CD = a, BA = = b, AD = c, and the angles at A and B respectively =a and 8; then prove that

a b sine 8+ b c sine aa c sine (a +

=} {ab

A =

B)}

VII. In a pleasure ground A B C D, known to be rectangular, only the following dimensions could be taken, on account of obstructions from buildings, shrubberies, and ponds, viz., the distance A E to the point where the perpendicular B E falls on the diagonal A C, and the prolongation E F of BE till it meet the side CD; it is required to find the area of the pleasure ground, when A E = 32, and EF = 4 chains.

Ans. 64 acres.

VIII. A large building is known to be of a square form, but no one of its sides could be measured on account of obstructions from other buildings; however, from a point P three streets diverge directly to its three nearest angles A, B, and C. Now PA 60, PB = 40 and PC = 70 yards; required the side of the building.

=

IX. Within a rectangular garden, the length of which is 4 and its breadth 3 chains, there is a piece of water in the form of a trapezium, the opposite angles of which are in a direct line with those of the garden, the distances of these opposite angles, taken in succession, are 20, 25, 40 and 45 yards; required the area of the water. Ans. 960 square yards.

X. Given the three perpendiculars of a triangle, from the angular points to their opposite sides, 10, 11, and 12 chains, to find the area of the triangle. Ans. 7a. Or. 4 p.

XI. Three objects, A, B, C, are observed at a point D, exterior to them; the distances of the objects are known to be as follows, AB = 8, BC = 12, and AC = 7 miles: the angles ADC, ABD are respectively 25° and 19°; required the distances A D, BD, and C D.

XII. A gentleman intends to make an elliptical garden, the principal axes of which are to be as 3 to 1, and the fence of which is to include three trees, one at the end of the transverse axis, the second 6 poles from it, and the third the same distance from the second, the three trees forming a right angle at the second; required the axes and area of the garden. (Gentleman's Mathematical Companion, Quest. 449.)

Ans. Minor axis 55.87 poles, area 73·55 sq. poles. XIII. In an elliptical enclosure of one acre the principal axes are as 5 to 4; required the length of a chord, which.

fastened to the end graze half an acre.

of the longer axis, will allow a horse to (G. M. Comp. Quest. 345.)

Ans. 48-45473 yards.

XIV. Given the diagonals and two opposite sides of a trapezium to construct it, when the area is a maximum or a minimum. (Prize Quest. G. M. Comp. for 1809.)

=

=e to

XV. In a trapezium ABDC are given all the sides AB = a, BD= b, DC c, CA = d, and the diagonal B C = find the diagonal A D, without constructing the figure. (B. Gompertz's Principles of Imaginary Quantities, page 29. Prob. V.)

XVI. There are given the four sides of a trapezium, and a line joining given points in two of its opposite sides, from which it is required to construct the figure.

XVII. The five sides a, b, c, d, and e of an irregular field are given, in which the angles between a and b, b and c, c and d, are equal but not given; from these data it is required to lay down the field.

NOTE. The solutions of the two last Problems are obtainable by cubic equations. The former was proposed by the author, in the Gentleman's Diary for 1838, it having occurred in his practice of town surveying for railway purposes. The latter was proposed by B. Gompertz, Esq., F.R.S., &c., in the Gentleman's Mathematical Companion; to which he gave a solution in a concise, novel, and ingenious manner by his Principles of Imaginary Quantities: other solutions by the ordinary methods were also given to the same Problem.

XVIII. A gentleman has an elliptical garden, the principal axes of which are 50 and 40 yards, enclosed by a brick-wall 13 feet high. He ordered his gardener to place his seat at equal distances from the centre, one of the foci and the boundary of the garden, and around the seat to make a gravel walk of equal breadth taking up of the area of the garden, and to be of such a nature, that, while the gentleman is seated, and the gardener moving along the middle of the walk, the gentleman's eye, the gardener's utmost height, and the top of the wall, may be in the same right line; the height of the gentleman's eye, (when seated), and that of the gardener, being 4 and 6 feet respectively. Required the position of the seat, and the nature and breadth of the walk. (G. M. Comp. for 1806, Quest. 14.) Ans. The equal distances of the seat from the centre, focus,

and curve = 10-7812 yards; the walk is elliptical, its axes being 11 and 91 yards, its breadth 2 yards, and the position of the walk with respect to the seat being the same as the fence of the garden.

