Question 1

6 different crews were assigned to measure a portion of a single township line. All the section corners were found along the line and it was determined that the corners were all within 0° 00’ 05” of being on a straight line. Each crew performed a series of measurements for their portion of the line. They all used the same equipment and provided the office the following information;

Crew	Mean Distance	Standard Deviation
1	5280.12’	±0.05’
2	5279.58’	±0.08’
3	5281.07′	±0.12’
4	5280.79’	±0.02’
5	5282.03′	±0.15’
6	5279.68′	±0.03′

Determine the most probable value for the distance of the township line, and the 2-sigma probability for the distance.

This can be considered a simple addition of errors. This is a set of measurements of a portion of the same “Township” line. So each crew went to their assigned section and somehow measured the distance with enough redundancy to establish a standard deviation.

Most Probable value

The most probable value is the sum of the mean distances or 31,683.27 feet

2-sigma probability for the distance

First find the standard deviation for the line using the formula:

\sf \sigma_{sum} = \pm\sqrt{0.05^2 + 0.08^2 + 0.12^2 + 0.02^2 + 0.15^2 + 0.03^2} = 0.217

Now to get 2 sigma, just mulitply the standard deviation by 2

\sf 2\sigma = 0.217 \times 2 = 0.434

Question 2

For a triangulation survey, 12 measurements were taken between point “A” and Point “B”, for clarity only the seconds are listed. They are 52.4”, 52.8”, 51.6”, 51.2”, 50.6”, 52.7”, 49.8”, 50.3”, 51.8”, 50.0”, 53.0” and 51.7”. The FGCC standards for second order class II triangulation survey requires the standard deviation of the mean to be within 0.8”. Determine if this angle meets that requirement.

Most Probable Value

This is a basic standard deviation calculation. The first thing that needs to be done is the most probable value (in seconds) needs to be determined. This is just the average in this case, so that comes out to be 51.492 seconds. I am carrying the average out far enough to eliminate a rounding error in my final value.

Residuals

The reason for the most probable value is so the residuals can be determined. The residual is the difference from the most probable value and the reading. After the residuals are calculated, they are squared for the standard deviation calculation. Here is the table of residuals:

Count	Reading (Seconds only)	Residuals (v)	v2
1	52.4	-0.908	0.825
2	52.8	-1.308	1.712
3	51.6	-0.108	0.012
4	51.2	0.292	0.085
5	50.6	0.892	0.795
6	52.7	-1.208	1.460
7	49.8	1.692	2.862
8	50.3	1.192	1.420
9	51.8	-0.308	0.095
10	50.0	1.492	2.225
11	53.0	-1.508	1.275
12	51.7	-0.208	0.043

Standard Deviation

The formula to use for the standard deviation is

\Large\sf \sigma_{s} = \sqrt{\frac{\sum\it{v}^2}{n - 1}}

The sum of the residuals squared = 13.809

So the standard deviation is

\large\sf \sigma_{s} = \sqrt{\frac{13.809}{11}} = 1.12 seconds

So, the survey does not meet the “FGCC standards for second order class II triangulation survey” and a resurvey will have to be done. That really ruins your day!

Question 3

The area of a rectangular field is determined using a 100 foot steel tape that has an accuracy of 0.03’ (2-sigma). One side of the field is measured as 2304.56 feet, and the other side is 996.32 feet. Assuming it is a perfect rectangle, determine the most probable value for the area and the values error (2-sigma).

This is an interesting question in that it addresses an issue that many do not understand, which is the concept of propagation of errors when calculating an area. It is hard to convince people that the area of a parcel of land has a large variance from the calculated area based on fairly precise measurements of the sides of the parcel.

This is a propagation of error based on multiplication. The formula used is :

\large\sf 2\sigma_{Area} = \sqrt{ (2\sigma_{A} \times B)^2 + (2\sigma_{B} \times A)^2}

The question is kind enough to give the 2 sigma value for the tape already, but even if just the standard deviation was given, the standard deviation for the area could be calculated using the formula and then just mulitply that number by 2.

2 sigma for the lengths

First the 2sigma for each line must be determined. This uses a propagation by sum. The first line is approximately 2300 feet long so it would be 23 measurements with the 100′ tape. The second line is approximately 1000 feet long so it would have 10 measurements with the tape. Since the 2sigma value is the same for each measurement, the formulas for the lines can be written

\sf 2\sigma_{Length} = \sqrt{n \times 2\sigma_{tape}^2}

Where n = number of measurements. So for each line:

\sf 2\sigma_{Length 1} = \sqrt{23 \times 0.03^2} = 0.144'

\sf 2\sigma_{Length 2} = \sqrt{10 \times 0.03^2} = 0.095'

2 sigma for the area

Now the propagation of error in multiplications is used

\sf 2\sigma_{Area} = \sqrt{ (0.144 \times 996.32)^2 + (0.095 \times 2304.56)^2} = 261 sq. feet

Final Answer

So the solution is:

\sf Area = 2,296,100 sq. ft. \pm 261 sq. ft.