Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 2 (Q.No. 15)
15.
A dumpy level was set up at the midpoint between two pegs A and B, 50 m apart and the staff readings at A and B were 1.22 and 1.06. With the level set up a A, the readings at A and B were 1.55 and 1.37. The collimation error per 100 m length of sight is
Discussion:
20 comments Page 1 of 2.
Goutom saikia said:
6 years ago
True level difference get when the level set midway between A & B.
True level difference=1.22-1.06=.16 downward (Greater value give lower elevation so A is lower elevation than B).
And when an instrument set near to a point collimation error eliminate.So we get true reading for A.
True reading at A=1.55.
When instrument set near A then true reading at B=1.55-.16=1.39.
Error=1.37-1.39=-.02.
Error for 50 m=-0.02.
Error for 100 m=2*(-.02)=-0.04.
True level difference=1.22-1.06=.16 downward (Greater value give lower elevation so A is lower elevation than B).
And when an instrument set near to a point collimation error eliminate.So we get true reading for A.
True reading at A=1.55.
When instrument set near A then true reading at B=1.55-.16=1.39.
Error=1.37-1.39=-.02.
Error for 50 m=-0.02.
Error for 100 m=2*(-.02)=-0.04.
(17)
DARSHAN said:
9 years ago
Here when you put level at centre all errors are eliminated .
Hence difference in RL is 1.22 - 1.06 is 0.16 but the observed staff difference is 1.55 - 1.37 = 0.180.
Therefore the error is 0.180 - 0.160 = 0.02 for 50 m, for 100m it is 0.04. It is positive value hence line of sight is upwards.
Hence difference in RL is 1.22 - 1.06 is 0.16 but the observed staff difference is 1.55 - 1.37 = 0.180.
Therefore the error is 0.180 - 0.160 = 0.02 for 50 m, for 100m it is 0.04. It is positive value hence line of sight is upwards.
(1)
M.Pagavan doss said:
1 decade ago
Distance between a and b = 50 m.
When instrument at center=1.22 - 1.06 = .160.
When instrument near A = 1.55-1.37 = .180.
DIFF = .160-.180 = -.020(- INDICATE DOWNWARDS).
COLLIMATION ERROR FOR 100M length = (.020/50)*100 = .04m inclined downward.
When instrument at center=1.22 - 1.06 = .160.
When instrument near A = 1.55-1.37 = .180.
DIFF = .160-.180 = -.020(- INDICATE DOWNWARDS).
COLLIMATION ERROR FOR 100M length = (.020/50)*100 = .04m inclined downward.
(1)
Karyeija Felex said:
4 years ago
@Ankit.
The 1st setting is a two peg test which eliminates error thus giving true difference in level.
We usually divide by two when the instrument is not at mid point so that errors on both BS and FS are eliminated.
The 1st setting is a two peg test which eliminates error thus giving true difference in level.
We usually divide by two when the instrument is not at mid point so that errors on both BS and FS are eliminated.
(2)
Prince said:
4 years ago
@Abhijeet,
This is the formula to be used in reciprocal levelling. Here the staff is kept at the centre first so the true difference between the elevations is the actual difference between them.
This is the formula to be used in reciprocal levelling. Here the staff is kept at the centre first so the true difference between the elevations is the actual difference between them.
Deep said:
7 years ago
Formula of errors =(b1-a1)-(b2-a2)/2.
Now, e = (1.06-1.22)-(1.37-1.55)/2 =0.01for 50 m and for 100 m error = 2 *0.01 = 0.02 upward.
Now, e = (1.06-1.22)-(1.37-1.55)/2 =0.01for 50 m and for 100 m error = 2 *0.01 = 0.02 upward.
Ashok said:
9 years ago
When the level is not set at the center what distance can I take instead of 50m?
Abhijitth said:
5 years ago
Why are they not dividing by 2?
(ha-hb)-(ha'-hb')/2.
Isn't this the formula?
(ha-hb)-(ha'-hb')/2.
Isn't this the formula?
(1)
Ankit said:
4 years ago
Agree @Abhijeet,
why are they not dividing it by 2? Explain, please.
why are they not dividing it by 2? Explain, please.
(1)
MD SADDAM said:
9 years ago
Please help about B.O.D & C.O.D formula.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers