Transit and Theodolite

The temporary adjustments are the operations done at every set-up of the theodolite, before observing any angle. They are done in this order:

1. Setting up the theodolite over the station

• Spread the tripod legs so that the head is about at chest height and roughly level, with the plumb bob (or optical plummet) near the station mark.
• Fix the instrument on the tripod head with the central fixing screw.
• Adjust the legs (move one leg in or out, or shift all legs together) until the plumb bob is vertically above the station mark. This is centering. Fine centering is done by loosening the fixing screw and sliding the instrument on the tripod head.

2. Levelling up

The vertical axis is made truly vertical with the plate levels and the three foot screws.

1. Turn the upper plate until the plate level is parallel to a line joining any two foot screws.
2. Turn those two screws together, in opposite directions (thumbs moving towards or away from each other), until the bubble is central.
3. Turn the instrument through 90 degrees so that the level is over the third foot screw; bring the bubble central using that third screw only.
4. Return to the first position and repeat. Continue until the bubble stays central in all positions (a full rotation through 180 degrees confirms it).

Levelling can disturb centering slightly, so check the plumb bob again.

   Foot screws: A, B, C         Step 1: level || AB
                                Step 2: turn 90 deg,
                                use C only
        A ---- B                Repeat until bubble stays
          \  /                  central everywhere
            C

3. Elimination of parallax

Parallax is the apparent movement of the cross hairs against the image when the eye is moved. It is removed in two steps:

1. Focusing the eyepiece: point the telescope to the sky or a white paper and turn the eyepiece until the cross hairs appear sharp and black.
2. Focusing the objective: sight the object and turn the focusing screw until the image is sharp and lies exactly in the plane of the cross hairs.

If the cross hairs and image do not move relative to each other when the eye is moved slightly, parallax is gone.

After these three adjustments the instrument is ready for observation.

The fundamental lines of a transit theodolite are the imaginary lines and axes about which the instrument is built. They are:

1. Vertical axis - the axis about which the whole upper part of the instrument rotates horizontally.
2. Horizontal (trunnion) axis - the axis about which the telescope rotates in the vertical plane.
3. Line of collimation (line of sight) - the line joining the intersection of the cross hairs to the optical centre of the objective and its extension.
4. Axis of the plate level (bubble line) - the tangent to the plate bubble at its centre; it is horizontal when the bubble is central.
5. Axis of the telescope level (altitude bubble) - the tangent to the altitude bubble at its centre, used for levelling the line of sight.

Relationship in perfect adjustment

• The axis of the plate level is perpendicular to the vertical axis. (So the vertical axis is vertical when the plate bubble is central.)
• The line of collimation is perpendicular to the horizontal axis. (So the telescope sweeps a plane, not a cone.)
• The horizontal axis is perpendicular to the vertical axis. (So the plane swept is vertical.)
• The axis of the telescope level is parallel to the line of collimation.
• The vertical axis passes through the centre of the graduated horizontal circle, and the horizontal axis through the centre of the vertical circle.

            line of collimation
         ---------------------->
   telescope  \
            ---O--- horizontal (trunnion) axis
               |
               |   vertical axis
               |
           plate level (axis perpendicular
           to vertical axis)

Method. The mean direction of each target is found from the two faces: mean = (Direct + (Reversed - 180°)) / 2. The directions are referred to the first reading on A (taken as 0°00′00″). The second reading on A closes the horizon, so the difference from 360° is the closing error, which is distributed equally.

Mean horizontal directions

Mean angles and adjustment

Closing error = 0°00′10.0″ (second reading on A larger than the first). The sum of the four angles at O must be 360°, and it is 360°00′10.0″; the error of 0°00′10.0″ is small and is therefore distributed equally: correction = -0°00′02.50″ per angle.

Missing vertical circle readings

For a theodolite with a zenith-angle vertical circle, FL + FR = 360° for a correct instrument. So the missing reading = 360° - the other face. The vertical angle is $90° - Z$ (positive = elevation, negative = depression).

Answer: adjusted angles AOB = 60°54′57.50″, BOC = 79°55′47.50″, COD = 129°29′07.50″, DOA = 89°40′07.50″; missing VCR: A (reversed) 239°44′50″, B (direct) 51°08′20″, C (direct) 90°44′50″, D (reversed) 182°39′50″; A closing (reversed) 270°59′50″.

Method. The mean direction of each target is found from the two faces: mean = (Direct + (Reversed - 180°)) / 2. The directions are referred to the first reading on A (taken as 0°00′00″). The second reading on A closes the horizon, so the difference from 360° is the closing error, which is distributed equally.

Mean horizontal directions

Mean angles and adjustment

Closing error = 0°00′35.0″ (second reading on A larger than the first). The sum of the four angles at O must be 360°, and it is 360°00′35.0″; the error of 0°00′35.0″ is small and is therefore distributed equally: correction = -0°00′08.75″ per angle.

