MISCELLANEOUS NOTES AND EXAMPLES. SUPPOSE a curve to contain 57° 24' curvature; distance between centres of inner and outer track 5ft. between inside and outside of track. Required difference in length To find the length of any circular arc, multiply tabular are of given number of degrees by the radius. Half of this tabular length gives the tabular area of a section of some number of degrees, and this tabular area multiplied by the square of radius, gives the required area of sector; or this tabular area, multiplied by the difference of the squares of the two radii, gives the area of a ring. Thus if inner radius = 3 ft., outer = 4, thickness being 1, we have 42327, which multiplied by tabular area gives area required. Suppose the radius of the intrados of an arch containing 134° 46′ is 6.5 ft., the thickness of voussoirs = 1.5. When the span and rise are given to find the curvature of arc, 18 ft., rise = 6 ft., then EXAMPLE.-Suppose span =nat. tang. 33° 41′, and 33° 41′ × 4 = 134° 46′ of curvature. Let it be required to find radius, we would then have, Had it been a 12 ft. span and 4 ft. rise, radius would have been 6.5 feet. Analogous to this last example, and derived from the same proposition of geometry, is an easy method of determining the distance across a river or ravine. Let the instrument be at B with a foresight upon C across river; from B lay off a right angle to D Set the instrument over D and lay off from DC a right angle D A meeting C B produced in A. Then by similar triangles, Es Now suppose the The instrumental To Triangulate round an Obstruction on a Curve. EXAMPLE-Suppose in running a 3° curve, I find the point for sta. 2645 to be occupied by a house; I find, however, that 2644 +75 and 2645 + 25 are clear of the house; also, that I have sufficient room for an equilateral triangle whose sides are 50 ft. each. tablish 2644 + 75 and set the instrument over it. last reliable point on curve to be at sta. 2640. deflection from 2640 to 2645 + 25 = 525 ft. is 7° 52'. Set the vernier to this reading, and clamp the instrument with a backsight on 2640, so that, when the vernier is at 0, the telescope may point towards 2645 + 25. Unclamp wernier, set the reading at 60°, and measure 50 ft. in line of telescope. Set instrument over this point, and turn the interior angle 60°, measuring 50 ft. as before. Set the transit over this last point, sta. 2645 + 25, with the vernier at 60° so that the zero line shall coincide with the chord from 2644 + 75 to 2045 + 25. Clamp the instrument with a sight on the second point or vertex of triangle. Then set the vernier at 1° 52', the instrumental deflection for 125 ft., and the telescope will point in direction of sta. 2646, from whence continue the curve, if required, as before. This was an expedient applied to advantage by a former associate in making the final location of the Ohio and Mississippi R. R., Ripley County, Indiana. Similar examples and corollaries to previous propositions might be added indefinitely, but this would transcend the proper limits of the work. To an adept practitioner possessing ordinary faculties of generalization, it is believed the rules and formulas already given will be suggestive of the means of solving most of the other problems which may occur in practice. |