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. chord for 800 ft. 1st distance, 2nd distance, etc. Intermediate Distance. =92.385 2 593.36 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, 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. -mid. ordinate 800 ft. curve ordinate at 1st station, Mid. ordinate 1000 ft. mid. ordinate 600 ft, = ordinate at 2nd station. For this purpose we have calculated a table of middle ordinates corresponding to that of long chords. From this we have, 107.39 - 0.00107.39 = 5th or middle ordinate. Were the chord for 900 ft. of curve we should have by tables, 87.25 53.05 34 20 1st ordinate. ordinates can be The same principle After passing the inversely; as will Then from end of This will sufficiently demonstrate how the obtained for any other length of chord or curve. obtains in regard to any other rate of curvature. middle ordinate, their lengths will be repeated also be the intermediate lengths of abscissas. first abscissa erect first ordinate, and so on in regular rotation. TABLE Of Middle Ordinates from Chords subtending Curves of from 100 to 1000 feet in length; calculated to every 15' of Curvature from 15' to 8°. Radius of 1° being 5730 feet. LENGTHS OF ARCS. 100 200 800 400 500 600 700 800 900 1000 6.32 14.20 25.19 89.24 6.54 14.69 26.65 40.57 99 15 124 78 1:867 162:42 128.84 157 98 6.76 15.18 26.91 41.90 60-08 81:37 105-68 132-88 162-86 6.98 15.66 27-77 43.23 61.97 83.90 108 92 136-89, 167-70 71-18 78.74 56:30 76:30 58.19 78.84 92.57 116.58 143.13 95.87 120-69 148.12 On the principles by which the following tables are calculated. Let m linear opening of switch rail, 8 = angular opening of |rail, ƒ = 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: fs; 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 of frog will be = f. ƒ=". But if the backsight be taken on a point 2 (say 5 inches distant) parallel with the main track, the deflection will Calling s 1° 15', f= 6° 45', g = 4'70, m = = 0·42, and g· 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, f+2 s sin. 2 : 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. = = |