b. To find v the secant point in the curve, the formula are: (cosec B-1) R=B v, or (tan 2 (180-B) + 1) = B v, to the radius 1. Again (2 R+B v) Bv=B t2, a quadratic equation. And 2 Bv=(Bt + R 2) 1 — R. From the foregoing we have 2 log Bt. =2·6622850, and 459·5 = B t2 By the formula for the cosec we have by reduction— c. To find the secondary tangents vd. The formula is R tan(-B)=vd (p. 111). Taking B as in the foregoing. Supplying the radius and taking logarithms— 16. For tracing circular curves with the chain (fig. 37). The formula for finding the perpendicular to the tangent is : ct2 2 R dc= (p. 111). Here R is the radius of the curve, and dc the required M perpendicular. Let R80 and c t2 = 2. Then d c (= 2 × 2÷80 × 2)=2÷80. If the unit of measure be a chain, then d c=0·025 chains, or 2 links. Again, the perpendicular to the chord produced may be found from the formula d' c'=! _t.d'xdc (p. 111). In most cases R will be large as compared with t d', then d'c' 2 dc. The application of the formula is so simple, that it does not require an example to assist in making the computation clear. 17. Tracing circular curves by perpendiculars to the chord (fig. 38). The formula for finding the perpendicular from a given reference point in the chord are :— og = r sin e (2-t, g2), = 2 r cos 0, or og = tate тс = (r2-ng2), the cosine for half the angle at the centre the chord for the radius r and angle 20. the perpendicular to the diameter parallel to the chord, which perpendicular shall pass through the given reference point in the chord. And mc-og-n c, the required perpendicular (p. 112). It may be seen that og and t te are computed only once for each curve, and that it is necessary to compute m c for each point to be determined in the curve. Let R 80 and 0-75° 30'. radius and taking logarithms Then supplying the To find the perpendicular at the reference point n— Let gn=10. Then (6400-100)=mc or mc=79-372. And nc (=79·372–77·272)=2·125, the required perpendicular. 18. Tracing circular curves. To find the perpendicular from a given reference point in the tangent. This is solved by Ex. 16 if the corresponding chord be given. By another formula R (1-cos ) = z t1 (=df) =zd (=tc) (p. 112). Let the radius of curvature, the angle made by tangents, &c., be taken as above, and let te f=1.396. Taking the radius 1— 1.39680 (sin )=0.01745. Hence =1°, and cos p=0.99985. Then 80 × (1-0-99985) (=c′ƒ)=0·012 the required perpendicular. 19 Tracing circular curves by tangents. To find the reference point in the extreme tangents for a tangent to the curve at given angles from the radius to the point of tangency of the extreme tangents (fig. 40). In these formula m is the number of equal subdivisions of the arc, and ø the angle at the centre subtended by one of the subdivisions. Let R=80, B = 60°, and 2 m=120. Supplying the radius, and taking logarithms— The line sq may be computed for an angle, as already shown for B v, Ex. 15. 20. To find the angle contained between tangents to the curve when the point of intersection of the tangents shall be inaccessible (fig.43). The formula is— } {{Am ƒ− n m ƒ − ) + (Cƒm—nƒm)} = n (p. 117). In this formula let the angles Cfm=135°, and Amf =145°. Now 180°—135° (=nfm)=45°, and 180° 145° (=nmƒ)=35°. Substituting these values in the formula we have {(145°—35°)+(135° — 45°)} (=n)=100°. 21. Division of lands (fig. 44). Let A represent area of the parcel for determination, a the area of the approximate parcel, and BD the line Let A=86972, a 86201, and B D=450. tution and reduction 2 2 (171) = +p. Here p 450 22. Proportional division of lands (fig. 45). By substi 3-4266. Let the parts be a, b, c, d, and let the proportions be Total of proportional parts 9.1 Let the total area to be divided be 1697854 square Then by formula (p. 122)— units of area. 23. Proportional division of lands, according to value (fig. 46). Let the number of parts, the proportionals, and the total area be as in Ex. 22. Let P1, P2, P3, P4 be the divisions; and let pi's pro 2 portion be of the value of the whole area for division; 9.1 also let the total units of value be £9900. Then the value of p's portion is 2175-8243 units of value (=2x.9900). Now let a =697000, and a'=1000854 units of area. areas of the approximate parcels c and c' be 200000 and 170000 units of area respectively. We have by substitution in the formula 2 xc the value of the value of Pia, i.e.: α v' Plar and x c' Let mn=13500, mn=6000, and m n=7500 lineal units, Then 13500 61-947:: 7500: 34-415 (=d。), and 13500 61.947:: 6000: 27-532 (=dc). This gives the deficiency of value on each division of v a quality. We have quality of a, and × 34-415=area α v on 7500 of m n. Also x 27.532 area on 6000 of m n. |