SCALING LOGS 43 



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Pif.ch Ring, Pitch Seam and Rot Tables. — These tables have been com- 

 puted by use of the defect formulae explained in the first part of this problem. 

 In the pitch ring table the first column gives the diameter of the ring, the 

 figures below "Waste Thickness" give the inches waste necessary to saw out 

 the ring and the figures under columns 3, 4, 5 show the number of feet B.M. 

 of defect per lineal foot. The pitch seam table similarly shows the defect 

 for different waste thicknesses and widths. The rot table shows the 

 defect for different average diameters. In each case the defect per lineal 

 foot should be multiphed by the total length of the defect and the result 

 subtracted from the full scale. 



Example: If a log has a pitch ring 20 inches in diameter and a waste 

 thickness of 4 inches, the defect in board feet per lineal foot wiU be found in 

 the column headed "4" under "Pitch Rings" opposite 20 in the "Diam." 

 column, which in this case is 16 board feet. Multiply this defect per lineal 

 foot by the distance the pitch ring extends into log and the result will be the 

 total number of feet B.M. defect caused by the ring. 



Example: If a log has a pitch seam 16 inches in width across the end of 

 the log and a waste thickness of 2 inches the defect in board feet per lineal 

 foot will be indicated in the column headed "2" under "Pitch Seam" oppo- 

 site 16 in the "Diam." column, which in this case is 2 board feet. Multiply 

 this defect per lineal foot by the distance the seam extends into the log, and 

 the result will be the total number of feet B.M. defect caused by the seam. 



Example: If a log has a uniform rot 16 inches in diameter, the defect in 

 board feet per lineal foot will be indicated in the column headed "Rot" 

 opposite 16 in the "Diam." column, which in this case is 17 board feet. 

 Multiply this defect per lineal foot by the distance the rot extends into the 

 log and the result will be the total number of feet B.M. defect caused by the 

 rot. 



Shingle Bolt Table. — By means of the formula for securing the contents 

 of shingle bolts the table above has been computed to show the contents in 

 board feet of bolts 52 inches long with triangular ends 18X18X18, 

 16X16X16, 14X14X14, and 12X12X12 inches. Under each column 

 heading giving the size of the bolt is showm a figure representing the number 

 of bolts of that size contained in a cord and in the table below is shown the 

 board foot contents of any number of bolts from one to ten inclusive. In 

 the last column under "length" is shown the length in feet corresponding 

 to the number of bolts indicated in the first column. 



In using this table for scaling slabs the scaler should first ascertain what 

 size bolts the cross section of the piece is equivalent to and then estimate 

 the number of that size bolts contained in the piece. The number of board 

 feet in the piece may then be secured by referring to the shingle bolt table. 



Should the cross section of the slab be larger than an 18X18X18 bolt it 

 may be divided into bolts of two sizes {i.e., 18X18X18 and 12X12X12, etc.), 

 and the number of such bolts and their equivalent board-foot contents 

 secured. 



