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now called common logarithms. If the expeditious mea thods for calculating hyperbolic logarithms explained in the foregoing articles*, had been known to Mr. Briggs, his trouble would bave been comparatively trivial with that yhịch he must have experienced in his operations.
16. It has been already determined that the hyperbolie logarithm of 5 is 1.6094379127, and that of 2 is 0.69314718054, and therelore the sum of these logarithms, viz. 2.30258509324 is the byperbolic logarithm of 19. If, therefore, tor the sake of illustration, as in article 14, we supa pose 17 al = 10, and alluw, in addition to the hypothesis there formed, that
&c. denote common lo
garithmus, then 6l = 2.30258509394, and l; and the ratio for reducing the hyperbolic logarithm of any number to the common logarithm of the same number, is that of 2.30258509324 to 1. Thus in order to find the common logarithm of 2, 2.30258509324 : 1 :: 0. 693147 18054 : 0.3010299956, the common logarithm of 2. The common logarithms of 10 and 2 being known, we obtain the common logarithm of 5, by subtracting the common logarithm of 2 from 1, the common logarithm of 10; for 10 being divided by 2, the quotient is 5. Hence the common logarithm of 5 is 0.6989700044. Again, to find the common logarithm of 3, 2.30258509324 : 1::1.0986122 8861: 0.4771212546 the common logarithm of 3.
17. As the constant ratio, for the reduction of hyperbolic to common logarithms, is that of 2.30258509324 to 1, it is evident that the reduction may be made by multiplying the hyperbolic logarithm, of the number whose com
1 mon logarithm is sought, by
Thus 1.94591014899, the hyperbolic logarithm of 7, being multiplied by 4342944818, the product, viz. .8450980378, &c. is the common logarithm of 7.
The commop logarithms of prime numbers being derived from the hyperbolic, the common logarithms of other numbers may be obtained from those so derived, merely by ad
Some of the principal particulars of the foregoing methods were discovered by the celebrated Thomas Simpson. See also Mr. Hellins Mathematical Sssays, published in 1788.
dition or subtraction. For addition of logarithms, in any set or kind, answers to the multiplication of the natural nurnbers to which they belong, and consequently subtraction of logarithms to the division of the natural nuinbers. Hyperbolic logarithms are not only useful as a medium through which common logarithms may be obtained: they are absolutely necessary for finding the fluents of many fluxional expressions of the highest importance.
It is deemed unnecessary, in this place, to show the uti. lity of logarithms by examples. Being once calculated and arranged in tables, not only for common numbers, but also for natural sines, tangents, and secants, it is manifest that a computor may save himself much time, and a great deal of labour, by means of their assistance; as otherwise multiplications and divisions of high numbers, or of deci. mals to a considerable number of places, would enter into his inquiries.
The writer of the foregoing articles now considers the design with which he set out as completed. He has endeavoured to explain, with perspicuity, the first principles of logarithms, and their relations to one another when of different sets or kinds; and he has laid before the young mathemnatical student the most improved and expeditious methods by which they may be calculated. If the reader should be desirous of further information on the subject, he may meet with full gratification by a perusal of the history of discoveries and writings relating to logarithms, prefixed to Dr. Hutton's Mathematical Tables. He will also find the Tables of Logarithms, contained in that volume, the most useful for calculation, if in his computations he does not go beyond degrees and minutes : if he aims at a higher degree of aceuracy, he will have reçourse to Taylor's Tables, in which the Logarithmic Sines and Tangents are calculated to every second of the Quadrant.
A. ROBERTSON, Savilian Professor of Astronomy, 0.cford.
A Sist: DEFINITIONS.
I. The pole of a circle of the sphere is a point in the super, а
through the centre of the sphere, and, whose centre : therefore is the same with that of the spbere.
sphere comprehended by three arches of three great cir"cles, each of which is less than a semicircle.
sphere is contained by two arches of great circles, and is
As they have a common centre, their common section will be a dianieter of each which will bisect them.
PROP. II. · Fig. 1. The arch of a great circle betwixt the pole and the circumference of another is a quadrant.
Let ABC be a great circle, and D its pole; if a great circle DC pass through D, and meet ABC in C, the arch DC will be a quadrant.
