connecting rod will be represnted by the formula ƒ cos 0 sin f.

The inertia of a point in the axis of the connecting rod, and in the line of that axis, will therefore be a maximum when cos 0 sin & is a maximum. The position of the connecting rod which corresponds to its maximum inertia in the line of its axis, depends, therefore, upon the relative length of the crank arm and the connecting rod.-ED.]

48. A cylindrical tower, radius r, is surrounded by a walk, width a. Two persons are on the walk; what is the probability that they can see each other?"

SOLUTION BY ARTEMAS MARTIN, ERIE, PA.

Let P be the position of one of the persons. Through P draw the diameter AP OB, and from P draw PG, PH tangent to the tower at C and D; then if the other G person be on the surface AEG CIDHFA they can see each other.

Let OP =x, then arc EA = arc AF

arc CI arc ID

== COS

GC

arc GE arc FH - (ra) cos

r

-1

Ar + a

= DH = √(2ar + a3); area ECG area FDH =

( r + a)2

rv (2ar + a2), area EA CI= area AFID= (r + a)3

Sta [(r+a)3 cos-1 (1a) + (2ar+a2)cos ̄1 (—) — rv (2ar + a2

π(2ar + a2) S***2лxdx

2(a+r)2 cos-1[r÷(a +r)]

π(2аr + a2)

2r

πV(2ar + a2)*

49.-"How far will a man travel in unwinding an inch rope from a frustrum of a cone whose upper diameter is 2 ft., lower 15 ft. and height 35 ft. the rope to be closely wound around the frustrum from top to bottom?"

Solution by Prof. J. Scheffer, Gambier, Ohio.

The area of the surface of the frustrum is π(R + r)√[h2 + (R +r)2]; R denoting the radius of the lower base, r that of the upper and h the axis, of the cone. The question arises now; what must be the length of a rectangular strip whose breadth is one inch, in order that its area may be equal to the above stated surface? Thus we find for the required distance:

12π(R + r) √[h2 + (R + r)2] = 7616.6π = 23928.3 ft., or nearly 4 miles.

This solution is not mathematically accurate. The line which the rope forms is not a curve (spiral or screw line) of double curvature, in fact it is no mathematical curve at all whose length we could find by means of the formulas of Calculus. The distance, however, which has been found in the above simple and elementary way, will not differ by a great deal from the actual one.

[It will be seen that, in the above solution, Prof. Scheffer has estimated approximately, the length of the rope, (or, rather, double the length, as, in substituting for R and r he has probably used the diameters of the bases instead of their radii,) but he has not attempted the more difficult part of the question, viz; to find the length of the involute curve traced on the ground in unwinding the rope.

Let r represent the distance from the apex of the cone to any point P, on its surface where the distance between the centers of two consecutive coils is x; and let represent the angle traversed by the radius vector r in describing that part of the curve which lies above the point P, then is r = fdxb0 the polar equation of the curve. If a were constant this equation would represent the Spiral of Archimedes. For the given frustrum æ is nearly constant and nearly equal to one inch, but not quite. Determine x in functions of ; then is r = Fron (1) determine the length of the curve between the limits r = h and r = k. (h representing the slant height of the complete cone and k the distance, on the slant side, from the apex of the cone to any point P, to which the rope may have been unwound). Let L represent the length between the above named limits and let p represent the perpendicular height of the point P; then will √(L2 +p2) =the radius of curvature of the involute curve, from which its length may be found.

Sd.&(0).

(1)

As a first step in the actual solution of prob. 49 we propose the question: To find the equation between x and as involved in (1) above.-ED.]

50.-"Assuming the earth's orbit to be a circle, if a comet move in a parabola around the sun and in the plane of the earth's orbit, show that the comet cannot remain within the earth's orbit longer than 78 days."

Solution by Prof. A. B. Evans, Lockport, N. Y.

Let MAN be the orbit of a comet whose time within the earth's orbit BEDF is a maximum. Let A be the comet's position when nearest the sun S, and C any other position of the comet in its orbit.

