Page images
PDF
EPUB

By substituting these values of the exponential in the equations (31), (32), (33), they take the following forms :

[subsumed][subsumed][ocr errors][subsumed][ocr errors][ocr errors][subsumed][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small]

Now 2 expresses the circumference, of which one half is on the positive side, and the other half on the negative side: in respect to the same diameter, — 27 expresses the same things taken in contrary directions; consequently 2 and 2 are two different expressions of the same circumference, when the diameters are the same; they then represent the same rings; and the last three equations give by reduction the following:

26 × 38

1 (26 x 3) = ± 2 % √ = 1 × & = ± 2′′ √ = I

±2% √ × 16

.(13)

[ocr errors][ocr errors][ocr errors][merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][ocr errors][merged small]

42

± 2 % √ − 1 ( 12 + 30 ) = ± 2% √ − 1 × 3 = ± 2 % √ − 1

[ocr errors]

x 18.....

72

-

..(45)

These last equations are evidently identical, and the problem proposed is now completely resolved by the roots (25), (26), (27), which, expressed aritlimetically, are

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small]

.(47)

.(48)

Unity being the area of any circle, which is an essential remark; for if unity were an abstract number, this solution would be absurd,

土 πν

It is only true when unity is expressed by e±2V -1 (= the area of any ring or central circle.)

Remark.

Wallis, who devoted much time and attention to the proposed problem, resolved the equations (1), (2), (3), by approximation. (Wallis's Algebra, Chap. 62.) His roots are,

[merged small][ocr errors][merged small][merged small][ocr errors][merged small]

Comparing these roots with the roots (46), (47), (48), it is easily perceived that they have not the least relation to each other. In order to prove how nearly the roots (49), (50), (51), verify the

[ocr errors]

equations (1), (2), (3), Wallis has substituted them in the place of a, b, c, by which he obtained the following result

[blocks in formation]

However near Wallis's approximation may be, the equations which he obtained are not for that reason the less absurd, nor are they the less

Equations, whose two members are not equal and never can be

come so.

In order that the two members should be equal, it is necessary that the inequality of the units compensate the inequality of the numbers, as in the equation 27. = 40s.; this compensation takes place if each member and its unit are in an inverse ratio to each other. This is the case in my solution, where the height of each ring, multiplied by the mean circumference between the two extreme circumferences, give for the product the constant area of the central circle, as I thus demonstrate, for we have for each ring the following proportion :

The sum of the radii of the extreme circumferences of the rings : To the tangent drawn to the smallest of these circumferences :: This tangent, (which is the radius of the central circle,) The difference of the radii of the extreme circumferences of the rings. (This difference being the height of the ring, 185.) From whence we have the following theorem.

In a series of concentric rings, each of whose areas are equal to the central circle. The rectangle formed by the sum of the radii of the extreme circumferences of any one of the rings and its height, is equal to the square of the central circles. (56.) Now the area of each ring is equal the area of a trapezium, which has for its base the height of the ring, and for the mean height half the sum of the extreme circumferences. But this half is equal to a mean circumference between the two extreme circumferences. We can then transform theorem (56) into the following.

In a series of concentric rings whose areas are equal to the arca of the central circle, the rectangle formed of the height of each ring and the circumference, which is a mean between the two extreme circumferences, is equal the area of the central circle, which is the proposition I had to demonstrate.

General Corollary.

The preceding resolution of equations (1), (2), (3), gives a complete solution of Gauss's problem, viz. " To divide a circumference into 17 equal parts." The dotted isosceles triangle O11 (fig. 1)

17 16 17

whose summit is the centre, has for its base the continued line 1 1 1; it is this base whose extremities are 1, which divides the

17 16 17

17 17

circumference into 17 parts. The arc 1 1 1 is one of these parts;

17 17 17

i.

the continued line 1 1 1 is equal to the central diameter 1 O I.

[merged small][ocr errors]

Now this central diameter is a unit; for the central circle and its diameter are respectively units, which are heterogeneous with each other. All the parallel lines, 1 1 1, 1 1 1, &c. are equal to this

101 2 1 2

last unit. These lines are the greatest which can be drawn between the two extreme circumferences of each ring, and they may be considered as the diagonals of those rings, in the same manner as

the diameter 1O I can be considered as the diagonal of the central

[ocr errors]

circle. We have, then, between the diagonal of the central circle and the diagonal of the rings, the same analogy as between the central circle itself and these rings; the central circle and the rings are both superficial units, and maxima; the diagonals of the central circle and of the rings are both linear units, and also maxima. This geometrical construction of the problem, as presented by fig. 1, is the only one which can be right; because all the straight lines of this figure are either maxima or minima, in such a manner that if they be maxima as lines, they are minima as being opposite to an angle or an arch which is a minimum; and if they be minima as lines, they are maxima, as being opposite to an angle or an arch which is a maximum, as I shall demonstrate at length in another paper. These lines, then, cannot be greater or less than they are. M. le Gendre, at the end of his Geometry, has given an algebraical solution of this problem, but he has not given the geometrical construction of his formula that construction was in fact impossible, without destroying the law of continuity.

