The problem is:
Calculate the state plane coordinates for station King whose NAD 27 coordinates are
latitude N40Â° 43' 37.302"
longitude W88Â° 41' 35.208"
The station is located in the State of Illinois, state plane zone Illinois
East.
Figure 1 shows the
map from the U.S. Coast and Geodetic Survey manual for the state of Illinois, also
reproduced in Rayner and Schmidt1. Illinois uses the transverse Mercator projection with
two zones, east and west. Each zone has its own axis for y, although both axes passing
through the east and west zones are given an x-value of 500,000'. Both zones use
theÂ sameÂ x-axis, which is located well below the southern limit of the state and
has a value of zero feet.Â TheÂ central meridian of the East Zone is 88Â°20' west
longitude; along this line the scale of the projection is one part in 40,000 parts too
small. The lines of exact scale are parallelÂ toÂ theÂ central meridian and
situated approximately 28 miles east and west. Of course, to the east and west of these
lines, the scale is too large. The parallelÂ ofÂ latitude 36Â°40' defines the
x-axis; the origin of coordinates for the east zone
isÂ aÂ pointÂ onÂ theÂ 36Â°40'Â parallelÂ situated 500,000'
west of longitude 88Â°20'.
Let's perform the calculations. Unlike the Lambert projection, there isn't a
sketch that shows the geometric relations between latitude, longitude and x,y. The
equations necessary to perform these calculations are as follows:
x = x' + 500,000Â Â Â (1)
x' = H Dl" +/- a bÂ Â Â (2)
y = yo + V ("/100)2 +/- cÂ Â Â (3)
Where x' is the distance, the point is either east or west of the central meridian; yo, H, V and a are quantities based on the geodetic latitude; b and c are based on Dl" (the difference in longitude of the point from the longitude of the central meridian, in seconds-of-arc).
Tables are needed to get the values for H, V, a, b, yo and c. Fortunately, all values can be found in two tables, which are given in the publication for the state of Illinois; but for this article, Tables 1 and 2 (on page 18) from Rayner and Schmidt are abstracts of the original tables that cover the values needed to solve our problem.
Repeating the problem:
Given:Â Â Â
Station King
latitude N40Â° 43' 37.302"
longitude W88Â° 41' 35.208"
State - Illinois, East Zone
Central Meridian - W88Â° 20' 00Solution:
1) Solve for Dl. Since we are in the western hemisphere, all values of longitude are minus.
Dl" = longitude - central meridian longitude.
Dl = -88Â° 41' 35.208" - (-88Â° 20' 00")
Dl = -0Â° 21' 35.208" = -1,295.208 seconds-of-arc
2) CalculateÂ Â Â (Dl"/100)2
Â Â Â (Dl"/100)2 = 167.756
Table 1. Values of H and V - Illinois East Zone
Values of H and V - Illinois East Zone |
|||||||
Lat. |
Y0 (feet) |
Â³Y0 per |
H |
Â³H per |
V |
Â³V per |
a |
40Â°Â Â Â
35' |
1,426,385.98 |
101.19683 |
77.158010 |
319.28 |
1.216989 |
1.85 |
- 0.509 |
Table 2. Values of b and c - Illinois Zones
Values of b and c - Illinois Zones |
|||
DlÂ¿ | b | Db | c |