Missing vertical circle readings

For a theodolite with a zenith-angle vertical circle, FL + FR = 360° for a correct instrument. So the missing reading = 360° - the other face. The vertical angle is $90° - Z$ (positive = elevation, negative = depression).

Answer: adjusted angles AOB = 60°55′16.25″, BOC = 79°55′41.25″, COD = 129°29′01.25″, DOA = 89°40′01.25″; missing VCR: A (reversed) 239°44′50″, B (direct) 51°08′20″, C (direct) 90°44′50″, D (reversed) 182°39′50″.

Note: the last pair of readings on A has no vertical circle readings (shown as - in the table), so only the four targets are completed. In this question the first reversed reading on A is 179°59′40″, which gives a closing error of 35″.

Errors eliminated by face observations

Face observation means observing a point with the telescope in the face-left position (vertical circle on the left, FL) and again after transiting and rotating the telescope 180 degrees (face right, FR), and taking the mean.

Errors eliminated by taking both faces

Mechanism of elimination

1. Collimation error. On face left the line of sight deviates from the correct direction by a small angle $e$ to one side. After changing face, the deviation is on the opposite side, so one reading is too large by $e$ and the other too small by $e$. The mean is free of it.

  FL:   X-------- true line ---------> P
          \_ line of sight (+e)
  FR:   X-------- true line ---------> P
          /_ line of sight (-e)
        mean of FL and FR = true direction

2. Trunnion axis error. If one end of the horizontal axis is higher, the telescope sweeps an inclined plane. After changing face, the high end is on the opposite side, so the angular error is reversed in sign and is cancelled in the mean. (It is large only for steep sights.)

3. Index error of the vertical circle. If the vertical circle reads $+i$ on face left when the line of sight is horizontal, it reads the same error with the opposite effect on face right. The mean of the two vertical angles is free of $i$.

For horizontal angles, the readings of both verniers (A and B, 180 degrees apart) are also taken and averaged, which removes the error of eccentricity of the centre of the circle.

Principle of operation of an optical (micrometer) theodolite

In an optical theodolite the horizontal and vertical circles are made of glass and are graduated finely (up to 20 minutes or 1 degree divisions). The images of the graduations are carried by a system of prisms and lenses to a reading microscope placed near the eyepiece of the telescope. The user reads the angle from the microscope while looking at the same place.

• The circle is lit by a mirror (or lamp) and light goes through the circle, a prism system and a lens to the reading eyepiece.
• In a micrometer theodolite, the graduations of the opposite sides of the circle are brought into the same view (diametrically opposite readings), so the reading is automatically the mean of two sides. This removes the error of eccentricity.
• A micrometer screw turns a plane parallel optical plate (glass plate) in the optical path. Turning it shifts the image of the circle graduation until the index line coincides exactly with a main graduation line. The drum attached to the screw is graduated in minutes and seconds, and gives the fraction of the smallest circle division.

   mirror -> circle -> prisms
        -> micrometer plate -> eyepiece
      (plate turned by micrometer screw)

   reading = circle division + micrometer drum reading

Advantages are: no verniers, accurate and quick reading to 1 second or 0.1 second, lighter and sealed instrument, and elimination of eccentricity error.

A theodolite is an instrument for measuring horizontal and vertical angles. A telescope that can turn about a vertical axis and a horizontal axis is sighted on a point, and the angular movement is read on graduated circles.

Construction

        eyepiece   telescope   objective
         \______________/
            |    ||  vertical circle
         ---+----++---  horizontal axis
         |  standards (A-frame)  |
         ======= upper plate (verniers) ======
         ======= lower plate (graduated circle) ===
              levelling head, 3 foot screws
                   tripod

• Levelling head with three foot screws, resting on the tripod.
• Lower plate carrying the graduated horizontal circle (0 to 360 degrees), with a lower clamp and tangent screw.
• Upper plate carrying the verniers (or micrometer), plate levels, and the standards; upper clamp and tangent screw.
• Telescope (internal focusing) with cross hairs, fitted to the horizontal axis, with altitude level.
• Vertical circle fixed to the telescope, with its vernier on the T-frame.

Principle

The telescope turns in a horizontal plane about the vertical axis and in a vertical plane about the horizontal axis. The difference of the circle readings when the telescope is sighted on two points gives the angle between them.

Function of the micrometer screw (optical theodolite)

The micrometer screw turns an optical plane-parallel plate which shifts the image of the circle graduation. It is turned until the index mark falls exactly on the nearest main graduation. The graduated drum on the screw gives the fraction of the smallest division (minutes and seconds). It replaces the vernier, with a least count of up to 1 second or less.

Uses of a theodolite

Measuring horizontal and vertical angles; setting out angles; ranging a line; prolonging a line; measuring distances by stadia tacheometry; finding heights; running traverse and triangulation; setting out buildings, roads and bridges.

Repetition method. The same horizontal angle (say AOB) is added several times mechanically on the circle, and the total is read only at the start and at the end. The mean angle = total accumulated angle / number of repetitions.

