MEASUREMENT OF HORIZONTAL DISTANCES
Problem Set 2
1. PACING. In walking along a 75-m course, the pacer of a field party counted
43.50, 44.00, 43.50, 43.75, 44.50, and 43.25 strides. Then 105.50, 106.00,
105.75, and 106.25 strides were counted in walking from one marker to another
established along a straight and level course. Determine the distance between
the two markers.
Given:
43.50, 44.00, 43.50, 43.75, 44.50, 43.25 strides for 75-m course
105.50, 106.00, 105.75, 106.25 strides for x
Requirements:
Distance between the two markers, x
Illustration:
75 m
x
Solution:
Pace factor =
75m
(43.50 + 44.00 + 43.50 + 43.75 + 44.50 + 43.25) strides
(
)
6
Pace factor =
Pace factor =
75m
262.50 strides
(
)
6
75m
2 paces
43.75 strides( stride )
Pace factor =
75m
87.50 paces
Pace factor = 0.857
m
pace
x = (pace factor)(mean number of pace)
m
x = (0.857 pace) (
(105.50+106.00+105.75+106.25) strides
4
m
)
2 paces
x = [0.857 pace] [(105.875 strides) ( stride )]
m
x = [0.857 pace] [211.75 paces]
Pace factor =
75m
(43.50 + 44.00 + 43.50 + 43.75 + 44.50 + 43.25) strides
(
)
6
Pace factor =
Pace factor =
Pace factor =
75m
262.50 strides
(
)
6
75m
2 paces
43.75 strides( stride )
75m
87.50 paces
Pace factor = 0.857
m
pace
x = (pace factor)(mean number of pace)
m
x = (0.857 pace) (
(105.50+106.00+105.75+106.25) strides
m
4
)
2 paces
x = [0.857 pace] [(105.875 strides) ( stride )]
m
x = [0.857 pace] [211.75 paces]
x = 181.50 m
2. PACING. A student paces a 50-m length five times with the following results:
57.00, 56.75, 56.50, 58.00, and 56.25 paces. Determine how many paces he
must step off in order to establish a distance of 450 meters on level ground.
Given:
57.00, 56.75, 56.50, 58.00, and 56.25 paces for 50-m length
Required:
No. of paces to establish 450 m
Illustration:
50 m
450 m
Solution:
∑(𝑃𝑎𝑐𝑒𝑠 𝑀𝑎𝑑𝑒)
𝑀𝑒𝑎𝑛 𝑁𝑜. 𝑜𝑓 𝑃𝑎𝑐𝑒𝑠 =
=
𝑁𝑜.𝑜𝑓 𝑇𝑟𝑖𝑎𝑙𝑠
57.00+56.75+56.50+58.00+56.25
5
= 56.90 𝑝𝑎𝑐𝑒𝑠
𝑃𝑎𝑐𝑒 𝐹𝑎𝑡𝑜𝑟 =
= 0.8787
50 𝑚
56.90 𝑝𝑎𝑐𝑒𝑠
𝑚
𝑝𝑎𝑐𝑒
𝑁𝑜. 𝑜𝑓 𝑃𝑎𝑐𝑒𝑠 =
450 𝑚
𝑚
0.8787 𝑝𝑎𝑐𝑒
𝑵𝒐. 𝒐𝒇 𝑷𝒂𝒄𝒆𝒔 = 𝟓𝟏𝟐. 𝟏𝟎 𝑷𝒂𝒄𝒆𝒔
3. PACING. Determine the length of a line negotiated in 208 paces by a person
whose pace is 0.76 meter long.
Given:
𝑁𝑜. 𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 = 208
Required:
Length of a line, D
Illustration:
𝑃𝑎𝑐𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 = 0.76
𝑚
𝑝𝑎𝑐𝑒
0.76 m
Solution:
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑁𝑜. 𝑜𝑓 𝑃𝑎𝑐𝑒𝑠 × 𝑃𝑎𝑐𝑒 𝐹𝑎𝑐𝑡𝑜𝑟
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 208 𝑝𝑎𝑐𝑒𝑠 × 0.76 𝑚/𝑝𝑎𝑐𝑒
𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝟏𝟓𝟖. 𝟎𝟖 𝒎
4. DISTANCE BY SUBTENSE BAR. With the use of a 1-sec theodolite
positioned at the center of a six-sided lot, the following readings were taken on a
2-m subtense bar setup at each corner: 0°26’16”, 0°12’35”, 0°15’05”, 0°22’29”,
0°30’45”, and 0°09’50”. Determine the distance of each corner from the
instrument position.
Given:
2-m subtense bar
Θ1= 0°26’16”
Θ3=0°15’05”
Θ5=0°30’45”
Θ2= 0°12’35”
Θ4=0°22’29”
Θ6=0°09’50”
Required:
Distance of each corner from the instrument position, D1, D2, D3, D4, D5, D6,
Illustration:
D1
D2
D3
D6
D5
D4
Solution:
D1 = cot
0°26’16”
D2 = cot
2
D1 = 𝟐𝟔𝟏. 𝟕𝟔 𝒎
D4 = cot
0°12’35”
2
D2 = 𝟓𝟒𝟔. 𝟒𝟎 𝒎
0°22’29”
D5 = cot
2
D4 = 𝟑𝟎𝟓. 𝟖𝟎 𝒎
0°30’45”
2
D5 = 𝟐𝟐𝟑. 𝟓𝟗 𝒎
D3 = cot
0°15’05”
2
D3 = 𝟒𝟓𝟓. 𝟖𝟑 𝒎
D6 = cot
0°09’50”
2
D6 = 𝟔𝟗𝟗. 𝟐𝟎 𝒎
5. DISTANCE BY SUBTENSE BAR. A 2-m long subtended bar was first set up
at A and subsequently at B, and the subtended angles to the bar, as read from a
theodolite positioned somewhere along the middle of line AB, were recorded as
0° 24’ 15”, and 0° 20’ 30”, respectively. Determine the length of AB.