0 Â Â Â 100 Â Â Â 200 Â Â Â 300 Â Â Â 400 Â Â Â 500 Â Â Â 600 Â Â Â 700 Â Â Â 800 Â Â Â 900 Â Â Â 1,000 Â Â Â 1,100 Â Â Â 1,200 Â Â Â 1,300 Â Â Â 1,400 Â Â Â 1,500 Â Â Â 1,600 Â Â Â 1,700 Â Â Â 1,800 Â Â Â 1,900 Â Â Â 2,000 |
0.000 Â Â Â +0.212 Â Â Â +0.424 Â Â Â +0.634 Â Â Â +0.842 Â Â Â +1.049 Â Â Â +1.252 Â Â Â +1.453 Â Â Â +1.649 Â Â Â +1.841 Â Â Â +2.028 Â Â Â +2.209 Â Â Â +2.384 Â Â Â +2.553 Â Â Â +2.715 Â Â Â +2.868 Â Â Â +3.014 Â Â Â +3.151 Â Â Â +3.279 Â Â Â +3.397 Â Â Â +3.504 |
+0.212 Â Â Â +0.212 Â Â Â +0.210 Â Â Â +0.208 Â Â Â +0.207 Â Â Â +0.203 Â Â Â +0.201 Â Â Â +0.196 Â Â Â +0.192 Â Â Â +0.187 Â Â Â +0.181 Â Â Â +0.175 Â Â Â +0.169 Â Â Â +0.162 Â Â Â +0.153 Â Â Â +0.146 Â Â Â +0.137 Â Â Â +0.128 Â Â Â +0.118 Â Â Â +0.107 Â Â Â +0.097 |
0.000 Â Â Â 0.000 Â Â Â -0.001 Â Â Â -0.002 Â Â Â -0.003 Â Â Â -0.005 Â Â Â -0.007 Â Â Â -0.010 Â Â Â -0.014 Â Â Â -0.018 Â Â Â -0.022 Â Â Â -0.027 Â Â Â -0.032 Â Â Â -0.038 Â Â Â -0.043 Â Â Â -0.049 Â Â Â -0.055 Â Â Â -0.061 Â Â Â -0.067 Â Â Â -0.073 Â Â Â -0.079 |
Table 3. Values of Scale Factors - Illinois East Zone
Values
of Scale Factors - |
||
xÂ¿ (feet) | Scale in Units of 7th Place of Logs |
Scale Expressed as a Ratio |
0 |
-108.6 |
0.9999750 0.9999750 0.9999751 0.9999752 0.9999755 0.9999757 0.9999760 0.9999764 0.9999768 0.9999773 0.9999778 0.9999784 0.9999791 0.9999798 0.9999806 0.9999814 0.9999823 0.9999833 0.9999843 0.9999853 0.9999864 0.9999876 0.9999888 0.9999901 0.9999915 0.9999929 0.9999943 0.9999958 0.9999974 0.9999990 |
3) From Table 1, interpolate for H, V, yo and a.
The argument for all values is the latitude of the point, 40Â°43'37.202". From Table
1 we must interpolate between 40Â° 43' and 40Â° 44' by the amount 37.202"/60" =
0.6217.
By doing this we get the following:Â Â Â
H = 76.992654
V = 1.217932
a = -0.492
yo = 1,478,725.73.
4) From Table 2, interpolate for b and c.
Â Â Â b = +2.545
Â Â Â c = -0.04
The argument for both values is Dl in seconds-of-arc.Â Â Â Â
5) Solve for H Dl and a b, which are needed to solve equation (2), given above:
Â Â Â x' = HDl" +/- a b,
Â Â Â HDl" = -99,721.50
Â Â Â a b = -1.25
Â Â Â x' = HDl +/- a b = -99,720.25
Note
The "sign convention" is as follows: when a b is negative, decrease HDl"
numerically. If a b is positive, increase HDl" numerically. Since Dl" is
negative because the station is west of the central meridian, x' is also negative.
6) Solve equation (3),
Â Â Â y = yo + V ( Dl/100)2 +/- c
Â Â Â y = 1,478,930.01 ft.
7) Solve for the scale factor.
The argument for scale factor is x'. Table 3 gives the scale factor for different values
of x'. In our problem,
x' = -99,720.25. The "minus" sign is not needed for this calculation.
Solution
Scale factor = 0.9999863
As you can see, the calculations on the transverse Mercator grid are more complicated than
calculations on the Lambert grid. However, at most, two conversions are needed for
traversing a small area; after that all calculations are made using plane trigonometry.
That"s what we will discuss in the next column. We are getting close to the end of
this series, at most two more.