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similar to the smaller triangle Bde, we have BD : Bd::AD: ed. Thus, suppose the length of wall represented by AD=28 feet, and its height AB=12 feet, we shall have the length of the diagonal = 30 feet 54 inches; and, taking the ninth part of AB, or 16 inches, as the thickness to be transferred on the diagonal from B to d, we have 30 ft. 6 in. : 16 in. :: 28 ft. : 14 in. : 8 lines (ed). The calculation may also be made trigonometrically; into which there is no necessity to enter, inasmuch as the rules for obtaining the result may be referred to in the section “ Trigonometry,” and from thence here applied.

Method of enclosing a given Area in any regular Polygon. 1518. It is manifest that a polygon may be divided by lines from the centre to its angles into as many triangles as it has sides. In fig. 601., on one of these triangles let fall from C(which is the vertex of each triangle) a perpendicular CD on the base or side AB which is supposed horizontal. The area of this triangle is equal to the product of DB (half AB) by CD, or to the rectangle DCFB. Making DB=x, CD=y, and the area given =p, we shall have,

For the equilateral triangle, xxYx =p, or ry=;
For the square, ky x 4=p, or xy=;
For the pentagon, xy + 5 =p, or xy= };

For the hexagon, xy x6 =p, or cy=..
Each of these equations containing two unknown quantities, it becomes necessary to as-
certain the proportion of x to y, which is as the sines of the angles opposite to the sides
DB and CD.

1519. In the equilateral triangle this proportion is as the sine of 60 degrees to the sine of 30 degrees; that is, using a table of sines, as 86603 : 50000, or 83 : 5, or 26 : 15, whence

*:y:: 26 : 15, and 15x=26y; whence y= 26 Substituting this value in the equation zy=;

we have 15.02

1, which becomes 13 =25, and x=V2017 Supposing the area given to be 3600, we shall therefore have

3600 x 26 M

=45.6, and the side AB=91.2.

45 For the pentagon, m:y::sin. 36° : sin. 54°, or as 58779 : 80902, whence

80902.5

152

26

y = 58799

and I=

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Substituting this value in the equation xy=5, we have
8090262 3600

✓ 58779x 720
58779

80902 which makes x=22:87, and the side AB=45074. For the hexagon, x:y::sin. 30° : sin. 60°, or as 50000 : 86603::5: 83, whence the value of

This value, substituted in the equation xy= o, will give 600; whence -; lastly, therefore, r =v346·15=18.61, and the side AB=37•22.

263

15 600 x 15

26

26.2%
15

Geometrically. 1520. Suppose the case that of a pentagon (fig. 601.) one of whose equal triangles is ACB. Let fall the perpendicular CD, which divides it into two equal parts; whence its area is equal to the rectangle CDBF.

1521. Upon the side AB, prolonged, if necessary, make DE equal to CD, and from the middle of BE as a centre describe the semi-circumference cutting CD in G, and GD will be the side of a square of the same area as the rectangle CDBF. The sides of similar figures (Geometry, 961.) being as the square roots of their areas; find the square root of the given area and make Dg equal to it. From the point g draw parallels to GE and GB, which will determine on AB the points e and b, and give on one side Db equal to one half of the side of the polygon sought; and, on the other, the radius De of the circum. ference in which it is inscribed. This is manifest because of the similar triangles EGB and egb, from which BD: DE :: 6D: De.

1522. From the truth that the sides of similar figures are to each other as the square roots of their areas we arrive at a simple method of reducing any figure to a given area. Form an angle of reduction (fig. 603.) one of whose sides is equal to the square root of the greater area, and the chord of the are, which determines the size of the angle equal

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B

Fig. 603.

to the square root of the smaller area. Let, for instance, the larger area =1156, and that of the smaller, to which the figure is to be reduced, = 529. Draw an indefinite line, on which make AB=34, the square root of 1156. Lastly, from the point A, as a centre, having described an indefinite arc, with a length equal to the square root 23 of 529, set out Bg; through 9 draw Ag, which will be the angle of reduction gAB, by means of which the figure may be reduced, transferring all the measures of the larger area to the line AD, with which arcs are to be described whose chords will be the sides sought.

1523. If it be not required to reduce but to describe a figure whose area and form aro given, we must make a large diagram of any area larger than that sought, and then reduce it.

1524. The circle, as we have already observed in a previous subsection (933.), being but a polygon of an infinite number of sides, it would follow that a circular enclosure would be stable with an infinitely small thickness of wall. This property may be easily demonstrated by a very simple experiment. Take, for instance, a sheet of paper, which would not easily be made to stand while extended to its full length, but the moment it is bent into the form of a cylinder it acquires a stability, though its thickness be not a thousandth part of its height.

