1226. PROBLEM XIII. To find the area of an ellipsis. Rule. Multiply the longest and shortest diameter together, and their product by 7854, which will give the area required. This rule is founded on Theorem 3. Cor. 2. in Conic Sections. (1098, 1100.) Example. Required the area of an ellipse whose two axes are 70 and 50. Here 70 x 50 x 78542748.9. 1227. PROBLEM XIV. To find the area of any elliptic segment. Rule. Find the area of a circular segment having the same height and the same vertical axis or diameter; then, as the said vertical axis is to the other axis (parallel to the base of the segment), so is the area of the circular segment first found to the area of the elliptic segment sought. This rule is founded on the theorem alluded to in the previous problem. Or, divide the height of the segment by the vertical axis of the ellipse; and find in the table of circular segments appended to Prob. 12. (1224.) the circular segment which has the above quotient for its versed sine; then multiply together this segment and the two axes of the ellipse for the area. Example. Required the area of an elliptic segment whose height is 20, the vertical axis being 70, and the parallel axis 50. Here 20÷702857142, the quotient or versed sine to which in the table answers the segment 285714. Then 285714 x 70 x 50 = 648 342, the area required. 1228. PROBLEM XV. To find the area of a parabola or its segment. Rule. Multiply the base by the perpendicular height, and take two thirds of the product for the area. This rule is founded on the properties of the curve already described in conic sections, by which it is known that every parabola is of its circumscribing parallelogram. (See 1097.) Example. Required the area of a parabola whose height is 2 and its base 12. MENSURATION OF SOLIDS. 1229. The measure of every solid body is the capacity or content of that body, considered under the threefold dimensions of length, breadth, and thickness, and the measure of a solid is called its solidity, capacity, or content. Solids are measured by units which are cubes, whose sides are inches, feet, yards, &c. Whence the solidity of a body is said to be of so many cubic inches, feet, yards, &c. as will occupy its capacity or space, or another of equal magnitude. 1 1230. The smallest solid measure in use with the architect is the cubic inch, from which other cubes are taken by cubing the linear proportions, thus, 1728 cubic inches=1 cubic foot; 27 cubic feet 1 cubic yard. = 1231. PROBLEM I. To find the superficies of a prism. Multiply the perimeter of one end of the prism by its height, and the product will be the surface of its sides. To this, if wanted, add the area of the two ends of the prism. Or, compute the areas of the sides and ends separately, and add them together. Example 1. Required the surface of a cube whose sides are 20 feet. Here we have six sides; therefore 20 × 20 × 6=2400 feet, the area required. Example 2. Required the surface of a triangular prism whose length is 20 feet and each side of its end or base 18 inches. Here we have, for the area of the base, - 1·52-752 (2·25 — ·5625 = ) 1·68752 for the perpendicular of triangle of base; and 1.6875=1.299, which multiplied by 1·5=1·948 gives the area of the two bases; then, 3 x 20 x 1 ·5 + 1 ·948 = 91 948 is the area required. Example 3. Required the convex surface of a round prism or cylinder whose length is 20 feet and the diameter of whose base is 2 feet. Here, 2 x 3.1416=6.2832, and 3·1416 × 20-125-664, the convex surface required. 1232. PROBLEM II. To find the surface of a pyramid or cone. Rule. Multiply the perimeter of the base by the length of the slant side, and half the product will be the surface of the sides or the sum of the areas of all the sides, or of the areas of the triangles whereof it consists. To this sum add the area of the end or base. Example 1. Required the surface of the slant sides of a triangular pyramid whose slant height is 20 feet and each side of the base 3 feet. Here, 20 × 3 (the perimeter) × 3÷2=90 feet, the surface required. Example 2. Required the convex surface of a cone or circular pyramid whose slant height is 50 feet and the diameter of its base 8 feet. Here, 8.5 x 3·1416 × 50÷2=667 5, the convex surface required. 1233. PROBLEM III. To find the surface of the frustum of a pyramid or cone, being the lower part where the top is cut off by a plane parallel to the base. Rule. Add together the perimeters of the two ends and multiply their sum by the slant height. One half of the product is the surface sought. This is manifest, because the sides of the solid are trapezoids, having the opposite sides parallel. Example 1. Required the surface of the frustum of a square pyramid whose slant height is 10 feet, each side of the base being 3 feet 4 inches and each side of the top 2 feet 2 inches. Here, 3 feet 4 inches x 4=13 feet 4 inches, and 2 feet 2 inches x 4 =8 feet 8 inches; and 13 feet 4 inches + 8 feet 8 inches=22. Then 22÷2 × 10=110 feet, the surface required. Example 2. Required the convex surface of the frustum of a cone whose slant height is 12 feet and the circumference of the two ends 6 and 8·4 feet. Here, 6+8·4=14·4; and 14·4 × 12·5÷2=180÷2=90, the convex surface required. 