 Correction of Length & Area Due To Incorrect Scale or Chain | 7 Numericals

Here in this post, we are explaining numerical like length correction, area correction, etc which are caused due to incorrect scale or chain.

1. Useful Formulas For Incorrect Scale or Chain Numerical

a. Error due to use of the incorrect scale

If the length of a line existing on a plan or a map is determined utilizing measurement with a wrong scale, the length so obtained will be incorrect. The true (or correct length) of the line is given by the relation.

Correct length =  (R.F of the wrong scale / R.F of correct scale) × Measured length

Similarly, if the area of a map or plan is calculated with the help of the wrong scale i.e.

Correct area = (R.F of the wrong scale/R.F of correct scale)² ×Calculated area

b. Error due to Shrunk scale

Shrunk scale = Shrinkage factor × Original scale

The shrinkage factor is equal to the ratio of the shrunk length to the actual length.

c. Error due to incorrect chain

If the length of the chain used in measuring the length of the line is not equal to the true length or the designated length, the measured length of the line will not be correct and suitable correction will have to be applied.

In case the chain is too long, the measured distance will be less. Therefore the error will be negative and the correction is positive.

Similarly, if the chain is too short, the measured distance will be more, the error will positive and the correction will be negative.

Let,

L= true or designated length of the chain or tape

L’= Incorrect length of the chain or tape used.

i. Correction to a measured length:

Let,

l’= measured length of the line

l= actual or true length of the line

Then,

the true length of line =  measured length of line × ( L’ / L)

l = l’ × ( L’ / L )

ii. Correction to the area:

let

A’= measured (or computed) area of the ground

A= actual or true area of the ground

Then,

true area = measured area × ( L’ / L )²

A= A’ × ( L’ / L)²

iii. Correction to the volume:

Let,

V’= measured or computed volume

V= actual or true volume

Then,

True volume = measured volume× ( L’ / L )³

V = V’ × ( L’ / L )³

2. Numerical Problems

1.1. A surveyor measured the distance between two points on the plan drawn to a scale of 1 cm = 40 m and the result was 468 m. Later, however, he discovered that he used a scale of 1 cm = 20 m. Find the true distance between the points.

Solution,

Measured length = 468m

R.F of wrong scale  = 1cm / 20m = 1cm / 2000cm = 1/2000 or 1:2000

R.F of correct scale  = 1cm / 40m = 1cm / 4000cm = 1/4000 or 1:4000

Correct length = (R.F of wrong scale/R.F of correct scale)×Measured length

∴ Correct length = ( (1/2000) / (1/4000) ) × 468  = 936 m

1.2. The area of the plan of an old survey plotted to a scale of 10 meters to 1 cm measures now as 100.2 sq. cm as found by a planimeter. The plan is found to have shrunk so that a line originally 10cm long now measures 9.7cm only. Find

i. Shrunk scale

ii. true area of the survey.

Solution,

i.

Shrinkage factor = 9.7/10 = 0.97

True scale R.F. = 1cm/10m = 1cm/ 1000cm  = 1/1000 or 1:1000

R.F. of shrunk scale = 0.97×(1/1000) = 1/1030.93

ii.

Original area on plan = (10/9.7)² ×100.2 sq.cm = 106.49 sq.cm

Scale of plan is 1cm = 10m

Area of survey = 106.49 * (10)² = 10649 sq.m.

1.3. A rectangular plot of land measures 20cm×30cm on a village map drawn to a scale of 100m to 1cm. Calculate its area in hectares. If the plot is redrawn on a topo sheet to a scale of 1km to 1cm, what will be its area on the topo-sheet? Also, determine the R.F. of the scale of the village map as well as on the topo-sheet.

Solution,

i. Village map:

1 cm on map = 100m on the ground

∴ 1cm² on map = 100² m² on the ground

The plot measures 20cm×30cm i.e. 600cm² on the map.

∴ Area of plot = 600×10000 = 6×1000000 m² = 600 hectares.

ii. Topo sheet

1 km² is represented by 1cm² or (1000×1000)m² is represented by 1cm²

∴ 6×10ˆ6 m² is represented by (1/(1000×1000))×(6×10ˆ6)= 6 cm²

iii.

R.F. of the scale of village map = 1/(100×100)=1/10000

R.F. of the scale of topo sheet= 1/(1×1000×100) = 1/100000

1.4. The length of a line measured with a 20-meter chain was found to be 250 meters. Calculate the true length of the line if the chain was 10cm too long.

Solution,

Incorrect length of the chain (L’) = 20 + (10/100) = 20.1m

Correct length of the chain (L) = 20m

Measured length (l’) = 250m

Hence true length of the line = l’×(L’/L) = 250×(20.1/20) = 251.25 meters

1.5. The length of a survey line was measured with a 20m chain and was found to be equal to 1200 meters. As a check, the length was again measured with a 25m chain and was found to be 1212m. On comparing the 20m chain with the test gauge, it was found to be 1 decimeter too long. Find the actual length of the 25m chain used.

Solution,

With 20m chain

Incorrect length of the chain (L’) = 20+0.10=20.10m

Correct length of the chain (L) = 20m

Measured length (l’) = 1200 m

Hence true length of the line (l) = l’×(L’/L) =1200×(20.10/20) = 1206m

With 25m chain

Correct length of the chain (L) = 25m

Measured length (l’) = 1212 m

the true length of the line (l) = l’×(L’/L)

or 1206=(L’/25)×1212

L’=24.88m

Thus, the 25m chain was 12cm too short.

1.6. A 20m chain was found to be 10cm too long after chaining a distance of 1500m. It was found to be 18cm too long at the end of the day’s work after chaining a total distance of 2900m. Find the true distance if the chain was correct before the commencement of the work.

Solution,

For first 1500 meters:

Average error (e)= (0+10)/2 = 5cm=0.05m

Incorrect length of the chain (L’) =20+0.05= 20.05m

True distance ( l1 )= (20.05/20)×1500 = 1503.75m

For next 1400 meters:

Average error (e)= (10+18)/2 =14cm = 0.14m

Incorrect length of the chain (L’) = 20+0.14 = 20.14m

True distance ( l2 ) = (20.14/20)×1400  +  1409.80m

∴ Total length=l1 +l2 = 1503.75 + 1409.80= 2913.55m

1.7. The area of the plan of an old survey plotted to a scale of 10 meters to 1cm measures now as 100.2 sq. cm as found by planimeter. The plan is found to have shrunk so that a line originally 10cm long now measures 9.7cm only. There was also a note on the plan that the 20m chain used was 8cm too short. Find the true area of the survey.

Solution,

Present length of 9.7cm is equivalent of 10cm original length.

Original area on plan =  (10/9.7)²×100.2 sq.cm= 106.49 sq.cm

Scale of the plan is 1cm=10m

∴ Original area of survey= (106.49)(10²)=1.0649×10ˆ4 sq.m

Faulty length of chain used= 20-0.08=19.92m

Correct area=(19.92/20)²×1.0649×10ˆ4 sq.m= 10564.7 sq.m