Question

pr.1 The Observations were made from station P to signal at station Q. The distance
between Pand Qis 12.5 km and diameter of signal at station Q was 20 cm. The sun rays make an angle
of 50° with line PQ. Calculate the phase correction if observations were made:on the bright portion,on the bright line.. ,A base line was measured with a steel tape of designated length 30 m at 20°C ata pull of 100 N. The measured length of base line was 1543 m. The field temperature was 31.5°C and
the pull applied was 130 N. Find the correct length of base line. The cross-sectional area of tape is
2 mm2, coefficient of thermal expansion of steel is 2.5 x 10-6°C-1 and E = 2x 105 N/mm?.

note:i will rate positive if you provide complete solution and justify your answer ...

Answer 1

Solution:- the values given in the are as follows:

distance between station P and Q =12.5 km

diameter of signal at station Q=20 cm

sun rays make an angle with line PQ =50o

phase correction=?

designated length of tape(l)=30 m

satndard tepmperature of tape(To)=20oC

standard pull on tape(Po)=100 N

measured lenth of base line with the help of tape(LI)=1543 m

field temperature(Tf)=31.5o

pull applied on tape(Pm)=130 N

correct length of base line(L)=?

cross-section area of tape(a)=2 mm^2

coefficient of thermal expansion()=2.5*10^-6 /oC

young's modulus of elasticity(E)=2*10^5 N/mm^2

phase correction calculation:

phase correction calculation made on bright portion:

let =phase correction due to bright portion

={(206265*r)/L}*cos^2(/2) [where, =50o, r=10 cm=10*10^-5 km]

={(206265*10*10^-5)/12.5}*cos^2(25)=1.3554 seconds

phase correction ade on bright portion()=1.3554 seconds

phase correction made on bright line(I)={(206265*r)/L}*cos(/2)

phase correction made on bright line(I)={(206265*10*10^-5)/12.5}*cos(25)

phase correction made on bright line(I)=1.4955 seconds

Calculation correct lenth of base line(L):

correction due to temperature(CT)=measured length(LI)**

where, =Tf-To =31.5-20 =11.5oC

correction due to temperature(CT)=1543*2.5*10^-6*11.5=0.04436 m

correction due to temperature(CT) is positive because Tf>To

correction due to temperature(CT)=+0.04436 m

correction due to pull(Cp)={(Pm-Po)*L}/(A*E)

values put in above equation is mm and N

correction due to pull(Cp)={(130-100)*1543*1000}/(2*2*10^5)=115.725 mm

correction due to pull(Cp)=115.725 mm

correction due to pull(Cp)=0.115725 m

correction due to pull(Cp) is positive because Pm>Po

correction due to pull(Cp)=+0.115725 m

correct lenth of base line(L)=measured length+correction due to temperature(CT)+correction due to pull(Cp)

correct lenth of base line(L)=1543+0.04436+0.11572

correct lenth of base line(L)=1543.16 m   [Ans]