Procedure (angle AOB, three repetitions)

1. Set up and level the theodolite at O. Keep the instrument on face left. Set the vernier A to 0°00′00″ (or any convenient reading) with the lower clamp tight, and read both verniers (A and B).
2. Loosen the lower clamp, bisect A exactly using the lower tangent screw, then tighten the lower clamp.
3. Loosen the upper clamp and swing the telescope clockwise to B. Bisect B with the upper tangent screw. Read the verniers (the first reading of the angle may be booked).
4. Loosen the lower clamp, swing the telescope back to A, and bisect it with the lower tangent screw. The reading on the circle is not changed because the upper plate is clamped to the circle.
5. Loosen the upper clamp and sight B again; bisect with the upper tangent screw. The reading now shows the angle added twice.
6. Repeat steps 4 and 5 once more for the third repetition. Read both verniers; the reading now shows the angle added three times.
7. Change face (transit the telescope, face right) and repeat the same three repetitions, anticlockwise if you like; take the mean of both faces.

   A                       B
    \ (upper plate)       /
     \                  /
      -------- O --------
   First: set to A, swing to B (upper clamp)
   Next: lower clamp, back to A, then upper again to B

Computation (example)

For face left: accumulated angle = (187°15′50″ - 0°00′20″ + 367°15′58″ - 180°00′20″)/2 = 187°15′34.0″. Angle (face left) = 187°15′34.0″ / 3 = 62°25′11.33″.

For face right: accumulated angle = 187°15′20.0″, angle = 62°25′06.67″.

$$\text{Mean angle AOB} = \frac{62°25′11.33″ + 62°25′06.67″}{2} = 62°25′09.0″$$

Answer: mean angle AOB = 62°25′09.0″.

The reading error of the circle is divided by the number of repetitions (here 3), so the final angle is more accurate than a single reading.

Methods of measuring horizontal angles

1. Ordinary (single angle) method: both faces, one reading each on A and B; the angle = difference of the readings.
2. Repetition method: one angle is added several times on the circle and the total is divided by the number of repetitions. Used for very accurate work for a single angle.
3. Direction (reiteration) method: several directions measured from a reference object and angles found as differences; used when many angles are measured at one station.
4. Method of closing the horizon: a special case of reiteration; the sum of all angles around the station must be 360 degrees.

Repetition method

Repetition method. The same horizontal angle (say AOB) is added several times mechanically on the circle, and the total is read only at the start and at the end. The mean angle = total accumulated angle / number of repetitions.

Procedure (angle AOB, three repetitions)

1. Set up and level the theodolite at O. Keep the instrument on face left. Set the vernier A to 0°00′00″ (or any convenient reading) with the lower clamp tight, and read both verniers (A and B).
2. Loosen the lower clamp, bisect A exactly using the lower tangent screw, then tighten the lower clamp.
3. Loosen the upper clamp and swing the telescope clockwise to B. Bisect B with the upper tangent screw. Read the verniers (the first reading of the angle may be booked).
4. Loosen the lower clamp, swing the telescope back to A, and bisect it with the lower tangent screw. The reading on the circle is not changed because the upper plate is clamped to the circle.
5. Loosen the upper clamp and sight B again; bisect with the upper tangent screw. The reading now shows the angle added twice.
6. Repeat steps 4 and 5 once more for the third repetition. Read both verniers; the reading now shows the angle added three times.
7. Change face (transit the telescope, face right) and repeat the same three repetitions, anticlockwise if you like; take the mean of both faces.

   A                       B
    \                    /
     \                 /
      ------ O ------
  (upper plate clamp: A -> B ; lower plate: B -> A)

The error of a single reading is divided by the number of repetitions, the errors of graduation are reduced as different parts of the circle are used, and the observer makes fewer readings.

(i) Repetition method

Repetition method. The same horizontal angle (say AOB) is added several times mechanically on the circle, and the total is read only at the start and at the end. The mean angle = total accumulated angle / number of repetitions.

Procedure (angle AOB, three repetitions)

1. Set up and level the theodolite at O. Keep the instrument on face left. Set the vernier A to 0°00′00″ (or any convenient reading) with the lower clamp tight, and read both verniers (A and B).
2. Loosen the lower clamp, bisect A exactly using the lower tangent screw, then tighten the lower clamp.
3. Loosen the upper clamp and swing the telescope clockwise to B. Bisect B with the upper tangent screw. Read the verniers (the first reading of the angle may be booked).
4. Loosen the lower clamp, swing the telescope back to A, and bisect it with the lower tangent screw. The reading on the circle is not changed because the upper plate is clamped to the circle.
5. Loosen the upper clamp and sight B again; bisect with the upper tangent screw. The reading now shows the angle added twice.
6. Repeat steps 4 and 5 once more for the third repetition. Read both verniers; the reading now shows the angle added three times.
7. Change face (transit the telescope, face right) and repeat the same three repetitions, anticlockwise if you like; take the mean of both faces.

