Solution of a Simple Curve
To solve a simple curve, the surveyor must know three elements. The first two are the PI station value and the I angle. The third is the degree of curve, which is given in the project specifications or computed using one of the elements limited by the terrain (see section II). The surveyor normally determines the PI and I angle on the preliminary traverse for the road. This may also be done by tri-angulation when the PI is inaccessible.
Chord Definition
The six-place natural trigonometric functions from table A-1 were used in the example. When a calculator is used to obtain the trigonometric functions, the results may vary slightly. Assume that the following is known: PI = 18+00, I = 45, and D = 15°.
Chords. Since the degree of curve is 15 degrees, the chord length is 25 feet. The surveyor customarily places the first stake after the PC at a plus station divisible by the chord length. The surveyor stakes the centerline of the road at intervals of 10,25,50 or 100 feet between curves. Thus, the level party is not confused when profile levels are run on the centerline. The first stake after the PC for this curve will be at station 16+50. Therefore, the first chord length or subchord is 8.67 feet. Similarly, there will be a subchord at the end of the curve from station 19+25 to the PT. This subchord will be 16,33 feet. The surveyor designates the subchord at the beginning, C1 , and at the end, C2 (figure 2).
Deflection Angles. After the subchords have been determined, the surveyor computes the deflection angles using the formulas. Technically, the formulas for the arc definitions are not exact for the chord definition. However, when a one-minute instrument is used to stake the curve, the surveyor may use them for either definition. The deflection angles are
The number of full chords is computed by subtracting the first plus station divisible by the chord length from the last plus station divisible by the chord length and dividing the difference by the standard (std) chord length. Thus, we have (19+25 – 16+50)-25 equals 11 full chords. Since there are 11 chords of 25 feet, the sum of the deflection angles for 25- foot chords is 11 x 1°52.5’ = 20°37.5’.
The sum of d1, d2, and the deflections for the full chords is
The surveyor should note that the total of the deflection angles is equal to one half of the I angle. If the total deflection does not equal one half of I, a mistake has been made in the calculations. After the total deflection has been decided, the surveyor determines the angles for each station on the curve. In this step, they are rounded off to the smallest reading of the instrument to be used in the field. For this problem, the surveyor must assume that a one-minute instrument is to be used. The curve station deflection angles are listed on page 6.
Special Cases. The curve that is solved on page 6 had an I angle and degree of curve whose values were whole degrees. When the I angle and degree of curve consist of degrees and minutes, the procedure in solving the curve does not change, but the surveyor must take care in substituting these values into the formulas for length and deflection angles. For example, if I = 42° 15’ and D = 5° 37’, the surveyor must change the minutes in each angle to a decimal part of a degree, or D = 42.25000°, I = 5.61667°. To obtain the required accuracy, the surveyor should convert values to five decimal places.
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 into the length formula gives
This method gives an exact result. If the surveyor converts the minutes to a decimal part of a degree to the nearest five places, the same result is obtained.
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 one- minute instrument. Since the surveyor normally stakes the PT prior to running the curve, the total deflection will be a check on the PT. Therefore, the surveyor should compute to the nearest 0.5 degree. If the total deflection checks to the nearest minute in the field, it can be considered correct.
Curve Tables
The surveyor can simplify the computation of simple curves by using tables. Table A-5 lists long chords, middle ordinates, externals, and tangents for a l-degree curve with a radius of 5,730 feet for various angles of intersection. Table A-6 lists the tangent, external distance corrections (chord definition) for various angles of intersection and degrees of curve.
Arc Definition. Since the degree of curve by arc definition is inversely proportional to the other functions of the curve, the values for a one-degree curve are divided by the degree of curve to obtain the element desired. For example, table A-5 lists the tangent distance and external distance for an I angle of 75 degrees to be 4,396.7 feet and 1,492.5 feet, respectively. Dividing by 15 degrees, the degree of curve, the surveyor obtains a tangent distance of 293.11 feet and an external distance of 99.50 feet.
Chord Definition. To convert these values to the chord definition, the surveyor uses the values in table A-5. From table A-6, a correction of 0.83 feet is obtained for the tangent distance and for the external distance, 0.29 feet.
The surveyor adds the corrections to the tangent distance and external distance obtained from table A-5. This gives a tangent distance of 293.94 feet and an external distance of 99.79 feet for the chord definition.
After the tangent and external distances are extracted from the tables, the surveyor computes the remainder of the curve.