Put SE =r, SA = a, SC P, LASC =0: then the equation of the parabola is

=

cos 20'

and the area

=

= √ax(r

+

Since cos20 a÷p, the area of ASD is found by putting cos 0 (ar), or tan 10 = v[(ra)+a], in (1). The area of ASD is therefore equal to a2v[(ra) a] + ža2v [(r — a)3 ÷ a3] =1/aX(r+2a)v (ra); and the area of BSA + ASD = 2ASD 2a)/(ra). . . . (2) Now the areas described about a common center of force by two bodies moving in different orbits being in the subduplicate ratio of the parameters of those orbits, and the parameters of the orbits in this case being 2r and 4a, we have (2): (4a):: r2: Vr3× √(2a) = the area described by the As the comet describes equal areas in equal times, we number of days in the year by T,

comet in one year.

have, denoting the

√r3 √(2a): Va.(r + 2a)v (r—a) :: T: T' the number of days the comet is within the earth's orbit

The only variable in the value of T being a, T' will be a maximum when (r+2a) v 2r-2a) is a maximum. Differentiating this expression with respect to and placing the differential coefficient equal to zero, we find 21 (2r-2a). - (r+ 2a)÷v (2r2a) = 0; whence r2a. This value of r substituted in (3) gives

PROBLEMS.

51. By Dr. N. R. Oliver, Sacaton, Arizona.-Given

to find x and y by quadratics.

(x2 + y2)y = 39,
2+ y = 97,

(1)

(2)

52. By J. M. Curtis, Ovid, Mich.-A hare starts 10 rods north of a hound and runs at an angle of 45° with the meridian. The hound pursues in a straight line and in such a direction that he will intercept the hare without changing his coursc. Supposing the hound to run n times as fast as the hare, how far will he run before he catches the hare?

43. By request of A. W. Mason.-If a ball of 6 in. diameter is discharged from a canon at the rate of 1 mile in 7 seconds, how much greater force will be required to throw a ball of double the weight with the same velocity, taking into account the resistance of the air and the diameters of the balls?-Silliman's Physics, page 108.

54. By Prof. J. Scheffer, Gambier, O.—Let ABC represent a spherical triangle and M the centre of the sphere. Find the three distances of the Dentre of gravity of the spherical triangle ABC from the three planes ABM, ACM and BCM.

55. By request of Walter Siverly.-A very small bar of matter is moveable about one extremity which is fixed half way between two centers of force attracting inversely as the square of the distance; if I be the length of the bar, and 2a the distance between the centers of force, prove that there will be two positions of equilibrium for the bar, or four, according as the ratio of the absolute intensity of the more powerful force to that of the less powerful is or is not greater than (a + 2)+(a-21): and distinguish between the stable and unstable positions.-Cambridge Problems, 1845.

1

56. By G. W. Hill. Find the real value of (√/—1)1

57. By Prof. E. W. Hyde.-Find the nature of the curve represented by the equation

4b2 (x2 + y2) [(x — a)2 + y2 — b2] — a3y3[(x — a)2 + y2]=0, and show that when a certain relation exists between a and b, the locus reduces to a right line perpendicular to the axis of x, and a 3rd degree curve which is a trisectrix.

58. By Prof. W. W. Johnson.-Referring to prob. 33, and Mr. Stille's figure in No. 9, find the rect. coordinates of the doub. pt. not on the axis of x.

We regret that, owing to a new arrangement for printing the Analyst and a new compositor, and other sufficient reasons which need not be explained, the present No. appears with an unusually large number of typographical We apologise to the authors for the appearance of their articles and hope to be able to bestow such personal attention to the proof reading, in the future, as will obviate the necessity of a like apology hereafter.

errors.

In the above errata we have, with two or three exceptions, given only the corrections of erroneous formulas that have been observed, without noticing the errors in spelling and punctuation which are very numerous: the reader, however, will correct these at a glance, and the authors will pardon our apparent inattention in this matter, we hope, for once at least.-ED.