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Hence, by addition, the whole volume required is
(a + b + 4 (p2 + P3 + ... + P2m + 1) + 2 (P2 +P1•
which may be expressed as a rule as follows:
· P2m) }
Between the first and last sections make an odd number of sections at equal distances along the road, the planes of the section being perpendicular to the road; then one third part of the common distance multiplied by the sum of the first and last sections, with four times the sum of the odd sections, and twice the sum of the even sections, gives the volume between the first and last section.
The student will observe that this rule is the same as that for finding the area of a figure bounded by a curve, which has been already given (p. 379), excepting that the ordinates in the former rule are replaced by trapezoidal sections in the latter. The formula given in article 10 is called the Prismoidal formula. It will be observed that the material on each side of a cutting being generally the same, is the reason the inclination of the planes AG and BH to BD being generally the same as stated in the last article. Mr. Macneill, to whom the prismoidal formula is due, has constructed tables founded on that formula, by which the volume of a cutting is very readily calculated. The volume of an embankment is to be found by the same formula, since, as a question of mere figure, an embankment is only an inverted cutting. If the calculation is made directly from the formula it is very tedious; the value of the Tables above referred to is therefore very great, and is enhanced by the following circumstance :-In constructing a long line of railway, the earth taken out of the cutting should be sufficient to form the embankment, otherwise land must be purchased for the mere purpose of obtaining earth; to effect this end of making the volume of the embankments equal that of cuttings, the ascents and descents (gradients) of the line of road have to be properly chosen, and this can only be done by trial, so that the calculation may have to be performed two or three times before a right adjustment can be hit upon. It is worth adding that, as a general rule, a cutting is followed not by a long level, but by an embankment; if possible, these two are adjusted to each other to prevent the need of carrying earth from long distances.
12. To find the Solid Content of a Military Earth-work.
The form of a military earth-work will be understood from the following explanations :
The form of a section of the work made by a vertical plane perpendicular to the face of the work is such as ABCDEF from B,C,D,E draw perpendiculars to AF, viz., Bm, Cn, Dp, Eq, then Am, mn, np, pq, qF, are of known magnitudes, as also are Bm, Cn, Dp, Eq. The plan of the work will be of the accompanying kind, viz., F'ƒ (Fig. 47) is the line corresponding to F (Fig. 46) E'e the line corresponding to E, and so on for the others; the lengths of all these lines are known; we will call them a,b,c,d,e,f, respectively.
In fig. 46 join pE, pC, pB, dividing the section of the work into triangles ApB, BC, CpD, DpE, and EpF; call these areas respectively Â ̧Â ̧Â ̧Â ̧Ã ̧, then the contents of
the work are clearly equivalent to the frustums of five prisms (similar to that in the Corol. to article 7), which have the area of perpendicular section A, and edges f,e,d, section A, and edges e,d,d, section A,, and edges d,d,c, section A, and edges c,d,b, and section A1 and edges b,d,a, it being evident that the edge through p = d. Hence if the whole volume equals V, we have
In which formula it will be observed that each line in the plan is multiplied by the triangle, or by the sum of the triangles, which have an angular point in that line, and that the line dD' is also multiplied by the whole area of the section. Hence if these products are formed and added together, the required volume is one third of the
13. To find the Volume of the Frustum of a right Cone made by a Plane parallel
to the Base.
Let ABCD be the frustum of the cone PCD. Join POO, where 00, are the centres of the ends of the frustums. Let AO = r CO2 = r1 002
which is the same as the volume of three cones whose common height is h, and which have the radii of their bases respectively r, r, and a mean proportional between r and 3⁄41⁄2.