ARTICLE III.

Some Experiments on pure Nickel, its Magnetic Quality, and its Deportment when united to other Bodies.* By W. A. Lampadius.

1. AFTER having in 1796 discovered a method of obtaining pure malleable nickel by means of an oxygen gas fire, either from Freiberg Bleispeise, or from the common regulus of nickel obtained from copper nickel by the usual process, I occasionally made a number of accurate experiments on many of the properties of this metal, which had been hitherto examined only in a cursory manner. 124 grains of speise gave me 43 grains of nickel, and 123 grains of copper nickel ore gave me 63 grains of the pure metal.

2. Magnetic Power of Nickel.

The magnetometer described in the preceding paper gave the * Translated from Schweigger's Journal für Chemie und Physik, x. 174. 1814.

magnetic energy of nickel 35, and that of iron

=

55. The

magnetic energy of cobalt was likewise tried, and found = 25; but as this metal was not quite pure, this experiment, as well as the magnetism of an alloy of cobalt and nickel, will be hereafter repeated.

3. Alloy of Nickel and Platinum.

This, as well as all the subsequent alloys, was made upon chareoal kept intensely hot by a stream of oxygen gas, according to the method described in my Manual for the Analysis of Minerals. A grain of each of the two bodies, nickel and platinum, was put upon the charcoal. After they had been softened by the application of the heat for about half a minute, both bodies incorporated together in a very striking manner. They formed an alloy possessing nearly the fusibility of copper, although nickel by itself is nearly as infusible as platinum. The alloy was completely malleable, acquired a fine polish, and had a light yellowish-white colour, not unlike that of sterling silver. Its magnetic energy was still 35.

4. Gold and Nickel (equal parts).

Both metals very readily melt into one round button; pretty hard, harder than the preceding alloy; externally malleable; capable of a fine polish; colour yellowish-white, a little darker than the preceding alloy. The magnetism continued = 35.

5. Silver and Nickel (equal parts).

When I attempted to alloy these two metals, I made the following observations. The silver melted in two seconds, and the nickel remained for some time unmelted upon the silver. In about a minute the silver, it is true, took up the nickel, but did not dissolve it. When the heat was continued some time longer, the two metals appeared to unite; but at that instant the silver burnt away with a blue flame, and left the malleable nickel behind it; but about one half of the nickel likewise was burnt.

6. Copper and Nickel (equal parts).

Both metals melted together in four seconds. The alloy was brittle and granular; the colour reddish-white; and the fracture porous. It exhibited no trace of magnetism.

7. Nickel and Iron.

Iron and nickel easily melted together into a round bead. The nickel was first melted, and the iron added to it, to prevent the last metal from being burnt by the heat. By continuing the heat, the greater part of the iron separated from the nickel in the state of a black oxide, still attracted by the magnet. By weighing the alloy I found that it consisted of ten parts of nickel and four parts of iron, or the iron amounted to rather less than one-third of the alloy. This alloy was moderately hard, quite malleable, and had the colour of steel. Its magnetism was = 35.

8. Phosphorus and Nickel.

The bead of nickel was heated red-hot, and then a small piece of

phosphorus placed in contact with it. They melted together in a few seconds. 34 parts of nickel thus treated increased in weight five parts; so that 100 nickel had combined with 15 phosphorus. The button externally was tin-white, and had the metallic lustre. It was moderately hard, and very brittle. Its fracture was foliated and crystalline, partly dull, and partly with the metallic lustre. Its magnetism was gone.

9. Nickel and Sulphur

Easily united together, when treated in the same way as the nickel and phosphorus had been. Externally the button was dull, swelled, and grey in colour. Its magnetism likewise was gone. 20 parts of nickel had taken up two parts of sulphur; so that 100 of the metal combine with 10. The mass was elastic, not very hard, the fracture uneven, and the colour yellowish-white, similar to that

of native copper nickel ore.

10. From these experiments we learn,

a. The readiness with which nickel and platinum unite together. b. The little affinity between silver and nickel, as the silver rather combines with oxygen and is dissipated, than remains united to the nickel.

c. The singular effect of combining it with copper, in which we see two malleable metals produce a brittle alloy.

d. The permanence of the magnetism of nickel when it is alloyed with gold and platinum.

e. Its complete destruction when nickel is alloyed with copper. f. Its diminution when nickel is alloyed with iron.

Perhaps a farther prosecution of these experiments might have a tendency to throw some light upon magnetism. At present I lay aside all hypotheses, and satisfy myself with stating simple facts.

ARTICLE IV.

Magnetical Observations at Hackney Wick. By Col. Beaufoy. Latitude, 51° 32′ 40.3" North. Longitude West in Time 6"

[blocks in formation]
« PreviousContinue »