1234. PROBLEM IV. To find the solid content of any prism or cylinder. Rule. Find the area of the base according to its figure, and multiply it by the length of the prism or cylinder for the solid content. This rule is founded on Prop. 99. (Geometry, 980.). Let the rectangular parallelopipedon be the solid to be measured, the small cube P(fig. 518.) being the measuring unit, its side being 1 inch, 1 foot, &c. Let also the length and breadth of the base ABCD, and also let the height AH, be divided into spaces equal to the side of the base of the cube P; for instance, H here, in the length 3 and in the breadth 2, making 3 times 2 or 6 squares in the base AC each equal to the base of the cube P. It is manifest that the parallelopipedon will contain the cube P as many times as the base AC contains the base of the cube, repeated as often as the height AH contains the height of the cube. Or, in other words, the contents of a parallelopipedon is found by multiplying the area of the base by the altitude of the solid. And because all prisms and cylinders are equal to parallelopipedons of equal bases and altitudes, the rule is general for all such solids whatever the figure of their base. Example 1. Required the solid content of a cube whose side is 24 inches. C P A B Fig. 518. Here, 24 x 24 x 24=13824 cubic inches. Example 2. Required the solidity of a triangular prism whose length is 10 feet and the three sides of its triangular end are 3, 4, and 5 feet. Here, because (Prop. 32. Geometry, 907.) 32 + 42=52, it follows that the angle con 3x4 tained by the sides 3 and 4 is a right angle. Therefore × 10=60 cubic feet, the content required. Example 3. Required the content of a cylinder whose length is 20 feet and its diameter 5 feet 6 inches. Here, 5.5 x 5.5 × 7854=23·75835, area of base; and 23.75835 x 20=47'5167, content of cylinder required. 1235. PROBLEM V. To find the content of any pyramid or cone. Rule. Find the area of the base and multiply that area by the perpendicular height. One third of the product is the content. This rule is founded on Prop. 110. (Geometry, 991.) Example 1. Required the solidity of the square pyramid, the sides of whose base are 30, and its perpendicular height 25. Example 2. Required the content of a triangular pyramid whose perpendicular height is 30 and each side of the base 3. 3+3+3 Here, 2 ==4.5, half sum of the sides; and 4.5-3=1.5, one of the three equal remainders. (See Trigonometry, 1052.) but 4.5 x 1.5 × 1·5 × 1·5 × 30÷3=3·897117 x 10, or 38.97117, the solidity required. Example 3. Required the content of a pentagonal pyramid whose height is 12 feet and each side of its base 2 feet. Here, 1.7204774 (tabular area, Prob. 6. 1218.) x 4 (square of side) = 6.8819096 area of base; 6-68819096 Example 4. Required the content of a cone whose height is 10 feet and the circumference of its base 9 feet. Here, 07958 (Prob. 9. 1221.) × 81 =6·44598 area of base, And 3.5 being of 10 feet, 6·44598 × 3·5=22·56093 is the content required. 1236. PROBLEM VI. To find the solidity of the frustum of a cone or pyramid. Add together the areas of the ends and the mean proportional between them. Then taking one third of that sum for a mean area and multiplying it by the perpendicular height or length of the frustum, we shall have its content. This rule depends upon Prop. 110. (Geometry, 991.). It may be otherwise expressed when the ends of the frustum are circles or regular polygons. In respect of the last, square one side of each polygon, and also multiply one side by the other; add the three products together, and multiply their sum by the tabular area for the polygon. Take one third of the product for the mean area, which multiply by the length, and we have the product required. When the case of the frustum of a cone is to be treated, the ends being circles, square the diameter or the circumference at each end, and multiply the same two dimensions together. Take the sum of the three products, and multiply it by the proper tabular number, that is, by 7854, when the diameters are used, and 07958 when the circumferences are used, and, taking one third of the product, multiply it by the length for the content required. Example 1. Required the content of the frustum of a pyramid the sides of whose greater ends are 15 inches, and those of the lesser ends 6 inches, and its altitude 24 feet. and 8125 × 24 (altitude)=19.5 feet, content required. Example 2. Required the content of a conic frustum whose altitude is 18 feet, its greatest diameter 8, and its least diameter 4. Here, 64 (area gr. diam.) +16 (less. diam.) + (8 × 4) = 112, sum of the products; and 7854x112 x 18 Example 3. 