From these computed values, we can determine thestations of the PI, PC, and PT as follows:By studying figure 11-10 and rememberingthat our task is to stake half-station intervals, youcan see that the first half station after the PC is Station18 + 50 and the last half station before the PT is23+ 00; therefore, the distance from the PC to Station18 + 00 is 42.2 feet [(18 + 50) - (18 + 07.80)].Similarly, the distance from Station 23+ 00 to the PTis 7.8 feet. These distances are used to compute thedeflection angles for the subchords using the formulafor deflection angles (d= .3CD) as follows:A convenient method of determining the deflectionangle (d) for each full chord is to remember that d equals1/2D for 100-foot chords, 1/4D for 50-foot chords, 1/8Dfor 25-foot chords, and 1/20D for 10-foot chords. In thiscase, since we are staking 50-foot stations, d = 15/4, or3°45’.Previously, we discussed the difference in lengthbetween arcs and chords. In that discussion, youlearned that to be within allowable error, the recom-mended chord length for an 8- to 16-degree curve is25 feet. Since in this example we are using 50-footchords, the length of the chords must be adjusted. Theadjusted lengths are computed using a rearrangementof the formula for the sine of deflection angles asfollows:As you can see, in this case, there is little differencebetween the original and adjusted chord lengths;however, if we were using 100-foot stations rather than50-foot stations, the adjusted difference for each fullchord would be substantial (over 3 inches).Now, remembering our previous discussion ofdeflection angles and chords, you know that all of thedeflection angles are usually turned using a transit thatis set up at the PC. The deflection angles that we turnare found by cumulating the individual deflectionangles from the PC to the PT as shown below:Notice that the deflection angle at the PT is equalto one half of the I angle. That serves as a check ofyour computations. Had the deflection angle beenanything different than one half of the I angle, then amistake would have been made.Since the total of the deflection angles should beone-half of the I angle, a problem arises when the Iangle contains an odd number of minutes and theinstrument used is a 1-minute transit. Since the PT isnormally staked before the curve is run, the totaldeflection will be a check on the PC therefore, itshould be computed to the nearest 0.5 degree. If thetotal deflection checks to the nearest minute in thefield, it can be considered correct.The curve that was just solved had an I angle of75° and a degree of curve of 15°. When the I angle anddegree of curve consists of both degrees and minutes,the procedure in solving the curve does not change;but you must be careful in substituting these valuesinto the formulas for length and deflection angles; forexample I = 42°15’, D = 5°37’. The minutes in eachangle must be changed to a decimal part of a degree.To obtain the required accuracy, you should convertthem to five decimal places; but an alternate methodfor computing the length is to convert the I angle anddegree of curve to minutes; thus, 42°15’ = 2,535 min-utes and 5°37’ = 337 minutes. Substituting this infor-mation into the length formula gives the following:11-9

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