GC CON PROPOSITION HIL. FIG. 2. In case the preceding method should consume too much of the tangent CG, it is required to change the origin of the curve, also the length of radius, so that the required tangent may commence opposite to B, running parallel to B H. In this case the radiating point will be changed from N towards A and B, the radius shortened, and the point A moved towards K. Let the required distance between tangents, the given radius, and curvature be as in Proposition II., then we have by logarithms: As the external secant of 40° Is to radius: So is 30 feet = To difference of radii = 98.23 09484879 1.992242 43 305407 98 And 1146 98 = 1048 radius of a 5° 28' curve. Then, as 1146: 1048:: 800: 782 = length of 5° 28′ curve. 98 (natural tangent of 40° 83910) 82 feet. Produce tangent 82 feet from A to K, and curve from thence 73 feet of a 5° 28 curve. MA will be the new radius. PROPOSITION IV. VEURE DU AM solbe Having located a curve with a given radius, terminating in a given point, it is required to change the origin of the curve, also the radius, so as to pass through the same terminating point, with a different direction of tangent. Fig.3. A Let the given radius M B equal 2292 feet; the given are BD equal 1000 feet, containing 25° of curvature; the given tangents DF and D E make an angle of (say) 4°, DF being 400 feet, and EF 28 feet. 31. By tables 1714 feet = radius of 3° 20′ curve. PROPOSITION V. FIG. 4. Having produced the two tangents to their intersection, it is required to connect them by a curve passing a given distance from the vertical point. Given the angle LCB 31° 44′, and CE = 50 feet, to find the 15° 52′. radius MA. By geometry, the angle AMELCB our pidio-geges bayers a businool to To find the tangent AC, or CB; or point of curve. By logarithms: As R. Is to A Mockus = By natural tangents: 2010 000000 3.101181 9.453668 12:554849 1262 x (natural tangent 15° 52' 26546) 388 feet CA CB. г FORMULE FOR RUNNING LINES, PROPOSITION VI. FIG. 5. Having located a curve connecting two tangents, it is required to move the middle of the curve any given distance, either towards or from the vertex. Let the angle TLG = 36°= whole amount of curvature; the E Fig. 5. arc ABC = 1200 feet; the radius A N = CN= 1910 feet, and IB = BF: = = 10 feet. It is required to find the radii H M and EO. We have by logarithms: External secant, 18°: half of 36° ANL Is to 10 So is R. 8.737153 1.000000 10 000000 2.262847 To difference of radii 183 feet OE radius of a 3° 19' curve. By natural tangents: |