3 =527.7888, content required. Required the content of a pentagonal frustum whose height is 5 feet, each side of the base 18 inches, and each side of the upper end 6 inches. Here, 1.52 + 1·52 + (1·5 × •5) =2·5625, sum of the products; but, 1-7204774 (tab. area) × 2 5625 (sum of products) × 5 – 9.31925, content required. 3 = 1237. PROBLEM VII. To find the surface of a sphere or any segment. Rule 1. Multiply the circumference of the sphere by its diameter, and the product will be the surface thereof. This and the rules in the following problems depend on Props. 113. and 114. (Geometry, 994, 995.), to which the reader is referred. Square the diameter, and multiply that square by 3·1416 for the surface. Rule 3. Square the circumference, then either multiply that square by the decimal 3183, or divide it by 3.1416 for the surface. Rule 2. Remark. For the surface of a segment or frustum, multiply the whole circumference of the sphere by the height of the part required. Example 1. Required the convex superficies of a sphere whose diameter is 7 and circumference 22. Here, 22 x 7=154, the superficies required. Example 2. Required the superficies of a sphere whose diameter is 24 inches. Here, 24 x 24 x 3·1416=1809.5616 is the superficies required. Example 3. Required the convex superficies of a segment of a sphere whose axis is 42 inches and the height of the segment 9 inches. Here, 13-1416::42: 131 ·9472, the circumference of the sphere; but 1319472 × 9=1187 5248, the superficies required. Example 4. Required the convex surface of a spherical zone whose breadth or height is 2 feet and which forms part of a sphere whose diameter is 12 feet. Here, 13-1416::12·5; 39-27, the circumference of the sphere whereof the zone is a part; and 39:27 × 2=78.54, the area required. 1238. PROBLEM VIII. To find the solidity of a sphere or globe. Rule 1. Multiply the surface by the diameter, and take one sixth of the product for the content. Rule 2. Take the cube of the diameter and multiply it by the decimal ·5236 for the content. Example. Required the content of a sphere whose axis is 12. Here 12 x 12 12 x 5236=904 7808, content required. 1299. PROBLEM IX. To find the solidity of a spherical segment. Rule 1. From thrice the diameter of the sphere subtract double the height of the segment, and multiply the remainder by the square of the height. This product multiplied by 5236 will give the content. Rule 2. To thrice the square of the radius add the square of its height, multiply the sum thus found by the height, and the product thereof by 5236 for the content. Example 1. Required the solidity of a segment of a sphere whose height is 9, the diameter of its base being 20. Here, 3 times the square of the radius of the base =300; and the square of its height =81, and 300 +81=381; but 381 x 9=3429, which multiplied by 5236=1795-4244, the solidity required. Example 2. Required the solidity of a spherical segment whose height is 2 feet and the diameter of the sphere 8 feet. Here, 8 x 3-4=20, which multiplied by 4=80; and 80 x 5236=41.888, the solidity required. It is manifest that the difference between two segments in which the zone of a sphere is included will give the solidity of the zone. That is, where for instance the zone is included in a segment lying above the diameter, first consider the whole as the segment of a sphere terminated by the vertex and find its solidity; from which subtract the upper part or segment between the upper surface of the zone and the vertex of the sphere, and the difference is the solidity of the zone. The general rule to find the solidity of a frustum or zone of a sphere is to the sum of the squares of the radii of the two ends add one third of the square of their distance, or the breadth of the zone, and this sum multiplied by the said breadth, and that product again by 1-5708, is the solidity. SECT. VIII. MECHANICS AND STATICS. 1240. It is our intention in this section to address ourselves to the consideration of mechanics and statics as applicable more immediately to architecture. The former is the science of forces, and the effects they produce when applied to machines in the motion of bodies. The latter is the science of weight, especially when considered in a state of equilibrium. 1241. The centre of motion is a fixed point about which a body moves, the axis being the fixed line about which it moves. 1242. The centre of gravity is a certain point, upon which a body being freely suspended, it will rest in any position. 1243. So that weight and power, when opposed to each other, signify the body to be moved and the body that moves it, or the patient and agent. The power is the agent which moves or endeavours to move the patient or weight, whilst by the word equilibrium is meant an equality of action or force between two or more powers or weights acting against each other, and which by destroying each other's effects cause it to remain at rest. PARALLELOGRAM OF FORCES. 