now called common logarithms. If the expeditious methods for calculating hyperbolic logarithms explained in the foregoing articles*, had been known to Mr. Briggs, his trouble would have been comparatively trivial with that which 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 0.69314718054, and therefore the sum of these logarithms, viz. 2.30258509324 is the hyperbolic logarithm of 10. If therefore, for the sake of illustration, as in article 14, we suppose 1910, and allow, in addition to the hypothesis 2 3 4 there formed, that &c. denote common lo m m m m 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 8864: 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 common logarithm is sought, by 818. = .4342944 2.30258509324 Thus 1.94591014899, the hyperbolic logarithm of 7, being multiplied by .4342944818, the product, viz. .8450980378, &c. is the common logarithm of 7. The common 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 Essays, published in 1788. dition or subtraction. For addition of logarithms, in any set or kind, answers to the multiplication of the natural numbers to which they belong, and consequently subtraction of logarithms to the division of the natural numbers. 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 utility 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 decimals 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 mathematical 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 accuracy, he will have recourse 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, Oxford. THE pole of a circle of the sphere is a point in the superficies of the sphere, from which all straight lines drawn to the circumference of the circle are equal. II. A great circle of the sphere is any whose plane passes through the centre of the sphere, and whose centre therefore is the same with that of the sphere. III.. A spherical triangle is a figure upon the superficies of a sphere comprehended by three arches of three great circles, each of which is less than a semicircle. A spherical angle is that on which the superficies of a sphere is contained by two arches of great circles, and is the same with the inclination of the planes of these great circles. PROP. I. GREAT circles bisect one another. As they have a common centre, their common section will be a diameter 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. 2. 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 angle 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 common 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 circles 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. 3. 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 equal. 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 F 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 DE 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 plases FEG, FHG, is perpendicular to the plane ADB (19. 11.); and because the angle FHG is the inclination 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 angles DEF, DHF 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 sphe rical 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. L.) ED is equal to DH, and the angles EDF, HDF are therefore equal (4.1.), 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 ABC be a spherical triangle, any two sides AB, BC will be greater than the other side AC. Let D be the centre of the sphere: Join DA, DB, DC. The solid angle at D is contained by three plane angles, |