Civil Engineering - Surveying - Discussion

First of all, See the question this is telling us the main scale division 59 is equal's to 60 division of vernier it means we can understand from it is direct vernier condition((direct vernier condition (n-1)=n where as( n-1) is main scale division and (n) is division of vernier)).

So from direct vernier condition lest count will be (s/n) where (s) is the least count of the main scale and (n) is the division of the vernier.

So first we will find (s) so think that question does not have directly given but one hint is there which is 1080 division of transit means the main scale has 1080 division and we know transit is nothing but it is transit theodolite so total angle will be 360 degrees and now if we will 360/1080 then least count of the main scale can easily get but the question is ask in second not in degree.

So we will convert 360 degree into second 360*60*60 (1degree=60minutes, 1 minute=60sec)
now (360*60*60/1080)=1200 so (s)=1200.
and (n) is already given (n)=60.

Now:-
Least count = (s/n) =(1200/60) = 20.

Least count = S - V.

Where S is the length of one division of main scale which is nothing but least count of main scale and V is the length of one division of vernier scale.

From the above problem we can conclude that (n-1) division of main scale is divided into n division of vernier scale because 60 divisions of vernier coincide exactly with 59 divisions of the main scale.

Therefore, n*V=(n-1)*S ------- (A)

Now, S= L.C of main scale = 360/1080 = 1/3 degrees because in transit main plate there is a circle with 360 degrees.

S= (1/3) degrees = (1/3)* 3600 sec= 1200 seconds.

Now from (A) 60*V=59*1200
V=1080 seconds.
Therefore Least Count = S - V = 1200 - 1080 = 20 seconds.

An angle-measuring instrument reading up to one-sixth of a degree on the main scale is equipped with a vernier having 19 main scale divisions divided into 20 parts.
The correct least count for the instrument is 30 seconds.

The Least Count=(1/6 *)/20
=>1/6*60*60 sec/20 =30" ans.

Least Count = S/n.

S = Least count of main scale = 360/1080
n = no of divion on vernier scale = 60

In transit main plate their is circle scale with 360 degree
So S=360/1080 = 1/3

Now least count = 1/3 ÷ 60 = 1/180 degree
Convert in seconds
L.C = 1/180 *60*60 =20 seconds.

Least count = length of one division on the main scale/ no. of divisions of vernier.

Legth of one division on main scale= 360/1080.
= 0.3333333 degrees,
= 0.3333333 x 3600 sec,
= 1200 sec.

No of divisions of vernier = 60.

Therefore, Least count= 1200/60 = 20.

Main plate of transit =360°,divided into 1080 equal div.

So Angular value of one div= (360/1080)*60 (for min) *60 (for sec) =1200"=S.

60div of vernier coincide exacty with 59div of the main scale so, n=60.
Least count = s/n = 1200/60= 20"

A transit in Theodolite is such that here,

1080 divisions =360 degree.
1 div =360/1080 degrees which is S the smallest div on the main scale.
n=60.

So, LC=S/n in second gives 20".

Least count is equal to ratio of (length of one division on main scale to no. of divisions of vernier) i.e least count= s/n.

S = 360/1080.

N = 60.

Lc = 1/180 degrees = 20 seconds.

Total angle 360.
Dividing in 1080 part.
So 1 part in main scale = 1/3 degree or 20'.
1 part of main scale dividing in vernier 60 part
So 1 part in vernier = 20'/60 = 1/3' = 20"

L scale = 360/1080.
= 0.3333333*3600sec.
= 1200sec.
No. Of divisions = 59.
= 1200/59 = 20.33.
Therefore, their least count is 20".