Im mm Itfl 1111 Ell RUhI II fegesasesgase sosscoragagc § LIBRARY OF CONGRESS ^m Chap. ....SB.S_S.iL Shelf KA UNITED STATES OF AMERICA. SELF-INSTRUCTOR LUMBER SURVEYING-, FOE THE USE OF LUMBER MANUFACTURERS, SURVEYORS, AND TEACHERS. by / OHAELES KINSLEY, PRACTICAL SURVEYOR AND TEACHER OF SURVEYING. ASSIGNOR, JAMES KINSLEY. PUBLISHED BY THE AUTHOR. Calais, Me., and St. Stephen, N. B. 1870. Entered according to Act of Congress, in the year 1870, Br Charles Kinsley, In the Office of the Librarian of Congress, at Washington. RIVERSIDE, CAMBRIDGE: ELECTROTYPED AND PRINTED BY H. 0. HOUGHTON AND COMPANY. ^ PREFACE. This work combines the theoretical and practical parts of surveying, in such a manner as to enable the energetic and uninitiated student who applies himself to the study of this useful and interesting science for a short time, to survey all kinds of lumber with accu- racy and expertness. It contains tables for measuring boards, plank, deal, and timber by board measure, by which the Surveyor can dispense with the use of the Board Rule. It contains the rules generally adopted by Surveyors, and also a more concise rule than that in general use : for plank, deal, and timber, this rule alone is worth more than the price of the book to any Surveyor, as it requires less mental calcu- lation than by the other rules, enabling him to sur- vey faster and with less trouble than he could other- wise do. It contains tables for inch, inch and a quarter, and inch and a half boards for battens and joist. It also contains rules and tables for surveying logs by board and cubic measure, and rules for ton tim- ber. It also contains tables showing the number of feet in length, of any dimension, which will make 1,000 feet board measure or 1,000 feet cubic measure ; IV PREFACE. a new method of finding the solid contents of timber ; a rule for finding what a round log will square, by hav- ing the circumference or diameter given, or in other words, to find the inscribed square ; how to make out specifications, survey bills, etc. ; rule for measuring tapering timber ; table of quarter-girts for logs ; rule for finding how much in length, of any dimension, which will make a solid foot, or any other desired quantity; table showing the weight of twenty-five kinds of wood, with a rule for finding the weight of the same from the contents ; the English and Amer- ican Government rules for finding the tonnage of ves- sels, and rules for gauging and ullaging casks. It also contains a correct and extensive interest table. TABLE OF CONTESTS. Page Rule for measuring Rectangular Boards 7 Exercise on the Rule 7 To find the Contents of a Triangular Board 8 Exercise on the Rule 8 To find the Solidity of a Sphere or Globe 9 To find the Contents of a Triangular Solid 9 To find the Superficial Contents of a Globe 9 To find the Contents of a Circular Board 9 Rule for finding the Contents of a Circle 9 Rule for finding the Diameter or Circumference of a Circle . . 9 Table for measuring Inch Boards, without a Board Rule, from Two Inches to Thirty-six Inches wide 10 Exercise on the Table 10 Table for measuring Inch-and-a-Quarter Boards without a Board Rule 11 Exercise on the Table 11 Table for Inch-and-a-Half Boards without using the Board Rule . . 12 Exercise on the Rule 12 Table for Plank from Two to Thirty Inches wide . . . . .13 Exercise on the Table 13 Table for Three-inch Deals from Three to Twenty-four Inches wide . 14 Table for Four inch Deals from Four to Twelve Inches wide . . 15 Table for Five-inch Timber from Five to Twelve Inches wide . . .16 Table for Six-inch Timber from Six to Twelve Inches wide . . 16 Table for Seven -inch Timber from Seven to Twelve Inches wide . . 17 Table for Eight-inch Timber from Eight to Twelve Inches wide . 17 Table for Nine-inch Timber from Nine to Twelve Inches wide . . 17 Table for Ten-inch Timber from Ten to Twelve Inches wide . . 17 Table for Eleven-inch Timber from Eleven to Twelve Inches wide . 17 Table for Twelve-inch Timber from Twelve to Twenty Inches wide . 17 Plan of drawing a Plank Shingle, with Directions for dotting Plank, Deal, &c 18,19 Rule for Plank Specifications 20 How to keep a Joist or Scantling Shingle 22 New York Deals, Three-inch 23 Specification Rule for Three-inch 24 Specification Rules for Four-inch 25, 26 Five-inch Timber Shingle, with Rule for Specification . . . 