Reference no: EM131080918

Part 1:

1. Missing side of traverse: Find the missing latitude and departure

Side	Bearing			Length	+N	-S	+E	-W
AB	S	6.25	W	189.53		188.4		20.63
BC	S	29.63333	E	175.18		152.27	86.62
CD	N	81.3	W	197.78	29.92			195.5
DE	N	12.4	W	142.39	139.07			30.58
EA

2. What errors should you correct in the previous example?
a. Latitude
b. Departure
c. Length
d. Latitude and Departure
e. Latitude, Departure and Length
f. Latitude and Length
g. Departure and Length
h. None
i. All

3. The data used to plot Figure 1 is given in Table 1.

Fill in the following table, find the bearings, latitude, departure, coordinates  x and y) and obtain  the area  Figure 1) using the coordinate method  ignore the corrections)

Table 1 Information used to plot figure 1

                                                        Latitude I Departure Point     X      Y     +      -

Side AB BC CD DE EF FA

Length 100 150 172 100 110 68

Azimuth 35 290 210 135 80 352

Bering

+N

-S

+E

-W

A

100

100

B

C

D

E

F

4. A measuring tape of 99.98cm is to be used to lay off a measurement of 600m. If temperature is constant at 40 Celsius and tension has been consistently applied at 5Kg, estimate the actual distance required to be measure on the field.

5. Assuming your coordinate system follows south azimuths. Subtract the following angles

a. 30° 20' 55" minus 10° 20' 55"
b. 60° 20' 05"  minus 10° 22' 55"
c. 10° 20' 50"  minus 30° 40' 55"

6. Three Horizontal angles of a triangle were read one time each, how would you correct the error

a. Evenly distributing angular misclosure over the number of angles
b. Using a weight average based on standard deviation from the 3 angles
c. Correction = length/perimeter x EL or ED accordingly
d. Considering accumulated distance from the starting point
e. Options c. and d.
f. All of the above
g. None of the above

Part 2:

Question 1

(a) Define the following terms briefly:
(i) Accidental errors
(ii) Systematic errors
(iii) Azimuth
(iv) Bearing

(b) Explain step by step how to utilize a Theodolite and a Tape only in order to determine a horizontal distance between point A, located on the ground shore side to a point B, located on an off-shore structure in the sea. Support your answer with sketch and equations.

Question 2. A 30 m steel tape measured 29.900 m when supported throughout under a tension of 90 N at 20°C standard temperature. The tape has a cross-section area of 5 mm2, a coefficient of expansion of 0.0000116 per meter per °C, an elasticity modulus of 1.92*1011N/m2, and a total weight of 0.9 kg. The tape was used supported only at the ends with a constant tension of 100 N to measure a straight line from A to B in 5 segments, of which only the last one was less than the full tape length. The data given in the following table were recorded. Apply corrections for tape length, temperature, pull, sag, and tape not horizontal to determine the correct length of the line.

Section	Recorded Distance Ern]	Recorded Temp [°C]	Difference in elevation [m]	Temp correction	Tape size correction	Pull correction	Sag correction	Slope correction
A-1		25	0.60
1-2		24	0.48
2-3		24	-0.40
3-4		23	-0.61
4-B		23	-0.34
Total (A-B)	134.5

Question 3. Fill-in the missing numbers (i.e. highlighted columns) in the table below using the given level field notes. You know that elevation of BM1 is 100 ft. Determine the loop closing error. Don't forget to perform the arithmetic check of your calculations.

Point	Station (hundred of ft)	B. S. (+)	HI	FS (-)	Rod (-)	Elevation	Adjusted Elevation
BM1	0+00	2.95
TP1	0+50	1.07		0.44
A	1+00				2.18
B	1+50				2.75
TP2	2+00	2.48		2.98
TP3	2+50	0.36		0.94
BM1	3 + 00			2.46

Question 4. Complete the highlighted cells in the following table for the given traverse ABC. Compute the linear closing error and determine if this is an acceptable survey. (Hint: Draw the traverse in your answer booklet)

Station	Bearing	Azimuth from North	Length			Unadjusted		Correction		Adjusted coordinates
A				Cos	Sin	Lat	Dep	Lat	Dep	100	100
	N78°49'50"E	78049'50"		642.65    ...	0.98106	124.49	630.48	-0.0300	0.0459	124.5185	630.4314
										224.52	730.43
	N50°49'52"W		618.25	0.63161	0.77529	390.49	-479.32	-0.0289	0.0441	390.5208	-479.3657
										615.04
	516°20'32"W ................	.................................. 536.75		0.95960		-515.06 .........			.. 0.0383		-151.0658
A										100.00	100.00
Sums							.... 0.1283

Question 5. Solve the following vertical curve by stage method. Assume that each stage is 50 ft (e.g. distance between points marked by 2 and 3 on the curve). In solving the curve provide the following:

a) Stations and elevations of PVC and PVT and the high point on the curve

b) The length of the curve tangent

c) Station and elevations for stations denoted by 1 and 5 on the figure.