XII TIMBER MEASURING

PROBLEM I

To find the area or superficial content of a bourd or plank.

Multiply the length by the mean breadth.

Note. When the board is tapering, add the breadths at the two ends toge ther, and take half the sum for the mean breadth.

BY THE SLIDING RULE

Set 12 on B on the breadth in inches on A; then against the length in feet on B is the content on A, in feet and fractional parts.

Ex. II.

EXAMPLES.

Ex. 1.—What is the value of a plank, at låd. per foot, whose length is 12 feet 6 inches, and mean breadth 11 inches? Ans. 1s. 5d. —Required the content of a board, whose length is 11 feet 2 inches, and breadth 1 foot 10 inches. Ans. 20 feet, 5 inches, 8". Ex. III.—What is the value of a plank, which is 12 feet 9 inches long, and 1 foot 3 inches broad, at 24d. a foot? Ans. 3s. 33d.

Ex. iv.—Required the value of 5 oaken planks at 3d. per foot, each of them being 17 feet long, and their several breadths as follows; namely, two of 13 inches in the middle, one of 14 inches in the middle, and the two remaining ones, each 18 inches at the broader end, and 114 at the narrower.

Ans. £1, 5s. 94d.

PROBLEM II.

To find the solid content of squared or four-sided timber. Multiply the mean breadth by the mean thickness, and the product again by the length, and the last product will give the content.

As length: 12 or 10 :: quarter girt : solidity.

That is, as the length in feet on C, is to 12 on D when the quarter girt is in inches, or to 10 on D when it is in tenths of feet; so is the quarter girt on D, to the content on C.

Note 1.-If the tree taper regularly from the one end to the other; either take the mean breadth and thickness in the middle, or take the dimensions at the two ends, and half their sum will be the mean dimensions.

Note 2.-If the piece do not taper regularly, but is unequally thick in some parts and small in others, take several different dimensions, add them all together, and divide their sum by the number of them, for the mean dimensions.

Note 3.-The quarter girt is a geometrical mean proportional between the mean breadth and thickness, that is the square root of their product. Sometimes unskilful measurers use the arithmetical mean instead of it, that is, half their sum; but this is always attended with error, and the more so as the breadth and depth differ the more from each other.

EXAMPLES.

Ex. 1.-The length of a piece of timber is 18 feet 6 inches, the breadths at the greater and less end 1 foot 6 inches and 1 foot inches, and the thickness

at the greater and less end 1 foot 3 inches and 1 foot: required the solid content? Ans. 28 feet 7 inches. Ex. 11.-What is the content of the piece of timber whose length is 24 feet, and the mean breadth and thickness each 1.04 feet? Ans. 26 feet. Ex. ш-Required the content of a piece of timber, whose length is 20.38 feet, and its ends unequal squares, the side of the greater being 19, and the side of the less 97? Ans. 29-7562 feet.

Ex. IV. Required the content of the piece of timber whose length is 27.36 feet, -at the greater end the breadth is 178, and thickness 1.23; and at the less end the breadth is 104, and thickness 0·91? Ans. 41.278 feet.

PROBLEM III.

To find the solidity of round or unsquared timber.

Multiply the square of the quarter girt, or of of the mean circumference, by the length, for the content.

BY THE SLIDING RULE.

As the length upon C: 12 or 10 upon D ::

quarter girt, in 12ths or 10ths on D: content on C.

Note 1.-When the tree is tapering, take the mean dimensions as in the former problems, either by girting it in the middle, for the mean girt, or at thè two ends, and take half the sum of the two. But when the tree is very irregular, divide it into several lengths, and find the content of each part separately.

Note 2.—This rule, which is commonly used, gives the answer about less than the true quantity in the tree, or nearly what the quantity would be after the tree is hewed square in the usual way; so that it seems intended to make an allowance for the squaring of the tree.

EXAMPLES.

Ex. I.-A piece of round timber being 9 feet 6 inches long, and its mean quarter girt 42 inches; what is the content? Ans. 116 feet.

Ex. II.-The length of a tree is 24 feet, its girt at the thicker end 14 feet, and at the smaller end 2 feet; required the content? Ans. 96 feet.

Ex. III. -What is the content of a tree, whose mean girt is 3.15 feet, and length 14 feet 6 inches? Ans. 8-9922 feet.

Ex. IV.—Required the content of a tree, whose length is 174 feet, which girts in five different places as follows; namely, in the first place 9-43 feet, in the second 7.92, in the third 6-15, in the fourth 4-74, and in the fifth 3.16?

Ans. 42-519525

PRACTICAL QUESTIONS IN

MENSURATION.

1. A plank is 14 feet 3 inches long, and I would have just a square yard slit off it; at what distance from the edge must the line be struck?

Ans. 7 inches.

2. A wooden trough cost 3s. 2d. painting within, at 6d. per yard; the length of it is 102 inches, and the depth 21 inches; what is the width?

Ans. 274 inches.

3. The paving of a triangular court, at 18d. per foot, came to £100; the longest of the three sides is 88 feet; required the sum of the other two equal sides? Ans. 106-85 feet.

4. What is the side of that equilateral triangle whose area cost as much paving at 8d. a foot, as the palisading the three sides did at a guinea a yard? Ans. 72.746 feet.

5. Let a, b, c be the sides of a triangle respectively opposite to the angles A, B, C; then will the area of the triangle ABC be

a2 sin. B sin. C cosec. A.

6. Let a, b, c be the three sides of a triangle; put h=b+c and k=b—c; then will the area of the triangle be=√(h2—a2) (a2—k2).