28-30 VI TABLE OF CONTENTS. Rule for Six-inch Specification 31 Rule for Seven-inch Specification 33 Rule for Eight-inch Specification 35 Rule for Nine-inch Specification ........ 37 Rule for Ten-inch Specification 40 Rule for Eleven-inch Specification . 41 Rule for Twelve inch Specification 42 Rule for finding the Contents of Battens or Two-and-a-Half-inch Stuff 44 Specification of Batten Shingle and Rule 45 Random Shingle, Contents given in the Columns 45 Random Shingle, Running Lengths in the Columns .... 46 Table showing the Number of Feet in Length, of all Dimensions, that will make 1,000 feet of Board Measure 47 Table showing the Number of Feet in Length, of all Dimensions, from Five Inches by Five Inches to Twenty-two Inches by Twenty- Four Inches that will make 1,000 Cubic feet, with Rules showing how both Tables are computed 48 New Rules for finding the Contents in Cubic Feet of Timber, from Five by Five up . . 49 Second Method of making out Specification, and Rule . . .51, 52 Specification of Philadelphia Deal, and Rule 53, 54 How to use the Board R'ule, with Exercise 54 Rule for measuring Logs, with Example 55 To find the Largest Square Piece of Timber that can be sawed from a Round Log, by having the Circumference or Diameter given . 56/'"' Form of Bills of Lading 57, 58 Surveyor's Bills and Receipts 59 New Rules for finding the Superficial Contents of Plank, Deal, Joist, Battens, and Timber 59 Given the Dimensions of the End of a Plank to find what Length of it will make a Foot 61 To find the Solid Contents of a Piece of Tapering Timber ... 61 When a Board or Plank is wider at one End than the other, to find what Length of it will make a Foot, or any Desired Quantity . 62 To find how much in Length will make a Solid Foot, or any other De- sired Quantity, of Squared Timber of Equal Dimensions from End to End 62 Table for measuring Round Timber by the Quarter-girt Areas . . 63 Table and Rule for finding the Weight of Timber from a Survey of its Contents 64 English Government Rule for finding the Tonnage of Vessels . . 65 United States Government rule for finding the Tonnage of Vessels . 66 Gauging of Casks 67 . Ullaging of Casks 68 Questions for Exercise 69 Log Rule for Round Timber 72-76 Directions for using the Log Rule ....... 77 Interest Table and Rule 77-79 SELF-INSTRUCTOR ON LUMBEE SURVEYING. Rule for measuring Rectangular Boards. Multiply the length in feet by the width in inches, and divide the product by 12, to find the contents in superficial feet. Or multiply the length in inches by the width in inches, and divide by 144, the number of inches in a square foot, for the contents in superficial feet. P. S. — A Rectangle is a plain figure bounded by four straight lines, which are equal and parallel, and whose angles are right angles, as B. B. QUESTIONS FOR EXERCISE. 1. What are the contents in feet of a rectangular board 30 feet long and 20 inches wide ? Ans. 50 feet. 2. How many feet in a board 26 feet 6 inches long, 12 inches in width ? Ans. 26 J feet. 3. What will be the cost of a walnut board 32 feet long and 16 inches wide, at 8 cents per square foot. Ans. $3.41. 4. "What are the contents of a board 22 feet 8 inches long, and 1 foot 9 inches in width ? Ans. 39 feet 8 inches. When a Board is wider at one End than at the other. Rule. — Add the width of both ends together, and take half the sum for a mean width, and multiply the width thus found by the length, for the contents ; or take the width in 8 SELF-INSTRUCTOR the middle of the board and multiply by the length, for the contents. EXAMPLE. 1. What are the contents of a board 14 inches at one end and 20 inches at the other, and 24 feet in length. Ans. 34 feet. 14 _L. 20 = 34 -;- 2 = 17, mean width in inches, which multiplied by the length, 24 feet = 408; 408 -r- 12 = 34 feet = contents. 2. What are the contents of a board 26 feet long, which measures i6 inches in the middle ? Ans. 34 feet 8 inches. 26 feet X 16 = 416 ; 416 ~ 12 = 34 feet 8 inches = contents. . To find the Contents of a Triangular Board. Rule. — Multiply the length in feet by the width in inches, and take half the sum for the contents in inches, which being divided by 12 will give the contents in feet of board measure. EXAMPLE. 