Given:
2-m subtense bar
ΘA= 0° 24’ 15”
ΘB=0°20’ 30”
Required:
Length of AB, DAB
Illustration:
0° 24’ 15”
A
0°20’ 30”
B
Solution:
DAB = DA + DB
1m
1m
DAB = (tan 0˚20′30") + (tan 0˚24′55")
DAB = 275.9383m + 375.3889m
DAB = 611.327m
6. SLOPE MEASUREMENT. A traverse line was measured in three sections:
295.85 m at slope 8° 45’, 149.58 m at slope 4° 29’, and 373.48 m at slope 4°
25’.Determine the horizontal length of the line.
Given:
295.85 m at slope 8° 45’
149.58 m at slope 4° 29’
373.48 m at slope 4° 25’
Required:
The horizontal length of the line, LT
Illustration:
4° 25’
4° 29’
8° 45’
Solution:
L1 = 295.85𝑐𝑜𝑠(8° 45’)
L2 = 149.58𝑐𝑜𝑠(4° 29’)
L1 = 292.407 𝑚
L2 = 149.122 𝑚
L3 = 373.48 𝑐𝑜𝑠(4° 25’)
372.371 𝑚
LT =292.407 𝑚 + 149.122 𝑚 +
L3 = 372.371 𝑚
LT =𝟖𝟏𝟑. 𝟖𝟗𝟖 𝒎
7. SLOPE MEASUREMENT. A slope measurement of 545.38 m is made
between points A and B. The elevation of A is 424.25 m and that of B is 459.06
m. Determine the horizontal distance between the two points.
Given:
Elevation of A = 424.25 m
Elevation of B = 459.06 m
Slope Measurement = 545.38 m
Required:
The horizontal distance between the two points, L
Illustration:
545.38m
424.25m
459.06m
Solution:
𝐷𝐸 = 𝑒𝑙𝑒𝑣. 𝐵 – 𝑒𝑙𝑒𝑣. 𝐴.
𝐿 = √𝑠 2 − 𝐷𝐸 2
𝐷𝐸 = 459.06 𝑚 − 424.25 𝑚
𝐿 = √545.382 − 34.512
𝐷𝐸 = 34.81 𝑚
𝑳 = 𝟓𝟒𝟒. 𝟐𝟕 𝒎
8. MEASUREMENTS WITH TAPE. The sides of a rectangular parcel of property
were measured and recorded as 249.50 m and 496.85 m. It was determined,
however, that the 30-m tape used in measuring was actually 30.05 m long.
Determine the correct area of the rectangle in hectares.
Given:
MLL = 496.85 m
NL = 30 m
MLw = 249.50 m
c = +0.05 m
Required:
The correct area of the rectangle in hectares, AHectares
Illustration:
496.85 m
249.50 m
Solution:
𝑊′ = 𝑊 + 𝐶
𝐿′ = 𝐿 + 𝐶
𝑊
𝐿
𝐶 = 𝑐 𝑥 (𝑁𝐿)
𝐶 = 0.05 𝑚 𝑥 (
𝐶 = 𝑐 𝑥 (𝑁𝐿)
249.50 𝑚
30𝑚
)
𝐶 = 0.05 𝑚 𝑥 (
496.85
30𝑚
)
𝐶 = 0.4158 𝑚
𝐶 = 0.8280 𝑚
𝑊′ = 249.50𝑚 + 0.4158𝑚
𝐿′ = 496.85𝑚 + 0.8280𝑚
𝑊′ = 249.92 𝑚
𝐿′ = 497.68𝑚
𝐴 = 𝐿′ 𝑥 𝑊′
𝐴 = 249.92𝑚 𝑥 497.68𝑚
𝐴 = 124380.18 𝑠𝑞. 𝑚
𝑨 = 𝟏𝟐. 𝟒𝟒 𝒉𝒂
9. MEASUREMENTS WITH TAPE. A 30-m steel tape when compared with a
standard is actually 29.95 m long. Determine the correct length of a line
measured with this tape and found to be 466.55 m.
Given:
NL = 30 m
c = -0.05 m
ML = 466.55 m
Required:
The correct length of a line, CL
Illustration:
Solution:
𝐶𝐿 = 𝑀𝐿 − 𝐶
𝐶=𝑐𝑥
𝑀𝐿
𝐶𝐿 = 466.55𝑚 − 0.78𝑚
𝑁𝐿
𝐶 = 0.05 𝑥
𝐶 = 0.78 𝑚
466.55𝑚
30𝑚
𝑪𝑳 = 𝟒𝟔𝟓. 𝟕𝟕𝒎
10. LAYING OUT DISTANCES. A track and field coach wishes to lay out for his
team a 200-m straightway course. If he uses a 50-m tape known to be 50.20 m
long, determine the measurements to be made so that the course will have the
correct length.
