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Example 4. Required the content of a cone whose heighit is 104 feet and the circu

ference of its base 9 feet.
Here, *07958 (Prob. 9. 1222.) x 81 =6.44598 area of base,

And 3.5 being of 10 feet, 6-44598 x 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. TH

taking one third of that sum for a mean area and multiplying it by the p pendicular height or length of the frustum, we shall have its content. This r

depends upon Prop. 110. Geometry, 991.). It may be otherwise expressed when the ends of the frustum are circles or regu polygons. In respect of the last, square one side of each polygon, and also multiply side by the other; add the three products together, and multiply their sum by the tabu area for the polygon. Take one third of the product for the mean area, which multi by the length, and we have the product required. When the case of the frustum of a c is to be treated, the ends being circles, square the diameter or the circumference at e end, and multiply the same two dimensions together. Take the sum of the three p ducts, and multiply it by the proper tabular number, that is, by *7854, when the diamet are used, and .07958 when the circumferences are used, and, taking one third of the p duct, multiply it by the length for the content required. Example 1. Required the content of the frustum of a pyramid the sides of wh

greater ends are 15 inches, and those of the lesser ends 6 inches, and its altit
24 feet.

Here, -5x -5=25, area of the lesser end,
and 1.25 x 1.25=1.5625, area of the greater end :
The mean proportional therefore v-25 x 1.5625='625.
-25+625+1.5625_ 2.4375

3°=8125, mean area,
and •8125 24 (altitude)=19-5 feet, content required.
Example 2. Required the content of a conic frustum whose altitude is 18 feet,

greatest diameter 8, and its least diameter 4.

Flere, 64 (area gr. diam.) +16 (less. diam.) +(8 x 4)=112, sum of the produ and 7854 x 112x 18

$=527-7888, content required.
Example 3. Required the content of a pentagonal frustum whose height is 5 f
each side of the base 18 inches, and each side of the upper end 6 inches.

Here, 1.52 +1:52 +(1.5 x :5)=2.5625, sum of the products;

1-7204774 (tab. area) x 2 5625 (sum of products) *5 =9-31925, content requir
1237. Problem VII. To find the surface of a sphere or any segment of one.
Rule 1. Multiply the circumference of the sphere by its diameter, and the product

be the surface thereof. This and the rules in the following problems depend

Props. 113. and 114. (Geometry, 994, 995.), to which the reader is referred.
Rule 2. 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 deci

3183, or divide it by 3.1416 for the surface.
Remark. For the surface of a segment or frustum, multiply the whole circumfere

of the sphere by the height of the part required.
Example 1. Required the convex superficies of a sphere whose diameter is 7
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 axi
42 inches and the height of the segment 9 inches.

Here, 1 : 3:1416::42 : 131.9472, the circumference of the sphere;

but 131 •9472 x 9=1187·5248, the superficies required. Example 4. Required the convex surface of a spherical zone whose breadth or hei is 2 feet and which forms part of a sphere whose diameter is 124 feet. Here, 1 : 3.1416::12.5 : 39:27, the circumference of the sphere whe

the zone is a part;

and 39.27 x 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

Rule 2. Take the cube of the diameter and multiply it by the decinal •5236 for

Example. Required the content of a sphere whose axis is 12.

Here 12 x 12 x 12 x5236 = 9047808, content required.

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1939. 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

diaineter 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. Esample 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 ineluded will give the solidity of the zone. That is, where for instance the zone is inHluded in a segment lying above the diameter, first consider the whole as the segment of a phere terminated by the vertex and find its solidity ; from which subtract the upper part I segment between the upper surface of the zone and the vertex of the sphere, and the lifference 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 he squares of the radii of the two ends add one third of the square of their distance, or the readth of the zone, and this sum multiplied by the said breadth, and that product again by 1-5708, is the solidity.

Sect. VII.

MECHANICS AND STATICS. 1940. 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 rience 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 quilibrium.

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, uch body will rest in any position.

1243. So that weight and power, when opposed to cach other, signify the body to be moved, and the body that moves it, or the patient and agent. The power is the agent which DOTES 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 ach other, and which by destroying each other's effects cause it to remain at rest.


