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 3 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. I.-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. I.—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 1-78, 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 the 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. L-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 Ex. III. What is the length 14 feet 6 inches? content of a tree, whose mean girt is 3.15 feet, and 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 h=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 42(ab+be+ca)-(a2+b2+c2). 8. A beam is 84 inches deep and 3 inches broad; what is the depth of another twice as large, which is 4 inches broad? Ans. 12:5263 inches. 9. Supposing the expense of paving a semicircular plot, at 28. 4d. per foot, come to £10; what is its diameter ? Ans. 14.7787 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. 274 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 foot 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 Aus. 20737 incnes. ounces. 15. Let Couter circumference of a circular ring, b = its breadth, and ☛=3·1416; then the area of the ring will be = b(C—«b). 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 34 feet. Ans. 58. 94d. 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. —sin ø), where cos. p= a2+2b2-c2 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 +7√8449) 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 .823. 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 city of the cone be= 3r(R-r) 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 eone 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 & complete power, and then the cube root being taken, gives finally a simple equation for the determination of the part immersed. PLANE TRIGONOMETRY, as the name imports, was originally employed solely in determining, from certain data, the sides and angles of plane triangles. In modern analysis, however, its objects have been much extended, and the formula of this branch of Mathematics are extensively employed as instruments of calculation in almost every department of scientific investigation. From this circumstance, some writers wishing to change its designation to one which might more fully express its nature and applications, have proposed to term it the Arithmetic of Sines, others, the Calculus of Angular Functions, but the original appellation is still retained by the great majority of authors upon these subjects. In treating of angular magnitude, we have hitherto confined ourselves to the consideration of angles less than two right angles, but in trigonometry it is frequently necessary to introduce angles which are greater than two, than three, or even than four right angles. We may take the following method of illustrating the generation of angular magnitude. Let Aa be a fixed straight line, and let a line CP be supposed to revolve round the point C in Aa, and to assume in succession the different positions CP1, CP2, CP3, .......... When CP coincides with CA, there is no angle contained between CP and CA, or the angle CAP is 0. When CP begins to revolve round C, and comes into the position CP1, it forms with CA an angle P1CA less than a right angle. P Ps When CP has performed one-fourth part of an entire revolution, and has thus reached the position CP2, where CP, is perpendicular to CA, it forms with CA the angle PCA, which is a right angle. As CP continues its revolutions, it will assume the position CP, forming with CA the angle P, CA, greater than one, and less than two right angles. When CP coincides with Ca, it has performed one half of an entire revolution, and forms with CA the angle aCA, equal to two right angles. CP having passed Ca, will assume the position CP4, forming with CA the angle P,CA, greater than two, and less than three, right angles. The dotted space indicates the angle which we are Considering. |