Given:
L = 200m
c = 0.20 m
NL = 50 m
Required:
The measurements to be made, L’
Illustration:
Solution:
𝐿′ = 𝐿 − 𝐶
𝐶=𝑐𝑥
𝐿
𝐿′ = 200𝑚 − 0.80𝑚
𝑁𝐿
200𝑚
𝐶 = 0.20𝑚 𝑥 ( 50𝑚 )
𝑳′ = 𝟏𝟗𝟗. 𝟐𝟎 𝒎
𝐶 = 0.80 𝑚
11. LAYING OUT DISTANCES. It is required to lay out a building 80 m by 100 m
with 30-m long metallic tape which was found to be 0.15 m too short. Determine
the correct dimensions to be used in order that the building shall have the
desired measurements.
Given:
L = 100m
c = +0.15 m
W= 80m
NL = 30 m
Required:
The correct dimensions to be used, L’ and W’
Illustration:
100 m
80 m
Solution:
𝐿′ = 𝐿 + 𝐶
𝐶=𝑐𝑥
𝑊′ = 𝑊 + 𝐶
𝐿
𝐶=𝑐𝑥
𝑁𝐿
100𝑚
𝑊
𝑁𝐿
80𝑚
𝐶 = 0.15𝑚 𝑥 ( 30𝑚 )
𝐶 = 0.15𝑚 𝑥 (30𝑚)
𝐶 = 0.50 𝑚
𝐶 = 0.40 𝑚
𝐿′ = 100𝑚 + 0.50𝑚
𝑊′ = 80𝑚 + 0.40𝑚
𝑳′ = 𝟏𝟎𝟎. 𝟓𝟎 𝒎
𝑾′ = 𝟖𝟎. 𝟒𝟎 𝒎
12. LAYING OUT DISTANCES. A steel tape whose nominal length is supposed
to be 30 m long was found to be 30.02 m long when compared with an invar tape
during standardization. If the tape is to be used in laying out a 520 m by 850 m
rectangular parking lot, determine the actual dimensions to be laid out.
Given:
L = 850m
c = -0.02 m
W= 520m
NL = 30 m
Required:
The actual dimensions to be laid out, L’ and W’
Illustration:
850 m
520 m
Solution:
𝐿′ = 𝐿 − 𝐶
𝐶=𝑐𝑥
𝑊′ = 𝑊 − 𝐶
𝐿
𝐶=𝑐𝑥
𝑁𝐿
850𝑚
𝑊
𝑁𝐿
520𝑚
𝐶 = 0.02𝑚 𝑥 ( 30𝑚 )
𝐶 = 0.02𝑚 𝑥 ( 30𝑚 )
𝐶 = 0.567 𝑚
𝐶 = 0.346 𝑚
𝐿′ = 850𝑚 − 0.567𝑚
𝑊 ′ = 520𝑚 − 0.346𝑚
𝑳′ = 𝟖𝟒𝟗. 𝟒𝟑 𝒎
𝑾′ = 𝟓𝟏𝟗. 𝟔𝟓 𝒎
13. CORRECTION DUE TO TEMPERATURE. A 30-m steel tape is of standard
length at 20°C. If the coefficient of thermal expansion of steel is 0.0000116/1°C,
determine the distance to be laid out using this tape to establish two points
exactly1235.65 m apart when the temperature is 33°C.
Given:
α= 0.0000116/1°C
NL= 30m
L=1235.65m
To= 20°C
T=33°C
Required:
The distance to be laid out, L’
Illustration:
Solution:
CT = L × α × (t – t0).
L’ = L - CT
CT = 1235.65 m × 0.0000116/1°C × (33°C - 20°C)
0.186m
L’ = 1235.65m –
CT = 0.186 m
L’= 1235.46 m
14. CORRECTION DUE TO TEMPERATURE. A steel tape having a correct
length at 22°C was used to measure a baseline and the recorded readings gave
the total of 856.815 m. If the average temperature during the measurement was
18°C, determine the correct length of the line.
Given:
ML = 856.815 m
α = 0.0000116/1°C
T = 18°C
To = 22°C
Required:
Correct length of the line, CL
Illustration:
Solution:
CT = L × α × (t – t0).
CL = ML ± CT
CT = 856.815 m × 0.0000116/1°C × (18°C - 22°C)
0.040 m
CL = 856.815 m -
CT = -0.040 m
CL = 856.775 m
15. CORRECTION DUE TO TENSION. A heavy 30-m tape having a crosssectional area of 0.05 cm2 has been standardized at a tension of 5 kg. If E = 2.10
x 106 kg/cm2, calculate the elongation of the tape for an increase in tension from
5.5 kg to 20 kg.
Given:
P = 20 kg
A = 0.05 cm2
P0 = 5.5 kg
E = 2.10 x 106 kg/c
L = 30 m
Required:
Elongation due to pull, CP
Illustration:
Solution:
𝐶𝑃 =
𝐶𝑃 =
(𝑃−𝑃0 )𝐿
𝐴𝐸
.
(20 𝑘𝑔 − 5.5 𝑘𝑔)(30 𝑚)
𝑘𝑔⁄
(0.05 𝑐𝑚2 )(2.10 × 106
𝑐𝑚2 )
𝑪𝑷 = 𝟒. 𝟏𝟒𝟑 × 𝟏𝟎−𝟑 𝒎
16. CORRECTION DUE TO TENSION. A steel tape is 30.0-m long under a pull
of 6.0 kg when supported throughout. It has a cross-sectional area of 0.035 cm2
and is applied fully supported with a 12-kg pull to measure a line whose recorded
length is 308.32 m. Determine the correct length of the line if E = 2.1 x 106
kg/cm2.