1944. If a body D suspended by a thread is drawn out of its vertical direction by to horizontal thread DE (fiy. 519.), such power neither increases nur diminishes the effort

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of the weight of the body; but it may be easily imagined that the first thread, by being the direction AD, will, besides the weight itself, have to sustain the effort of the po that draws it out of the vertical AB.

1245. If the direction of the horizontal force be prolonged till it meets the vert which would be in the first thread if it were not drawn away by the second, we shall triangle ADB, whose sides will express the proportion of the weight to the forces of two threads in the case of equilibrium being established; that is, supposing AB to exp the weight, AD will express the effort of the thread attached to the point A, and BD 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 (fig.519.) any length of line DF to represent the weight of the body. If from the poir the parallels FI, FG be drawn in the direction of the threads, their forces will be indie by the lines ID, DG, so that the three sides of the triangle DGF, similar to the tria ADB, will express the proportion of the weight to the two forces applied to the thread

1247. Suppose the weight to be 30 lbs.: if from a scale of equal parts we set up of those parts from D to F (fig. 519.), we shall find DG equal to 21, or the pound 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 forces DG and ID by the proportions of 30: 21:100: x, where x expresses the force The value resulting from this proportion is x = 30= 70. The second propor 30 : 35::100 : y,where y represents the effort ID, whose value will be y = 35x100 =1166

1249. If the angle ADH formed by AD and DH be known, the same results ma obtained by taking DF for the radius, in which case IF=DG becomes the tangent, in 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 mentione
figure DIFG has been formed, which is called the parallelogram of the forces, because
diagonal DF always expresses a compounded force, which will place in equilibrio the
others FI, FG, represented by the two contiguous sides IF, FG.

1251. Instead of two forces which draw, we may suppose two others which act by p ing from E to D (fig. 522.) and from A to D. If we take the vertical DF to express weight, and we draw as before the parallels FG and FI in the e direction of the forces, the sides G D 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 po higher than the body (fig. 520.), it follows that the oblique forces AB and BC support, independent of any lateral efforts, a part of the weight of the body. In orde tind the proportion of these parts to the total weight, take any distance BD on a ver raised from the centre of the body B to express the weight, and complete the parall gram DEBF, whose sides EB, BF will express the obliqúe forces of the powers A C. These lines, being considered as the diagonals of the rectangular parallelograms LE BHFM, may each be resolved into two forces, whereof one of them, vertical, sustains body, and the other, horizontal, draws it away from the verticals AO, CQ, Hence IB express the vertical force, or that part of the weight sustained by the power EB, and that sustained by the other power BF: as these two forces act in the same direct when added together their sum will represent the weight DB. In short, IB b equal to HD, it follows that BH + BI= BI + ID.

1252. As to the horizontal forces indicated by the lines LB and BM, as they are ec and opposite they destroy one another.

1253. It follows, from what has been said, that all oblique forces may be resolved two others, one of which shall be vertical and the other horizontal, by taking their direc for the diagonal of a rectangular parallelogram.

1254. In respect of their ratio and value, those may be easily found by means of a sif the diagrain be drawn with accuracy; or by trigonometry, it we know the ang

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Fig. 523.


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ABD, DBC, which AB and BC form with the vertical BV, by taking successively for talius the diagonals BD, BE, and BF.

1255. In the accompanying diagram, the weight, instead of being suspended by strings ating by tension, is sustained by forces which are supposed to at by pushing. But as this arrangement makes no alteration 2 the system of forces, we may apply to this figure all that has been said with respect to the preceding one. The only differ. race is, that the parallelogram of the forces is below the weight instead of being above it. Thus ID+1B=BD exTesses the sum of the vertical forces which support the weight, und MB and BL the horizontal forces which counteract each ber.

1256. In the two preceding figures the direction of the forces which act by tension or compression in supporting the weight arm an acute angle. In those represented in fig. 521. and the igure at the side (524.), these directions make an obtuse angle; rience it follows that in fig. 521. the force C which draws the weight out of the vertical IL, instead of tending to support the weight B, increases its tfæet by its tendency to act in the same direction. In order to scertain the amount of this effect upon BD in figs. 521. and 94., which represents the vertical action of the weight, describe be parallelogram BADF, for the purpose of determining the blique forces BA, BF, and then take these sides for the diagoals of the two rectangles LAIB, BHFM, whose sides Bİ, BH rall express the vertical forces, and LB, BM the horizontal

1257. It must be observed that in fig. 521. the force AB eting upwards renders its vertical effect greater than the weight [ a quantity ID, which serves as a compensation to the part XH, that the other force BF adds to the weight by drawing bwowards. Similarly, the vertical effect of the force BE ( fig.

