12S56 Circle data Collected
11/03/2009
Data
AT TO Angles
(deg) Distance
(m)
A B 0.0001 34.371
C 67.5996 34.359
B C 0.0001 38.233
A 56.1876 34.371
C A 0.0001 38.362
B 56.2187 38.233
a 315.2607 8.027
b 338.1861 23.402
c 20.5806 40.271
d 43.8662 40.753
e 70.5921 33.013
f 93.4452 20.656
O 33.8481 20.702
Solution: |
Solution adjustment. The first step in the analysis is to
make the angles consistent (i.e., sum to 180 degrees). These adjustments are usually made by
distributing the "misclose" (the difference from 180 deg), into
each angle inversely proportional to the line lengths. In our case the lengths are all about
the same length so we subtract 0.002 deg to each angle. (This corresponds to
mis-pointing by ~1.0mm over the 33-39 meter distances). The distance measurements all agree
in the forward and back directions except for one 1 mm difference. The first measurement was adopted. |
(a) Using the geometry from the
figure above at site 00, we can write two equations for the radius: |
|
The division of these two equations results in the R being
canceled and using the expansion of we can write |
|
By expansion, this equation reduces to: |
|
Using the estimate of a1,
we can then solve for the radius R. |
For each corner point the results are: |
|
(b) To find the radius to each of the intermediate points,
we use the data from site C. The
cosine rule is used to solve for r
and the sine rule to solve for y. To solve these equations we use: |
|
|
(c) The position of the sprinkler at the center (CEN) and
computed by geometry. If the
spigot had been exactly at the center, the distance to it would have been
20.677 m (compared to the measured value of 20.702 m). The difference in position places the
spigot 0.028 m from the center at y
=-27 deg. |
|
The total results are shown in the figure below. (ÒSouthÓ is the direction from the
center of the circle to point A, ÒEastÓ at right angles to this direction. |
The residuals to the mean radius and a function of the
angle at the center are in the figure below: |
This project was solved using Matlab code Proj_3_09.m.
The output of the code (in addition to the figures above is: |
12S56 Project Number 3
Sum of angles in triangle
is 180.0057 deg, adding -0.0019 to each angle
-------12S56
2009------------------------
Results for each
angle/distance pair
Alpha 1 33.8125 Radius
1 20.677
Beta 1 33.7836 Radius 2 20.677
Gamma 1 22.4044 Radius
2 20.677
Mean radius 20.677
-----------------------------------
Point Radius Drad
Angle
B 20.677 0.000 67.5639
a 20.749 0.072 270.0690
b 20.637 -0.040
223.2940
c 20.689 0.011 138.9328
d 20.691 0.014 92.3788
e 20.645 -0.032
38.5476
f 20.608 -0.069
352.5300
Sprinkler Position 0.028 (m) at -27.39 deg