Given:
P = 12 kg
A =0.035cm2
P0 = 6.0 kg
ML = 308.32 m
E = 2.10 x 106 kg/cm
Required:
Correct length of the line, CL
Illustration:
Solution:
𝐶𝑃 =
𝐶𝑃 =
(𝑃−𝑃0 )𝐿
𝐴𝐸
.
(12 𝑘𝑔−6.0 𝑘𝑔)(308.32 𝑚)
𝑘𝑔
⁄ 2)
𝑐𝑚
(0.035 𝑐𝑚 2 )(2.10 × 106
𝐶𝑃 = 0.025 𝑚
CL = ML ± CP
CL = 308.32 m + 0.025 m
CL = 308.345 m
17. CORRECTION DUE TO TENSION. A 30-m steel tape weighing 1.75kg is of
standard length under a pull of 4.55 kg, supported for full length. This tape was
used in measuring a line (found to be1371.50 m) on smooth level ground under a
steady pull of 8 kg. Assuming E = 2.05 x 106 kg/cm2 and that the unit weight of
steel is 7.9 x 10-3 kg/cm3, determine the following: cross-sectional area of the
tape, correction for increase in tension for the whole length measured, and the
correct length of the measured line.
Given:
P = 8.0 kg
ML = 1371.50 m
P0 = 4.55 kg
E = 2.05 x 106 kg/cm2
W = 1.75 kg
NL = 30 m
ρ = 7.9 x 10-3 kg/cm3
Required:
Cross-sectional area of the tape, correction to be applied and correct
length of the line, A, C and CL
Illustration:
Solution:
V = A × L, and ρ × V = W,
𝐴=
𝐴=
𝑊
𝜌×𝐿
,
1.75 𝑘𝑔
𝑘𝑔
7.90 × 10−3 ⁄𝑐𝑚3 × 3000 𝑐𝑚
𝑨 = 𝟎. 𝟎𝟕𝟒 𝒄𝒎𝟐
𝐶𝑃 =
𝐶𝑃 =
(𝑃−𝑃0 )𝐿
𝐴𝐸
, and
(8.0 𝑘𝑔 − 4.55 𝑘𝑔)(1371.50 𝑚)
𝑘𝑔⁄
(0.074 𝑐𝑚2 )(2.05 × 106
𝑐𝑚2 )
𝑪𝑷 = 𝟎. 𝟎𝟐𝟔 𝒎
CL = ML ± CP
CL = 1371.50 m + 0.026 m
CL = 1371.526 m
18. CORRECTION DUE TO SAG. A 30-m steel tape weighs 1.5kg and is
supported at its end points and at the 5 and15-meter marks. If a pull of 8 kg is
applied, determine the correction due to sag between supports for one tape
length.
Given:
L1 = 5.0 m
L2 = 10.0 m
P = 8.0 kg
W = 1.5 kg
L3 = 15.0 m
Required:
Correction due to Sag, Cs
Illustration:
Solution:
𝑊2𝐿
𝐶𝑆 = − ∑( 24𝑃2).
𝐶𝑆 = −
(1.5 𝑘𝑔)2
(5 + 10 + 15)𝑚
24(8.0 𝑘𝑔)2
𝑪𝑺 = −𝟎. 𝟎𝟒𝟒 𝒎
19. CORRECTION DUE TO SAG. A 30-m steel tape weighing 0.04 kg/m is
constantly supported only at its endpoints, and used to measure a line with a
steady pull of 8.5kg. If the measured length of the line is 2465.18 m, determine
the correct length of the line.
Given:
ML = 2465.18 m
NL = 30.00 m
w = 0.04 kg/m
P = 8.5 kg
Required:
Correct length of the line, CL
Illustration:
Solution:
𝐶𝑆 = − ∑
𝑛=
2460 𝑚
30 𝑚
𝑛(𝑤 2 𝐿3 )
24𝑃2
= 82 𝑓𝑢𝑙𝑙 𝑡𝑎𝑝𝑒𝑙𝑒𝑛𝑔𝑡ℎ𝑠,
𝑘𝑔
𝑘𝑔
82 × (0.04 ⁄𝑚)2 × (30 𝑚)3 1 × (0.04 ⁄𝑚)2 × (5.18 𝑚)3
𝐶𝑆 = −
−
24 × (8.5 𝑘𝑔)2
24 × (8.5 𝑘𝑔)2
𝐶𝑆 = −0.025 𝑚
CL = ML ± CS
CL = 2465.18 m – 0.025 m
CL = 2465.155 m
20. NORMALTENSION. Determine the normal tension required to make a tape
exactly 30.0 m between its ends when used in an unsupported mode, if the tape
has a cross- sectional area of 0.045 cm2and weighs 0.90 kg. Assume that the
tape is exactly 30.0 m when supported throughout its length under a standard
pull of 6.0 kg, and its modulus of elasticity is 2.10x106 kg/cm2.