1525. But as walls must have a certain thickness to acquire stability, inasmuch as they are composed of particles susceptible of separation, we may consider a circular enclosure as a regular polygon of twelve sides, and determine its thickness by the preceding process. Or, to render the operation more simple, find the thickness of a straight wall whose length is equal to one half the radius.

1526. Suppose, for example, a circular space of 56 ft. diameter and 18 ft. high, and the thickness of the wall be required. Describe the rectangle ABCD (fig. 594.), whose base is equal to half the radius, that is, 14 ft., and whose height AB is 18 ft. ; then, drawing the diagonal BD, make Bd equal to the ninth part of the height, that is, 2 ft. Through d draw ad parallel to the base, and its length will represent the thickness sought, which is 144 inches.

1527. By calculation. Add the square of the height to that of half the radius, that is, of 18 =324, and of 14=196 (=520). Then extract the square root of 520, which will bc found = 22-8, and this will be the value of the diagonal BD. Then we have the follow. ing proportion : 22.8 : 14:: 2 ft. (f the height): 1474.

1528. The exterior wall of the church of St. Stefano Rotondo at Rome (Temple of Claudius) incloses a site 198 feet diameter. The wall

, which is contructed of rubble masonry faced with bricks, is 2 ft. 4 in. (French) thick, and 22; ft. high. In applying to it the preceding rule, we shall find the diagonal of the rectangle, whose base would be the side of a polygon, equal to half the radius and 224 ft. high, would be 7491 494 + 2222= 547%. Then, using the proportion 54.37 : 49.5 :: 2 : 2 ft. 3 in. and 4 lines, the thickness sought, instead of 2 ft. 4 in., the actual thickness. We may as well mention in this place that a circle encloses the greatest quantity of area with the least quantity of walling; and of polygons, those with a greater number of sides more than those with a lesser: the proportion of the wall in the circle being 31416 to an area of 78540000; whilst in a square, for the same area, a length of wall equal to 35448 would be required. As the square falls away to a flat parallelogram, say one whose sides are half as great, and the others double the length of those of the square, or 17724 by 4431, in which the area will be about 78540000, as before ; we have in such a case a length of walling = 44310.

On the Thickness of Walls in Buildings not vaulted. 1529. The walls of a building are usually connected and stiffened by the timbers of the roof, supposing that to be well constructed. Some of the larger edifices, such as the ancient basilicæ at Rome, have no other covering but the roof; others have only a simple ceiling under the roof; whereas, in palaces and other habitations, there are sometimes two or more floors introduced in the roof.

1530. We will begin with those edifices covered with merely a roof of carpentry, which are, after mere walls of enclosure, the most simple.

1531. Among edifices of this species, there are some with continued points of support, such as those wherein the walls are connected and mutually support each other ; others in which the points of support are not connected with each other, such as piers, columns, and pilasters, united only by arcades which spring from them.

1532. When the carpentry forming the roof of an edifice is of great extent, instead of being injurious to the stability of the walls or points of support, it is useful in keeping them together.

1533. Many edifices exist wherein the walls and points of support would not stand without the aid of the carpentry of the roofs that cover them.

1534. The old basilica of St. Paolo fuori le murà at Rome was divided into five naves formed by four ranks of columns connected by arcades, which carried the walls whereon the roof rested; the centre nave 73} ft. (French) wide, and 93 ft. 10 in. high. The walls of it are erected on columns 31 ft. 9 in. high, and their thickness is a little less than 3 ft., that is, only s part of their height.

1535. At Hadrian's Villa the most lofty walls, still standing, were but sixteen times their thickness in height, and 51 ft. 9 in. long. The walls were the enclosures of large halls with only a single story, but assisted at their ends by cross walls. And we may therefore conclude that if the walls of the basilica above mentioned were not kept in their places by the carpentry of the great roof they would not be safe. It is curious that this supposition, under the theory, was proved by the fire which destroyed the church of St. Paolo in 1823. The walls which form the nave of the church of Santa Sabina are raised on columns altogether 52 ft. high ; they are 145 ft. long, and somewhat less than 2 ft., that is, a part of their height, in thickness. They are, therefore, not in a condition of stability without the aid of the roof. In comparing, however, the thickness of these walls with the height only of the side aisles, in the basilica of St. Paolo the thickness is ', and at Santa Sabina is. In the other basilicæ or churches with columns, the least thickness of wall is ; of greater proportion unconnected with the nave, as at Santa Maria Maggiore, Santa Maria in Trastevere, St. Chrysogono, St. Pietro in Vincolo, in Rome ; St. Lorenzo and St. Spirito, in Florence ; St. Filippo Neri, at Naples; St. Giuseppe and St. Dominico, at Palermo.