NOTE. The solutions to all the Questions, taken from the works referred to, may be seen in those works.

(DIVISION OF Land.)

XIX. In a triangle ABC, AB = a, BC = b, CA = c, and from D, a given point in AB, the distance DB = d, DE a line meeting B C and dividing the triangle, so that ▲ ABC: ABDE:: m : n; then prove that

But if it be found from the calculation that BE is greater than B C, and that the divisional line will meet AC in some point F; then prove that

AF

ac (m n)
m (a — d)

XX. ABCD is a given trapezium, and T a given point in AD, from which a line TE is drawn to meet BC in E, so that trapezium A B C D : trapezium DCET :: m : n. Now, since the trapezium A B C D is given, if the sides A D, BC be prolonged till they meet in Z, the point Z will also be given; therefore, put AZ = a, BZ = b, DZ = c, CZ = d, and

TZ=ƒ; then prove that

EZ =

n ( a b − cả)
m f

cd

+ F
f

XXI. It is required to divide a given trapezium into four equal parts, by two right lines perpendicular to one another. NOTE. This Problem was proposed, the author believes, in one of the Diaries, hany years ago, where its solution may be seen.

(RAILWAY ENGINEERING.)

XXII. Let be distance between the tangent points of a serpentine curve, consisting of two circular arcs of equal radii r, a and ẞ the angles that makes with the tangents, and σ = arc to (cos. a+ cos. B); then prove that

XXIII. Find the distances of the tangent points to the point of contrary flexure of the curve, in the last Problem, and the

angles that these distances make with the tangents to the curve, and give the requisite formula.

XXIV. The widths of a laterally sloping railway cutting from the centre of the line are expressed generally by the following formula.

the positive sign being used for the width measure down the slope, and the negative one for that up the slope, in which formula b = width of the cutting, assuming its surface to be level, h = differences of level readings at the distances s and l on the slope and level, and r the ratio of the slopes. See figure to Prob. II., Chap. III.

bot

XXV. Let a and b be the depth of a railway cutting to the intersection of the slopes, = length of the cutting, w tom width, all in feet, and r = ratio of the slopes, the surface of the cutting being assumed to be horizontal; then prove that the content of the cutting in cubic yards is

XXVI. When a and b are the depths of a horizontal cutting to the formation level, and the other dimensions the same as in the last Problem; then prove that the content of the cutting in cubic yards is

XXVII. Let A and B be the areas of the cross sections of a cutting to the intersection of the slopes, d, the area of the end of the prism below the formation level, (see fig. p. 194) and 7 the length of the cutting; then prove that the content in cubic yards is

(A + B + √AB − 3 d).

XXVIII. Let the dimensions be as in Problem XXV then prove that the error in defect of the method of finding the content of a cutting by using the mean depths is

l r (a 324

b)2.

as in Problem XXVII.;

XXIX. Let the dimensions be

then prove that the error in excess of the method of finding the content of a cutting by using mean areas is

XXX. Prove the formula at p. 202, for finding the error per cent. of Mr. Bashforth's method of finding the contents of railway cuttings from sectional areas.

NOTE. The demonstrations of the nine last Problems are given in Baker's Railway Engineering.

CENTRIFUGAL FORCE OF TRAINS IN RAILWAY CURVES.

Since all moving bodies have a tendency to continue their motion in a direct line, from this cause the carriages of a railway train of great velocity are strongly urged towards the outer rail, and would ultimately be driven off the rails, were it not for the flanges of the wheels and the conical inclination of their tire.

Let F centrifugal force thus generated, W = weight of the train, V = its velocity, R = radius of the curve, in which the train moves, and g the force of gravity at the earth's surface; then by Dynamics

F=

W V2
9 R

that is, the force that urges the train to quit the curve is of its whole weight, in this case.

2. When V = 60 miles per hour = 88 feet per second, and R the same as in Example 1; then

* This great amount of centrifugal force, in curves of small radius, would be very much increased by the high velocities, which some are sanguine enough to expect as likely to be attained on railways; since this force varies as