Given:
NL = 30m
E = 2.0 x
kg
106 cm2
PS = 6.0 kg
W= 0.90 kg
A = 0.045 cm2
Required:
Normal tension, PN
Illustration:
Solution:
𝑃𝑛 =
𝑃𝑛 =
0.204𝑊 2 √𝐴𝐸
√𝑃𝑛 − 𝑃𝑠
0.204(0.90)2√(0.045)(2.0 × 106)
√𝑃𝑛 − 6
𝑷𝑵 = 𝟏𝟓. 𝟖𝟏𝟗 𝒌𝒈
21. NORMAL TENSION. A 30-m steel tape supported at its ends weighs 0.03
kg/m and is of standard length under a pull of 6.5 kg. If the elastic modulus of
steel is 2.0 x 106 kg/cm2 and its weight density is 7.9 x 10-3 kgm3, determine the
tension at which the effect of sag will be eliminated by the elongation of the tape
due to increased tension.
Given:
NL = 30m,
PS = 6.5 kg
wῥ = 7.90 x
W = 0.03
kg
10-3 3 ,
cm
kg
m
,
Required:
Normal Pull, PN
Illustration:
E = 2.0 x 106
kg
cm2
,
Solution:
mass
A = (density x length)
𝑃𝑛 =
W x NL
A = (density x length)
A=[
A=[
𝑃𝑛 =
kg
)(30m)
m
kg
100cm
3
(7.9 x 10
)(30m x m )
cm3
(0.03
0.09 kg
]
0.204𝑊 2 √𝐴𝐸
√𝑃𝑛−𝑃𝑠
0.204(0.09𝑘𝑔)2√0.03797468𝑐𝑚 2 𝑥 2.0𝑥106
𝑘𝑔
𝑐𝑚2
√𝑃𝑛−6.5𝑘𝑔
𝑷𝒏 = 𝟏𝟔. 𝟐𝟐𝟓 𝒌𝒈
]
kg
(7.9 x 10−3 3 )(3000cm)
cm
A = 0.037974684 cm2
22. COMBINED CORRECTIONS. A 30-m tape weighs 12.5 g/m and has a cross
section of 0.022 cm2. It measures correctly when supported throughout under a
tension of 8.0 kg and at a temperature of 20ºC. When used in the field, the tape
is only supported at its ends, under a pull of 9.0 kg and at an average
temperature of 28ºC. Determine the distance between the zero ad 30-m marks.
Given:
P = 9.0 kg
P0 = 8.0 kg
A = 0.022 cm2
E = 2.0 x 106
t = 28ºC
t0 = 20ºC
α = 11.6 x 10-6/ ºC
NL = 30.0 m
Required:
Corrected Length, CL
𝑘𝑔⁄
𝑐𝑚2
`
Illustration:
Solution:
(𝑃−𝑃0 )𝐿
CT = Lα(t - t0)
CP =
CT = 30 m(11.6 x 10-6 / ºC)(28ºC -20ºC)
CP =
CT = 2.784 x 10-3 m
CP = 6.818 x 10-4 m
𝐴𝐸
(9.0 𝑘𝑔−8.0 𝑘𝑔)(30 𝑚)
𝑘𝑔
(0.022 𝑐𝑚 2 )(2.0 × 106 ⁄ 2 )
𝑐𝑚
CL = L ± Cp ± CT
CL = 30.0 m + 6.818 x 10-4 m + 2.784 x 10-3 m
CL = 30.0034658 m
CL = 30.003 m
23. COMBINED CORRECTIONS. A line was found to be 2865.35 m long when measured with
a 30-m tape under a steady pull of 6.5 kg at a mean temperature of 30ºC. Determine the
correct length of the line if the tape used is of standard length at 20ºC under a pull of 5.5 kg.
Assume the cross-sectional area of tape to be 0.025 cm2, elastic modulus as 2.10 x 106
kg/cm2, and coefficient of thermal expansion to be 0.0000116/1ºC
Given:
ML = 2865.35 m
t = 30ºC
α = 0.0000116/1ºC
t0 = 20ºC
P = 6.5 kg
P0 = 5.5 kg
A = 0.025 cm2
E = 2.10 x 106 kg/cm2
Required:
Correct Length of the Line, CL
Illustration:
`
Solution:
CT = ML × α × (t - t0)
CT = 2865.35 m × 0.0000116/1ºC × (30ºC - 20ºC)
CT = 0.332 m
𝐶𝑃 =
(𝑃 − 𝑃0 )𝑀𝐿
𝐴𝐸
𝐶𝑃 =
(6.5 𝑘𝑔 − 5.5 𝑘𝑔)(2865.35 𝑚)
𝑘𝑔
(0.025 𝑐𝑚2 )(2.10 × 106 ⁄𝑐𝑚2 )
CP = 0.055 m
CL = ML ± CP ± CT
CL= 2865.35 m + 0.055 m + 0.332 m
CL = 2865.737 m
`
24. MEASURING ANGLES WITH TAPE. The sides of a triangle measure 1063.55, 1840.33,
and 1325.05 m. Determine the three angles in the triangle.
Given:
a = 1063.55 m,
b = 1840.33 m,
c = 1325.05 m
Required:
Angles A, B and C
Illustration:
Solution:
𝐴 = 𝑐𝑜𝑠 −1 (
𝐴 = 𝑐𝑜𝑠 −1 (
𝑏2 +𝑐 2 −𝑎2
2𝑏𝑐
1840.332 +1325.052−1063.552
2×1840.33×1325.05
)
𝑨 = 𝟑𝟒°𝟑𝟗′ 𝟒𝟐. 𝟏𝟖"
𝐵 = 𝑐𝑜𝑠 −1 (
𝐵 = 𝑐𝑜𝑠 −1 (
𝑎2 +𝑐 2 −𝑏2
2𝑎𝑐
)
−1840.332+1325.052 +1063.552
2×1063.55×1325.05
)
𝑩 = 𝟏𝟎𝟎°𝟏𝟐′ 𝟓𝟗. 𝟔𝟖"
𝐶 = 180° − (34°39′ 42.18"+100°12' 59.68")
𝑪 = 𝟒𝟓°𝟕′ 𝟏𝟖. 𝟏𝟒"
25. OBSTRUCTED DISTANCES. In the accompanying sketch it is required to determine the
distance between points A and B which spans a wide and deep river. Lines BD and CE, which
measure 385.75 m and 529.05 m, respectively, are established perpendicular to line ABC. If
points D and E are lined up with A and the length of BC = 210.38 m, determine the required
distance.
