operation. To calculate contents, or areas, in square inches, set the slide A to 10 in. (as here shewn), which means that the result multiplied by 10, gives the content, or area, in square inches. To obtain this result, place the point E, at a convenient distance from the figure to be measured, so that the tracer F may traverse the entire periphery of the figure. But if the figure is too large to allow this, it can be subdivided by drawing straight lines through it, and the contents, or areas, of the several parts computed separately, and added together. Then place B the point of the tracer on any convenient starting point in the periphery. When the instrument is thus adjusted, read off the division on the horizontal disc G; also that on the perpendicular wheel and vernier, H. Suppose that the horizontal disc gives 3, and the vertical wheel gives 905, namely 90 on the wheel and 5 on the vernier; this reading must be put down thus, 3.905. Then carry the tracing point round the figure in the direction of the hands of a watch; and when the whole circuit has been made, observe the readings again. Suppose them 5.763; then subtract the former reading from the latter; the result will be 1.858; multiply this by 10, and you will get the content of the figure in square inches, namely 18.58 square inches. Notice must be taken whether the disc G has made an entire circuit; if so, 10 must be added, for every revolution, to the whole number. If the disc, in the above case, had gone once round, the second reading would have been 15.763. If twice round, 25.763; and so on. If the result is required in acres, multiply the above result by the number of acres in the square inch, according to the scale used in drawing the plan. The other divisions on the arm A, may be easily applied to any scale, by finding the result of one square acre or chain, and using that number as the coefficient to give the content of the surface required. The proof of the principle of this instrument, which can only be understood by advanced students, will be found in a Note at the end of the book. THE THEODOLITE. 279. The Theodolite is the most complete and efficient instrument used by Surveyors. It measures, with great accuracy, the angular distance between any two visible objects, either in the horizontal, or in the vertical, plane; and it is not necessary, that, in the former case, the objects themselves be in the same horizontal plane, or in the latter case, that they be in the same vertical plane. But the Theodolite, in its best form, is so complicated, that a minute description here of its various parts would serve no good purpose. The only way really to understand it, is to see and handle the instrument itself. CURVED SURFACES AND SOLIDS. 280. Before proceeding to consider the mensuration of Solids generally, we may call attention to two cases of curved surfaces, where the surface can be unwrapped, so as to form a plane surface. Its area may then be obtained by methods already given. DEF. A Right Cone is a solid which may be supposed to be generated by the revolution of a right-angled triangle about one of its shorter sides as an axis, the conical surface being generated by the hypothenuse; the base of the cone is a circle. DEF. A Right cylinder is a solid which may be supposed to be generated by the revolution of a rectangle about one of its sides as an axis, the opposite side of the rectangle describing the cylindrical surface: the extremities of the cylinder are two equal parallel circles. Now the curved surface of a cone, when slit down in a straight line from the vertex of the cone to the base, and unwrapped, becomes the sector of a circle; and that of a cylinder, slit down by a straight line at right angles to the base of the cylinder, and unwrapped, becomes a rectangle. Therefore we can measure the surface of a right cone, or of a right cylinder, by a very simple process. 281. To measure the curved surface of a right cone. E B с cumference of the base Let ABCD be the cone, AE its altitude, and BE the radius of its base, which is a circle; then the area of its curved surface is equal to that of a sector, whose radius is AB, the slant side, and whose arc is the circumference of the circle BCDB. Let BD, the diameter of the base, be measured; then the cir=π × BD, (237). And, therefore, the curved surface of the cone, which is equal to the area of the above-mentioned sector, is equal to half the product of the slant side and the circumference BCDB, by (241), ABX T×BD Exs. If AB=4in., and BD=3in., then the curved surface = × 6 square inches, =3.1416×6 sq. in.-18.8496 sq. in. If AE and BE are known AB may be found without further measurement. For since ABE is a right Thus, if AE-8, and BE-6, then AB-64+36=100, .. AB=10. And the curved surface will be equal to 282. To measure the curved surface of a frustum of a right cone. If the upper part of the cone were cut off by a plane parallel to the base, the lower part would be called a frustum, and the surface of the frustum would evidently be found by subtracting the surface of the small cone so cut off from that of the complete cone. Suppose the section be made through the middle of AB, or AD. Then, the curved surface of the small cone Subtracting this from the whole surface, we have area of the curved surface of the frustum A method of finding the surface of a frustum, índependently of that of the complete cone, will be given in a future article. A 283. To measure the surface of a right cylinder. B C Let ABCD be the cylinder, of which D the height AB, and the diameter of the base, BC, are known. Then the area of the curved surface is equal to that of a rectangle, whose adjacent sides are respectively AB, and the circumference of the base. This circumference = =π× BC; therefore the area of the curved surface =T×AB× BC. Exs. If the height of the cylinder be 10 in., and the diameter of the base be 4 in.; then, the area of the curved surface =π×10×4 sq. in., = π × 40 sq. in., =125.664 sq. inches. In practice, we shall more often have the circumference, than the radius, of the base given; and if it be not given, it is readily measured. We have then nothing to do with π, since Curved surf. Circumf. × height. = Thus, if the circumference of a cylinder be 10 feet, and its height 5 feet, Curved surf.=104x5=52 sq. ft. (1) A cylinder has a base whose radius is 1.75 ft.; and the height is 5·26 ft.; find the whole surface, including the two ends. Ans. 77.11 sq. in. (2) The perpendicular height of a conical tent is 9 ft., and the radius of its base is 44 ft.; find the area of the canvas. Ans. 142.305 sq. ft. |