EXPLANATION OF THE USES AND APPLICATIONS OF THE TABLE OF LONG CHORDS. PROBLEM. Required to find the distances or abscissas on the chord from which, if ordinates or perpendiculars be drawn, they will pass through the station points on the curve. EXAMPLE. Let the given curve be 1000 ft. long of 5° curvature, or 1146 ft radius. For the first station from the beginning we have chord for 1000 ft. 2 chord for 800 ft. chord 800 - chord 600 1st distance, 2nd distance, etc. 484-435= half length == 2 Thus for any given station we take from the length of the whole chord the length of a chord of twice as many stations less than the one under consideration; that is, 1st station from beginning 2 less; 2 from beginning, 4 less, etc., and take half the difference. If the chord had been for 900 ft of curve, we should have, 877.32 689.39 93.9651st distance. MIDDLE ORDINATES. In like manner we may find the ordinates connecting these abscissas with their points on the curve. Let the length of chord and radius be as already given. Then we have, Mid. ordinate 1000 ft. nate at 1st station. Mid. ordinate 1000 ft. 2nd station. 107.39= 5th or middle ordinate. Were the chord for 900 ft. of curve we should have by tables, 87.2553.05 34 20 1st ordinate. 87 2527 17: 87.25 9.81 87.25 60:08 = 77.44 1·09= : 86.16 2nd, 66 : 4th 0.00 87 25 = middle This will sufficiently demonstrate how the ordinates can be obtained for any other length of chord or curve. The same principle obtains in regard to any other rate of curvature. After passing the middle ordinate, their lengths will be repeated inversely; as will also be the intermediate lengths of abscissas. Then from end of first abscissa erect first ordinate, and so on in regular rotation. angle D A meeting CB 7 BC. ces. point of +507 e curve rate per rom 30 to nes=42 ft., Le 18 to 20 ft. 506 of frog MISCELLANEOUS NOTES AND ANEOUS NOTES (say 5 incances one to contain frog, radi then be=" will be homolog d outer tr outsid 2° 00' 0.44 1.75 3.92 15 On the principles by which the following tables are calculated. Let m linear opening of switch rail, s = angular opening of rail, f= angle of frog, g = gauge of track. Let x = length of chord from opening of switch rail to point of frog. Then will the amount of curvature between the opening of rail where curve commences and point of frog = ƒ-8; therefore the instrument setting over the open end of switch rail with a backsight on the fixed end of it, the instrumental deflection to the point In like manner we may find_' MIDD abscissas with their points on tacksight be taken on a point Let the length of chord T we have, subi the main track, the deflection will Making the value of x, radius, g-m Mid. ordi f+s o the sine of: Then we have, 2nd station. For th corresp XAMPLE: Calling &= 1° 15', ƒ 6° 45', g = 4'70, m = 0.42, and g-m= 4.28, we have sin. 4° R:: 4.28 x 61.36ft. When a double opening of a switch rail for a double turnout occurs, we have, sin. (f +20): : Rg. -- 2 x m: x = distance to nearest frog. The linear and angular opening of rail being the same, this table may be adapted to any other gauge by increasing the value of x as given in this table, and the length of radius of turnout 2 per cent. for every additional inch in the gauge. This is a little too much; the correction for a 6 ft. gauge being about 30 per cent Thus 100 ft. chord of turnout on this track will give 130 ft. on 6 ft. gauge, and 1000 ft. radius will give 1300 ft. This is for a straight line. When on a curve going the same way as turnout, it is sufficiently accurate for practice to add rate of curve of main track to that of the table; but when going in opposite direction, subtract it; thus making relative departure from main track the same as on a straight line. EXAMPLE: Thus a 5° frog for a 4ft. 8 inch gauge gives a distance of 78.5 ft. curvature 4° 46'. If the main track were a 4° curve and going the same way, distance being the same, the rate of curvature would be 4° 46′+4° 8° 46', radius 653 ft.; but going the other way 4° 46' ·4° 0° 46′, radius 7473 ft. |