A strong axis D is fitted to the top of the frame; to the lower part of the axis any of the wheels: (as i) may be adapted; when screwed to it they are, immoveable. EG is an arm contrived to turn round upon the main axis D; two sliding boxes are fitted to this arm; to these boxes any of the wheels belonging to the geometric pen may be fixed, and then moved so that the wheels may take into each other, and the immoveable wheel i; it is evident, that by making the arm EG revolve round the axis D, these wheels will be made to revolve also, and that the number of their revolutions will depend on the proportion between the teeth. fg is an arm carrying the pencil; this arm slides backwards and forwards in the box c d, in order that the distance of the pencil from the centre of the wheel h may be easily varied; the box cd is fitted to the axis of the wheel h, and turns round with it, carrying the arin f g along with it; it is evident, therefore, that the revolutions will be fewer or greater, in proportion to the difference between the numbers of the teeth in the wheels h and i; this bar and socket are easily removed for changing the wheels. When two wheels only are used, the bar fg moves in the same direction with the bar EG; but if another wheel is introduced between them, they move in contrary directions. The number of teeth in the wheels, and consequently, the relative velocity of the epicycle, or arm fg, g, may be varied ad infinitum. The numbers we have used are, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96. The construction and application of this instrument is so evident from the figure, that nothing more need be pointed out than the combinations by which the figures here delineated may be produced. To render the description as concise as possible, Ishall in future describe the arm E G by the letter A, and fg by the letter B.. To describe fig. 1, plate 12. The radius of A must be to that of B, as 10 to 5 nearly, their velocities, or the numbers of teeth in the wheels, to be equal, the motion to be in the same direction.. If the length of B be varied, the looped figure, delineated at fig. 12, will be produced. A circle may be described by equal wheels, and any radius, but the bars must move in contrary direction. A To describe the two level figures, see fig. 11, plate 12. Let the radius of A to 3 be as 10 to 34, the velocities as 1 to 2, the motion in the same direction. To describe by this CIRCULAR MOTION, STRAIGHT LINE AND AN ELLIPSE. For a straight line, equal radii, the velocity is 1 to 2, the motion in a contrary direction; the same data will give a variety of ellipses, only the radii must be unequal; the ellipses may be described in any direction; sce fig. 10, plate 13. Fig. 13, plate 12, with seven leaves, is to be formed when the radii are as 7 to 2, velocity as '2 to 3, motion in contrary directions. The six triangular figures, seen at fig. 2, 4, 6, 8 9, 10, are all produced by the same wheels, by only varying the length of the arm B, the velocity should be as 1 to 3, the arms are to move in contrary directions. Fig. 3, plate 12, with eight leaves, is formed by equal radii, velocities as 5 to 8, A and B to move the same way; if an intermediate wheel is added, and thus a motion produced in a contrary direction, the pencil will delineate fig. 16, plate 12. The ten-leaved figure, fig. 15, plate 12, is produced by equal radii, velocity as 3 to 10, directions of the motions contrary to each other. Hitherto the velocity of the epicycle has been the greatest; in the three following figures the curves are produced when the velocity of the epicycle is less than that of the primum mobile. For fig. 7, the radius of A to B to be as 2 to 1, the velocity as 3 to 2; to be moved the same way. For fig. 14, the radius of A, somewhat less than the diameter given to B, the velocity as 3 to 1; to be moved in a contrary direction. For fig. 5, equal radii, velocity as 3 to 1; moved the same way. These instances are sufficient to shew how much may be performed by this instrument; with a few additional pieces, it may be made to describe a cycloid, with a circular base, spirals, and particularly the spiral of Archimedes, &c. OF THE DIVISION OF LAND. To know how to divide land into any number of equal, or unequal parts, according to any assigned proportion, and to make proper allowances for the different qualities of the land to be divided, form a material and useful branch of surveying. In dividing of land, numerous cases arise; in some it is to be divided by lines parallel to each other, and to a given fence, or road; sometimes, they are to intersect a given line; the division is often to be made according to the particular directions of the parties concerned. In a subject which has been treated on so often, novelty is, perhaps, not to be desired, and scarcely expected. No considerable improvement has been made in this branch of surveying since the time of Speidell. Mr. Talbot, whom we shall chiefly follow, has arranged the subject better than those who preceded him, and added thereto two or three problems; his work is well worth the surveyor's perusal. Some problems also in the foregoing part of this work should be considered in this place. PROBLEM 1. To divide a triangle in a given ratio by right lines drawn from any angle to the opposite side thereof. 1. Divide the opposite side in the proposed ratio. 2. Draw lines from the several points of division to the given angle, and then divide the triangle as required. Thus, to divide the triangle AB C, fig. 13, plate &, containing 26 acres, into three parts, in proportion to the numbers 40, 20, 10, the lines of division to proceed from the angle C to A B, whose length is 28 chains; now, as the ratio of 40, 20, 10, is the same as 4, 2, 1, whose sum is 7, divide A B into seven equal parts; draw C a at four of these parts, Cb at six of them, and the triangle is divided as required. Arithmetically. As 7, the sum of the ratios, is to AB 28 chains; so is 4, 2, 1, to 16, 8, and 4 chains respectively; therefore A a 16, a b = 8, and bB 4 chains. To know how many acres in each part, say, as the sum of the ratios is to the whole quantity of land, so is each ratio to the quantity of acres ; { 7:26,5 4: 15,142857 = triangle A Ca 7: 26,5 2: 7,571428 = triangle a C b PROBLEM 2. To divide a triangular field into any number of parts, and in any given proportion, from agiven point in one of the sides. 1. Divide the triangle into the given proportion, from the angle opposite the given point. 2. Reduce this triangle by problem 51, so as to pass through the given point. Thus, to divide the field A BC, fig. 14, plate 8, of seven acres, into two parts, in the proportion of 2 to 5, for two different tenants, from a pond b, in BC, but so that both may have the benefit of the pond. * Talbot's Complete Art of Land Measuring. 1. Divide BC into seven equal parts, make Ba 5, then Ca= 2; draw A a, and the field is divided in the given ratio. 2. To reduce this to the point b, draw A b and a c parallel thereto, join c b, and it will be the required dividing line. Operation in the field. Divide BC in the ratio required, and set up a mark at the point a, and also at the pond b; at A, with the theodolite, or other instrument, measure the angle b Aa; at a lay off the same angle A ac, which will give the point c in the side Ac, from whence the fence must go to the pond. 2. To divide AB C, fig. 15, plate 8, into three equal parts from the pond c. 1. Divide AB into three equal parts, A a, a b, b B, and C a Cb will divide it, as required, from the angle A; reduce thesc as above directed to c d and ce, and they will be the true dividing lines. PROBLEM 3. To divide a triangular field in any required ratio, by lines drawn parallel to one side, and cutting the others. Let ABC, fig. 16, plate 8, be the given triangle to be divided into three equal parts, by lines parallel to AB, and cutting AC, BC. Rule 1. Divide one of the sides that is to be cut by the parallel lines, into the given ratio. 2. Find a mean proportional between this side, and the first division next the parallel side. 3. Draw a line parallel to the given side through the mean proportional. 4. Proceed in the same manner with the remaining triangle. Example 1. Divide BC into three equal parts BD, DP, PC. 2. Find a mean proportional between BC and DC. 3. Make CG equal to this mean proportional, and draw G H parallel to AB. Proceed in the same manner with the remaining triangle CHG, dividing GC into two equal parts at I, finding a mean proportional between C G and CI; and then making CL equal to this mean proportional, |