1. What are the contents of the board ABC, whose base B C is 26 inches, and perpendicular height A D is 18 feet. Ans. 19 feet 6 inches. 18 X 26 = 468 -r- \ = 234 -f- 12 = 19 feet 6 inches. b 2. What are the contents of the trian- gular board ABC, whose base B C is 2 feet 6 inches, and perpendicular A C, 24 feet. Ans. 30 feet. 24 feet X 2J = 60 feet ; 60 feet + 2 = 30 feet. Or— - 2 feet 6 inches = 30 inches ; 30 inches X 24 feet = 720 inches; 720 -f- 2 = 360 inches = contents; 360 -7- 12 = 30 feets = contents in feet. ON LUMBER SURVEYING. 9 The contents of a triangular solid can be found in the same manner by the foregoing rule, by multiplying the con- tents thus found by the thickness of the solid. How many feet of boards in a triangular piece of timber, ABC, whose length A B is 24 feet, breadth B C 18 inches, and thickness C E 2 feet 6 inches ? 24 feet X 18 inches = 432 ; 432 -^ 2 = 216 inches; 216 inches -f- 12 = 18 feet = contents of superficial triangle A B C, which being multiplied by the thickness, C E, 2 feet 6 inches, will give the contents of the solid triangle A B C D E F, 18 feet X 2 J feet = Ans. 45 cubic feet, or 540 board measure. For Measurement of a Globe. Rule. — To find the solidity of a globe, cube the diame- ter, and multiply the product by 5,236 ; and to find the sur- face of a globe, multiply the diameter by the circumference. To find the cir- cumference by having the diameter given, say as 7 is to 22, so is the diame- ter to the circumference, or as 22 is to 7, so is the circumference to the diameter. To find the Contents of a Circle. Rule 1. — Multiply half the circum- ference by half the diameter, for the contents. Rule 2. — Square the diameter, and multiply it by .7854 for the contents, or square the circumference, and multiply it by .07958 for the contents. P. S. — The square of a number is found by multiplying the number by itself. 10 SELF-INSTKUCTOR Table for measuring Inch Boards without a Rule, from 2 Inches to 36 Inches wide. Inches. Feet. Inches. Feet. Inches, Feet. Inches. Feet. 2X1=| 3X1 = | nXi = H 20 X 1 = if 29 X l = 2fV 12x1 = 1 2lXl = l| 30 X 1 = 2 J 4X1=| 13X1 = 1 T V 22Xl = lf 31 X 1 = 2 T 7 2 6X1 = A i4Xi = ii 23 X 1 = Hh 32 X l = 2f exi=i i5Xi = ii 24 X 1 = 2 33 X l=2j 7X1 = T 7 2 16Xl = lJ 25 X 1 = 2 T V 34X1=2| 8X1 = | 17 X 1 = IfV 26 X 1 = H 35X1 = 2H 9X1 = | 18 x i = ii 27X1=2^ 36 X 1 = 3 10X1 = 1 19 X 1 = 1 T 7 2 28 X 1 = 2 J In order to survey boards by the Table of Board Meas- ure, the Surveyor must commit the table to memory, and by a little practice, he will become expert at surveying by this method. Questions for JExercise done by the Table of Board Measure. 1. What are the contents of a board 24 feet long and 18 inches wide? Ans. 24 X 1^ = 36 feet. 2. How many feet in a board 32 feet long and 17 inches wide ? Ans. 45 J feet. By the table, 17 inches wide is lj\ the length, for the contents ; therefore 32 feet X 1-& = 45^ feet. 3. What are the contents of a board 21 feet 6 inches long and 6 inches wide ? Ans. 10 feet 9 inches. By the table, 6 inches wide is half the length, for the con- tents; therefore 21 feet 6 inches — 2 = 10 feet 9 inches == contents. 4. Required the contents of a board 36 feet long and 3 inches wide? Ans. 36 -4- 4 = 9 feet. 5. Find the contents of a board 24 feet 8 inches long and 14 inches wide ? Ans. 24 feet 8 inches X H = 28 feet 9 inches 4". ON LUMBER SURVEYING. 11 6. Required the contents of a board 27 feet long and 30 inches wide ? Arts. 67 J feet. 7. What is the value of a walnut board 23 feet 6 inches long, and 36 inches wide, @ 12J- cents per square foot? Ans. $8.81J. 8. Required the contents of a board 16 feet long and 27 inches wide ? Ans. 36 feet. 9. How many feet in a board 38 feet long and 28 inches wide ? Ans. 88 feet 8 inches. 10. Required the contents of a board 16 feet long and 19 inches in width ? Ans. 25 feet 4 inches. Table for Inch-and-a- Quarter Boards, from 2 Inches to 36 Inches wide. Inches. Feet. Inches. Feet. Inches. Feet. 2xii=A i4Xii=Hi 26Xli = 2if BXii=tV 15Xli = l T 9 ^ 27Xli = 2|f 4Xii=A i6xii=if 28Xli = 2H 5Xl± = i* i7xii==ifi 29X1± = 3A 6Xii=f 18Xli = lJ 30Xli = 3l 7X1± = H 19Xli = l|i sixii=s« 8Xii=f 20Xli = 2 T L 32Xli = 3i 9Xli = |f 2ixii=aft 33 X li ~ 3 T V ioxii=i^ 22 X l£ = 2& 34Xli = 3i| iiXiJ = iA 23 X 1± = 2|! 