. s obtained by the table of sines, is 767*54, onlyy^Q- of a foot difference. That is, sine 28° 40 x radius 800 x 2 = now it be required to find the length of chord corres-ponding to 950 feet of a 6° curve. 900 feet gives length of chord . 867451000 . 955-37 Sum 1822-82 Mean 950 91141 Now sine 28° 30 x radius 955-37 x 2 = length of chord = 911-7],being only j^ ^^ ^ ^^^^ difference, so that this table will be sufficient for ordinary purposes. For common rates of curvature for aless distance, say 650 feet, the variations
. s obtained by the table of sines, is 767*54, onlyy^Q- of a foot difference. That is, sine 28° 40 x radius 800 x 2 = now it be required to find the length of chord corres-ponding to 950 feet of a 6° curve. 900 feet gives length of chord . 867451000 . 955-37 Sum 1822-82 Mean 950 91141 Now sine 28° 30 x radius 955-37 x 2 = length of chord = 911-7],being only j^ ^^ ^ ^^^^ difference, so that this table will be sufficient for ordinary purposes. For common rates of curvature for aless distance, say 650 feet, the variations from the true length wouldbe scarcely perceptible. 403 FOR RlNNLXG LlXZ>, ifco. Problem.—L-t A and C br two fxfd iangcnt points, tlie positions Iof irhose tangents are deterniined by the angles DA C = m =. 18, ;B C jE =^ n =z G^, and the perpendicular distance D C ^= p ^= A^^oi//.* Required the amount of cnrvature in the arc A B, its reversionB C ajid the laujh of the comnion radius 0 £ =^ Jf B by whichthe arcs A B a?.d B C .irc Let m = nrit vers, sine DAC, and n = nat. vers, sine B CE. Let .r = nat vers, sine (A O B — -rn) = (B M C — n). (>r curvature A B = jn -f ^, and onrvatiire BG = n -{- xA T ^ J , yi - V V. s. IS -^ V. s. 6io nna x we have, .>• = == — 0-048944 -h 0-005478 o =0 020 x sine IS- = 15iX) > 0-3}9*>2 = 463-53. t D G E being equal to
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Keywords: ., bookcentury1800, bookdecade1850, booksubjectenginee, bookyear1856