`
Given:
BD = 385.75 m
CE = 529.05 m
BC = 210.38 m
Required:
Length of AB
Illustration:
Solution:
∆𝐴𝐶𝐸 ~ ∆𝐴𝐵𝐷
𝐴𝐶
𝐶𝐸
=
𝐴𝐵−𝐵𝐶
𝐶𝐸
𝐴𝐵
𝐵𝐷
=
𝐴𝐵−210.38
529.05
𝐴𝐵
𝐵𝐷
=
𝐴𝐵
385.75
𝑨𝑩 = 566.32
`
MEASUREMENT OF HORIZONTAL DISTANCES
Unit Exam 2
1.
a)
b)
c)
d)
e)
A pace is defined as the length of a step in walking. It may be measured from
Heel to heel
Toe to heel
heel to heel
mid-heel to mid-toe
tip of toe to tip of heel
2. The method of measuring or laying out horizontal distance by stretching a
calibrated tape between two points and reading the distance indicated on the
tape is referred to as
a) Taping
b) Pacing
c) Tacheometry
d) stadia measurement
e) range finding
3. The subtense bar is a convenient and practical device used for quick and
accurate measurement of horizontal distances. It consists of a rounded steel tube
through which runs a thin invar rod and at each end of the frame the target marks
are house exactly
a) 1.00 m apart
b) 1.50 m apart
c) 3.00 m apart
d) 2.00 m apart
e) 4.00 m apart
4. The first electronic distance measuring instrument was the geodetic distance
meter (geodimeter) which was developed in1948 by Swedish physicist named
a) Dr. T. L. Wadley
b) Erik Bergstrand
c) Sir Edmund Gunter
d) Pierre Vernier
e) Hipparchus
`
5. A special tape made of an alloy of nickel (35%) and steel (65%) with a very low
coefficient of thermal expansion, and used only for precise measurements in
geodetic work as well as for checking the lengths of other kinds of tape is the
a) engineering tape
b) fiberglasstape
c) invar tape
d) nylon-coated tape
e) builder’s tape
6. The standard practice of measuring short distance on uneven sloping ground to
accumulate a full tape length wherein the tape is held horizontally above ground
and plumbed at one or both ends is referred to as
a) slope taping
b) horizontal taping
c) incremental taping
d) breaking tape
e) partial taping
7. Normal tension is defined as the applied pull which will lengthen the tape to equal
the
a) decrease in standard pull
b) shortening due to temperature
c) increase in length due to absence of intermediate support
d) shortening caused by sag
e) increase in gravitational forces
8. A surveyor counted 50, 52, 53, 51, 53, and 51paces in walking along a 45-m
course laid out on a concrete pavement. He then took 768,771,772,770, 769,
and 770 paces in walking an unknown distance XY. His pace factor should be
equal to
𝑚
a) 1.148𝑝𝑎𝑐𝑒
𝑚
b) 0.001𝑝𝑎𝑐𝑒
𝑚
c) 14.904𝑝𝑎𝑐𝑒
d) 0.067
𝑚
𝑝𝑎𝑐𝑒
𝒎
e) 0.871𝒑𝒂𝒄𝒆
`
Given:
L= 45 m
Paces = 50, 52, 53, 51, 53, 51
= 768, 771, 772, 770, 769, 770
Required:
Pace Factor, PF
Illustration:
45m
Solution:
𝑃̅ =
∑𝑃
𝑃𝑓 =
𝑛
=
=
50+52+53+51+53+51 𝑝𝑎𝑐𝑒𝑠
6
310 𝑝𝑎𝑐𝑒𝑠
6
= 51.66666667 𝑝𝑎𝑐𝑒𝑠
𝑁𝐿
𝑃̅
=
45 𝑚
51.66666667 𝑝𝑎𝑐𝑒𝑠
= 0.8709677419
𝒎
𝑚
𝑝𝑎𝑐𝑒
≈ 𝟎. 𝟖𝟕𝟏 𝒑𝒂𝒄𝒆
9. In question 8, the length of XY based on the pace factor of the surveyor is equal
to
a) 670.67 m
b) 883.96 m
c) 11476.08 m
d) 51.59 m
e) 715.67 m
Given:
𝑚
Pf= 0.871𝑝𝑎𝑐𝑒
