ACET108 Revision PDF
ACET108 Revision PDF
ACET108 Revision PDF
Question 1:
Use either the Rise and Fall method or the Height of Plane of Collimation (HPC) method to reduce
the data.
2 B 2.311 5
3 D 1.220 15
3 E 3.675 20
4 G 0.339 30
4 H 0.157 35
1
Answer:
Arithmetic checks:
THEREFORE OK
2
- Height of Plane of Collimation method
7.561 7.719
Arithmetic checks:
THEREFORE OK
3
(b) Plot the longitudinal profile using the height against the distance
Solution:
45.000
44.500
44.000
43.500
Reduced Level (m)
43.000
42.500
42.000
41.500
41.000
40.500
40.000
0 5 10 15 20 25 30 35 40 45
Chainage (m)
(c) On the same graph you plotted at (b), plot the profile of a proposed road which will start at
Chainage 0 meters with Reduced Level of 41.100m. The road will end at Chainage 50m and
Reduced Level of 42.300m. Indicate which areas will need to be excavated and which areas will
need to be filled.
Answer:
45.000
44.500
44.000 excavation
43.500
Reduced Level (m)
43.000
42.500
(50, 42.300)
42.000
41.500
41.000
40.500
fill
40.000
0 5 10 15 20 25 30 35 40 45 50
(0, 41.100)
Chainage (m)
4
(d) Find the slope of the above road.
Answer:
45.000
44.500
44.000 excavation
43.500
Reduced Level (m)
42.000
(50, 42.300)
41.500
41.000
40.500
fill
40.000
0 5 10 15 20 25 30 35 40 45 50 55 60
(0, 41.100)
Chainage (m)
Answer:
5
(f) We plan to build a different road which would start at a Chainage of 0 meters at an elevation
of 42.000m. The road would have a slope of + 5°. What would the elevation of the road be at a
chainage of 80m?
Answer:
Question 2:
An Electronic Distance Measurement (EDM) device measured a slope distance of 50.874m. The
EDM is 1.60m above its station (point A) and the prism is at height 2.10m above point B. If the
reduced levels of A and B are RLA=+27.000m RLB=+22.700m, calculate the horizontal distance from
A to B.
Answer:
H = 50.732m
Question 3:
A 30m steel tape, standardized at 20°C using a tensile force of 60N. Measured against a standard
tape, the tested tape had a length of 29,994m. The 30m tape has a weight of 0,195N/m and a cross
sectional area of 1,41mm2. A field measurement (Lf) of 29,716m found at a temperature of 15°C
using a tensile force of 50N.
The coefficient of linear thermal expansion = 1.17x10-5 °C-1
The Young’s Modulus of Elasticity = 2.068x1011N/m2
− w2 L f ( P − Ps ) L f
3
∆m
Cs = 2
Cp = CT = a(T − Ts ) L f CL = Lf
24P AE m
Answer:
CT = - 0.001738m
Cs = - 0.0166m
Cp = - 0.0010191m
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CL = - 0.00594m
LT = 29.691m
Question 4:
Give answers to the following using the correct number of significant figures:
Answer:
3.36 – 2.2 = 1.2 (the answer 1.16 is rounded to 1.2 so that it has 1 decimal place)
2218.26 x 7.2 = 1.6 x104 (the answer 15971.472 is rounded so that it has only 2 significant
figures)
12.0009 + 0.000278 + 312 = 324 (the answer 324.001178 is rounded so that it has no decimal
places)
2.45 + 4.567 = 11.2 (the answer 11.18915 is rounded so that it has only 3 significant figures)
Question 5:
Answer:
Systematic errors: they are defined as the errors whose magnitude and algebraic sign can be
determined. Systematic errors can be determined and therefore corrected. They result from factors
that comprise the ‘measuring system’ and include the environment, the instruments and the
observer. As long as the ‘measuring system’ remains constant, the systematic errors remain
constant. If the conditions change, the magnitudes of the systematic errors will change. An example
is the error due to the temperature effects which changes the length of a steel tape. If the
temperature is known, the change in length of the steel tape (shortening or lengthening) can be
determined.
Because systematic errors tend to accumulate, they are also called cumulative errors.
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Random errors: they are the errors remaining after mistakes and systematic errors are eliminated.
They are due to the eliminations of the observer and the instruments. Smaller errors occur more
frequently than larger ones, while +ve and –ve errors are equally likely to occur and balance each
other.
Answer:
A common mistake is when you record a measured distance as 682.38 instead of 862.38. Another
mistake is when you transport a number maybe from a field book to a ‘clean’ paper and instead of
the correct number which is 85.96 you write 85.69.
Levelling
Question 6:
Answer:
Accuracy is the absolute proximity of the observed value to the ‘true’ measurement. In other words,
it shows how close a measurement is, to the true value.
Precision is the degree of refinement of the measuring process and the ability to repeat the same
measurement with consistently small variations i.e. how close is one measurement to another.
(b) Three groups carried out a levelling survey. The measurements of the elevation of the same
point are shown in the following tables for each group. The true elevation of the point measured is +
50.000m. Describe the results of each group in terms of accuracy and precision.
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Answer:
Question 7:
Describe how you will bring a perpendicular (offset) from a survey line. Use a sketch to support your
answer.
Answer: Q
Question 8:
Describe how you will bring a perpendicular (offset) from a point to a survey line. Use a sketch to
support your answer.
Answer:
OR
• With the free end of the tape (zero reading) at
point P, cross the survey line at any point A.
Bring a line from point P to point A. Find the
centre of line AP and mark it as B. From B, strike
an arc with radius BA. Where it cuts survey line
mark point Q. Make a line from point P to point Q.
Angle PQA =90°.
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Question 9:
Points A and B are visible to each other and are located on opposite sides of a lake. Explain how
you will measure the distance between these two points using a tape. Use a diagram to support
your answer.
OR
OR
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Question 10:
Answer:
The purpose of the Peg Test is to check that the line of sight – Plane of Collimation – through the
level is horizontal (i.e parallel to the axis of the bubble tube).
To perform the peg test, the surveyor first places two staffs at a distance of 60 to 90 m apart. The
level is set up midway (paced) between the two staffs (stadia) and readings are taken at both
locations.
A
Reading Reading
on the on the
staff a1 staff b1
B
30m 30m
If the line of sight through the level is not horizontal, the errors in readings ∆e1 at both point A and B
will be identical because the level is halfway between the points. Because the errors are identical,
the calculated difference in elevation between points A and B (difference in readings) will be the true
difference in elevation.
The level is then moved to one of the points (A) and set up so that the eyepiece of the telescope just
touches the staff as it is being held at point A. The staff reading (a2) can be determined by sighting
backward through the objective lens at a pencil point that is being moved slowly up and down the
staff. The pencil can be centered precisely, even though the cross hairs are not visible, because the
circular field of view is relatively small. Once that reading has been determined, the staff is held at
B. If the surveyor cannot look backward through the telescope, the instrument is set up at 2m away
from A with the staff read normally. Any error generated over that short distance will be relatively
insignificant.
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Question 11:
There are two points on a road axis, A and B. The length between these two points was measured
to be 40.226m. If the slope of the road is -4% from point A dropping to point B, what is the horizontal
distance between these two points? What is the difference in height between A and B?
Answer:
12