Let the great circle CD meet ABC again in A, and let AC be the common section of the great circles, which will pass through E, the centre of the sphere: Join DE, DA, DC: By def. 1. DA, DC are equal, and AE, EC are also equal, and DE is common; therefore (8. 1), the angles DEA, DEC are equal; wherefore the arches DA, DC are equal, and consequently each of them is a quadrant. Q.E.D.
"!! PROP. III. Fig. 9. Ir a great circle be described meeting two great circles AB, AC passing through its pole A in B, C, the angle at the centre of the sphere upon the circumference BC, is the same with the spherical an.. gle BAC, and the arch BC is called the measure of the spherical angle BAC..
Let the planes of the great circles AB, AC, intersect one another in the straight line AD, passing through D their coinmon centre: join DB, DC.
Since A is the pole of BC, AB, AC will be quadrants, and the angles ADB, ADC right angles; therefore (6 def. 11.) the angle CDB is the inelination of the planes of the cireles AB, AC; that is, (def. 4.) the spherical angle BAC: Q.E.D.
Cor. If through the point A, two quadrants AB, AC, be drawn, the point A will be the pole of the great circle BC, passing through their extremities BC.
Join AC, and draw AE, a straight line to any other point E, in BC; join DE: Since AC, AB are quadrants, the angles ADB, ADC are right angles, and AD will be perpendicular to the plane of BC: Therefore the angle ADE is a right angle, and AD, DC are equal to AD, DE, each to each; therefore AE, AC are equal, and A is the pole of BC, by def. 1. Q: E. D.
PROP. IV. Fig. 9. In isosceles spherical triangles, the angles at the base are equal.
Let ABC be an isosceles triangle, and AC, CB, the equal sides; the angles BAC, ABC, at the base AB, are equa).
Let D be the centre of the sphere, and join DA, DB, DC; in DA take any point E, from which draw, in the plane ADC, the straight line EF at right angles to ED, meeting CD in F, and draw in the plane ADB, EG at right angles to the same ED; therefore the rectilineal angle FEG is (g. def. 11.) the inclination of the planes ADC, ADB, and therefore is the same with the spherical angle BAC From P draw FH perpendicular to DB, and from H draw, in the plane ADB, the straight line HG at right angles to HD, meeting EG in G, and join GF. Because DĚ is at right angles to EF and EG, it is perpendicular to the plane
FEG (4. 11.) and therefore the plane FEG is perpendicular to the plane ADB, in which DE is (18. 11.): In the same manner, the plane FHG is perpendicular to the plane ADB; and therefore GF, the common section of the planes FEG, FHG, is perpendicular to the plane ADB (19. 11.); and because the angle FHG is the inclipation of the planes BDC, BDA, it is the same with the spherical angle ABC; and the sides AC, CB of the spherical triangle being equal, the angles EDF, HDF, which stand upon them at the
centre of the sphere, are equal; and in the triangles EDF, HDF, the side DF is common, and the apgles DEF, DHE are right angles ; therefore EF, FH are equal: and in the triangles FEG, FHG the side. GF is common, and the sides EG, GH, will be equal by the 47. 1. and therefore the angle FEG is equal to FHG (8. 1.); that is, the spherical angle BAC is equal to the spherical angle ABC.
PROP. V. FIG. 3.
IF, in a spherical triangle ABC, two of the angles BAC, ABC be equal, the sides BC, AC opposite to them are equal.
Read the construction and demonstration of the preceding proposition, unto the words, “and the sides AC, CB," &c. and the rest of the demonstration will be as follows, viz.
And the spherical angles BAC, ABC being equal, the rectilineal angles FEG, FHG, which are the same with them are equal; and in the triangles FGE, FGH, the angles at G are right angles, and the side FG opposite to two of the equal angles is common; therefore (26. 1.) EF is equal to FH: And in the right angled triangles DEF, DHF, the side DF is common; wherefore (47.. 1.) ED is equal to DH, and the angles EDF, HDF are therefore equal (4. k.), and consequently the sides AC, BC of the spherical triangle are equal.
PROP. VI. FIG. 4. Any two sides of a spherical triangle are greater than the third.
Let ABE be a spherical triangle, any two sides AB, BC will be ater than the other side AC.
Let D be the centre of the sphere: Join DA, DB, DC.