35Xli = 3fi i2Xii=ii 24Xl± = 2j . 36Xli = 3| 13X1± = 1« 25 X li = 2|| Examples of \\-inch Board Measure done by the Table. 1. What are the contents of a board 1J inches thick, 32 inches wide, and 30 feet long ? Ans. 100 feet. By the table 32 inches is 3^ times the length ; for the con- tents, therefore, 30 feet X H = 100 feet. 2. What are the contents of a board 1 J inches by 18 inches, and 36 feet in length? Ans. 67 feet 6 inches. 12 SELF-INSTRUCTOR 3. Required the contents of a board 1J inches by 24 inches, and 32 feet 8 inches in length? Ans. 81 feet 8 inches. 4. How many feet in a 1^-inch board 16 inches wide and 24 feet long ? Ans. 40 feet. 5. What will be the cost of a piece of mahogany 1 J inches by 12 inches, and 36 feet long, @ 6 cents per foot ? Ans. $2.70. Table for One-and-a- Half -inch Boards, from 2 to 24 Inches wide. Inches. Feet. Inches. Feet. Inches. Feet. Inches. Feet. 2Xl* = i sx4=i 1.4X4= If 20x4=4 2lX4=2f 3X4 = 1 9X4 = 4 15 x 4=4 4xii=i 5X4=* 10x4=4 16X4 = 2 22X4 = 2} nxii=i| 17X4 = 21 23X4 = 2| 24 X 4 = 3* ex4=f 12 x 4=4 isx4 = 2 i 7X4 = | 13 x 4=4 i9x4= 2 l 1. What are the contents of a lj-inch board 32 feet long and 24 inches wide ? Ans. 32 feet X 3 feet = 96 feet. 2. Required the contents of a lj-inch board 18 feet long and 18 inches wide ? Ans. 40 J- feet. 3. Find the contents of a board \\ X 10 inches and 28 feet 8 inches in length ? Ans. 35 feet 10 inches, By the table lj X 10 is 1J the length, for the contents. 28 feet 8 inches X 1J = 35 feet 10 inches. 4. What are the contents of a board 24 feet long, 20 inches wide, and 1-J inches thick? Ans. 60 feet. 5. Required the contents of a board 16 inches wide, 1J- inches thick, and 27 feet long. Ans. 54 feet. 6. What is the value of a board 17 inches wide, and 4 inches thick, and 20 feet long, at 6 cents per foot ? Ans. $2.55. * Equal three times the length, for contents. ON LUMBER SURVEYING. 13 Table for Two-inch or Plank, from 2 to 30 Inches wide. Inches. Feet. Inches. Feet. Inches. Feet. Inches. Feet. 2X2 = i 2X3 = | 2X10 = 1$ 2Xl7 = 2f 2 X 24 = 4 2XH = lf 2 X 18 = 3 2 X 25 = 4^ 2X4 = f 2 X 12 = 2 2Xl9 = 3j 2 X 26 = 4 j- 2X5 = 1 2X13 = 21- 2X20 = 3j 2 X 27 = 4^ 2X6=1 2X14 = 2j 2X15 = 21? 2X21 = 3^ 2X28 = 4§ 2X8 = lJ 2X9 = lJ 2X22 = 3f 2X29 = 4f 2X16 = 2§ 2X23 = 3f 2 X 30 = 5 EXERCISE. 1. Required the contents of a plank 18 feet long and 15 inches in width ? Ans. 45 feet. By the table 15 inches wide is 2 \ times the length, for the contents in feet of board measure; therefore 18 feet X 2^ = 45 feet. 2. Required the contents of a plank 36 feet long and 12 inches wide at one end, and 16 inches at the other end? Ans. 84 feet. 12 inches -f- 16 inches = 28 inches ; 28 inches —- 2 = mean width 14 inches. By the table 14 inches is 2^ times the length ; therefore 36 feet X 2^ = 84 feet. 3. What is the value of a plank 24 feet long and 27 inches wide @ 3 J cents per foot ? Ans. $3.92. 4. Required the contents of a plank 18 feet long and 4 inches wide ? Ans. ^ X § = ¥ = 12 feet - 5. What are the contents of 1,860 feet running lengths of 2 inches X 2 inches ? Ans. 620 feet. Solution. — 1,860 -~ J= 620 feet. 6. In 2,500 feet running lengths how many feet contents of 2 inches X 12 inches ? Ans. 5,000 feet or 5 M. 2,500 feet X 2 = 5,000 feet, or 5 M. 14 SELF-INSTRUCTOR Table for Three-inch Deals, from 3 to 24 inches wide. Inches. Feet. Inches. Feet. Inches. Feet. Inches. Feet. 3X3 = | 3X 9==2i 3X15 = 3J 3 X 20 = 5 3X4=1 3 X 10 = 2j 3 X 16 = 4 3X21 = 51 3X5=lJ 3X6=lJ 3X H = 2j 3X 17 = 4j 3 X 22 = 5j 3 X 12 = 3 3 X 18 = 4j 3X23 = 5| 3X7 = 1| 3X 13 = 3j 3X 19 = 4j 3 X 24 = 6 3X 8 = 2 3 X 14 = 3j EXERCISE. 1. What are the contents of a deal 3 inches thick, 6 inches wide, and 30 feet long ? Ans. 45 feet. By the table 3X6 is 1^- times the length, for the con- tents ; therefore 30 feet X l£ == 45 = contents. 