet. As there are 4 returns, with 2 studs for each, as shown at _a a_, the total number is 46 + (4 × 2) = 54 studs. As a general rule, when (as is customary) the studs are set at _16-inch_ centers, _1 stud for each foot_ in length of partition will be a sufficient allowance to include sills, plates, and double studs. Thus, if the total length of partitions is 75 feet, 75 studs will be sufficient for sills, double studs, etc. If the studs are set at _12-inch_ centers, the number required will be equal to the _number of feet in length of partition plus one-fourth_. Thus, if the length of partitions is 72 feet, 72 + 18, or 90, studs will include those required for sills, plates, etc.