1244. If a body D suspended by a thread is drawn out of its vertical direction by an horizontal thread DE (fig. 519.), such power neither increases nor diminishes the effort KM M G E L H H М of the weight of the body; but it may be easily imagined that the first thread, by being in the direction AD, will, besides the weight itself, have to sustain the effort of the power that draws it out of the vertical AB. 1245. If the direction of the horizontal force be prolonged till it meets the vertical, which would be in the first thread if it were not drawn away by the second, we shall have triangle ADB, whose sides will express the proportion of the weight to the forces of the two threads in the case of equilibrium being established; that is, supposing AB to express the weight, AD will express the effort of the thread attached to the point A, and BD that of the horizontal power which pulls the body away from the vertical AB. 1246. These different forces may also be found by transferring to the vertical DH (fig. 520.) any length of line DF to represent the weight of the body. If from the point F the parallels FI, FG be drawn in the direction of the threads, their forces will be indicated by the lines ID, DG, so that the three sides of the triangle DGF, similar to the triangle ADB, will express the proportion of the weight to the two forces applied to the threads. 1247. Suppose the weight to be 30 lbs.: if from a scale of equal parts we set up 30 of those parts from D to F (fig. 519.), we shall find DG equal to 21, or the pounds of force of the horizontal line DE, and 35 for the oblique power ID. 1248. If the weight, instead of 30 lbs., were 100, we should find the value of the forces DG and ID by the proportions of 30: 21::100; x, where x expresses the force DG. The value resulting from this proportion is r = = 70. The second proportion 21 x 100 35x100 30 =116-666. 30: 35:100 y,where y represents the effort ID, whose value will be y 1249. If the angle ADH formed by AD and DH be known, the same results may be obtained by taking DF for the radius, in which case IF=DG becomes the tangent, in this instance, of 35 degrees, and ID the secant; whence DF: DI: IF:: radius: tang. 35 sec. 35. If ID be taken for the radius, we have ID: IF: FD:: radius: tang. 35; sin. 35. 1250. We have here to observe, that in conducting the operation above mentioned a figure DIFG has been formed, which is called the parallelogram of the forces, because the diagonal DF always expresses a compounded force, which will place in equilibrio the two others FI, FG, represented by the two contiguous sides IF, FG. G Fig. 522. 1251. Instead of two forces which draw, we may suppose two others which act by pushing from E to D (fig. 522.) and from A to D. If we take the vertical DF to express the weight, and we draw as before the parallels FG and FI in the E direction of the forces, the sides GD and DI of the parallelogram DGFI (fig. 519.) will express the forces with which the powers act relatively to DF to support the body: thus FI= GD the weight and two powers which support it will, in case of equilibrium, be represented by the three sides of a rectangular triangle DFI; so that if the weight be designated by H, the power which pushes from G to D by E, and that which acts from I to D by P, we shall have the proportion H: E: P:: DF: FI: ID, wherein, if we take DF for radius, it will be as radius is to the tangent of the angle FDI and to its secant. As a body in suspension is drawn away from the vertical line in which it hangs by a power higher than the body (fig. 520.), it follows that the oblique forces AB and BC each support, independent of any lateral efforts, a part of the weight of the body. In order to find the proportion of these parts to the total weight, take any distance BD on a vertical raised from the centre of the body B to express the weight, and complete the parallelogram DEBF, whose sides EB, BF will express the oblique forces of the powers A and C. These lines, being considered as the diagonals of the rectangular parallelograms LEIB, BHFM, may each be resolved into two forces, whereof one of them, vertical, sustains the body, and the other, horizontal, draws it away from the verticals AO, CQ. Hence IB will express the vertical force, or that part of the weight sustained by the power EB, and HB that sustained by the other power BF: as these two forces act in the same direction, when added together their sum will represent the weight DB. In short, IB being equal to HD, it follows that BH + BIBI+ ID. 1252. As to the horizontal forces indicated by the lines LB and BM, as they are equal and opposite they destroy one another. 1253. It follows, from what has been said, that all oblique forces may be resolved into two others, one of which shall be vertical and the other horizontal, by taking their direction for the diagonal of a rectangular parallelogram. 1254. In respect of their ratio and value, those may be easily found by means of a scale if the diagram be drawn with accuracy; or by trigonometry, if we know the angles |