2. What are the contents of a deal 3 inches X 12 inches and 331 feet long? Ans. 100 feet. 3. In 2,700 feet of running lengths of 3 inches X 20 inches, how many feet ? Ans. 13,500 feet. By the table 3 X 20 is 5 times the length, for the con- tents ; 2,700 X 5 = 13,500 feet. 4. Required the number of feet running lengths of 3 X 4 that will be equal to 2,000 feet running lengths of 3 inches X 10 inches ? Ans. 5,000 feet. 5. What number of feet of running lengths of 2 X 3 will be equivalent to 24,000 feet running lengths of 3 X 12 inches. Ans. 144,000 feet. Solution. — By the table 3 X 12 is 3 times the length, for the contents ; therefore 24,000 feet X 3 = 72,000 feet = contents of 3 X 12 inches, and by the table 2 X 3 is = to half the length, for the contents ; therefore 2 X 3 is 2 times the contents for the running lengths, consequently 72,000 feet X 2= 144,000 feet running length. ON LUMBER SURVEYING. 15 Table for Four-inch Deals, from 4 to 12 Inches wide. Inches. Feet. Inches. Feet. Inches. Feet. Inches. Feet. 4X4 = 1| 4X7 = 2^ 4X 9 = 3 4XH = 3f 4X5=lf 4X8 = 2f 4X10 = 31 4 X 12 = 4 4X6 = 2 EXERCISE. 1. What are the contents of a deal 4X4 inches, and 20 feet long? Ans. 26f feet. 2. What are the contents of a deal 4x5 and 24 feet long? 3. Required the contents of a deal long? Ans. 40 feet. 4X6 and 26 feet Ans. 52 feet. 4. Required the contents of a deal 4 inches X^ inches and SO feet long? Ans. 120 feet. 5. What is the value of a piece of oak 36 feet long, 4 inches thick, and 11 inches wide, @ 4^- cents per square foot? 6. In 2,800 feet of running lengths of 4 inches X 12 inches, how many feet of superficial measurement are there ? Ans. 11,200 feet. 7. How many feet running lengths of 4 inches X 12 inches deals are equal to 3,000 feet running lengths of 2 X 6 ? • Ans. 750 feet. 8. What is the amount of lumber in the following cargo, and its value @ $15.00 per M ? Surveyed from Bennett & Co., of Boston, Mass., to Ship Aurora, Capt. Jones, — 2,758 pieces 2 X 8 and 16 feet long. 3,800 pieces 4 X 12 and 30 feet long. 2,600 nieces 4 X 10 and 16 feet long. 250 M of Mer. spruce laths @ $2.50 per M. Ans. 653,497 feet of lumber. 250 M laths. Value of lumber, $9,802,451 Value of laths, 625.00 $10,427.45i 16 SELF-INSTRUCTOR Table of Five-inch Timber, from 5 to 12 Inches wide. Inches. Feet. 5X6 = 2^ 5X7 = 2H 5 X 8 = 3 J Inches. Feet. 5X 9 = 3f 5X10 = 4 5XH = 4j^ 5 X 12 = 5 Table of Six-inch Timber, from 6 to 12 Inches wide. Inches. Feet. Inches. Feet. 6X 6 = 3 6 X 10 = 5 6X7 = 3j 6 x H = 5j 6X8 = 4 6 X 12 = 6 6X9 = 4 EXERCISE. 1. What are the contents of a piece of timber 5 inches X 5 inches and 24 feet long ? Ans. 50 feet. By the table 5 X 5 is 2^ times the length, for the con- tents ; therefore 24 feet X 2 T V = ¥ X ff =W = 50 feet in board measure. 2. Required the contents of a joist 5x8 and 30 feet long? 30 feet X H = 100 feet. Ans. 100 feet. 3. Find the contents of a beam 6 inches X 8 inches and 36 feet in length ? Ans. 144 feet. 36 feet X 4 =144 feet. 4. How many running feet of 6-inch X 8-inch timber are equal to 3,500 feet running lengths of 5 X 1 2 inches ? Ans. 4,375 feet. By the table 5 X 12 is 5 times the length, for the con- tents, and 6X8 = 4 times the length ; therefore 3,500 feet X 5= 17,500 feet = contents of 5 X 12 ; then 17,500 -r- 4 =s 4,375 feet = the number of feet in length of 6 X 8 = 3,500 feet of 5 X 12. ON LUMBER SURVEYING. IT 5. What will a beam cost 48 feet long, 6 inches by 11 inches, @ 3J cents per foot ? Ans. $9.24. 48 X 5 J feet = 264 feet = contents ; 264 X H cents = $9.24. Table of Timber from 7 X 7 to 12 X 20. Seven-inch Timber. Eight-inch Timber. Nine-inch Timber. Inches. Feet. Inches. Feet. Inches. Feet. 7X 7=4^ 7X 8 = 4§ 8 X 8 = 5^ 9X 9 = 6} 8 X 9 = 6 9X 10 = 7j 7X 9 = 5^ 8 X 10 = 6§ 8 X H=7^ 9XH = 8^ 7 X 10 = 5f 9 X 12 = 9 7X11 = 6^ 8 X 12 = 8 7 X 12 = 7 Ten-inch Timber. Eleven-inch Timber. Twelve-inch Timber. Inches. • Feet. Inches. Feet. Inches. Feet. 10 X 10= 8± 10x11= 9 00 a o i— i 00 Ci © s C 00 cs o Lengths. | § X X X X X X X X Contents. 5* 7 2 5 16 2 1 LO 5 5 20 2,941 21 8 6 5 8 5 5 4 4 3,167 22 6 5 5 2 3 5 5 5 2,778 23 6 2 5 5 8 2 8 3,191 24 20 7 3 8 3 3 15 9 5,570 25 2 1 1 2 4 4 5 1,865 26 7 3 2 1 4 4 2 1 2,004 30 12 6 3 3 10 3 2 4,025 31 7 5 3 3 3 3 6 9 4,405 32 7 3 2 4 3 2 1 2,387 33 10 4 2 1 1 3 3 5 9 4,345 Total, 36,678 feet. Example, showing how to compute a 5-inch Specification. No. Br. No. Br. No. Br. No. Br. 7 X 5= 35 8 X 5 = 40 6 X 5 = 30 6 X 6 = 36 2 X 6= 12 6X 6=36 5X 6= 30 2 X 7=14 5 X 7= 35 5X 7 = 35 5 X 7 = 35 5 X 8 = 40 8 X 16 = 128 8 X 8 = 64 2X 8 = 16 5X 9 = 45 2 X 9= 18 5X 9 = 45 3 X 9 = 27 8 X 10 = 80 1 x 10= 10 5 X 10 = 50 5 X 10 = 50 2 X H = 22 5 X 11= 55 4 X 11=44 5 X H = 55 8 X 12 = 96 5 X 12 = 60 4 X 12=48 5 X 12 = 60 353 362 303 333 *20 — 2f 3= 8 J 214-2|==8| Contents, 3,167 22 -5- 2-g- = 9g- 23-=-2f = 9 T V Contents, 2,941 Contents, 2,778 Contents, 3,191 * 20 feet, the length of the pieces, divided by 2§, and the result, 8£, multiplied by 353 = 2,941 feet = contents of 20 ft pieces. Invert J£ = ft X ^ = Iff. = 8J ON LUMBER SURVEYING. Timber Shingle, Six-inch, No. 5. 