Suppose we have a bar of 12-inch steel and want to find the distance across the corners, and the angle it will make with the base. The 1-inch side is the radius, the diagonal is the secant, and the 2-inch side is the tangent of the angle. Reducing these to a basis of one inch we have a bar 1 inch by inch and the inch is the tangent of the angle. Looking in the table we find this to be almost exactly the tangent of 26 degrees and 45 minutes. With this angle a 3 1926 FIG. 6 the circumference by dividing 360 degrees by the number of divisions, but what we want is to find the chord or the distance from one point to the next in a straight line as a pair of dividers would step it off. First divide 360 by the number of divisions - say 9 - and get 40 degrees in each part. Fig. 5 shows this and we want the distance shown or the chord of the angle. This equals twice the sine of half the angle. Half the angle is 20 degrees and the sine for this is .342. Twice this or 0.684 is the chord of the 40-degree angle for every inch 1.026 the secant or diagonal is 1.1198 for a radius of I inch and 1 times this gives 1.6797 as the dis tance across corners. A very practical use for this kind of calculation is in spacing bolt holes or otherwise dividing a circle into any number of equal parts. It is easy enough to get the length of each arc of FIG. 7 of radius. If the circle is 14 inches in diameter the distance between the holes will be 7 times 0.684 or 4.788 inches. This is very quick and the most accurate method known. Draftsmen often lay out jigs with the angles marked in degrees as in Fig. 6, overlooking the fact that the toolmaker has no convenient or accurate protractor for measuring the angle. Assume that a drawing shows three holes as a, b, and c, with b and c 20 degrees apart. The distance from a to b is 3 inches, what is the distance from b to c or from a to c? As the known radius is from a to b, the distance b c is the tangent of the angle and the tangent for a one-inch radius is .36397, so for a 3-inch radius it is 3 X .36397 1.09191 inches from b to c and at right angles to it. = But we need not depend on the accuracy of the square or of the way we use it, as we can find the distance from a to c just as easily and just as accurately as we did bc. This distance is the secant, and is 1.0642 for a one-inch radius. Multiplying this by 3 as the distance which can be accurately measured. = 3.1926 If the distance between a and c had been 3 inches, then b c would have been the sine and a b the cosine of the angle, both of which can be easily found from the tables. It often happens that we want to find the angle of a roller or other piece of work as Fig. 7. Always work from the center line and continue the lines to complete the angle. Every triangle has the sides and they are called the side "opposite," "side adjacent," and "hypotenuse," the first being opposite the angle, the second the base line, and the third the slant line. The following rules are very useful in this kind of work: FIG. 8 (9) Side Adj. = Co-Tan. X Side Opp. Side Adj. If we have the dimensions shown in Fig. 7, the side opposite, and the hypotenuse, we use formula No. 1, and dividing 2 by 4 we get or .5 as the sine of the angle. The table shows this to be the sine of the angle of 30 degrees, consequently this is a 30-degree angle. or .5 use = If we have the side opposite and the side adjacent we formula No. 3, and find that the tangent of the angle. The table shows this to be the tangent of 26 degrees and 34 minutes. Should it happen that we only knew the hypotenuse and the angle we use formula No. 6 and multiply 4 X 5 = 2, the side opposite. In the same way we can find the side adjacent by using formula No. 7. The cosine of 30 degrees in .866 and 4 × .866 = 3.464 inches as the side adjacent. Having a bar of steel 2 by inches, Fig. 8, what is the distance across the corners? Either formulas 3 or 4 will answer for this. Number of Sides Taking No. 4 we have 2 as the side opposite, 3 as the side adjacent. Dividing 3 by 2 gives 1.5. Looking under co-tangents for this we find 1.5108 after 33 degrees 30 minutes, which is nearly the correct angle. Then look for the secant of this and find 1.1958. Multiply this by 3 and get 3.5874 as the distance across the corners. Complete tables of sines, tangents, secants, etc., will be on pages 371 to 405. USING THE TABLE OF REGULAR POLYGONS THE easiest way to lay out figures of this kind is to draw a circle and space it off, but it saves lots of time to know what spacing to use or how large a circle to draw to get a figure of the right size. Suppose we wish to lay out any regular figure, such as pentagon