Therefore, 4 = 4 (2 r and 4 x = 4 X (2 r .ix') the fluxion of the solid. The 4 x g2 fluent of 4 ,y8 3 = 4 r for the solidity of the se ment of the dome. Now, therefore, if the solidity of the whole dome be required, it is only supposing x to become equal to r; 4 x3 3 3 8 we then find 4 r = 4 = Sup 3 pose d = the diameter, equal to 2 r, and a = r, the altitude, 2 2 d'a 8r then will = , that is, the solidity of the dome is equal to two-thirds of the circumscribing rectangular prism. If in the above, dome all the horizontal sections had been circles instead of squares, the dome would have been sphe rical ; let us suppose that p = .7854, the area of a circle, the diameter of which is unity, then in the case of the segment 2 of the sphere we have p r 4 .r for the solidity of the dome when less than a hemisphere ; or, putting 2 r = d, and a equal to the altitude of the segment, then 43 4 xs (4 r = p d and the hemispheric 3 2 a solidity is p . The same form of expression may be 3 shown for all polygonal domes whatever : it is only using a proper multiplier for p, when the radius of its circumscribing circle is unity.
Suppose it were required to find the solidity of a trun cited hemispheric dome, let .v = the altitude of the dome, as before, then, by the nature of the circle, we have = equal to one quarter of the generating square. 4 = 4 4 for the generating square.
4 x = 4 x 4 x9 x the fluxion of the solid.
a x The fluent of 4 x = 4 r2x 43 : now let d = the To find the solidity of a truncated square dome, the axal section being that part of a semi-circle left by cutting of a segment parallel to the diameter.
From the square ot' the side of the base, multiplied into the altitude of the dome, subtract the third part of four times the cube of the altitude, and the remainder is the solidity required.
Example.Suppose the side of the base 20 feet, and the altitude 6 feet, what is the solidity ? 20 6 20 6 400 36 6 6 2400 216 288 42112 solidity of the dome required. 3)864 2S8 But if the horizontal sections had been circles instead of squares, we should then have the solidity of the circular dome as follows : 211" .7654 8443 10560 16896 14784 1658.76-IS the solidity of the dome when the horizontal sections are circular.
It may here be remarked, that the segment and truncated domes make together a complete square dome, each side of the base being 20 feet, and the altitude 10 feet.
Now the solidity of the segment dome is 5543 feet. and that of the truncated dome . . . 2112 Therefore the whole square doine whose vertical section is a semi-circle, is . 26663
Now, the rule for measuring a square dome with semi-circu lar vertical sections parallel to the sides of the base, is to take two-thirds of the area of the base, multiplied into the height: 20 20 400 area of the base 10 3)4000 13333 13=1 26663 the solidity, as before.
But as it may be objected by many, that, when a square or circular dome, whose vertical section is the segment of a cir cle, is required to find its solidity, it is difficult to find the diameter of the vertical section, and that it would be more eligible to find the solidity from the side or diameter of its base, and the altitude of the section : in order to save the trouble of finding the diameter, we shall here show the investigation of another rule, independent of any foreign or adventitious dimension.
Let s equal the side of the square base, then -- is equal to half the side of the base; and if d be the diameter of the section, and a its altitude; then, by the property of the circle, 4 =a(da) therefore, d a = + consequently, d = + a 4a s' and 2 d 2 a 2a Therefore, by substituting + 2 a for 2 d in the formula 2 as' 2 d we obtain + 3 for the solidity of the 2 segmental dome, independent of the diameter of the section. This rule may be expressed thus: To half the area of the square base, multiplied into the altitude, add two-thirds of the cube of the altitude, and the sum will be the solidity of the dome.
Let us take the same example as at first, and we shall find the side of the square of the base to be 10 feet : therefore, 16 4 p 16 4 96 1616 4 256 square of the base. 64 cube of the altitude.
4 altitude.
2)1024 3)128 512 42-3423 5543 cubic feet, as before, in the segmental dome. The solidity of any dome whatever may be found by the following general rule : To the areas of the two ends, add four times the area in the middle ; then, one-sixth of the sum multiplied by the altitude, gives the solid contents. This rule applies to all domes whose vertical section is contained between any two opposite arcs of the same circle, and two parallel lines, and will even apply to those domes which are the segment of a sphere. In the seccond example, the side of the square of the base being 20, and of the top 16, the middle area will be found to be 364.
20 16 364 20 16 4 400 area of base. 96 1456 16 256 400 256 area of top.
2112 solidity.
Because multiplying and dividing by 6, gives the same number.
To find the solidity of a hollow square truncated dome, the shell being of equal thickness, supposing/ each edge of the base equal to the diameter of the circle of which the section is a part.