Paces: 768, 771, 772, 770, 769, 770
Required:
Length of XY,
`
Illustration:
XY
Solution:
𝑃̅ =
∑𝑃
𝑛
=
768+771+772+770+769+770 𝑝𝑎𝑐𝑒𝑠
4620 𝑝𝑎𝑐𝑒𝑠
6
=
6
= 770 𝑝𝑎𝑐𝑒𝑠
𝑿𝒀 = 𝑃𝑓 × 𝑃̅
𝑚
= 0.871 𝑝𝑎𝑐𝑒 × 770 𝑝𝑎𝑐𝑒𝑠
= 𝟔𝟕𝟎. 𝟔𝟕 𝒎
10. Two points, A and B, are established along the same direction from a theodolite
station. Of the subtended angle read on a subtense bar held at A and B are
0°55'20" and 0°23'44", respectively, the horizontal distance between the two
points is
a) 82.73 m
b) 165.45 m
c) 206.98 m
d) 289.70 m
e) 124.25 m
Given:
𝜃𝐴 = 0°55′20"
𝜃 = 0°23′44"
Required:
HD
Illustration:
A
B
`
Solution:
∝
𝐷𝐴 = 𝑐𝑜𝑡 ( 𝐴)
∝
𝐷𝐵 = 𝑐𝑜𝑡 ( 𝐵)
2
2
0°55′20"
= 𝑐𝑜𝑡 ( 2 )
= 124.25 𝑚
𝑯𝑫 = 𝐷𝐵 − 𝐷𝐴
= 289.70 𝑚 − 124.25 𝑚
= 𝟏𝟔𝟓. 𝟒𝟓 𝒎
= 𝑐𝑜𝑡 (
= 289.70 𝑚
0°23′44"
2
)
11. A slope distance of 465.82 m is measured between two points with a slope angle
of 12°35'. The corresponding horizontal distance between the points is
a) 101.48 m
b) 454.63 m
c) 103.98 m
d) 358.70 m
e) 207.14 m
Given:
𝐷𝑆 = 465.82 𝑚
𝜃𝑆 = 12°35′
Required:
HD
Illustration:
12°35’
SOLUTION:
𝐻𝐷
Cos 𝜃 = 𝐷
𝑆
𝑯𝑫 = cos 𝜃 (𝐷𝑆 )
= cos 12°35′ (465.82 𝑚)
= 454.63108 𝑚
≈ 𝟒𝟓𝟒. 𝟔𝟑 𝒎
12. A line measured with a 30-m steel tape was recorded as 325.70 m. If the tape is
found to be 30.05 m long during standardization, the correct length of the line is
a) 325.16 m
b) 325.70 m
`
c) 327.45 m
d) 325.44 m
e) 326.24 m
Given:
NL= 30 m
ML= 325.70 m
TL= 30.05 m
REQUIRED:
CL
ILLUSTRATION:
SOLUTION:
𝑐 = 𝑇𝐿 − 𝑁𝐿
= 30.05 𝑚 − 30 𝑚
= 0.05 𝑚
𝑀𝐿
𝑪𝑳 = 𝑀𝐿 + 𝑐 ×
𝑁𝐿
325.70 𝑚
= 325.70 𝑚 + 0.05 𝑚 × 30 𝑚
= 326.2428333 𝑚
≈ 𝟑𝟐𝟔. 𝟐𝟒 𝒎
13. A rectangular building 250.00 m by 130.00 m is to be laid out with a 30-m long
steel tape. If during standardization the tape is found to be 30.03 m, the correct
length and width to be laid out should be
a) 249.75 m by 129.87 m
b) 250.25 m by 130.13 m
c) 249.87 m by 129.75 m
d) 250.00 m by 130.00 m
e) 249.97 m by 129.97 m
Given:
Tape Material = Steel
L= 250.00 m
W= 130.00 m
NL= 30 m
TL= 30.03 m
REQUIRED:
`
the correct length and width to be laid out should be, L’ and W’
ILLUSTRATION:
250 m
130 m
Solution:
𝑐 = 𝑇𝐿 − 𝑁𝐿
= 30.03 𝑚 − 30 𝑚
= 0.03 𝑚
𝐿′ = 𝐿 + 𝐶
250𝑚
𝑳′ = 250 𝑚 − 0.03𝑚
30𝑚
= 𝟐𝟒𝟗. 𝟐𝟓 𝒎
𝑊′ = 𝑊 + 𝐶
130𝑚
= 130 𝑚 + 0.03𝑚 30𝑚
= 𝟏𝟐𝟗. 𝟖𝟕 𝒎
14. A line measured with a 50-m long steel tape was determined to be 645.22 m
when the average temperature during taping was 15.75°C. If the tape is of
standard length at 20°C and the coefficient of thermal expansion of steel is
0.0000116/1°C, the correct length of the measured line is
a) 645.23 m
b) 645.22 m
c) 645.24 m
d) 645.19 m
e) 645. 21 m
Given:
NL= 50 m
ML= 645.22 m
T= 15.75°C
𝑇𝑂 = 20°𝐶
∝= 0.0000116/1°C
Required:
the correct length of the measured line, CL
Illustration:
`
Solution:
𝑪𝑳 = 𝑀𝐿 + 𝐶𝑇
0.0000116
= 645.22 𝑚 +
°C (645.22 m)
1
= 645.1881907 𝑚
≈ 𝟔𝟒𝟓. 𝟏𝟗 𝒎
15. A steel tape with a cross-sectional area of 0.03cm2 is 30.00 m long under a pull
of 5 kg when supported throughout. It is used in measuring a line 875.63 m long
