PROBLEM VI.

To find the volume of a wedge.

Def. A wedge is a solid, bounded by five planes; the rectangular base, ABCD, two trapezoids, ABFE, DGCFE, intersecting in the edge EF,

and two triangular ends, D.

ADE, BCF.

The altitude of the wedge is

the perpendicular drawn from

any point in the edge to the plane of the base, as EH.

Put

1=AB=length of the base.

B

Now if the length of the base and that of the edge are equal to each other, it is plain that the wedge is half a parallelopiped of the same base and altitude.

If the length of the base is greater than that of the edge, let a plane, EGI, be drawn parallel to BCF. The wedge will then be divided into two parts, viz., the triangular prism BCG-F, and the pyramid AIGD-E.

The vol. of prism=bhl'

(Vol. of Solids, Prob. I.) vol. of pyr. =3bh(l—l'′), (Vol. of Solids, Prob. II.)

.. vol. of wedge-bh(21+1').

If the length of the base is less than that of the edge, then the wedge will be equal to the difference between the prism and pyramid.

.. vol. of wedge={bh(21+1'),

the same as above.

Hence we derive the following

RULE.

To twice the length of the base add the length of the edge, and multiply the sum by one sixth of the product of the height of the wedge and breadth of the base.

EXAMPLE.

What is the volume of a wedge whose base is 30 inches long and 5 inches broad, its altitude 12 inches, and the length of the edge 24 inches? Ans. 840 cubic inches.

PROBLEM VII.

H

To find the volume of a rectangular prismoid. Def. A rectangular prismoid is a solid bounded by six planes, of which the two bases are rectangles, having their corre- G sponding sides parallel, and the four oth- 72 er sides are trapezoids.

Put

l'=AD=BC;

b' AB CD;

1=HE=GF;

b=HG=EF;

halt. of the prismoid;

M=the length of the mid section=pq;

m the breadth

66

If a plane be passed through the opposite edges of the upper and lower bases, the prismoid will be divided into two wedges, whose bases are the bases of the prismoid, and whose edges are l' and l.

The volume of these wedges, by Prob. VI., is

fbh(21+1') + b'h (2 l'+1)

=h(2b1+bl'+2b'l'+b')=vol. of the prismoid.

But, since M and m are the length and breadth of the mid sections, we have

2M=1+l', and 2m=b+b'

.. 4Mm=(l+1') (b+b')=bl+bl′+b'l+b'l'. Substituting 4Mm for its value in the preceding expression, we have for the value of the prismoid

th(bl+4Mm+b'l'),

from which we derive the following

RULE.

To the sum of the areas of the two bases add four times the area of a middle section, and multiply the sum by one sixth of the altitude.

EXAMPLE.

What is the volume of a block of marble in the form of a rectangular prismoid, the length and breadth of one end being 16 and 12 inches, and of the other 7 and 4 inches, the length of the block being 24 feet?

Ans. 163 cubit feet.

PROBLEM VIII.

To determine the volume of a railroad cutting.

In Prob. VII. the figure is very nearly that of a portion of a railroad cutting in which AC is the road, and AG and DF the sloping sides of the embankment; hence the solid contents required can be found by means of the rule given in the last problem.

It is to be observed that the rule requires that HG and EF be straight lines, or, as an approximation, to be very nearly straight lines; or AG to be a plane, which is not true if the cutting is a long one. For this reason we shall derive the following:

Let ABCD represent a section, made by a vertical D plane, of the hill to be cut through; AB the level of the road; and suppose sections of the cutting to be Al

made by planes perpendic

PPR P

ular to AB, at equal distances along that line, viz., at M1,

M2

Man+1

.....

Let the terminal sections at A and B be denoted by a and b respectively, and the sections at P1, M1, P2, M1⁄2 . . . &C., P2n+19 M2n+1, be denoted by P1, P2 ·

•P2n+19 the number of sections being odd. And let the common distance between the sections equal h. Then (by Prob. VII.),

vol. of AM,P,D=‡AM,(a+4p1+P2); or

=(a+4p1+P); and

vol. of PMMP1=}h(p2+4P3+P1), and so on ;
vol. of PMBC=h(P2n+4P2n+1)+b.

·2n

IIence, by addition, the whole required volume

=}h{(a+b)+4(P1+P2+, &c. .... P2n+1)+2(P2+P1+

Between the first and last sections, make an odd number of sections at equal distances along the road, the planes of the sections 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 required.

SPHERICAL TRIGONOMETRY.

CHAPTER I.

1. SPHERICAL TRIGONOMETRY treats of the relations subsisting between the parts of a solid angle formed by the inclination of three planes; or, supposing the angular point the centre of a sphere, the relations between the sides and angles of the triangular surface of the sphere inclosed by the three planes.

2. A spherical triangle is the portion of the surface of a sphere inclosed by the three planes which form the solid angle at the centre: it is bounded by three arcs of great circles.

3. A spherical angle is the angle contained by any two of the planes passing through the centre of the sphere.

4. The sides of a spherical triangle are the three arcs which measure the rectilinear angles at the centre respectively. The sense in which the spherical triangle is employed being once understood, we may transfer our attention from the solid angle to the triangle, and the solid angle need not be represented in our diagrams.

The first formula which we will establish is to spherical trigonometry what the expression for the sine of two angles is to plane trigonometry; that is, it serves as a key to all the rest.

*PROBLEM I.

To express the cosine of an angle of a spherical triangle in terms of the sines and cosines of the sides.

Let ABC be a spherical triangle, O the centre of the sphere. Let the angles of the triangle be denoted by the large letters A, B, C, and the sides opposite to them by the corresponding small letters a, b, c. At the point A draw AT, a tangent to the arc AB; and AT' a tangent to the arc AC.

A