7. Let the three sides be a, √b, √c; then prove that the area of the triangle is

1/2(ab+be+ca)—(a2+b2+c2).

8. A beam is 8 inches deep and 3 inches broad; what is the depth of another twice as large, which is 44 inches broad? Ans. 12.5263 inches.

9. Supposing the expense of paving a semicircular plot, at 2s. 4d. per foot, come to £10; what is its diameter ? Ans. 14.7737 feet.

10. Two sides of an obtuse angled triangle are 20 and 40 poles; required the third side, that the triangle may contain just an acre of land?

Ans. 58.876, or 23.099.

11. A circular fish-pond is to be made in a garden that shall enclose just half an acre; what must be the length of the chord that strikes the circle ? Ans. 27 yards.

12. Having a rectangular marble slab, 58 inches by 27, I would have a square foot cut off parallel to the shorter edge; I would then have the like quantity divided from the remainder, parallel to the longer side; and this alternately

repeated till there shall not be the quantity of a foòt left; what will be the dimensions of the remaining piece? * Ans. 20-702 inches by 6.086.

13. If a round pillar, 7 inches across, have 4 feet of stone in it, what is the diameter of a column of equal length, that contains 10 times as much?

Ans. 22.136 inches.

14. Find the thickness of the lead in a pipe of an inch and a quarter bore, which weighs 14lb. per yard in length, the cubic foot of lead weighing 11325 Ans. 20737 inches.

ounces.

15. Let Couter circumference of a circular ring, b = its breadth, and T=3·1416; then the area of the ring will be = = b(С—πb).

16. What will be the expense of a curb to a round well, at 8d. per foot square, the breadth of the curb being 8 inches, and the interior diameter 3 feet. Ans. 5s. 9ąd.

17. A garden is 100 feet long and 80 feet broad, and a gravel walk is to be made of an equal width half round it, so as to occupy half the garden; find both by construction and calculation, the breadth of the walk.

Ans. 25.968. 18. If the sides of a triangle are 28, 25, and 17, what is the area of its greatest inscribed square? Ans. 101-67124. 19. Let a, b, c be the distances between a tree and three corners of a square field in a successive order; then will the area of the field be

=a2+b2—ab√/2 (cos. p—sin ø), where cos. =!

a2+2b2-c2
2ab/2

20. The four sides of a field, whose diagonals are equal to each other, are known to be 25, 35, 31, and 19 poles, in a successive order; required the content of the field.

A. R. P.

Ans. (793+78449) sq. poles, or 4 1 38.

21. The length and breadth of a vessel in the form of a rectangular parallelopepid are respectively 6 and 3 feet; what must be its depth, to contain exactly 200 imperial gallons? Ans. 1 foot 9.4 inches.

22. Seven men bought a grinding stone of 60 inches diameter, each paying ✈ part of the expense; what part of the diameter must each grind down for his share?

Ans. The 1st, 4·4508; 2d, 4·8400; 3d, 5·3535; 4th, 6·0765; 5th, 7-2079; 6th, 9.3935; 7th, 22.6778 inches.

23. Divide a cone into three equal parts by sections parallel to the base, and find the altitudes of the three parts, the height of the cone being 20 inches.

Ans. The upper part, 13.8672; the middle, 3.6044; the lower, 2-5284. 24. What quantity of canvass is necessary for a conical tent whose perpendicular height is 8 feet, and the radius at bottom 6+ feet?

Ans. 210 sq. feet.

This question admits of an elegant general solution, as may be seen in the "Ladies' Diary" for 1823. In this example, the student will find that the slab contains more than 10, but less than 11 feet; hence the operation must be performed 10 times, and the dimensions of the remaining part of a foot will be the answer.

25. A cable 3 feet long, and 9 inches in compass, weighs 22lb.; what will be the weight of a fathom of that cable whose circumference is a foot?

Ans. 783lb.

26. If R and r be the radii of two spheres inscribed in a cone, so that the greater may touch the less, and also the base of the cone; then will the capa2π R5 3r(R-r)

city of the cone be=

27. If a heavy sphere whose diameter is 4 inches be let fall into a conical glass full of water, whose diameter is 5, and altitude 6 inches; it is required to determine how much water will run over. Ans. 26-272 cubic inches.

28. The dimensions of a sphere and cone being as in last question, and the cone only full of water; what part of the axis of the sphere is immersed in the water ?* Ans. 5459 inches.

29. Let A represent the number of degrees in the arc of a segment of a circle whose radius is r; it is required to prove that

area of segment = r2 {arc A° (radius 1) — sin A°}.

30. What is the length of a chord which cuts off of the area from a circle of which the diameter is 289? Ans. 278-6716.

31. How high above the surface of the earth must a person be raised to see of its surface? Ans. The height of its diameter.

32. If a cubic foot of brass be drawn into wire of inch diameter, what will be the length of the wire, supposing no loss of metal in working? Ans. 97784-5684 yards, or 55 miles 984.5684 yards.

33. Supposing the diameter of an iron 91b. ball to be 4 inches, as it is very nearly; it is required to find the diameters of the several balls weighing 12, 18, 24, 32, and 36lb., and the calibre of their guns, allowing of the calibre, or of the ball's diameter for windage.

This question admits of a beautiful algebraical solution, by putting = the part of the axis immersed. The resulting cubic equation is easily rendered a complete power, and then the cube root being taken, gives finally a simple equation for the determination of x the part immersed.