right triangle IKO, KO KI, cos. IKO (R — r) cos. 6. In the right triangle FIE, EF : EI, cos. IEF= (R — r) cos. A. Also, EH=LN=D, since each is equal to R – KL. Then EF=EH + HF=EH + K O; i.e., (R – ?) cos. l=(R — r) cos. b + D. Hence nat. cos. (=nat. cos. b + ID = (R — r)]. Were the curve CM located, and the curve B N to be substituted for it, that is to say, were a given and b required, we should have, by transposition, nat. cos. I nat. cos. aID - (R --r)]. E.cample. A 50 compounding into a 3o curve at B, which consumes an angle of 44° 20', and terminates at N, 28 feet too far to the left. Here D 28, R 44° 20; and, by the solution, nat. cos. a = nat. cos. 44° 20' + (28 • 764). The nat. cos. 41° 20' 0.69883; 28 : 764 log. 1.447158 log. 2.883093 = log. 2.564065, corresponding to the decimal 0.03665, which, being added to nat. cos. 44° 20', gives 0.73548, the nat. cos. 42° 29'. Then BKN – CEM : : 44° 20' - 42° 29' = 1° 51' = angle BIC, equivalent on a 5o curve to 37 feet, which therefore is the distance around the arc from B, the erroneous P.C.C., to C, the correct one. 3. From these formulas the following general rule may be drawn: Divide the distance between terminal tangents by the difference of the radii, and call the quotient Q. Find the natural cosine of the terminal arc already located, and call it C. The sum or the difference of Q and C will be the natural cosine of the terminal arc to be substituted for that already located. With radii in the order R, r, should the terminal tangent located strike { outside } the proposed tangent; or, with radii in the order r, R, should the terminal tangent outside located strike { } the proposed tangent, — take the inside { difference} of Q and C for the required cosine. XXXII. HAVING LOCATED A TANGENT, A B, INTERSECTING A CURVE, CD, FROM THE CONCAVE SIDE, TO FIND THE POINT E ON SAID CURVE AT WHICH TO BEGIN A CURVE OF GIVEN RADIUS WHICH SHALL MERGE IN THE LOCATED TANGENT. В. 1. Place the transit at the intersection point B. Set points at equal distances therefrom in both directions on the curve already located, by means of which the direction of a tangent to that curve at B may be fixed, and the angle F B A measured. Call that angle (; and, as shown in the figure, suppose the located curve to be prolonged in10 a terminal tangent, parallel with the newly located tan gent A B. Complete the diagram. Call the larger radius R; the proposed radius, r; the central angle of the proposed curve, x. Then, obviously, the line AG = Rcos, (1. It is also equal to (R — r) cos. x + r. That is to say, R cos. a = - (R — 9) cos. x+r. Hence cos. x Rcos. at p• (R — 7); and x angle BGE, subtended by the arc B E, from which the length of the arc may be deduced, and the point E ascertained. E a = curve. Example. DC, a 1o curve; angle a = 64° 32': to connect with a 4° Here cos. X = (5,730 X 0.43) 1,433 = (5,7:30 1,433) 0.24 = cos. 76° 06'; and X 11° 31', equivalent to a distance from B around the 1° curve of 1,157 feet to E, (l= XXXIII. HAVING LOCATED A TANGENT, A B, INTERSECTING A CURVE, C I), FROM THE CONVEX SIDE, TO FIND THE POINT E ON SAID CURVE AT WHICH TO BEGIN A CURVE OF GIVEN RADIUS WHICH SHALL MERGE IN THE LOCATED TANGENT. D B E COS. (L= 1. This problem is analogous to the preceding one. The preparatory steps are the same in both. Having found the angle a, however, it will be manifest to the altentive reader, that, in this case, R (R + r) cos. . t r. Hence cos. X = (R cos. (l – - T) Ć (R + ). Example. 2. DC, a 1o curve; angle (1 = 64° 32': to connect with a 40 Here cos. x = (5,730 X 0.43) — 1,433 : (5,730 +1,433) 0.1439 = cos. 81° 43'; and x 17° 11', equivalent to a distance from B around the 1° curve of 1,718 feet to E, the point at which to start the 4o curve. G curve. a = XXXIV. TO LOCATE A Y. A 1. The processes of the two former problems may be adopted. In this case the angle al vanishes, and the cos. x clearly is equal to (R — r) + (R + ). 2. Another solution of the Y problem is as follows: Draw the tangent E D intersecting the tangent BA. Then is BD= D A E с G son that each is equal to D E. Make GF=R+ , the diameter of a semicircle. Said semicircle touches tangent B A at D, its middle point; and D E being perpendicular to G F, we have by geometry GE:DE::DE: EF; i.e., GE X EF, or R X 1,= D E. Hence DE=BD=D A=VR Xv=R tan. } x, and we are thus enabled to fix the points E and A. 3. In the two foregoing problems, the angle consumed by curve E A is 180° -X. Example. BE, a 21° curve located; BA, a tangent: to complete the Y with a 6o curve, E A. By the first method, cos. x = (R – 7) = (Rtg) = (2,292 955) = (2,292 + 955) 1,337 = 3,247 - log. 3.126131 log. 3.511482 1.614649, which corresponds to log. cos. 9.614649, or to the decimal number 0.4118, indicating in either case the angle 65° 41' = x. DE=BD=DA= R tan. { x = 2,292 X 0.6455 1,479.4. DE may be found also by reference to Table XVI., where the tangent of a 1o curve for 65° 41' is seen to be 3,698.6. Dividing this number by 23, we have 1,479.4, as above. Or, by the second method, Ilaving thus the means of fixing points E, D, and A, the curve E A can be laid down. 4. If B A is curved convex to the Y, construct the figure as in margin, and reason thus: In the triangle EGF, formed by lines connecting the curve-centres, the sides are respectively equal to the sums of the contiguous radii: the angles may therefore be found as in Case Ill., Trigonometry. A F с curves will pass through the points of intersection of the tangents to those curves severally. But lines so drawn in this case bisect also the angles of a triangle, and, demonstrably by geometry, meet in one point equidistant from the three sides of the triangle. That point, therefore, must be a common P. I. for all the curves, and that equidistance the “tangent” length common to them all. E.cample. Given B A, a 3o, and BC, a 4o curve: to complete the Y with a 5o curve, CA. EF=1,910 + 1,146 = 3,056. Adding half the difference to half the sum of the segments of the base EG, we shall have the greater of them; i.e., (3,343 + 804) • 2 = 2,073.5, which is the cos. E, E F being radius. Hence 2,073.5 ; 3,056 log. 3.316704 log. 3.485153 = 9.831551 = = cos. 47° 16' = E. By Table XVI., the tangent of a 1o curve corresponding to this angle is 2,507.3: that of a 3o curve, therefore, is 835.8 = the common tangent BD or D A. Multiplying the common tangent by 4, we shall find opposite the product in Table XVI. the central angle of the 4o curve to be 60° 32'; multiplying it by 5, we find, in like manner, the central angle of the 5o curve to be 72° 12'. Arc = 47° 16', is equivalent to 1,575 feet on the 3o curve; arc Ᏼ Ꮯ, 60° 32', is equivalent to 1,513 feet on the 4o curve. Points being thus fixed at A and C, curve CA can be laid on the ground. 5. If curve B A is concave to the Y, the radii being given, construct the figure as follows: First draw the triangle GFE, the sides of which are obviously derived from the given radii. Prolong the sides E G and BA, |