and trouble gained by using the rules and principles given in this book, the Author submits it, without further preface, to the profession, fully confident that its use will be practical proof of its merits. The tables and examples have been prepared with great care, and their accuracy may be relied upon. While the Author claims a fair share of originality in the following work, he would acknowledge many valuable suggestions derived from Mifflin's Diagrams, as also from Henck on Compound and Reversed Curves, authors to whom he would refer those wishing to follow the subject at greater length. On the manner of working an instrument Mifflin is very clear and concise. This work is designed especially for practical field engineers, already familiar with minor details. Cincinnati, 1855. C. H THE ENGINEER'S FIELD BOOK. FORMULÆ FOR RUNNING LINES, LOCATING SIDE-TRACKS, &c. PROPOSITION I. FIG. 1.* To change the origin of a curve so that it shall terminate in a tangent parallel to a given tangent. Suppose the curve A C to have been described containing 60° of curvature, and that the distance G D equal 50 feet. We have by logarithms: Sine 60° (total amount of curvature), Is to R.. So is G D, 50 feet, To AB = 57*73 feet, Or by nat. sines = 9.937531 10.000000 1.698970 1.761439 Produce the tangent from A to B *The diagrams in this work are not drawn to any exact scale, but are designed to represent merely the abstract geométrical relation of lines, curve BD equal A C; that is AMC-BND; then the tangents will be parallel. This rule will apply to the origin of a compound curve, using the total amount of curvature run. PROPOSITION II. FIG. 2. Having a curve A B terminating in a tangent D F, it is required to find the radius of a curve that will give a tangent CG parallel to DF at any given distance therefrom, as at D C say 30 feet. = Let AM be the given radius = 1146 feet, the arc AB 800 feet, containing 40°, and D C perpendicular distance 30 feet. By logarithms: Then we have 1146 + 128 = 1272 radius of a 4° 30' curve. Then say 1146: 1272:: 800: 888 = arc A C. This case is equally applicable to changing the last radius used in a compound curve terminating in a parallel tangent. PROPOSITION III. FIG. 2. In case the preceding method should consume too much of the tangent O G, 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 Í. N Fig. 2. M F H A In this case the radiating point will be changed from M 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° 9484879 10:000000 1:477121 1.992242 =98. And 1146- 98 1048 radius of a 5° 28′ curve. Then, as 1146: 1048::800: : 732 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 732 feet of a 5° 28′ curve. PROPOSITION IV. F. 3. 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. Let the given radius M B equal 2292 feet; the given are BD equal 1000 feet, containing 25° of curvature; the given tangents D ̊F and D E make an angle of (say) 4°, DF being 400 feet, and EF 28 feet. We have 28 4° angle E D F, consequently the angle 29. CLD 25° + 4° By logarithms: As versed sine 29° Is to versed sine 25° So is radius given B M = 2292 To radius required C L= 1714 feet. By tables 1714 feet = radius of 3° 20′ curve. PROPOSITION V. FIG. 4. 9.098229 8.971703 3.360217 3.233991 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 |