In the last column I gave a sketch of the sample problem,(Figure 2, page 18, Oct. POB), that will be used to
demonstrate traversing using state plane coordinates. There is a change, as shown in
Figure 1 of this column, because now a building is under construction that obstructs
Wakeman from being observed from Reilly. Do the following:
Occupy Reilly with a total station and back-sight to Bromilow. State
plane coordinates are known for both stations; the grid azimuth from Reilly to Bromilow
can be calculated.
Establish a temporary station called Temp in a position that's
visible from both Reilly and Wakeman.
With the total station at Reilly, back-sight to Bromilow, fore-sight to
Temp and record horizontal angle, zenith angle and slope distance to Temp. Observe the
horizontal and zenith angles in both direct and reverse positions.
Move the total station to Temp, back-sight to Reilly, fore-sight to
Wakeman and record horizontal angle, zenith angle and slope distance. As before, observe
all angles direct and reverse.
| Station
Name |
Latitude
(North) |
Longitude
(West) |
| Bromilow |
32 16 52.33969 |
106 45 15.77636 |
| Reilly |
32 16 55.93458 |
106 45 15.16429 |
| Wakeman |
32 17 0.10142 |
106 45 29.49809 |
 |
| Figure 1. Sketch of sample problem. |
|
|
As we said in earlier columns, state plane coordinates are
based on conformal map projections. This means all horizontal angles as measured are
correct, but all measured distances must be reduced to grid distances. Since the observed
angles are correct, let¿s start by calculating the grid azimuths from Reilly to Temp
and from Temp to Wakeman.
Calculate the grid azimuth from Reilly to Bromilow using state plane
coordinates determined earlier:
| Name |
Northing
(meters) |
Easting
(meters) |
| (2) Bromilow |
142158.262 |
452489.852 |
| (1) Reilly |
142268.912 |
452506.387 |
| 1 |
DN(2-1) = 110.650m |
DE(2-1) = 16.535m |
Figure 2, below shows the quadrant of the azimuth from Reilly to Bromilow.
As can be seen, the azimuth from is greater than 180° but less than 270°. The equation
for azimuth is
a = tan-1 (E2 - E1) = tan-1 DE
(N2 - N1)DN
Solving this equation gives
a = 8° 29' 56.8"
We know the azimuth is in the third quadrant (greater than 180°), so we must add 180° to a to get grid azimuth.
Azimuth Reilly to Bromilow = 188° 29' 56.8"
= 188° 29' 57"
 |
| Figure 3. Reduction to the ellipsoid. |
|
|
Reduce horizontal distances to grid distances.
This is a two-step process. First, horizontal distances at the surface must be reduced to
the ellipsoid (geodetic distances) by using the elevation factor, then reduced to the grid
(grid distances) by using the scale factor.
I'm going to borrow two figures from the NGS state plane coordinate manual1 (Figures
4.1b and 4.1c on pages 47 and 48). These will be combined as Figure 3 in this column. As
can be seen from the equation in the figure,
S = D ( R/ (R + N + H) )
Where S = Geodetic Distance
D = Horizontal Distance
H = Mean Elevation
N = Mean Geoid Height
R = Mean Radius of Earth
NGS recommends using R = 6,372,000m. In Las Cruces, N.M.,
N = -25m and H = 1188.720m (3,900 ft).
Putting these numbers into the above equation gives
Geodetic Distance = Horizontal Distance (0.99982)
Elevation factor = 0.99982
The scale factors are given in Figure 4 of this column. We can average the scale factor.
Since the traverse is short, let's use 0.9999278 = 0.99993. Multiplying the scale
factor times the elevation factor gives the "combined factor."
Combined Factor = (0.99982) (0.99993) = 0.99975
Because state plane coordinates are given in meters, I'm going to use the horizontal
distances in meters to calculate grid distances, also in meters.
Grid Distance =
Horizontal Distance x Combined Factor.
Grid Distance Reilly to Temp = (338.697m) (0.99975)
= 338.612m
Grid Distance Temp to Wakeman = (213.894m) (0.99975)
= 213.840m
STATION
NAME |
LATITUDE
(NORTH) |
LONGITUDE
(WEST) |
NORTHING
(Y) METER |
EASTING
(X) METER |
ZONE |
CONVERGENCE |
SCALE
FACTOR |
ELEV
(M) |
GEOID
HT (M) |
| D |
M |
S |
| Bromilow |
32 16 52.33969 |
106 45 15.77636 |
142158.262 |
452489.852 |
NM C |
-0 |
16 |
9.78 |
0.99992783 |
1 |
1 |
| Reilly |
32 16 55.93458 |
106 45 15.16429 |
142268.912 |
452506.387 |
NM C |
-0 |
16 |
9.48 |
0.99992781 |
1 |
1 |
| Wakeman |
32 17 0.10142 |
106 45 29.49809 |
142399.023 |
452131.948 |
NM C |
-0 |
16 |
17.17 |
0.99992825 |
1 |
1 |
Calculate state plane coordinates of stations Temp and Wakeman.
The following equations are needed to calculate coordinates of station Temp:
E Temp = E Reilly + (Grid Distance Reilly to Temp ) sin (Azimuth Reilly to Temp )
N Temp = N Reilly + (Grid Distance Reilly to Temp ) cos (Azimuth Reilly to Temp )
E Temp = 452506.387m + (338.612m) sin (256° 32' 21") = 452177.077m
N Temp = 142268.912m + (338.612m) cos (256° 32' 21") = 142190.090m
Repeating the process going from Temp to Wakeman,
E Wakeman = 452177.077m + (213.840m) sin (347° 48' 03") = 452131.890m
N Wakeman = 142190.090m + (213.840m) cos (347° 48' 03") = 142399.101m
How close did I get to the published coordinates?
Using a COGO package, you may find some minor errors in my calculations; I
used an HP 48 calculator (Hewlett Packard, Palo Alto, Calif.).
| 1 |
Northing
(meters) |
Easting
(meters) |
| Published Values |
142399.023 |
452131.948 |
| Calculated Values |
142399.101 |
452131.890 |
| 1 |
0.078m |
0.058m |
Now I have to tell what instrument I used. When I went to
my office to get the instrument for this survey, only an old manual total station was
available. Even with that, the error of closure is about 1:5000. With modern instruments
available in your company, you should get much better results. The point of this series of
articles was to show how easy and convenient it is to work with state plane coordinates.
Long article, and I haven¿t started on surface coordinates. I will
have to continue this in the next column to make a fair comparison of state plane
coordinates vs. surface coordinates. I'll also explain convergence of the meridian.