Chain
the survey is the simplest method of surveying. It is the exercise of physically
measuring horizontal distances. In this method, the lengths of lines marked on
the field are measured, while the details are measured by offsets and ties from
these lines. This fieldwork will continue for 3 field hours. This is most
suitable adapted to small plane areas with very few details.
Errors
in chain survey
In
general, the distance measurement obtained in the field will be in error.
Errors in distance measurement can arise from several sources:
1. Instrument errors:
A
tape may be faulty due to a defect in its manufacturing or from kinking.
2. Natural errors.
The actual horizontal distance between the ends of the tape can vary due to the
effects of
·
temperature,
·
elongation due to
tension
·
Sagging.
3. Personal
errors.
Errors will arise from carelessness by the survey crew:
1.
poor alignment
2.
tape not horizontal
3.
improper plumbing
4.
faulty reading of the
tape
Errors
in Chaining: - The errors that occur in
chaining is classified as (i) Compensating, (ii) Cumulative. These
errors may be due to natural causes such as say variation in temperature,
defects in construction and adjustment of the instrument, personal defects in
vision etc.
Compensating
Errors:- The compensating errors are those
which are liable to occur in either direction and hence tend to compensate i.e.
they are not likely to make the apparent result too large or too small.
In
chaining, these may be caused by the following: -
Incorrect
holding of the chain:-
The follower may not bring his handle of the chain to the arrow but may hold it to
one or other side of the arrow.
Fractional
parts of the chain or tape may not be correct if the total length of the chain
is adjusted by insertion or removal of a few connections rings from one portion
of the chain, or tape is not calibrated uniformly throughout its length.
During
stepping operation crude method of plumbing (such as dropping of stone from the end of the chain) is adopted.
When chain angles are set out with a chain which is not uniformly adjusted or with a combination of chain and tape.
Cumulative
Errors: - The cumulative errors are
those which occur in the same direction and tend to add up or accumulate i.e. either to make the apparent measurement always too long or too short.
Positive
errors (making the measured lengths more than the actual) are caused by the
following:-
The length
of the chain or tape is shorter than the standard, because of bending of links, removal of too many links in adjusting the length, ‘knots’ in the connecting links, cloggings of rings with clay, temperature lower than that at which the tape was calibrated, shrinkage of tape when becoming wet.
The slope
correction is not applied to the length measured along the sloping ground.
The sag
correction is not applied when the tape or the chain is suspended in the air.
Measurements
are made along the incorrectly aligned line.
The tape bellies out during offsetting when working in the windy weather.
Negative
errors (making the measured lengths less than the actual) may be caused because the length of the tape or chain may be greater than the standard because of the wear or flattening of the connecting rings, the opening of ring joints, a temperature higher than the one at which it was calibrated.
The final
error in a linear measurement is composed of two portions:
Cumulative errors which are proportional to L and compensating errors which are proportional to √L, where L is the length of the line.
Illustration: - Suppose a line 1280 m in length is measured with a 20 m
chain which is 0.02 m too long, and error in marking a chain length is say
±0.03 m.
Compensating error of marking
The latter
error though smaller has a greater effect than the former though it is larger.
Mistakes
in Chaining: - The mistakes are generally
avoidable and cannot be classed under any law of probability. The
following mistakes are commonly made by inexperienced chainmen.
Displacement
of arrows: - When the arrow is displaced,
it may not be replaced accurately. To guard against this mistake, the end
of each chain length should be marked both by the arrow and by a cross (+) scratched on the ground.
Failure
to observe the position of the zero points of the tape: - The chainmen should see whether it is at the end of the ring or on the tape.
Adding or omitting a full chain or tape length (due to wrong counting or loss of arrows): - This is the most serious mistake and
should be guarded against. This is not likely to occur if the leader has
the full number (ten) of arrows at the commencement of chaining and both the leader and follower count them at each transfer. A whole tape length may be dropped if the follower fails to pick up the arrow at the point of beginning.
Reading
from the wrong end of the chain: - e.g.
reading 10 m for 20 m in a 30 m chain, or reading in the wrong direction from a tally, e.g. reading 9.6 m for 10.4 m. The common mistake in reading a chain is to confuse 10 m tag with 20 m tags. It should be avoided by noticing the 15 m tag.
Reading numbers incorrectly: - Transposing figures
e.g.37.24 for 37.42 or reading tape upside down, e.g. 6 for 9, or 36 for 98.
Calling
number wrongly: - e.g. calling 40.2 as “forty-two”.
reading the wrong metre marks: - e.g. 58.29 for 57.29.
Wrong booking: - e.g. 345 for 354.
