SOLVING AND LAYING OUT A
Now let’s solve and lay out a simple curve usingthe arc definition, which is the definition you will more often use as an EA. In figure 11-10, let’s assume that the directions of the back and forward tangents and the location of the PI have previously been staked, but the tangent distances have not been measured. Let’s also assume that stations have been set as far as Station 18 + 00. The specified degree of curve (D) is 15°, arc definition. Our job is to stake half-stations on the curve.
Solving a Simple Curve
We will begin by first determining the distancefrom Station 18 + 00 to the location of the PI. Since these points have been staked, we can determine the distance by field measurement. Let’s assume we have measured this distance and found it to be 300.89 feet. Next, we set up a transit at the PI and determine that deflection angle I is 75°. Since I always equals D, then D is also 75°, Now we can compute the radius of the curve, the tangent distance, and the length of curve as follows:
From these computed values, we can determine thestations of the PI, PC, and PT as follows:
By studying figure 11-10 and remembering that our task is to stake half-station intervals, you can see that the first half station after the PC is Station 18 + 50 and the last half station before the PT is 23+ 00; therefore, the distance from the PC to Station 18 + 00 is 42.2 feet [(18 + 50) - (18 + 07.80)]. Similarly, the distance from Station 23+ 00 to the PT is 7.8 feet. These distances are used to compute the deflection angles for the subchords using the formula for deflection angles (d= .3CD) as follows:
A convenient method of determining the deflection angle (d) for each full chord is to remember that d equals 1/2D for 100-foot chords, 1/4D for 50-foot chords, 1/8D for 25-foot chords, and 1/20D for 10-foot chords. In this case, since we are staking 50-foot stations, d = 15/4, or 3°45’.
Previously, we discussed the difference in length between arcs and chords. In that discussion, you learned that to be within allowable error, the recommended chord length for an 8- to 16-degree curve is 25 feet. Since in this example we are using 50-foot chords, the length of the chords must be adjusted. The adjusted lengths are computed using a rearrangement of the formula for the sine of deflection angles as follows:
As you can see, in this case, there is little difference between the original and adjusted chord lengths; however, if we were using 100-foot stations rather than 50-foot stations, the adjusted difference for each full chord would be substantial (over 3 inches).
Now, remembering our previous discussion of deflection angles and chords, you know that all of the deflection angles are usually turned using a transit that is set up at the PC. The deflection angles that we turn are found by cumulating the individual deflection angles from the PC to the PT as shown below:
Notice that the deflection angle at the PT is equal to one half of the I angle. That serves as a check of your computations. Had the deflection angle been anything different than one half of the I angle, then a mistake would have been made.
Since the total of the deflection angles should be one-half of the I angle, a problem arises when the I angle contains an odd number of minutes and the instrument used is a 1-minute transit. Since the PT is normally staked before the curve is run, the total deflection will be a check on the PC therefore, it should be computed to the nearest 0.5 degree. If the total deflection checks to the nearest minute in the field, it can be considered correct.
The curve that was just solved had an I angle of 75° and a degree of curve of 15°. When the I angle and degree 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 values into the formulas for length and deflection angles; for example I = 42°15’, D = 5°37’. The minutes in each angle must be changed to a decimal part of a degree. To obtain the required accuracy, you should convert them to five decimal places; but an alternate method for computing the length is to convert the I angle and degree of curve to minutes; thus, 42°15’ = 2,535 minutes and 5°37’ = 337 minutes. Substituting this information into the length formula gives the following:
This method gives an exact result. By converting the minutes to a decimal part of a degree to the nearest fives places, you obtain the same result.
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