   A                       B
    \                    /
     \                 /
      ------ O ------

Field notes

Computation (example)

For face left: accumulated angle = (187°15′50″ - 0°00′20″ + 367°15′58″ - 180°00′20″)/2 = 187°15′34.0″. Angle (face left) = 187°15′34.0″ / 3 = 62°25′11.33″.

For face right: accumulated angle = 187°15′20.0″, angle = 62°25′06.67″.

$$\text{Mean angle AOB} = \frac{62°25′11.33″ + 62°25′06.67″}{2} = 62°25′09.0″$$

Answer: mean angle AOB = 62°25′09.0″.

The reading error of the circle is divided by the number of repetitions (here 3), so the final angle is more accurate than a single reading.

(ii) Direction (reiteration) method

Direction (reiteration) method. Used when several angles are to be measured at one station. All the directions are observed from one reference object (RO), and the angles are obtained as differences of directions. The circle is not moved between objects, so many angles are obtained in one round.

Procedure (stations A, B, C observed from O)

1. Set up at O, level, and centre. Face left.
2. Bisect the reference object A (a well-defined point). Set the vernier A at 0°00′00″ (or any value) with the upper clamp, and read both verniers.
3. Loosen the upper clamp and swing clockwise to B; bisect, read both verniers. Do the same for C, and finally close the horizon by sighting A again. The second reading on A should be nearly the same as the first. If it differs, the difference is the closing error.
4. Change face (face right). Sight A again, and swing anticlockwise through C, B, back to A, reading both verniers.
5. Mean the two faces for each station. The angle between any two stations = difference of their mean directions.
6. For more accuracy, repeat the whole round (a set) with different initial readings (for 2 sets, start at 0° and 90°), and take the mean of the angles of the different sets.

         B       C
          \     /
           \   /
   A ------ O          RO = A (reference object)

Field notes (example)

The angles are AOB = 100°40′20.0″, BOC = 60°40′10.0″. Closing reading on A differs from the first by 0°00′05.0″, which is small enough to be accepted, so no adjustment is needed.

Assumption: the title says ABCDEA, but only four instrument stations (A, B, C, D) are given, so the traverse is taken as the closed quadrilateral ABCDA with angle sum 360°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 360°02′50″. Angular error = 360°02′50″ - 360°00′00″ = +0°02′50″. Correction per angle = -0°00′42.5″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 110°24′52.5″, B = 90°15′37.5″, C = 89°07′37.5″, D = 70°11′52.5″.

Assumption: closed traverse ABCDA with four stations; angle sum must be 360°. Readings are booked in the standard field book form (instrument station, target station, face L, face R, mean direction, angle).

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 360°02′55″. Angular error = 360°02′55″ - 360°00′00″ = +0°02′55″. Correction per angle = -0°00′43.75″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 110°24′56.25″, B = 90°15′31.25″, C = 89°07′36.25″, D = 70°11′56.25″.

Assumption: closed traverse ABCDA; angle sum must be 360°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 369°23′20″. Angular error = 369°23′20″ - 360°00′00″ = +9°23′20″. Correction per angle = -2°20′50″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 112°05′00″, B = 98°15′00″, C = 76°48′15″, D = 72°51′45″.

Note on the data: the angular error computed from the readings as printed is about 9°23′, which is far larger than any acceptable error for a four-sided traverse (a few tens of seconds). So there is a recording or printing mistake in the readings (for example in the readings 204°25′40″ and 190°36′10″). The method and the equal-correction adjustment are shown above on the values as printed; with correct data the same steps give a small error that is adjusted in the same way.

Assumption: closed traverse ABCDA; angle sum must be 360°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 369°03′00″. Angular error = 369°03′00″ - 360°00′00″ = +9°03′00″. Correction per angle = -2°15′45″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 117°09′50″, B = 88°00′30″, C = 86°52′50″, D = 67°56′50″.

Note on the data: with the face-left reading 209°25′40″ on B (face right 29°25′30″, as printed) the angle at A is 119°25′35″ and the sum is 369°03′00″, an error of about 9°, which is far too large for adjustment. The printed reading is probably a misprint for 200°25′40″ (face right 20°25′30″), the value used in the other papers with the same remaining readings. With 200°25′40″ and 20°25′30″:

• Angle at A = 110°25′35″; sum of angles = 360°03′00″; error = 0°03′00″; correction per angle = -0°00′45″.
• Corrected angles: A = 110°24′50″, B = 90°15′30″, C = 89°07′50″, D = 70°11′50″.

The fundamental lines of a transit theodolite are the imaginary lines and axes about which the instrument is built. They are:

1. Vertical axis - the axis about which the whole upper part of the instrument rotates horizontally.
2. Horizontal (trunnion) axis - the axis about which the telescope rotates in the vertical plane.
3. Line of collimation (line of sight) - the line joining the intersection of the cross hairs to the optical centre of the objective and its extension.
4. Axis of the plate level (bubble line) - the tangent to the plate bubble at its centre; it is horizontal when the bubble is central.
5. Axis of the telescope level (altitude bubble) - the tangent to the altitude bubble at its centre, used for levelling the line of sight.