Fig. 521. 24.) exceeds the expression BD of the weight by a quantity DI, b counterpoise the effect BH of the other power BF, which acts downwards; so that in oth cases we have BD only for the vertical effect of the weight. As to the horizontal facts LB and BM, they being equal and in oppoite directions in both figures, of course counteract nch other.

1958. For the same reason that a force can be reolved into two others, those two others may be reolved into one, by making that one the diagonal of a arallelogram whose forces are represented by two mitiguous sides. It is clear, then, that whatever x be number of forces which affect any point, they may be reduced into a single one. It is only necestry to discover the results of the forces two by two nd to combine these results similarly two by two, ill we come to the principal ones, which may be ulimately reduced to one, as we have seen above. By ucb a process we shall find that PY (fig. 525.) is be result of the forces PA, PB, PC, PD, which feet the point P.

1259. This method of resolving forces is often of grcat utility in the science of building br the purpose of providing a force to resist several others acting in different directious but meeting in one point.




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Fig. 525


1260. The lever is an inflexible rod, bar, or beam serving to raise weights, whilst it is kapported at a point by a fulcrum, or prop, which is the centre of motion. To render the Heinonstrations relative to it easier and simpler, it is supposed to be void of gravity or bright. The different positions in which the power applied to it, and the weight to be afected, may be applied in respect of the fulcrum, have given rise to the distinction of thræ sorts of levers.

1. That represented in fig. 526., in which the fulcrum () between the power applied Pand the weight Q.

II. That represented in fig 527., in which the weight Q is placed between tiie fulerum

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Fig. 526.
Fig. 527.

Fig. 528.
111. That represented in fig. 528., wherein the power P is placed between the wei
and the fulcrum, in which case the power and the weight act in contrary directions.

1261. In considering the fulerum of t. ese three sorts of levers, we must notice, a third species of power introduced for creating an equilibrium between the others, That in which the directions of the weight and of the powers concur in the point R (fig. 529.).

That in which they are parallel.

1262. In the first case, if from
the point R (figs. 529. and 530.)
we draw parallel to these directions
Om Rn, the ratio of these three
forces, that is, the power, the weight,
and the fulcrum, will be as the three ez
sides of the triangle Om R, or its
equal On R; thus we shall have P

:Q:R::mR: Rn : OR; and as
the sides of a triangle are as the
sines of their opposite angles, by

Fig. 529.

Fig. 530
taking OR as the radius we shall have

P:Q::sin. ORn : sin. O Rm.
And if from the point O two perpendiculars be let fall, Odof, on the directions RQ, R.

Sin. O Rn : sin. ORm:: 00 : Of;
from which two proportions we obtain

P:Q::00 : Of; whence P x Of=Q - Od. This last expression gives equal products, which are called the momenta, moments, or qua tities of motion of the force in respect of the fulcrum 0. This property is the same for t straight as for the angular levers (figs. 529. and 530.). As this proportion exists, howev large the angles mRO and ORn of the directions RQ RP in respect to RO, it follows th when it becomes nothing, these directions become parallel without the proportion bei changed ; whence is derived the following general theorem, found in all works on mechanic If two forces applied to a straight or angular lever are in equilibrio, they are in an inve ratio to the perpendiculars let fall from the fulcrum on their lines of direction : or in other wor In order that two forces applied to a straight or angular lever may be in equilibrio, their momer in respect of the fulcrum must be equal.

1263. Since, in order to place the lever in equilibrio, it is sufficient to obtain equal m menta, it follows that if we could go on increasing or diminishing the force, we might pla it at any distance we please from the fulcrum, or load it without destroying the equilibriu This results from the formula Px Of=Q x Od, whence we have Of=bxod. Hence the distance Of is easily found, to which by applying the known force P, it may counterpoise the weight Q applied at the distance Od. In respect of the other points, we have only to know the perpendiculars Of and Od, for Oa and oh, which are the arms of the real levers, are deduced from the triangles Ofb, Oda, to which they belong.

1264. Suppose two levers (figs. 531, 532.), whereof

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