SOLVING AND LAYING OUT A
SIMPLE CURVE Now let’s solve and lay out a simple curve using the
arc definition, which is the definition you will more often use as an EA. In
figure 1110, 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 halfstations on the curve.
Solving a Simple Curve
We will begin by first determining the distance from
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 the stations
of the PI, PC, and
PT as
follows:
By studying figure 1110 and remembering that
our task is to stake halfstation 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 100foot
chords, 1/4D for
50foot chords, 1/8D
for 25foot chords, and 1/20D
for 10foot chords. In
this case, since we are staking 50foot 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 16degree curve is 25 feet. Since in this example we
are using 50foot 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 100foot stations rather
than 50foot 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 onehalf of the I
angle, a problem arises
when the I
angle contains an odd number of minutes and the instrument used is a 1minute
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.