31 ST D f Dimen- sions. CO X C© X CO QO X SO C5 X CO o X CO i—i X CO X CO 20 3 7 10 5 10 ... 14 18 3 21 7 4 5 9 •••• 26 22 7 6 6 5 10 11 .... 15 23 7 5 3 6 11 22 .... 15 24 3 4 4 3 3 5 25 6 5 4 10 20 9 26 7 3 4 5 4 11 ... 14 27 3 3 5 5 6 10 ... 14 28 10 3 4 4 5 6 6 29 7 6 4 5 9 3 5 30 7 4 2 6 5 5 Rule for finding the Contents of 6-inch Timber. Multiply the number of pieces or dots by the width of said pieces, and then multiply the product by half the length of one of the pieces, for the contents. 32 SELF-INSTRUCTOR What are the contents of 18 pieces of 6 X 6, and 20 feet long? 18X6 = 108; 20 h- 2 = 10,108 X 10 = 1,080 feet. By the Table 6 X 6 is three times the length for the con- tents, therefore 20 X 18 = 360 feet running length, 360 feet X 3 feet = 1,080. Am. 1,080. So we find the same result by both rules. Specification of Timber Shingle, No. 5. O f-i H <~ ' rH Lengths. f § X X X X X X X Contents. gco CO CO co CO CO CO CO 20 18 3 3 10 5 10 14 5,710 21 7 7 4 5 9 26 6,321 22 7 6 6 5 10 11 15 6.358 23 7 5 3 6 11 22 15 7,900 24 3 4 4 3 3 5 2,544 25 9 6 5 4 10 20 6,712 26 7 3 4 5 4 11 14 6,097 27 3 3 5 5 6 10 14 6,237 28 10 3 4 4 5 6 6 4,718 29 7 6 4 5 9 3 5 4,988 30 7 4 2 6 5 5 6 4,755 Total, 62,340 feet. Examples showing how to compute the Specification No. 5 of 6-inch Timber. Br. No. Br. No. Br. No. Br. No. 6X 18=108 7 X 7 = 49 6 X 7= 42 7 X 6= 42 7X 3= 21 8 X 7 = 56 7 X 6= 42 7 X 5= 35 8 X 3 = 24 9 X 4= 36 8X 6= 48 8X 3= 24 9 X 10 = 90 10 X 5 = 50 9 X 5= 45 9 X 6= 54 10 X 5 = 50 11 X 9= 99 10 X 10=100 10 X 11 = 110 11 x io = no 12 X 26 = 312 11 X 11 = 121 11 X 22 = 242 12 X 14 = 168 12 X 15=180 12 X 15 = 180 571 602 578 687 20-7-2= 10 21-7-2= 10J 22-7-2= 11 Contents, 6,358 23-7-2= llj Contents, 5,710 Contents, 6,321 Contents, 7,900 ON LUMBER SURVEYING. 33 What is the cost of a piece of pine timber 6 inches X 10 inches, and 38 feet in length @ 3 J cts. per foot ? Ans. $6.65. Solution. — Length 38 -r- 2 = 19 ; 19 X by the breadth 10 = 190 feet, contents. 190 feet @ 3J = $6.65. By the Second Rule. 6 inches X 10 inches = 5 times the length, for the contents, therefore 38X5 = 190 feet. 190 feet XH cts. = $6.65. Rule for finding the Contents of 1-inch Timber. Multiply the width by the length, divided by If. Required the contents of a piece of timber 7X7 and 20 feet long ? Divide the length, 20 feet, by If (20 -r- If = llf), and multiply the breadth, 7 inches, by the quotient, llf. 1 If = 3^ ; 3^> x 7 — - 2|& = 81 f feet = contents in su- perficial feet. 2d Operation. — By the table 7X7 is = to 4 T ^ times the length, for the contents, therefore 20 feet X 4 I J j = 81f feet = contents. Timber is often surveyed and the contents marked on each piece, and then put down on a shingle for contents in its proper column. 34 SELF-INSTRUCTOR Timber Shingle , Seven-inch. iVb. 6, erne? Specification, 13 Dimen- sions. X 00 X 16 C5 X o r-H X 5 7 X *>• i— t X 1> Contents. 20 24 7 18 8 8 16 9,321 21 9 6 8 6 5,059 22 4 4 4 4 6 8 10,062 23 9 7 7 6 18 9 27 10,062 7,420 24 21 7 8 8 25 4 5 6 6 4 3,807 26 14 5 2 6 5 7 5 1,493 27 6 4 4 5 7 4 5,197 28 8 15 4 7 8 5,390 29 7 „ 4 6 2 5,988 30 8 4 5 8 7 8 6,755 31 16 12 4 2 1 3 5,624 ON LUMBER SURVEYING. Timber Shingle, Eight-inch. No. 7. 35 a,: Lengths. g = 5* 00 X 00 X CD o X 00 pi X 00 X 00 7 26 12 5 12 9 18 27 5 3 ... 3 8 4 28 2 3 6 5 29 5 4 3 7 5 30 3 2 2 8 10 10 31 5 2 8 32 4 1 2 5 3 4 7 33 5 3 4 7 34 2 6 3 5 2 35 5 9 36 6 7 12 9 37 7 15 32 24 21 i?w/e for finding the Contents of 8 by 8 Timber. Divide the length by 1J, and multiply the quotient by the width of the timber for the contents in feet of boajd measure. 36 SELF-INSTRUCTOR Example showing how the first column of 8-inch specifi- cation is done. Br. No. pieces each 26 feet long. 8X 12= 96 9 X 18 = 162 10 X 12 = 120 11 X 9= 99 12 X 7= 84 561 26-5-H= 14 26 11 3 x 2 2* Invert the divisor, fX 2 T 6 =¥ = 14. 