or five-sided figure, having sides 1 inches long. Looking in the third column we find "Diameter of circle that will just enclose it," and opposite pentagon we find 1.7012 as the circle that will just enclose a pentagon having a side equal to 1. This may be 1 inch or 1 anything else, so as we are dealing in inches we call it inches. As the side of the pentagon is to be 1 inches we multiply 1.7012 by 1 and get 2.5518 as the diameter of circle to draw, and take half of this or the radius 1.2759 in the compass to draw the circle. Then with 1 inches in the dividers we space round circle, and if the work has been carefully done it will just divide it into five equal parts. Connect these points by straight lines, and you have a pentagon with sides 1 inches long. If the pentagon is to go inside a circle of given diameter, say 2 inches, look under column 5 which gives "Length of side when diameter of enclosing circle equals 1," and find 5878. Multiply by 2 as this is for a 2-inch circle, and the side will be 2 X .5878 = 1.1756. Take this distance in the dividers and step around the 2-inch circle. Assume that it is necessary to have a triangular end on a round shaft, how large must the shaft be to give a triangle 1.5 inches on a side? Look in the table under column 3, and opposite triangle find 1.1546, meaning that where the side of a triangle is 1, the diameter of a circle that will just enclose it is 1.1546. As the side is 1.5, we have 1.5 X 1.1546 1.7318, the diameter of the shaft required. If the corners need not be sharp probably a shaft 1.625 would be ample. Reversing this to find the size of a bearing that can be turned on a triangular bar of this size, look in column 4, which gives the largest circle that will go inside a triangle with a side equal to 1. This gives .5774. Multiply this by 1.5 .8661. = A square taper reamer is to be used which must ream 1 inch at the small end and 1.5 at the back, what size must this be across the flats at both places? Under column 5 find .7071 as the length of the side of a square when the diameter of the enclosing circle is 1, so this will be the side of the small end of the reamer and 1.5 X .7071 side of the reamer at the large end. = 1.0606 is the FINDING THE RADIUS WITHOUT THE CENTER IT sometimes happens in measuring up a machine that we need to know the radius of curves when the center is not accessible. Three such cases are shown in Figs. 9, 10, and 11, the first two being a machine and the last a broken pulley. In Fig. 9 the rule is short enough to go in the curve while in Fig. 10 it has one end touching and the other across the sides. It makes no difference which is used so long as the distances are measured correctly, the short distance or versed sine being taken at the exact center of the chord and at right angles to it. It is easier figuring when the chord or the hight are even inches, so in measuring slip the rule until one or the other comes even; sometimes it is better to make the hight come I inch and let the chord go as it will, while at others the reverse may be true. The rule for finding the diameter is: Square half the chord, add to this the square of the hight, and divide the whole thing by the hight. If the chord is 6 inches, as in Fig. 9, and the hight 1 inches we have Or as shown in Fig. 10 the chord is 10 inches and the hight 1 inch, then the figures are In Fig. 11 we have a piece of a broken pulley, and find the chord B to be 24 inches, and the hight A to be 2 inches. This becomes 122 + 22 2 74 inches. 144 + 4 2 148 74, so that the diameter of the pulley is 2 A circle is a continuous curved line having every point at an equal distance from the center. Its perimeter or circumference is always 3.14159265359 times the diameter, although 3.1416 is generally used and 34 is a very close approximation. Area equals the diameter squared X .7854, or half the diameter squared X 3.1416, or half the diameter X half the circumference. Diameter of a square having equal area diameter of circle times .89 very nearly. Triangle = Equilateral triangle is a regular figure having three equal sides and three equal angles of 60 degrees each. The side equals .866 times the diameter of enclosing circle. Distance from one side to opposite point equals the side times .866 or diameter of enclosing circle X .75 or inside circle X 1. Diameter of enclosing circle equals the side times 1.1546 or 1 times distance from side to point or twice inside circle. Diameter of inside circle equals side times .5774 or the enclosing circle. The area equals one side multiplied by itself and by .433013. Diameter of circle having equal area equals side of triangle times .73. |