reduce the division of an angle in any fashion to the like division of a line.

Now the division of a line in any fashion, that is, into a set of segments in any given ratio, is at once solved so soon as we have learnt by the aid of a pair of compasses or a' set square' to draw parallel lines. Thus suppose we require to divide the line c'p into segments in the ratio of 3 to 5 to 4; we have only to mark off along any line through c', say c'q, steps c'R, RS, S T placed end to end and containing 3, 5, and 4 units of any kind respectively. If the finish of the last step T be joined to P

R

S

FIG. 68.

and the parallels Rr, ss to T P through R and s be drawn to meet c'p in r and s, then c'p will be divided in r and s into segments in the required ratio of 3 to 5 to 4. This follows at once from our theory of triangles of the same shape (see p. 106). For, since R c'r, s c's, and To'p are such triangles, they have their corresponding sides proportional, and the truth of the proposition is obvious.

A spiral of Archimedes accurately cut in a metal or ivory plate is an extremely useful addition to the ordinary contents of a box of so-called mathematical instruments.

§ 9. The Equiangular Spiral.

Another important spiral was invented by Descartes, and is termed from two of its chief properties either the equiangular or the logarithmic spiral.

Let BOA be a triangle with a small angle at o, and whose sides o A and O B are of any not very greatly different lengths. Upon o в and upon the opposite side of it to a construct a triangle B OC of the same shape as the triangle A O B, and in such wise that the angles at B and A are equal. Then upon o c place a triangle COD of the same shape as either BOC or AO B; upon O D a fourth triangle D O E, again of the same shape; upon o E a fifth triangle, and so on. We thus ultimately form a figure consisting of a number of triangles A O B, BO C, CO D,

DO E, &c., of the same shape, all placed with one of their equal angles at o, and in such fashion that each pair has a common side consisting of two non-corresponding sides (that is, of sides not opposite to equal angles). The points A B C D E, &c., will form the angles of a polygonal line, and if the angles at o are only taken small enough, the sides of this polygon will appear to form a continuous curved line. This curved line, to which we can approach as closely as we please by taking the angles at o smaller and smaller, is termed an equiangular spiral. It derives its name from the following property,—A B, B C, C D, &c.,

being corresponding sides of triangles of the same shape, make equal angles O B A, O C B, OD C, &c., with the corresponding sides o B, O C, O D, &c. ; but when the angles at o are taken very small A B, B C, C D, &c., will appear as successive elements of the curved line or spiral. Hence the arc of the spiral meets all rays from the pole o at the same constant angle.

Let us now endeavour to find the relation between any radius vector o P (=r) and the vectorial angle A O P (=0).

Since all our triangles AO B, BO C, COD, &c., are of the same shape, their corresponding sides must be proportional (see p. 106); or,

Each of these equal ratios will therefore have the same scalar value; let us denote that value by the symbol μ. Then we must have

OB = μ.OA; OC = μ. Oв; OD = μ.OC; &c.

Or, O B = μ. OA; OC = μ2.0 A; OD = μ3. O A, and so on. Hence if oN be the radius vector which occurs after n equal angles are taken at o, we must have

ON μ.0 A.

Now let the very small angles at o be each taken equal to some small part of the unit angle; thus we might take them or of the unit angle. We will represent this fraction of the unit angle by 1/b, where we may suppose b a whole simplicity. Further let us use power of μ, or λ=. p. 144 we then term

number for greater to denote the bth With the notation explained on μ a bth root of λ, and write

Hence finally we have ON OA. Anx1/b, or in words: The base of the (n+1)th equal-shaped triangle placed about o is equal to the base of the first multiplied by a certain quantity λ raised to the power of n-times the quantity 1/6 which expresses the magnitude of the equal angles at o in units of angle.

Now let the spoke or ray o P fall within the angle which is formed by the successive rays on and oQ of the system of equal-shaped triangles round o. Then ON makes an angle n-times 1/b, and oQ an angle (n+1)times 1/b with o A. Hence the angle A o P, or 0, must lie in magnitude between n/b and (n+1)/b. Similarly the magnitude of o P must lie between those of o N and o Q. Now by sufficiently decreasing the angles at o we can approach nearer and nearer to the form of the spiral, and the ray o P must always lie between two successive rays of our system of triangles. The angle 0, which will thus always lie between n/b and (n+1)/b, can only differ from either of them by a quantity less than 1/b. If then b be taken large enough, or the equal angles at o small enough fractions of the unit angle, this difference 1/b can be made vanishingly small. In this case we may say that in the limit the angle becomes equal to n/b and the ray o P equal to ON or o Q, which will thus be ultimately equal. Hence o P=0A.X"0 A. λo, or in words: If a ray OP of the equiangular spiral make an angle A OP with another ray o A, the ratio of op to o A is equal to a certain number λ raised to the power of the quantity 0 which expresses the magnitude of the angle A O P in units of angle.

This is

If a and r be the numbers which express the magnitudes of OA and O P, we have r=a λo. termed the polar equation of the spiral.

We proceed to draw some important results from a

consideration of this spiral. The reader will at once observe that the ratio of any pair of rays o P and o q is equal to the ratio of any other pair which include an equal angle, for the ratio of any pair of rays depends only on the included angle. Further, if we wanted to multiply the ratio of any two quantities p and q by the ratio of two other quantities r and s we might proceed as follows: Find rays of the equiangular spiral o P, o Q, O R, O S containing the same number of linear units as P, q, r, s contain units of quantity (see p. 99), and let

R

T

FIG. 70.

◊ be the angle between the first pair, the angle

ratio

whence it follows that

or is equal to the include an angle 0+.

of any pair of rays which Thus if the angle QOT be taken equal to 4, and or be the corresponding ray of

the spiral,

от
ОР

=

x+, and is a ratio equal to the pro

duct of the given ratios. Hence to find the product of ratios we have only to add the angles between pairs of rays in the given ratios, and the ratio of any two rays including an angle equal to the sum will be equal