7854 feet= contents. Specification Shingle, Eight-inch. No. 7. S - • Lengths. | § X X X X Contents. 5 " « OS OS C5 26 6 14 6 5 6,240 27 18 12 5 5 8,039 28 2 3 4 2 2,436 29 4 3 2 4 2,958 30 6 2 5 3,195 31 8 3 4 4 4,510 32 5 4 2 15 6,888 33 2 1 5 3 2,945 34 2 2 4 3 3,009 35 2 4 5 5 4,541 36 6 2 1 3 3,267 Contents, 48,028 feet. Specification of Timber Shingli ?, Ten-inch. No. 9. a k Lengths. | § o X i— i X X Contents. a* o o o 5 26 36 13 12,198 27 4 5 4 3,297 28 11 5 11 6,930 29 8 3 4 3,891 30 4 6 4 3,850 31 6 2 1 2,428 32 27 3 3 9,040 33 5 5 5 4,587 34 3 2 6 3,513 35 1 2 5 2,683 36 12 25 24 20,490 Contents, 72,907 feet. 40 SELF-INSTRUCTOR Rule for Ten-inch Timber. Divide the length by 1£ and multiply the quotient by the breadth, for the contents in feet of board measure. Required the contents of a stick 36 feet long 10 inches by 11 inches? 36-^-1^ = 30, and 30 X 11 = 330 feet = contents. 2d Solution. — By the table 10X H is 9£ times the length, for the contents ; therefore, 36 feet X $k — 330 feet = contents. Examples showing how 9 and 10 inch specifications are made out. Nine-inch. Br. Pieces. Pro. 9 X 6= 54 10 X 14 = 140 11 X 6= 66 12 X 5= 60 320 26~li= 191 2880 320 Ten-inch. Br. Pieces. Pro. 10 X 36 = 360 11 X 13 = 143 12 X 5= 60 563 563 2 21\ 3)1126 563 1126 375 375 12,198 feet Contents = 6240 Length, 26 -f- 1£ ; If = Length, 26 -h 1 J ; 1 i = f = inverted to g ; £ X ¥ i . Inverted = f ; j X ¥ = H~ = 21f . = ¥=19*. ' P. S. — All the specifications in this book are done in a manner similar to the specification of the Plank Shingle .No. 1. ON LUMBER SURVEYING. 41 Eleven-inch Shingle No. 10. Dimen- sions. X i—i X 20 24 36 21 6 4 22 ' 3 9 23 4 2 24 5 1 25 5 3 26 1 2 27 5 3 28 5 2 29 6 4 30 5 2 d o © CV. © CO r s^ © -t-» r£3 u o & .d (M rH 5^ 'S X i-C © § rO r-l CM 03 'S *s +3 § ^2 d — - O *> © ^ O J3 e» V cr 1 o *; ««9 5a 1!! s 13 .2* S :43 © ° o § ^ H3 d c3 •all <3 s^ «r-i rH J? -m i—i © GO rd I— t © © 8 II © d 2* > •-* A X js + © O 42 SELF-INSTRUCTOR Timber Shingle, Twelve-inch, No. 11. Lengths. g § X X CO i—i X 00 X I— t o X 20 4 4 25 16 16 21 3 2 1 1 2 22 4' 1 1 2 2 3 23 2 2 1 24 8 3 8 1 2 25 4 1 1 3 1 26 1 2 2 1 4 27 2 3 3 2 3 28 1 2 3 4 4 29 4 3 1 1 2 30 2 2 3 2 5 i?w/e /or Twelve-inch Timber. Multiply the length by the width for the contents in feet. Required, the contents of 16 pieces of 12 X 20 inch timber, and 20 feet long ? 16 X 20 = 320. 320 X 20 = 6,400 feet = contents in feet of board measure. ON LUMBER SURVEYING. Specification of Shingle No. 10. 43 i« Lengths, s § 11 X 11 11 X 12 Contents. 5 » 20 24 36 12,760 21 6 4 2,194 22 9 3 2.722 23 4 2 1,434 24 5 1 1,774 25 5 3 2,085 26 1 2 834 27 5 3 2,252 28 5 2 2,028 29 6 4 3,030 30 5 2 2,272 Total, 33,285 Specification of Shingle No. 11. 00 a © I— ( r-l A 3 *A H 6§ 10 Total . 42,673 ft. 1700 697 3140 2970 1052 1.710 3525 2740 1065 9420 10500 L133 348 1046 44 427 57 53 1046 2833 1045 4186 1096 2137 2797 1118 10466 ON LUMBER SURVEYING. 47 oooooooooooooooooooooooo oooooooooooooooooooooooo oooooooooooooooooooooooo No. of feet in Length = to 1,000 feet of Contents. _Ji-i _>-< "sH it- lO|f r-<|<» ~rH *-|o»r-<(coWl|eO0C^aae^|O9 »h|M*H»-<*4« |i- , H««>l l = ! H6» l«- ooiooeooo©i^cos r H|3» HrH H^rH:-,H<»e3|t-,-<|e*0|ca O O t-(iO(Mt-?OOcr.iCCOCOi-HrHOC50asOCCi— iG^iOOOO NLTTfOOCOiOCOINiiOJINr! 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OS CN CN i— i m CO CO O CN CO CO co cn m os m o CO 00 o CO CO"* O ■<* CC CN CO r— CO CN HCO^CD ^f ** "* T* i g OlflCOON'C* (M N (M N rH CO O — < CO -* CO N m cn o m CM CO CN CN O N -* 00 o, o CN CO oo m cn CN t» CN in co oo CO CO CO OS CO CD O — ■ CN CN i— i in os CO OS o cn co m CN CN CN CO CO CN CO cn co m co CC N 00 OJ CN CN CN CN O H- OS CN CO OS O CM OrHCO-* CO CO CO CO COOOOOlOCOfO-tCMOCOCO-^fMOCOcOTfOlOCOcOTfCMO Tinot-oiooooo)OHC'iwn' rf| incDcoNODa>ooH(McoTji co CONGO CO O i- ilMijiiniflNOSClOi- ifMCO^inNOOOiOi- 1 Co H rtHHHHHHHCN(NIM(NM(N(N(SWCOCOm !NT|i©ooO(M't(Oooo(M'*cooooninMnNaHnmN ^^^ Jr _,r>icNoir>iCNcococococO'+'*-*-*TtncOCNOOSNcO - '+CO'— i OCONCO % co OK3c0N00fflC)HH(N«'tmtCNN000)OH(NWM'l | in OOOOOOOOOOOOr- i CN CO Tt< in — i CN CN CO -* m CO t>- CO 0CtO-rflMO00IOitiMO0tltO-tlMO00)(M(NN(N co n co as o ON LUMBEE SUEVEYING. 