Relationship (condition) between them

• Plate level axis $\perp$ vertical axis
• Line of collimation $\perp$ horizontal axis
• Horizontal axis $\perp$ vertical axis
• Altitude bubble axis $\parallel$ line of collimation

Minimising errors when these relations are violated

If the errors are large, the adjustments are corrected by the permanent adjustment procedure.

Temporary adjustments

These are done at every set-up before observation:

1. Setting up (centering): fix the theodolite on the tripod and bring the plumb bob (or optical plummet) exactly over the station mark.
2. Levelling up: with the foot screws and the plate levels, make the vertical axis truly vertical. Set the plate level parallel to two foot screws, centre the bubble; turn 90 degrees and centre it with the third screw; repeat until the bubble stays central in all positions.
3. Elimination of parallax: focus the eyepiece until the cross hairs are sharp, then focus the objective on the object so that there is no apparent movement between the image and cross hairs when the eye moves.

Fundamental lines of a theodolite

• Vertical axis - about which the instrument rotates horizontally.
• Horizontal (trunnion) axis - about which the telescope rotates vertically.
• Line of collimation - the line of sight through the intersection of the cross hairs and the optical centre of the objective.
• Axis of the plate level and axis of the altitude (telescope) level.

            line of collimation
         ----------------------->
   telescope \
           ---O--- horizontal axis
              |
              |  vertical axis

In adjustment: plate level axis $\perp$ vertical axis; line of collimation $\perp$ horizontal axis; horizontal axis $\perp$ vertical axis; altitude level axis $\parallel$ line of collimation.

Errors eliminated by face observations

Face observation means observing a point with the telescope in the face-left position (vertical circle on the left, FL) and again after transiting and rotating the telescope 180 degrees (face right, FR), and taking the mean.

Errors eliminated by taking both faces

Mechanism of elimination

1. Collimation error. On face left the line of sight deviates from the correct direction by a small angle $e$ to one side. After changing face, the deviation is on the opposite side, so one reading is too large by $e$ and the other too small by $e$. The mean is free of it.

  FL:   X-------- true line ---------> P
          \_ line of sight (+e)
  FR:   X-------- true line ---------> P
          /_ line of sight (-e)
        mean of FL and FR = true direction

2. Trunnion axis error. If one end of the horizontal axis is higher, the telescope sweeps an inclined plane. After changing face, the high end is on the opposite side, so the angular error is reversed in sign and is cancelled in the mean. (It is large only for steep sights.)

3. Index error of the vertical circle. If the vertical circle reads $+i$ on face left when the line of sight is horizontal, it reads the same error with the opposite effect on face right. The mean of the two vertical angles is free of $i$.

For horizontal angles, the readings of both verniers (A and B, 180 degrees apart) are also taken and averaged, which removes the error of eccentricity of the centre of the circle.

Temporary adjustments

1. Setting up and centering over the station mark with the plumb bob or optical plummet.
2. Levelling up with the three foot screws and the plate level: bring the bubble central with the level parallel to two screws, turn 90 degrees, use the third screw, and repeat.
3. Elimination of parallax: focus the eyepiece for sharp cross hairs, then focus the objective on the target, so that there is no movement between the image and cross hairs.

In an optical-reading theodolite the glass circles are read through a microscope; there are no open verniers. Two types are used.

(i) Optical-reading repeating theodolite

• It has an upper plate and a lower plate, each with its own clamp and tangent screw, as in a vernier repeating theodolite.
• The graduated circle can be clamped to the alidade (upper part) or to the lower part. So an angle can be repeated (added) several times on the circle without resetting the reading.
• The circle is read through a micrometer or optical-scale microscope; both sides of the circle are brought together in the field of view (reducing eccentricity).
• It is used for the repetition method as well as the ordinary method.

(ii) Optical-reading directional theodolite

• The horizontal circle is fixed relative to the base during observation; it has only one clamp and tangent screw for the alidade (no separate lower motion), so repetition is not possible.
• Directions to each station are read from the fixed circle, and angles are found from differences of directions. This is the direction (reiteration) method.
• The circle is turned by a circle-setting screw to change the zero between sets.
• It gives more accurate results (reading to 1 second or 0.1 second) and it is quick, so it is used in triangulation.

In the direction method with two sets, the zero setting of the circle is changed between the sets (e.g. 0° for set 1 and 90° for set 2) so that different parts of the circle are used. Each set has face left and face right observations.

Booking format (angle AOB, 2 sets)

Mean direction = [Face L + (Face R - 180°)] / 2.

Set I: A = (0°00′00″ + 0°00′20″)/2 = 0°00′10.0″, B = (62°24′40″ + 62°25′00″)/2 = 62°24′50.0″, angle = 62°24′40.0″.

Set II: A = (90°00′20″ + 90°00′00″)/2 = 90°00′10.0″, B = (152°24′50″ + 152°25′10″)/2 = 152°25′00.0″, angle = 62°24′50.0″.

$$\text{Mean angle AOB} = \frac{62°24′40.0″ + 62°24′50.0″}{2} = 62°24′45.0″$$

Answer: mean horizontal angle AOB = 62°24′45.0″. The two sets agree within 10 seconds, so the result is accepted.

