Explain briefly the Subtense bar method.

Subtense bar is an instrument used for measuring horizontal distances in the areas where direct chaining becomes difficult due to undulation or other obstructions.

It consists of a metal tube of length varying from 3m to 4 m. Two disc 20 cm in diameter painted either black or red on one side and white on the other, each with a 7.5m white or black center are placed 3m apart. Red or black faces of discs are kept towards the theodolite.

At the center of the bar,an alidade perpendicular to the axis of the bar is attached. The bar is mounted on special tripod.

Here, from triangle `OAC` and `OCB`,

`D*tan (theta/2)=S/2`

`D=S/2 cot(theta/2)`

Or, `D=S/(2tan(theta/2))`

Now, for an small angle `theta`,

`tan(theta/2)=theta/2`, where `theta` is in radian.

      `=1/2*theta/(206265)`, where `theta` is in second..

 Thus, we have,

 `D=S/(2*(1/2*theta/206265))`
Or,`=(S*206265)/thet....

Subtense bar is an instrument used for measuring horizontal distances in the areas where direct chaining becomes difficult due to undulation or other obstructions.

It consists of a metal tube of length varying from 3m to 4 m. Two disc 20 cm in diameter painted either black or red on one side and white on the other, each with a 7.5m white or black center are placed 3m apart. Red or black faces of discs are kept towards the theodolite.

At the center of the bar,an alidade perpendicular to the axis of the bar is attached. The bar is mounted on special tripod.

COMPUTATION OF SUBTENSE BAR DISTANCE:

Here, from triangle `OAC` and `OCB`,

`D*tan (theta/2)=S/2`

`D=S/2 cot(theta/2)`

Or, `D=S/(2tan(theta/2))`

Now, for an small angle `theta`,

`tan(theta/2)=theta/2`, where `theta` is in radian.

      `=1/2*theta/(206265)`, where `theta` is in second..

 Thus, we have,

 `D=S/(2*(1/2*theta/206265))`
Or,`=(S*206265)/theta`, Where `theta` is in second.

 EFFECT OF ANGULAR ERROR ON HORIZONTAL DISTANCE:

 From figure,

 `tan(theta/2)=((S/2)/D)`

 `S/2=D*tan(theta/2)=D*theta/2`

Or,`S=D*theta`.......(i)

 We know,

 `D=(S*206265)/theta`

 Let the negative error in `theta` be `delta theta` and the positive error in `D` be `delta D`then,

 `S=(D+delta D)(theta- delta theta)`......(ii)

 Or,`(D+delta D)/D=theta/(theta+delta theta)`

 On cross multiplication,

 `D theta-D delta theta-theta*D delta theta=D theta`

Or,`delta D(theta - delta theta)=D*delta theta`

Or,`delta D=(D delta theta)/(theta + delta theta)`

Similarly, if `delta theta` is positive error and `delta D` is the negative error.,

`delta D=(D delta theta)/(theta+delta theta)`

However,`delta theta` is too small as compared to `theta`,so,

`delta theta=(D delta theta)/theta`