wbw, bww; or three white, www. Hence, before drawing from the bag, the chances that the balls are one white, two white, three white are respectively,,.

16

6,

Now, if there is only one white ball in the bag, the chances that it will be drawn four successive times are (1); if two of the balls are white, the chances are (); if three of the balls are white the chances are (). Combining, we get (1)=16, }(})* = 21%, (3), as the a priori probabilities that a white ball was drawn from the bag four times in succession. But since a white ball is drawn four successive times, the respective probabilities are

=

1. Notes on the Inclined Plane and the Wedge.-Let A B C represent any inclined plane; W the weight, placed at 0; P the power, which is constant and may be denoted by Oa, the radius of a circle; and R the reaction in the line OD perpendicular to the plane. The condition of equilibrium will be represented by the three sides of a triangle. R must always act in the line OD, W must be perpendicular to the horizon, and P may vary in direction, and will give different values for R and W in different positions.

When P acts in the line Oa,

W must be in the same line, and therefore PW and R = 0. When P acts in the line Og, R must act in the same line, and therefore PR and W= 0. All the positions including the limits

are then represented as follows:

In general W increases from an equality with P to its maximum value when P acts parallel to the plane, and afterwards decreases again to its minimum value, or zero, when P acts perpendicular to the plane. R increases from its minimum value, or zero, to its maximum, when P acts parallel to the base of the plane, then decreases again to equality with P, when P acts perpendicularly to the plane. When P acts in directions beyond the limits Oa and Og, negative results for W and R are obtained, of an interesting character.

Let A B C represent any wedge whatever; also RR and TT any resistances whatever, acting at any angles, and '. Resolve RR' into RM and RM; and RM into RN and MN. Also in like manner resolve TT" into TV and T'V; and TV into VS and S T Produce RM and TV until they meet in 0, and beyond O take OX= VT, and OY

0 Y= RM. Next, complete the

parallelogram OXPY; then draw the diagonal OP, and lastly,

MN represents the effective force

of RR', and VS that of TT. Therefore when there is an equilibrium, MN+VS=PP.

From Y draw YK perpendicular to PP, and complete the parallelogram YKLO. The triangles RMN and LKP are equal, since the sides are respectively parallel, and L K=0 Y= RM by construction; therefore P' K= MN. The triangles PKY and TVS are equal, for a similar reason, and PK = VS. Consequently MN+VS=P' K+KP=PP'. This method of resolution also gives the place E, where the power PP' must be applied to produce equilibrium. When the prolongation of RM and TV meet in a point 0 beyond B C, either to the right or left, negative results are obtained.-Prof. JOHN L. CAMPBELL, Wabash College, Crawfordsville, Ind.

2. Law of Gravity. Solution of the problem on page 204, Vol. II. -Two similar systems of bodies have all corresponding linear dimensions in the same ratio to each other; that is, the linear dimensions and the distances apart of the bodies are in the one system in a fixed ratio to those of the other.

In order that two similar systems may remain similar for successive instants of time, it is necessary that the motions of corresponding bodies be in the same relative directions inter se, and that the velocities of corresponding motions be proportional to the linear dimensions of the two systems.

If, now, these motions be continually modified by the action of central fixed forces, or by forces dependent directly upon the masses of the bodies, and upon some function of their distances apart, then, since the changes in the motions must also be proportional to the dimensions of the systems and to the motions themselves, the values of these central forces will be proportional to the same dimensions. But the forces, so far as they are dependent upon the masses, are proportional to the cubes of the linear dimensions; hence, so far as

they depend upon the distances, they must be inversely proportional to the squares of these dimensions, in order that on the whole, from the masses and distances combined, they may be simply dependent on the first power of the distances, dimensions, and velocities of the bodies. Hence the law of gravity, and conversely corresponding motions of revolution or oscillation in two similar systems must, by the law of gravity, have the same periods, since the dimensions of the paths or orbits of these motions, their velocities, and their changes of velocity, are all proportional to the dimensions of the systems.

Hence all measures of time, whether by periods of orbital, of rotary, or of oscillatory motions, are by the law of gravity independent of the dimensions of the material universe; and if the solar sytem had been constructed on the scale of a common planetarium, it would still have moved, by virtue of the forces inherent in matter, pari passu, through the same phases of motion and configuration, with the same periods as now. - - W.

3. Develop the Naperian logarithm of x into a series.

Put xy+1; then dx dy, and

=

dx dy

=

By integration

But by division

=dy—ydy + y2dy—y3dy+y^dy &c.

log xy-y2 + } y3 — { y2 + fy0 &c.

Restoring the value of y=x-1, we get

log x = (x − 1) — 1 (x − 1)2 + } (x − 1)3 — † (x — 1)* &c. -ARTEMAS MARTIN, Franklin, Pa.

4. Note on Right-angled Triangles.—I have for many years kept on hand for my own convenience a list of the fifty two right-angled triangles described by Professor HOYT in the May number of the Monthly. I prepared the series by using the formula

(a2 + b2)2 = (a2 — b2)2 + (2 a b)2,

which is easily seen to be true, and in which any numbers whatever may take the places of a and b. If for these letters we substitute the natural numbers in succession we shall obtain thirty sets of numbers no one of which will exceed a hundred; fourteen of these however are equimultiples of some of the others, and twenty-two other multiple sets may be found within the same limit. We thus find, as Professor HOYT has done, fifty-two right-angled triangles whose sides are expressed by integral numbers not exceeding one hundred, and sixteen of which are dissimilar in form. I cannot now call to mind where I found the formula given above. - Prof. E. S. SNELL, Amherst College, Mass.

—

5. Note on Equal Temperaments. It is assumed that the number of vibrations in a given time, producing a musical tone, is to the number producing its octave as 1 is to 2; that the numbers in like manner corresponding to a note and its "fifth" are to each other as 1 to 1.5; and that in a scale of equal temperament the numbers corresponding to the successive tones are in geometrical progression. Required, the number of equal intervals into which an octave must be divided, so as to have one of the tones approximate nearly to a "fifth." Let z = the ratio in the geometrical progression, y = the number of intervals approximating the fifth. x = the number of intervals in the octave. Then

which, by the method of continued fractions, gives the approximate