Horizontal angles

1. Ordinary method (single angle). The theodolite is set at O and levelled. Set zero (or a reading) on A, bisect B with the upper clamp, read both verniers, change the face and repeat; mean the two. The angle AOB = reading on B - reading on A.

2. Repetition method - a single angle is added several times on the circle with the upper and lower clamps used alternately; the angle = total / number of repetitions.

3. Reiteration (direction) method - several directions from a reference object, angles from differences of directions, sets with different zeros.

        B
        /
   A --- O          horizontal angle AOB at O

Vertical angles

A vertical angle is the angle between the line of sight and the horizontal plane (angle of elevation or depression).

1. Set up the theodolite and level it. Set the altitude bubble central (or clamp vertical circle to zero if the instrument has an index bubble).
2. With the face left, bisect the target by using the vertical clamp and tangent screws. Centre the altitude bubble and read both verniers of the vertical circle.
3. Change face (transit) and repeat. The mean of the two results is the vertical angle; the index error is removed.

$$\text{Index error} = \frac{(FL + FR) - 360^\circ}{2}, \quad \text{Vertical angle} = 90^\circ - Z$$

where $Z = (FL - FR + 360^\circ)/2$ is the mean zenith distance, for a zenith-type circle.

                    target P
                  /
      horizontal /  alpha (angle of elevation)
   O ------------+

Methods of plotting a theodolite traverse

1. Angle and distance (polar) method: the stations are plotted one after another from the measured angles and lengths using a protractor or a set of tangents (the tangent method) and a scale. Easy, but errors accumulate; each angle is plotted from the previous line.
2. Coordinate method: the latitudes and departures of all the lines are computed, adjusted (Bowditch's rule), and the total coordinates of every station from a chosen origin are found. The stations are then plotted on the sheet by their coordinates, so the error does not accumulate. This is the more accurate method and is used for large traverses.

There is also the tangent method, where the angle is plotted using the tangent of the angle for greater accuracy than the protractor.

Errors eliminated by the repetition method

• Errors of reading (least count error), since the angle is added several times and the single reading error is divided by the number of repetitions.
• Errors due to imperfect graduation of the horizontal circle, since different parts of the circle are used.
• Error of eccentricity of the verniers (if both verniers are read and averaged).
• The usual errors eliminated by changing face (line of collimation, trunnion axis error) are eliminated when the repetitions are done on both faces.

It does not remove errors of centering, sighting (bisection) or levelling.

Assumption: the traverse ABCDA is a closed four-sided figure, so the sum of the interior angles must be 360°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 360°03′20″. Angular error = 360°03′20″ - 360°00′00″ = +0°03′20″. Correction per angle = -0°00′50″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 70°11′50″, B = 110°24′55″, C = 89°07′55″, D = 90°15′20″.

Missing VCR readings

Missing VCR. For a vertical circle graduated as zenith angle, the sum of the face-left and face-right readings of the same target is 360° when the instrument has no index error. So the missing reading = 360° - the given reading. The vertical angle is $90^\circ - Z$ (+ elevation, - depression).

Assumption: a closed traverse ABCDA (four stations), so the interior angles must add up to 360°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 360°01′10″. Angular error = 360°01′10″ - 360°00′00″ = +0°01′10″. Correction per angle = -0°00′17.5″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 96°24′12.5″, B = 98°08′22.5″, C = 81°03′42.5″, D = 84°23′42.5″.

Missing VCR readings

Missing VCR. For a vertical circle graduated as zenith angle, the sum of the face-left and face-right readings of the same target is 360° when the instrument has no index error. So the missing reading = 360° - the given reading. The vertical angle is $90^\circ - Z$ (+ elevation, - depression).

Method. The mean direction of each target is found from the two faces, mean direction = [Face L + (Face R - 180°)]/2.

The readings decrease from P to Q to R and back to P, i.e. the telescope was turned anticlockwise (the circle is graduated clockwise). So each angle = (previous mean direction) - (next mean direction). The three angles at O must add up to 360°, and the second observation of P closes the horizon.

Mean angles

$$\begin{aligned} \angle POQ &= 359°59′50″ - 294°30′05″ \;(+360^\circ) = 65°29′45″\\ \angle QOR &= 294°30′05″ - 137°54′30″ = 156°35′35″\\ \angle ROP &= 137°54′30″ - 0°00′30″ = 137°54′00″ \end{aligned}$$

Sum = 359°59′20″; required = 360°00′00″; error = -0°00′40″ (sum is less than 360° by 0°00′40″). The error is small, so it is distributed equally: correction = +0°00′13.33″ per angle.

Answer: adjusted angles POQ = 65°29′58.33″, QOR = 156°35′48.33″, ROP = 137°54′13.33″.

Method. The reiteration (direction) method: all directions are observed from the reference object A, and the horizon is closed on A again. Mean direction = [Face L + (Face R ∓ 180°)]/2. The angles are differences of the mean directions.