77 How to use the Log or Timber Rule. If the timber is tapering, the girt should be taken about one third the distance from the larger to the smaller end. Some take the girt in the middle. Girt the log to be meas- ured, and take the quarter of it, and measure the length of the log. Then look along the top of the table till you come to the corresponding quarter girt ; then run down the column underneath the quarter girt till you get opposite the length, - where you will find the contents. Or, you can find the con- tents by taking the diameter of the small end and the length. Then find the corresponding diameter at the foot of the table, and ascend the line perpendicularly till you come opposite the length, where you will find the contents. P. S. — This table allows one fourth of the true contents of the log for bark, saw kerf, and waste slab. It has been extensively used by timber merchants, and is just about as fair a rule to go by as any I have seen. There are many allowances to be made which are left to the scaler's judg- ment, and for which it would be almost impossible to make due allowance in the table. INTEREST. Rule for finding the interest at 6 per cent. — Multiply the sum by the number of days, divide the product by 6, then strike off the right-hand figure. EXAMPLE. $200 12 days. 6)2400 400 = 40 cents is the interest. 78 SELF-INSTRUCTOR. M Q. < *3 Q 8 c • |H g£ e r- 3 H £p tf O q h "« s P3 e ^ pq <1 £ H g o H Q r/> *> H J* Pi ES H Q H 5 S i— i o eo 1 CM - rHi-lrHOO <£ OrinnuN^w - *-r ,_< ,_, ,_, ^q eg M co CO L~ O t*i CO CO rHr-ir-icO - <& 1 ©T^^r-^CMClCOCOCOCO^^CMgCOCD^CMOC* r-JrHN to g£ Ot-lrHiHTHINNWeNOS^OOrHcgjDOg^^gegWgggfg ira o 1 m 1 r-1 oi t* j w 1 €£ 1 OOrMrHrHrMCNlSS|cNC G» r-1 S o 6© OOrlHHiHrtrtNM^iOl-OJHMTtiOODm^NOH I— ItHi— lr-lT-HCOiOt~C»CO o 6 OOHHrlHriHrtlMC0i0t-C0O'S)Mi0^C001©IMTri r-ir-HrHT-irHCOTfrlCOCOCO r-i O 1£ OOOHHHHHrtHOJ^mi^OSOMCOiOO^ffi^OO CO OOOr-IHHHrtHHM^iOl-ooOHIMWtPrBCO'XilM th r-i th cq co >o cd co t- OOOHHHHHHHIMCCIlO«Ot--affiO(MMO;DoniO |H tH CM CO ■>* l« r-j CO 6 OOOOOHHHHHNCOiHOfflt-COfflOOOOffla T-H (M CO •* Tjt 05 U3 6 OOOOOOHHHHNNM^iOfflt-NOOffliOMHN tH CM CO ■* CO tH 6 OOOOOOOHHHHNMM^iOiO^l-MOCOmtO i— t CM C-l CO CO CO 6 OOOOOOOOOOH(MNNM»^^iOOin~iPOi t— 1 1— 1 CM CM ■* CI 6 O O O O O O O O O O tH rH rH CM CM CM CO CO CO t- O CO CO CO HH •— ' CO I-H O OOOOOOOOOOOOHHriHHHCqMiOStOM 00 1 CO O A H«BTi(iij»hooaiooooooooooooeoo rlNOJ^iOtONOOOOOOOOO rlMMttHOO First Example at Qper cent. — Required, 50 days interest on $100. Interest on $100 for 30 days = 49 cents. Interest on $100 for 20 days= 33 " Ans. 82 cents. INTEREST. 79 ©3 L- O •* CO rHlOOOCflCOIMOOTn O W M CO -tfi O O O S O O r^iHi^eqNCQedc6t^o^odi^u5c»c0 IIMCO^iOO HHrlNflMM^TjuociSCUCiOCCCCCCOO OrHrHC'lC^COeO-^-^iOO-^Ol-^CiCCcrcoCCiCCO^^OOL^ OrHrHC^COCO^^C^OOMCCg^^OglCC^O© r-ti-KM-. o o C £ fe » ^ o ° s o — I ^ "=> l±i +j e a o « s 99 o m 11 ' = s% O B C3 || 2 CD* X o "*° >> | ^J o S>S,0 S3 p g S 3 *^ +J ,fi r> 5 t« M a) a, g W H & « '> ^ *X ■■§ cJD — II S ® > «V^ ? "s ^ aj ft? £ a> r-4 CM CO ^ tO < /Second Example at 6 ;?er cercf. — Required, the interest of $50 for 3 years, 2 months, and 10 days. Interest on $50 for 3 years = $9.00 Interest on $50 for 2 mos. = 50 Interest on $50 for 10 days = 8 Arts. $9.58 WANTED. Agents to sell this Book, throughout the United States, the Dominion of Canada, California, and Oregon. Exclu- sive territory given. Good inducements to agents. The book will be sent to any address, free of postage, on receipt of Two Dollars. Send Post-Office orders, or by express. Address CHARLES KINSLEY, Calais, Me., or St. Stephen, N. B. NOTE. All Lumber Manufacturers, Lumber Dealers, Millmen, Carpenters, Carriage Makers, Shipbuilders, Cabinet Mak- ers, Ship Brokers, Ship Carpenters, Railroad Conductors, Engineers, Machinists, Freight Agents, Teachers, Students, Architects, Merchants, Accountants, and others, will find it to their advantage to procure a copy of this book, as the knowledge it imparts may save them in a few years' prac- tice hundreds of dollars. The book contains twelve new rules for finding the superficial contents of lumber, which do the same work as one hundred and fifty of the rules generally used. CHARLES KINSLEY. VfliiN' H