To guard against this mistake, the chainmen should call out the measurements loudly and distinctly, and the surveyor should repeat them as he books them.
Tape
Corrections: - Precise measurements of
distance is made using a steel tape 30 m or 50 m in length. Before
use, it is desirable to ascertain its actual length (absolute length) by
comparing it with the standard of known length, which can be done for a small fee by the Survey and Standards department. It is well to note here the distinction between the nominal or designated length and absolute length of a tape. By the former is meant it’s designated length, e.g. 30 m, or 100 m, while by the latter is meant it’s actual length under specified
conditions. The tape may be standardized when supported horizontally
throughout its full length or in catenary. The expression that “tape is
standard at a certain temperature and under a certain pull” means that under these conditions the actual length of the tape is exactly equal to its nominal length. Since the tape is not used in the field under standard conditions it is necessary to apply the following corrections to the measured length of a line to obtain its true length:
Correction for absolute length,
(ii) Correction for temperature,
(iii) Correction for tension or pull,
(iv) Correction for sag, and
(v) Correction for slope or vertical alignment.
A correction is said to be plus or positive when the uncorrected length is to be increased, and minus or negative when it is to be decreased to obtain true length.
Correction
for Absolute Length: - It is the usual
practice to express the absolute length of tape as its nominal or designated length plus or minus a correction. The correction for the measured length is given by the formula,
Ca = Lc / l ------------------- (1)
Where Ca =
the correction for absolute length.
L = the measured length of a line.
l = the nominal length of a tape.
C =
the correction to a tape.
The sign
of the correction (Ca) will be the same as that of c.
it may be noted that L and l must be expressed in the same units and the
unit of Ca is the same as that of c.
Correction for Temperature: - It is necessary to apply this correction, since the length of a tape is increased as its temperature is raised, and consequently, the measured distance is too small. It is given by the formula,
Ct = a (Tm – To) L----------- (2)
in which Ct =
the correction for temperature, in m.
a = the coefficient of thermal expansion.
Tm = the mean temperature during
measurement.
To = the temperature at which the tape is
standardized.
L = the measured length in m.
The sign
of the correction is plus or minus according to Tm is greater
or less
than To. The coefficient of expansion for steel varies from
10.6 x 10-6 to 12.2 x 10-6 per degree centigrade and that for invar from 5.4 x 10-7 to 7.2 x 10-7. If the coefficient of expansion of tape is not known, an average value of 11.4 x 10-6 for steel and
6.3 x 10-7 for invar may be assumed. For very precise
work, the coefficient of expansion for the tape in question must be carefully determined.
Correction
for Pull (or Tension): - The correction is necessary when the pull used during the measurement is different from that at which the tape is standardized. It is given by the formula,
Cp =
(P-Po) L / AE ---------- (3)
Where Cp = the correction for pull in metres.
P = the pull applied during measurement, in newton (N).
Po= the pull under which the tape is standardized in newton
(N).
L = the measured length in metres.
A = the cross-sectional area of the tape, in sq.cm.
E = the modulus of elasticity of steel.
The value
of E for steel may be taken as 19.3 to 20.7 x 1010 N/m2 and
that for invar 13.8 to 15.2 x 1010 N/m2. For
every precise work, its value must be ascertained. The sign of the
correction is plus, as the effect of the pull is to increase the length of the
tape and consequently, to decrease the measured length of the line.
Correction
for Sag: - (Fig.1). When a tape is
stretched over points of support, it takes the form of a catenary. In
actual practice, however, the catenary curve is assumed to be a parabola.
The correction for sag (or sag correction) is the difference in length between the arc and the subtending chord (i.e., the difference between the horizontal distance between supports and the length measured along the curve). It is required only when the tape is suspended during measurement. Since the effect of the set on the tapes is to make the measured length too great this correction is always subtractive. It is given by the formula,
Cs = l1 (mgl1)2 /
24P2 = l1 (Mg) 2 / 24P2 ………………
(4)
in which Cs = the sag correction for a single span, in metres.
l1 = the distance between supports in
metres.
m = the mass of the tape, in kilograms per metre.
M = Total mass of the tape in kilograms.
P = the applied pull, in newton (N).
If there are n equal spans per tape length, the sag corrections per tape length
is given, by
Cs = nl1 (mgl1)2 / 24P2 =
l1 (mgl1)2 / 24P2 = l (mgl)
2 / 24n2P2 …………. (4a)
in which l
= the length of the tape = nl1, and l1= l/n.