1536. We must take into account, moreover, that the thickness of walls depends as much on the manner in which they are constructed, as on their height and the weight with which they are loaded. A wall of rough or squared stone 12 inches thick, wherein all the stones run right through the walls in one piece, is sometimes stronger than one of 18 or 20 inches in thickness, in which the depth of the stones is not more than half or a third of the thick. ness, and the inner part filled in with rubble in a bad careless way. We are also to recollect that stability more than strength is ofttimes the safeguard of a building; for it is certain that a wall of hard stone 4 inches thick would be stronger than would be necessary to bear a load equal to four or five stories, where a thickness of 18 inches is used; and yet it is manifest that such a wall would be very unstable, because of the narrowness of the base.

1537. From an examination which Rondelet made of 280 buildings in France and Italy, ancient as well as modern, he found that in those covered with roofs of two inclined sides and constructed in framed carpentry, with and without ceilings, and so trussed as not to act at all horizontally upon the walls, the least thickness in brick or rough stones was

of the width in the clear.

1538. In private houses, divided into several stories by floors, it was observed, generally, that the exterior walls ran from 15 to 24 inches, party walls 16 to 20 inches, and partition walls 12 to 18 inches.

1539. In buildings on a larger scale, exterior walls 2 to 3 feet thick, party walls 20 to 24 inches, partition walls 15 to 20 inches.

1540. In palaces and buildings of great importance, whose ground floors are vaulted, the exterior walls varied from 4 to 9 feet, and the partition walls from 2 to 6 feet. In many of the examples which underwent examination, the thicknesses of the walls and points of support were not always well proportioned to their position, to the space they enclosed, nor to the loads they bore. In some, great voids occur, and considerable loads were supplied with but slender walls and points of support ; and in others, very thick walls enclosed very small spaces, and strong points of support had but little to carry.

1541. For the purpose of establishing some method which in a sure and simple manner would determine the thickness of walls in buildings which are not arched, we have considered, says Rondelet, that the tie-beams of the trusses of carpentry whereof the roofs are composed, being always placed in the direction of the width, as well as the girders and leading timbers of Moors, serve rather to steady and connect the opposite walls; but, considering the elasticity and flexibility of timber, it is found that they strain the walls which support them in proportion to the widths of the spaces enclosed, whence it becomes often the better plan to determine the thickness of the walls from the width and height of the apartments requisite. Hence the following rules.

First Rule. 1542. In buildings covered with a simple roof, if the walls are insulated throughout, their height up to the under side of the tie-beams of the trusses, being as shown in fig. 604. Having drawn the diagonal B D and thereon made Bb and Dil, equal to the twelfth part of the height AB, then through the points b and d, draw lines parallel to BA and DC, which will bound the thickness of the walls required.

1543. If the height AB and width AD be known, the thick ness Ac may be calculated

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32. O

24x2} = 1ft., or

First Example. 1544. Supposing the width AD=24 ft., and the height AB=32, we shall have VAB* + AD= 24 x 24 + 32 x 32; whence BD=576 + 1024=11600 = 40 ft. Bb, which is the twelfth part of AB, or of 32 ft. = 2 ft. 8 in. ; the thickness of the wall

ADx Bb expressed by

will be BD

40 i ft. 7 in. 2 lines, for the thickness sought.

1545. If the walls supporting the roof were stiffened by extra means, such as lower roofs at an intermediate height, as in churches with a nave and side aisles, we may make Be in the diagonal BD (fig. 605.) equal to one twelfth of the height above the springing of the side roofs, and ef a twenty-fourth part of that height below it, and draw through the point f a line

Fig. 604. parallel to AB, which will determine the thickness Af' sought; or, which amounts to the same thing, add together the total beight AB of the interior, and that of E B above the point of support, E, whereof take the twenty-fourth part, which will be equal to Be + ef.

2.0

Second Example. 1546. Fig. 605. is a section of St. Paolo fuori le murà, near Rome, as it was in 1816.

