« НазадПродовжити »
A C, G F parallel to BC, FH parallel to B D, H I parallel to DE, and so on for as many convolutions as may be desired, and we obtain four points on each convolution.
To find intermediate points, divide each quadrant into four equal angles by the lines Sa, Sb, &c.; then the line Sb is a mean-proportional between SA and SC; Sa is a mean-proportional between SA and Sb; Sc is a meanproportional between Sb and SC; Se is a mean-proportional between SC and SB; Sd is a-mean-proportional between SC and Se; Sf is a mean-proportional between Se and SB. Having determined eight points on the first half of the convolution, the other points are found from them; for the chord of Bg is parallel to the chord of A a, the chord of gh is parallel to the chord of ab, and so on all round the spiral. The points where the other convolutions cut the lines drawn from S are found in the same manner by drawing chords parallel to the chords in the first convolution; thus En is parallel to A a, n p to ab, and so forth.
Divide the given length AS by the given length BS, and take the common logarithm of the quotient; divide 1:3643 by that quotient, and call the result a; or, in algebraical formula,
Next divide 1705 by the value of a found above, and subtract the quotient from the common logarithm of SA; the difference will be the common logarithm of Sa; or, in algebraical formula,
log (Sa)=log (SA) – 1705
Having thus determined the length of Sa, draw two lines 0 A and O A', making an acute angle with each Fig. 155.
other at 0 (fig. 155), let O A equal SA, O A' less than 0 A; take 0 a equal to Sa (fig. 154), determined as above, join A A', A'a; then draw a a' parallel to A A', a'b parallel to A'a, 6 b' parallel to A A', Vc parallel to A'a, and so on; then the lengths O a, 06, 0c, &c., thus found on 0 A will represent the length Sa, Sb, Sc, &c., on the spiral (fig. 154); and in this way all the points on the contcur from A to B can be found. Also if we require to draw another spiral similar to the first, as shown by the dotted line A'a' (fig. 154), we have
only to take OA' (fig. 155) equal to SA', and the lengths 0 a', 06, 0 c', &c., will represent Sa', SU', Sc', dic., on the second curve.
190. To draw tingents and normals at various points on the equiangular spiral.—Let the lengths SA, S B (fig. 154) be given ; then the trigonometrical tangent o of the angle between the radius-vector and the geometrical tangent at any point of the spiral is found by dividing SA by SB, taking the logarithm of the quotient and dividing the number 1:3643 by the quantity obtained ; or, algebraically expressed,
Suppose, for example, that SA=20, SB= 14, as in figure 154 ; then we find from this formula that a=9. Upon S A measure Sy equal to 9 on any scale, draw y z at
right-angles, and equal to 1 on the same scale, or to oneninth of Sy; join Sz, then Sy divided by y z represents the trigonometrical tangent (@) of the angle Szy or Sxy (yx being taken equal to y2); therefore Sxy or Szy is the angle which the tangent and radius-vector make with each other at every point of the spiral; and if A Kis drawn parallel to Sx, A K is the normal or perpendicular to the tangent at A ; and the tangent itself can be drawn by simply drawing a line at A perpendicular to AK. If A z is produced to meet the curve at P, then P is the point where the tangent is parallel to the axis SC, and is the highest point on the convolution. The normals at a, b, c, &c., can all be found by drawing lines making an angle equal to KAS with Sa, Sb, Sc, &c., and the tangents at those points will be perpendicular to those normals. If the curve is drawn in by hand the tangents should all be drawn at the several points so as to give the direction of the spiral. If it is desired to imitate the spiral by drawing arcs of circles from point to point, the centres of those circles will be found in the intersection of the normals drawn from any two consecutive points.
191. To draw the equiangular spiral when the depth and the angle between the tangent and radius are given.—Let A B (fig. 156) be the given depth, a the trigonometrical tangent of the given angle, the value of which is found from a table of tangents when the angle is given. Then the ratio of A S to BS is found by the following formula :
or, divide 1.3643 by the value of a and the quotient is the logarithm of the ratio of A S to BS, the value of which is found from a logarithmic table. We have then to divide A B in the ratio thus obtained and find the pole S. The spiral can now be drawn by one of the methods above described (189).
For example, let the given angle be 80°, as in figure 156; the tangent of 80° is given in the tables as 5.67,
which is the valne of a in this case. Dividing 1.3643 by 5.67, we get 2406, which is the logarithm of 1.74, or very nearly 1. Divide the given length A B into eleven equal parts, and measure four of those parts from B towards A, and we obtain the pole S. Draw DSC at right-angles to A B, and we find the point C by taking SC a mean-pro
portional between AS and BS, either by the geometrical process (17), or by extracting the square root of the product of A S and B S, for
SCP=AS ~ BS = 7 * 4= 28. This gives us SC = 5-3 very nearly.
The point b is found by taking Sb a mean-proportional between SA and SC, or, Sb=v7 x 5:3= 6.1.
The point a is found by taking Sa a mean-proportional between S A and Sb, or,
Sa= v7 x 6.1 = 6-53. We can also obtain the value of Sa by the formula previously given (189), namely, log (Sa) = log (SA) – 1705= log (7) -57605
= -8451 – 03 = -8151 = log. (6:53) or, Sa=6:53 as found above.
Now draw 0 A, O A' (fig. 157), making an acute angle at 0, and take 0 A' less than 0 A, as, for
Fig. 157. example, let O A' equal S A' (fig. 156). Take 0 A equal to SA (fig. 156), and O a equal to Sa just found, namely 6.53, the length of SA being 7. Join A A' (fig. 157) and A’a; draw a a' parallel to A A', a' b parallel to A'a, bŲ parallel to A A’, b'c parallel to A'a, and so on; then Oa, Ob, Oc, &c, represent the lengths of Sa, Sb, Sc, &c. (fig. 156). Also O a', Ob', &c., represent the lengths Sa', S6, &c., in the dotted spiral.
On SA (fig. 156) take Sy equal to 5.67 on any scale, draw zy x perpendicular to Sy, making y z and yx each