Normal-Tension: - The normal tension is a tension at which the effects of pull and sag are neutralized, i.e. the elongation due to an increase in tension is balanced by the shortening due to sag. It may be obtained by equating the corrections for pull and sag. Thus we have,
(Pn-Po) l1 / AE = l1(mgl1)2 /
24Pn2 or (Pn-Po) Pn2 =
W2AE / 24
~ Pn =
0.204 W √AE / √ (Pn-Po) ………………………………………….. (5)
In which Pn =
the normal tension in newton (N).
W = the weight of the length of tape between supports
in newton (N).
The value
of Pn may be determined by trial
Correction for Slope or
Vertical Alignment: - (Fig 2) This correction is required when the points of support are not exactly at the same level.
Let l1 l2,
etc. = the lengths of successive uniform slopes.
lt1, lt2 etc.
= the differences in height between the extremities of each of these
slopes.
Cs = the total correction for slope.
If l is the length of anyone slope, and h the difference in height between the ends of the slope,
The slope correction = l - √ l2-h2
= l – l (1
– h2 / 2l2 – h4 /
3l4 – etc..)
= (h2 / 2l + h4 /
3l3 + etc.) = h2 / 2l ------------------------
(6)
Hence,
Cs = (h12 / 2l1 + h22 /
2l2 + ….. + hn2 /
2ln) ------------------------------- (6a)
When the slopes are of uniform length l we have
Cs = l / 2l (h12 + h22 +
……… + hn2) = ∑h2 / 2l -------------------------
(6b)
This correction is
always subtractive from the measured length. If the slopes are given in terms of vertical angles (plus or minus angles), the following formula may be used:
The correction for the slope
= l – l cos 0 = 2l sin2 0
/ 2
= l versin 0 (-ve) -------------------------- (7)
In which l =
the length of the slope : 0 = the angle of the slope.
Examples
on Tape Corrections
Examples
1: - A line was measured with a steel tape which was exactly
30m long at 18oC and found to be 452.343 m. The temperature
during measurement was 32oC. Find the true length of the line. Take
the coefficient of expansion of the tape per oC=0.0000117.
Temperature correction per tape length = Ct
= α (Tm -
To) l
Here
l = 30 m: To =18oC; Tm =
32oC;
α = 0.0000117
~ Ct =
0.0000117 (32-18) 30
= 0.004914 m (+ ve)
Hence the
length of the tape at 32oC = 30 + Ct
= 30 + 0.004914 = 30.004914 m.
Now true length of a line = L’ / L x its measured length.
L = 30 m:
L’ = 30.004914 m; measured length = 452.343 m.
~ True
length = 30.004914 / 30 x 452.343 = 452.417 m.
Example
2: - A line was measured with a steel rape which was exactly 30 m
at 18oC and a pull of 50 N and the measured length was 459.242
m. The temperature during measurement was 28oC and the pull
applied was 100 N. The tape was uniformly supported during the
measurement. Find the true length of the line if the cross-sectional area
of the tape was 0.02 cm2, the coefficient of expansion per oC = 0.0000117 and the modulus of elasticity = 21 x 106 N per cm2.
Temperature
Correction per tape length
= α (Tm – To) L
= 0.0000117 x (28 -18) 30
= 0.00351 m (+ ve)
Sag correction per tape
length
= 0
Pull correction per tape
length
= (Pm - Po) L / AE
= (100 – 50)30 / 0.02 x 21 x 106
= 0.00357 m (+ve)
~ Combined
correction
= 0.00351 + 0.00357 m.
= 0.00708 m
True length of
tape
= 30.00708 m
True length of the
line
= 30.00708 / 30 x 459.242
= 459.350 m.
Example
3: - A 50 m tape is suspended between
the ends under a pull of
150 N. The mass of the tape is 1.52 kilograms. Find the corrected
length of the tape.
Correction for sag
= Cs = l1 (Mg)2 / 24 P2
l1 = 50 m; M = 1.52 kilograms; P = 150
N.
~ Cs = 50 x (1.52 x 9.81)2 / 24 x 1502 =
0.0206 m.
~ Corrected length of the
tape
= l1 – Cs
= 50 – 0.0206
= 49.9794 m.
Example
4: - The downhill end of the 30 m
tape is held 80 cm too low. What is the horizontal length?
Correction for slope = h2 / 2l
Here h = 0.8 m; l = 30 m
~ The required correction
= 0.82 / 2 x 30 = 0.0167 m.
Hence the
horizontal
length
= 30 – 0.0167
= 29.9833 m
Example
5: - A 100 m tape is held 1.5 m out
of line. What is the true length?
Correction for incorrect alignment
= d2 / 2l (- ve)
Here d = 1.5 m; l = 100 m.
~ Correction = 1.52 / 2 x 100 = 0.011 m.
~ True length = 100 – 0.011 = 99.989 m.