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The interior height to the under side of the tie-beams is 93 ft. 10 in. (French), whereof 26 ft. 2 in. is the exterior height above the roofs of the side aisles. These two dimensions together make 120 ft., whose twenty-fourth part is 5 ft., to which, on the diagonal BD make Bj' equal; then from the point f letting fall a vertical line, the horizontal line Be will determine the thickness, which will be 3 ft., the width of the nave being 73 ft. 6 in. In figures, as follows:

BD=793 ft. 10 in. x 93 ft. 10 in. + 73 ft. 6 in. x 78 ft. 6 in. = ~14207=119 ft. 2 in. 1547. For the thickness, eB, as before, BD: AD::Bf: Ag"; whence, Af' =A

ADXB, 73} x5_=3 ft. 1 in., instead of 2 ft. 11 in. 9 lines, the actual thickness of the walls. 119 ft. 2 in. 1548. The same calculation being applied to the walls of the nave of Santa Sabina

Dd

BD

(Rome), whose beight of nave is 51 ft. 2 in, and width 42 ft. 2 in., with a height of 16 ft. of wall above the side aisles, gives 21 in. 4 lines, and they are actually a little less than 24 in.

1549. In the church of Santa Maria Maggiore, the width is 52 ft. 7} in., and 56 ft. 6 in. and 4 lines high, to the ceiling under the roof. The height of the wall above the side aisles is 19 ft. 8 in., and the calculation requires the thickness of the walls to be 264 in. instead of 28fin., their actual thickness.

1550. In the church of St. Lorenzo, at Florence, the internal width of the nave is 37 ft. 9 in., and the height 69 ft. to the wooden ceiling; from the side aisles the wall is 18 ft. high. The result of the calculation is 21 in., and the actual execution 21 in, and 6 lines.

1551. The church of Santo Spirito, in the same city, which has a wooden ceiling sus. pended to the trusses of the roof, is 76 ft. high and 37 ft. 4 in. wide in the nave the walls rise 19 ft. above the side aisles. From an application of the rule the thickness should be 21 in. 3 lines, and their thickness is 224 in.

1552. In the church of St. Philippo Neri, at Naples, the calculation requires a thickness of 21 in., their actual thickness being 224 in.

1553. In the churches here cited, the external walls are much thicker ; which was necessary, from the lower roofs being applied as leantoes, and hence having a tendency, in case of defective framing of them, to thrust out the external walls. Thus, in the church of St. Paolo, the walls are 7 ft. thick, their height 40 ft. ; 3 ft. 4 in. only being the thickness required by the rule. A resistance is thus given capable of assisting the walls of the aisles, which are raised on isolated columns, and one which they require.

1554. In the church of Santa Sabina, the exterior wall, which is 26 ft. high, is, as the rule indicates, 26 in. thick ; but the nave is flanked with a single aisle only on each side, and the walls of the nave are thicker in proportion to the height, and are not so high.

For at St. Paolo the thickness of the walls is only at of the interior width, whilst at Santa Sabina it is at At San Lorenzo and San Spirito the introduction of the side chapels affords great assistance to the external walls.

Second RULE. For the Thickness of Walls of Houses of more than one Story. 1555. As in the preceding case, the rules wbich Rondelet gives are the result of observations on a vast number of buildings that have been executed, so that the method proposed is founded on practice as well as on theory,

1556. In ordinary houses, wherein the height of the floors rarely exceeds 12 to 15 ft., in order to apportion the proper thickness to the interior or partition walls, we must be guided by the widths of the spaces they separate, and the number of foors they have to carry. With respect to the external walls, their thickness will depend on the depth and height of the building. Thus a single house, as the phrase is, that is, only one set of apartments in depth, requires thicker external walls than a double house, that is, more than one apartment . in depth, of the same sort and height ; because the stability is in the inverse ratio of the width.

1557. Let us take the first of the two cases (fig. 606.), whose depth is 24 ft. and height

12

18

wwws

24 30

36 Ft.

Fiz. 606. to the under side of the roof 36 ft. Add to 24 ft. the half of the height, 18, and take i part of the sum 42, that is, 21 in., for the least thickness of each of the external walls above the set-off on the ground floor. For a mean stability add an inch, and for one still more solid add two inches.

1558. In the case of a double house (fig. 607.) with a depth of 42 ft., and of the same height as the preceding example, add half the height to the width of the building; that is, 21 to 18, and of the sum =194 is the thickness of the walls. To determine the thickness of the partition walls, add to their distance from each other the height of the story, and take za of the sum. Thus, to find the thickness of the wall IK, which divides the space LM into two parts and is 32 ft., add the height of the story, which we will take at 10 ft., making in all 42 ft., and take g or 14 in. Half an inch may be added for each story above The ground floor. Thus, where three stories occur above the ground floor, the thickness in

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