kg
under a steady pull of 10 kg. Assuming E= 2.0x106 cm2 , the elongation of the
tape due to increase in tension is
a) 0.0730 m
b) 0.730 m
c) 0.50 m
d) 0.043 m
e) 0.0025 m
Given:
Tape Material = Steel
A= 0.03cm2
NL= 30 m
𝑃𝑜 = 5 𝑘𝑔
ML= 875.63 m
P= 10 kg
kg
E= 2.0x106 cm2
Required:
the elongation of the tape due to increase in tension, Cp
Illustration:
`
Solution:
(𝑃 − 𝑃𝑜 )𝑁𝐿
𝑪𝑷 =
𝐴𝐸
(10 𝑘𝑔−5 𝑘𝑔)30 𝑚
=
2
6 kg
0.03cm (2.0x10
)
cm2
−3
= 2.5 × 10 𝑚
≈ 𝟎. 𝟎𝟎𝟐𝟓 𝒎
16. In question 15, the correct length of the measured line is
a) 875.56 m
b) 875.63 m
c) 875.68 m
d) 875.60 m
e) 875.70 m
Given:
ML= 875.63 m
A= 0.03cm2
𝑃𝑜 = 5 𝑘𝑔
P= 10 kg
kg
E= 2.0x106 cm2
Required:
The correct length, CL
ILLUSTRATION:
SOLUTION:
(𝑃−𝑃𝑜 )𝑀𝐿
𝐶𝑃 =
𝐴𝐸
=
(10 𝑘𝑔−5 𝑘𝑔)875.63 𝑚
0.03cm2 (2.0x106
kg
)
cm2
= 0.07296916667 𝑚
𝑪𝑳 = 𝑀𝐿 + 𝐶𝑃
= 875.63 𝑚 + 0.07296916667 𝑚
= 875.7029692 𝑚
`
≈ 𝟖𝟕𝟓. 𝟕𝟎 𝒎
17. A 30-m steel tape weighs 1.05 kg and is supported at its end and at the 10-m
and 25-m marks. If a pull of 6.0 kg is applied at the ends of the tape, the
correction due to sag for a full tape length is
a) 0.038 m
b) 0.006 m
c) 0.050 m
d) 0.45 m
e) 0.06 m
Given:
Tape Material = Steel
NL= 30 m
W= 1.05 kg
Supported: @ its ends (0 m & 30 m – marks)
: (10 m & 25 m – marks)
P= 6 kg
Required:
the correction due to sag for a full tape length, Cs
ILLUSTRATION:
SOLUTION:
𝐿1
𝑤1 = 𝑊 ( )
𝑁𝐿
10 𝑚
= 1.05 𝑘𝑔 (
)
30 𝑚
= 0.25 𝑘𝑔
𝐿
𝑤2 = 𝑊 (𝑁𝐿2 )
10 𝑚
= 1.5 𝑘𝑔 (30 𝑚)
= 0.50 𝑘𝑔
𝐿
𝑤3 = 𝑊 (𝑁𝐿3 )
15 𝑚
= 1.5 𝑘𝑔 (30 𝑚)
= 0.75 𝑘𝑔
𝑪𝑺 = 𝑐𝑆1 + 𝑐𝑆2 + 𝑐𝑆3
`
=
=
𝑊12 𝐿1
+
𝑊22 𝐿2
+
𝑊32 𝐿3
24𝑃 2
24𝑃 2
24𝑃 2
(0.35 𝑘𝑔)2 (10 𝑚)
(0.525 𝑘𝑔)2 (15 𝑚)
+
24 (6 𝑘𝑔)2
−3
24 (6 𝑘𝑔)2
+
(0.175 𝑘𝑔)2 (5 𝑚)
24 (6 𝑘𝑔)2
−3
= 1.417824074 × 10 𝑚 + 4.78515625 × 10
= 6.380208333 × 10−3 𝑚
≈ 𝟎. 𝟎𝟎𝟔 𝒎
𝑚 + 1.772280093 × 10−4 𝑚
18. In a triangular-shaped lot ABC, the two sides and the included angle are: CA=
90.95 m, BC= 73.80 m, and angle C= 43°15'. The length of the remaining side
AB is
a) 62.77 m
b) 117.13 m
c) 153.28 m
d) 82.38 m
e) 81.93 m
GIVEN:
CA= 90.95 m
BC= 73.80 m
𝜃𝐶 = 43°15
REQUIRED:
Length of AB
ILLUSTRATION:
SOLUTION:
𝐴𝐵 = √𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶
= √(90.95 𝑚)2 + (73.80 𝑚)2 − 2(90.95 𝑚)(73.80 𝑚) cos 43°15′
= 62.77364382 𝑚
≈ 𝟔𝟐. 𝟕𝟕 𝒎
19. In question 18, the relationship between angle C and the two remaining angles, A
and B, of the triangle could be expressed correctly as
a) A < C > B
b) A > C > B
c) C= A – B
d) C= A + B
`
e) A > C < B
20. In the accompanying sketch it is desired to determine the length of AB across a
wide and deep river.
C
471.48m
D
a)
b)
c)
d)
e)
A
B
Line AC, which measures 471.48 m, is established perpendicular to AB; CD is
similarly established perpendicular to BC with point D on the prolongation of line
AB. If the length of AD is 322.35 m, the length of AB is equal to
689.60 m
220.39 m
389.85 m
453.40 m
517.23 m
REQUIRED:
Length of AB
SOLUTION:
𝐴𝐵
𝐴𝐶
=
𝐴𝐶
𝐴𝐷
𝑨𝑩 =
𝐴𝐶 2
𝐴𝐷
(471.48 𝑚)2
= 322.35 𝑚
= 689.6025761 𝑚
≈ 𝟔𝟖𝟗. 𝟔𝟎 𝒎