its attached level, is truly horizontal. This is called the true 0 of the limb.

If the instrument be accurately constructed, and the parts have not been disarranged, this point is the 0 point of the limb. This, however, is easily ascertained by turning the limb till the o's correspond, and then examining if the upper level be truly horizontal. If not, direct the telescope to a distant and elevated object, and read the degrees on the vertical limb. Turn the vernier plate 180o, reverse the telescope, direct it a second time to the same point, and read the arc on the vertical limb. The half difference of these two readings, counted from the 0 point of the limb, in the direction of the greater arc read, gives the true 0 point of the vertical limb; that is, the point at which the o of the vernier stands when the line of collimation is horizontal.

Suppose for example, that we had directed the telescope to a point and found the 0 of the vernier to stand at 10° of elevation. If we now reverse the telescope, it ought to incline at an equal angle of depression. If then we turn the whole 180°, and then raise the depressed end of the telescope with the thumb-screw Z, until it is directed to the same point as before, the o ought to stand at 10°. If it shows a less arc, the true 0 is between the 0 of the limb and the first arc read; if a greater, it is on the other side, and the difference divided by two will indicate the exact 0 point. The half difference thus found is called the correction. When the true 0 falls between the marked 0 and the eyeglass, the correction is to be subtracted from the arc read, for angles of elevation, and added, for angles of depression; and the reverse when it falls on the other side. The eyeglass is supposed to be over the thumb-screw Z, as in the plate.

These preparatory steps being taken, let the axis of the telescope be directed to any point either above or below the plane of the limb, and read the arc indicated by the o of the vernier. To the arc so read apply the proper correction, if any, and the result will be the true angle of elevation or depression.

87. Having explained the preliminary principles, it only remains to apply them to the measurement of Heights and

PROBLEM I.

To determine the horizontal distance to a point which is inaccessible by reason of an intervening river.

88. Let C be the point. Measure along the bank of the river a hori

zontal base line AB, and select the stations A and B, in such a manner that each can be seen from the other, and the point C from both of them. Then measure the horizontal angles CAB and CBA.

B

Let us suppose that we have found AB=600 yards, CAB= 57° 35′ and CBA=64° 51'.

To determine the altitude of an inaccessible object above a given

horizontal plane.

FIRST METHOD.

89. Suppose D to be the inaccessible object, and BC the horizontal plane from which the altitude is to B be estimated: then, if we suppose DC to be a vertical line, it will re

C

Measure any horizontal base line, as BA; and at the extremities B and A, measure the horizontal angles CBA and CAB. Measure also, the angle of elevation DBC.

Then in the triangle CBA there will be known, two angles and the side AB; the side BC can therefore be determined. Having found BC, we shall have, in the right-angled triangle DBC, the base BC and the angle at the base, to find the perpendicular DC, which measures the altitude of the point D above the horizontal plane BC.

Let us suppose that we have found

BA=780 yards, the horizontal angle CBA=410 24',

the horizontal angle CAB=96° 28', and the angle of elevation DBC=10° 43'.

In the triangle BAC, to find the horizontal distance BC.

The angle BCA 180°-(41° 24′+96° 28')=42° 08′ = C.

In the right-angled triangle DBC, to find DC.

REMARK I. It might, at first, appear that the solution which we have given, requires that the points B and A should be in the same horizontal plane, but it is entirely independent of

For, the horizontal distance, which is represented by BA, is the same, whether the station A is on the same level with B, above it, or below it (Art. 74). The horizontal angles CAB and CBA are also the same, so long as the point C is in the vertical line DC (Art. 75). Therefore, if the horizontal line through A should cut the vertical line DC, at any point as E, above or below C, AB would still be the horizontal distance between B and A, and AE which is equal to AC, would be the horizontal distance between A and C.

If at A, we measure the angle of elevation of the point D, we shall know in the right angled DAE, the base AE, and the angle at the base; from which the perpendicular DE can be determined.

Let us suppose that we had measured the angle of elevation DAE, and found it equal to 20° 15′.

First: In the triangle BAC, to find AC or its equal AE.

In the right-angled triangle DAE, to find DE.

Now, since DC is less than DE, it follows that the station B is above the station A. That is,

DE-DC-283.66-218.64=65.02=EC,

which expresses the vertical distance that the station B is above the station A.

REMARK II. It should be remembered, that the vertical distance which is obtained by the calculation, is estimated from a horizontal line passing through the eye at the time of observation. Hence, the height of the instrument is to be

SECOND METHOD.

90. When the nature of the ground will admit of it, measure a base line AB in the direction of the object D. To do this, it will be well to A

place the theodolite at A, and range the chain staves by means of the upper telescope. Having measured the base, measure with the instrument the angles of elevation at A and B.

Then, since the outward angle DBC is equal to the sum of the angles A and ADB, it follows, that the angle ADB is equal to the difference of the angles of elevation at A and B. Hence, we can find all the parts of the triangle ADB. Having found DB, and knowing the angle DBC, we can find the altitude DC.

This method supposes that the stations A and B are on the same horizontal plane; and therefore can only be used when the line AB is nearly horizontal.

Let us suppose that we have measured the base line, and the two angles of elevation, and

AB=975 yards

found A=15° 36′

DBC 27° 29';

required the altitude DC.

First: ADB=DBC-A=27° 29'-15° 36'11° 53'.