Check on the observations

For each target the difference between Face L and (Face R - 180°) should not exceed the least count (1′ = 60″). All differences are below 60″, so the observations are accepted. The closing reading on A must agree with the first reading within the least count.

Closing error on A = 0°00′10″, which is less than 1′, so the round is accepted.

Mean angles and adjustment

Sum of the three angles AOB + BOC + COA should be 360°. Distributing the closing error 0°00′10″ equally (correction -0°00′03.33″ per angle):

Missing VCR

FL + FR = 360° for each target.

Answer: AOB = 100°40′00″, BOC = 60°40′00″, COA = 198°40′00″ (to the least count of 1′); missing VCR: A (face R) = 294°15′00″, B (face L) = 91°05′00″, C (face L) = 59°20′00″.

Assumption: closed pentagon ABCDEA, so the sum of the interior angles is $(5-2) \times 180^\circ = 540^\circ$.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(5-2) \times 180^\circ$ = 540°00′00″. Observed sum = 539°57′40″. Angular error = 539°57′40″ - 540°00′00″ = -0°02′20″. Correction per angle = +0°00′28″ (equal for all 5 angles).

Answer: corrected horizontal angles: A = 128°47′38″, B = 102°06′48″, C = 108°52′58″, D = 91°00′28″, E = 109°12′08″.

VCR and vertical angles from A

For a zenith-angle vertical circle, FL + FR = 360°.

Answer: VCR (FR) to E = 294°49′30″; VCR (FL) to B = 62°34′20″; VA to E = 24°49′30″ (elevation); VA to B = 27°25′40″ (elevation).

Method. In each set the mean direction of each target is found from the two faces, mean direction = [Face L + (Face R - 180°)]/2. The angle AOB in the set is the difference of the mean directions of B and A. The mean of the two sets is the final angle.

(Face R - 180° for A in set I is 359°59′30″, i.e. -0°00′30″, and the mean of 0°00′00″ and -0°00′30″ is -0°00′15″ = 359°59′45″. For B in set II, 31°00′20″ + 180° = 211°00′20″.)

$$\text{Mean angle AOB} = \frac{121°00′25″ + 121°00′35″}{2} = 121°00′30″$$

The two sets differ by 0°00′10″, which is small, so the result is accepted.

Answer: mean horizontal angle AOB = 121°00′30″.

Assumption: closed traverse ABCDA (four stations), angle sum 360°. The readings are entered in the standard booking format below.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 360°02′55″. Angular error = 360°02′55″ - 360°00′00″ = +0°02′55″. Correction per angle = -0°00′43.75″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 70°11′51.25″, B = 89°08′01.25″, C = 110°24′51.25″, D = 90°15′16.25″.

Missing VCR readings

Missing VCR. For a vertical circle graduated as zenith angle, the sum of the face-left and face-right readings of the same target is 360° when the instrument has no index error. So the missing reading = 360° - the given reading. The vertical angle is $90^\circ - Z$ (+ elevation, - depression).

Method. Mean directions are obtained from the two faces, angles from differences of directions, the triangle sum is checked against 180°, the error is distributed equally, and the side AB is then found by the sine rule.

Check and correction

Sum of angles = 179°58′35″; required = 180°00′00″; error = -0°01′25″ (sum is less than 180°). The permissible error is about the least count times $\sqrt{3}$ = 1′ × 1.73 ≈ 1′44″, so the error is acceptable and is distributed equally: correction = +0°00′28.33″ per angle.

Span AB

In triangle ABC, the side AB is opposite to the angle C and BC is opposite to angle A.

$$\begin{aligned} \frac{AB}{\sin C} &= \frac{BC}{\sin A}\\ AB &= BC\,\frac{\sin C}{\sin A} = 58.232 \times \frac{\sin 36°00′08.33″}{\sin 54°38′28.33″} = 41.972\ \text{m} \end{aligned}$$

Answer: corrected angles A = 54°38′28.33″, B = 89°21′23.33″, C = 36°00′08.33″; span of the bridge AB = 41.972 m (about 41.97 m, since the instrument reads only to 1′).

Assumption: the title says ABCDEA, but only four instrument stations (A, B, C, D) are given, so the traverse is taken as the closed quadrilateral ABCDA with angle sum 360°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 360°03′10″. Angular error = 360°03′10″ - 360°00′00″ = +0°03′10″. Correction per angle = -0°00′47.5″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 110°24′52.5″, B = 90°15′32.5″, C = 89°07′47.5″, D = 70°11′47.5″.

Method. A direction theodolite (least count 6″) gives directions from a fixed circle. At D, the directions to C, B and A are observed on both faces. The mean direction of each = [Direct + (Reversed - 180°)]/2. All directions are then referred to C (the reference object) by subtracting the mean direction of C, which gives the abstract of average directions. Angles are the differences of the directions.

Abstract of average directions (at D)

The reduced average directions are: DC = 0°00′00″, DB = 83°06′48″, DA = 242°35′24″.