14. To find the Volume of a Portion of a Sphere.
Let ABO be a quadrant of a circle, draw BC, AC, tangents at A, and B, join OC, and take any point P in AB and draw QPN parallel to BO; then if the whole figure revolve round AO, the quadrant ABO will clearly describe a hemisphere, the square AOBE will describe a cylinder, on a base whose radius is OB, and height AO, and the triangle AEO will describe a cone on a base whose radius is AC and height AO. Also APN will describe a portion of a sphere, ACQN a cylinder whose height is AN and radius AC, and ACMN a frustrum of a cone, the radii of whose ends are AC and MN, and height AN. Divide AN into any number of equal parts, of which let m m be one, through m, draw mk parallel to BO, meeting AB in p, and OC in k, and through m
draw a line m1 k, parallel to mk, and from p and h draw pp, and hh, perpendicular to m, k. Join Op. Now Op2 mp2 + m2 since Omp is a right angle. But since triangle Omh is similar OAC and ACAO .. Om mh also km во Op .. mk2 mp2 + mh2. π × mk2 = π × mp2 + × mh2 or (Mensuration of Areas, Art. 8) the circle, the radius of which is mk is equal to the sum of those whose radii are mp, and mh; also since kk, = pp1 = hh, we have
or the cylinder, the radius of whose base is mk and height kk, is equal to the cylinder, the radius of whose base is mp, and height pp1, together with that the radius of whose base is mk, and height hh,; but these cylinders are those which will be described by mk, mp, and mh,, when the whole figure revolves round AO. And the same is true of the cylinders corresponding to any other one of the equal parts into which AN is divided, and therefore is true of all of them. Now all the cylinders corresponding to km, make up the cylinder described by AQ, and all those described by mp, make up the series of cylinders inside the sphere, and those described by m1 h make up those described about the cone. Hence,
Cylinder AQ= sum of cylinders inside portion of sphere + sum of cylinders outside portion of cone.
This being true, however small m m, may be, is true in the limit, but the portion of the sphere is the limit of the inscribed cylinders, and the frustrum of cone is the limit of the cylinders outside the cone (compare note, p. 376). Hence,
Cylinder described by AQ= portion of sphere described by APN + portion of cone described by ACMN.
Now column of cylinder described by AQ Trh. And volume of frustum of
Hence if V = volume of the whole sphere, in which case h
COR. 1.-The above proposition may be demonstrated in the following manner. Suppose a solid having any number of plane faces to be described in the sphere, and let A, A, A, &c., be the areas of these faces, and P1 P2 P3 &c., be the perpendicular distances of these faces from the centre of the sphere; now this inscribed solid may be conceived to be made up of pyramids, the bases of which are the faces of the solids, having the centre of the sphere for their common vertex. Hence, if the volume of the solid is V1
Now this is true, however great the number of faces may be, and.. is true in the limit, but in the limit P, P2 P3... become equal to one another and to the radius of sphere. Hence,
+ A, x Ag...)
Limit of A, p, + A2 P2 + A3 P3 + ... = x (limit of A, Now the limit of A, + A + A + ... = surface of sphere of Areas, Art. 19). Also the limit of V1 IV the volume of sphere.
4 92 (Mensuration
COR. 2.-To find the volume of the portion of the sphere corresponding to BPNO. Let the volume be called V, and let h
= ON which is
Now volume of cylinder BN
Volume of cone MNO
15. To find the Volume of a Spheroid.
DEF. A spheroid is a figure formed by the revolution of an ellipse about one of its axes; if about the major axis it is called a prolate spheroid, if about its minor axis it is called an oblate spheroid.
Let ABa be a semi-ellipse, OA its semi-major axis, OB its semi-minor axis. With centre O and radius OA describe a semi-circle ACa, draw QPN through any point in the ellipse parallel to OC,
draw Mnm parallel to NPQ and complete the rectangles MP, MQ (Mensuration of Areas, Art. 12).
Now if the figure revolve round Aa the semi-ellipse will describe a prolate spheroid and the circle a sphere, also MP and MQ will describe cylinders with altitudes MN and having the radii of the bases NP and NQ respectively, and.. having volumes TMNx PN2 and T. MN x NQ2. Now by a property of the ellipse, PN a: b
:: QN2: PN2: a2: b2
Cylind. MQ cylind. MP :: MQ2 : MP2
.. cylind. MQ cylind. MP :: a2 : b2
and the same is true of any other cylinders described in like manner within the sphere and spheroid, and .. is true of their sum
.. all cylinders within sphere all within spheroid :: a2: b2
and this being true, however many cylinders there may be, i. e. however small we suppose MN to be, is true in the limit, but the sphere is the limit of its inscribed cylinders, and the spheroid the limit of its inscribed cylinder,
Cor: In like manner the volume of an oblate spheroid is
16. To find the Volume of a portion of a Spheroid, cut off by a plane Perpendicular to the Axis of Revolution.
In figure 50, suppose we wish to find the volume of the portion of the spheroid corresponding to the portion APN of the generating ellipse; then, as in Art. 15, we shall have
Now (Mensuration of Solids, Art. 14.) if AN = h the volume of the portion of the Th2 sphere is equal to
-(3a-h.) .. If V, is the volume required,
as is evident from corol. 2. Art. 14. Mensuration of Solids.
17. To find the Volume of a Cask.
We may consider a cask to be either the middle portion of a spheroid, or two frustums of equal cones joined together at their bases, though it will not coincide