Average angles

$$\begin{aligned} \angle CDB &= 83°06′48″ - 0 = 83°06′48″\\ \angle BDA &= 242°35′24″ - 83°06′48″ = 159°28′36″\\ \angle ADC &= 360^\circ - 242°35′24″ = 117°24′36″ \end{aligned}$$

Check: 83°06′48″ + 159°28′36″ + 117°24′36″ = 360°00′00″ = 360°, so the horizon is closed.

Answer: average directions DC = 0°00′00″, DB = 83°06′48″, DA = 242°35′24″; average angles CDB = 83°06′48″, BDA = 159°28′36″, ADC = 117°24′36″.

Assumption: closed traverse ABCDA (interior angles), so the sum must be 360°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 360°00′40″. Angular error = 360°00′40″ - 360°00′00″ = +0°00′40″. Correction per angle = -0°00′10″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 60°42′10″, B = 98°55′30″, C = 82°12′10″, D = 118°10′10″.

Check. The three stations A, B, C form a triangle, so the sum of the three angles must be 180°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(3-2) \times 180^\circ$ = 180°00′00″. Observed sum = 180°00′02.5″. Angular error = 180°00′02.5″ - 180°00′00″ = +0°00′02.5″. Correction per angle = -0°00′00.83″ (equal for all 3 angles).

Answer: corrected horizontal angles: A = 30°00′04.17″, B = 59°59′54.17″, C = 90°00′01.67″.

(a) Horizontal angle BAC (instrument at A)

Mean direction of each line = [FL + (FR - 180°)]/2.

$$\begin{aligned} \text{Mean direction AB} &= \frac{0°00′00″ + 0°00′10″}{2} = 0°00′05″\\ \text{Mean direction AC} &= \frac{80°40′00″ + 80°40′20″}{2} = 80°40′10″\\ \angle BAC &= 80°40′10″ - 0°00′05″ = 80°40′05″ \end{aligned}$$

(b) Vertical angle P to Q (instrument at P)

Readings FL = 68°30′30″ and FR = 291°29′00″. The vertical circle is read as zenith angle (FL + FR should be 360°).

$$\begin{aligned} FL + FR &= 359°59′30″ \;(\text{index error present})\\ \text{Index error}\ i &= \frac{(FL + FR) - 360°}{2} = -0°00′15″\\ \text{Mean zenith distance}\ Z &= \frac{FL - FR + 360°}{2} = \frac{68°30′30″ - 291°29′00″ + 360°}{2} = 68°30′45″\\ \text{Vertical angle} &= 90° - Z = 21°29′15″ \;(\text{angle of elevation}) \end{aligned}$$

Answer: horizontal angle BAC = 80°40′05″; vertical angle of Q from P = 21°29′15″ (elevation), index error -0°00′15″.

Note: the vertical circle is taken as a zenith-angle circle (reading 0° at the zenith), which is the usual IOE convention.

(a) Horizontal angle ABC (instrument at B)

Mean direction = [FL + (FR - 180°)]/2.

$$\begin{aligned} \text{BA}: &\; FL = 0°00′00″,\; FR = 179°59′20″ \Rightarrow \text{mean} = 359°59′40″\ (= -0°00′20″)\\ \text{BC}: &\; FL = 160°20′30″,\; FR = 340°20′00″ \Rightarrow \text{mean} = \frac{160°20′30″ + 160°20′00″}{2} = 160°20′15″\\ \angle ABC &= 160°20′15″ - (-0°00′20″) = 160°20′35″ \end{aligned}$$

(b) Vertical angle P to Q (instrument at P)

FL = 61°30′00″, FR = 298°30′30″ (zenith-angle circle).

$$\begin{aligned} \text{Index error} &= \frac{(FL+FR) - 360°}{2} = \frac{360°00′30″ - 360°}{2} = 0°00′15″\\ Z &= \frac{FL - FR + 360°}{2} = \frac{61°30′00″ - 298°30′30″ + 360°}{2} = 61°29′45″\\ \text{Vertical angle} &= 90° - Z = 28°30′15″ \end{aligned}$$

Answer: horizontal angle ABC = 160°20′35″; vertical angle from P to Q = 28°30′15″ (elevation).

Note: a zenith-type vertical circle is assumed.

Method. The instrument is at A and four targets B, C, D and E are observed on both faces. Mean direction of each target = [F.L. + (F.R. - 180°)]/2. The angle between two targets = difference of their mean directions.

$$\begin{aligned} \angle BAC &= 62°10′20″ - 0°10′15″ = 62°00′05″\\ \angle CAD &= 135°20′25″ - 62°10′20″ = 73°10′05″\\ \angle DAE &= 250°30′30″ - 135°20′25″ = 115°10′05″ \end{aligned}$$

(Check: BAE = 250°20′15″ = 62°00′05″ + 73°10′05″ + 115°10′05″.)

Answer: BAC = 62°00′05″, CAD = 73°10′05″, DAE = 115°10′05″.