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The Theodolite is a semi-circle divided into 180°, with an index which turns about on its centre, and retains any situation given it, on which are two fights, called the moveable sights; there are also two other fights fixed on the diameter of the theodolite, which are called the fixed fights. Fig. 2. plate 4.
Sights are small pieces of wood or brass, having small holes or slits in them, to view the object through ;---They are fixed perpendicular to the plane of the theodolite, but parallel to the plane of the quadrant.
The geometrical square may be made of brass, wood, or any solid body, having equal fides and angles; from one of the angles, a thread is suspended, with a small weight at the end, so as to point always to the centre. The two fides opposite to the centre of suspension, are divided each of them into 100 equal parts; there is also an index, which, (when occasion serves), may be fixed to the centre of suspension, and is made fo as to turn round, and retain any fituation; on this index, are two fights. See fig. 3. plate 4.
Heights and distances are of two kinds, viz. accessible and inaccessible: accessible objects are houses, growing trees, &c. inaccessible ones are all mountains, celestial bodies, also houses and trees, in certain situations.
PROBLEM I. See Plate 4. fig. 58.
To measure accessible heights.
Let AB be a horizontal plane and BC a tower, whose height is required : From B, the foot of the tower, measure any convenient distance, 80 feet upon the horizontal plane AB. Suppose the tower to subtend an angle of 39° 49' from A. What is its height?
As cosine ang. elev. 39° 49'
11.70950 To the height of the tower 66.69 = 1.82408
A tower, surrounded by a ditch 40 feet broad: from the other side of the ditch, the tower subtends an angle of 53° 13'. Required the height of the tower, also the length of a ladder sufficient to scale the tower. See fig. 58. plate 4.
To find the height of the tower. Tofind the length of the ladder.
10.0000 is to the breadth of
is to the br. of ditch 40 1.60206 the ditch 40
1.60206 So is sec. elev. 53° 13' 10.22256 So is tan. el. 53° 13' 10.1 2631
To ladder 66.78 = 1.82462 To the height of the tower 53.5
EXAMPLE III. Plate 4. fig. 59.
From the top of a ship-mast 100 feet above the level of the water, I took an angle of depression of another ship's hull, 74° 15'; required the distance of the other fhip.
To the dift. 354.6
To measure inaccessible heights and diffances.
EXAMPLE I. Plate 4. fig. 60.
At the foot of a hill, I took an angle of elevation of its top, and found it to be 50° 42'. I then measured back 120 yards on the horizontal plane, and observed the angle to be 40° 12'. Required the perpendicular height of the hill.
N. B. When any side AB of the triangle ADB is produced,
the exterior angle DBC is equal to both the interior and opposite angles DAB, ADB ; therefore the angle ADB will
be 10° 30':
To find BD.
To find DC the height. Assine ADB=10°30' 9.26063 | As rad. 90
10.00000 is to AB 120 2.07918
is to BD 425
2.62839 So is fine an. A 40°12' 9.80987 So is fine DBC 50°42' 9.88865
To BD 425
11.88905 To the height 328.9 2.5:704 2.62842
EXAMPLE II. Plate 4. fig. 67.
I observed an object on the other side of a river, on a level with the place where I stood ; behind me was a regular declivity, which I might reckon a straight line.
I marked my station by the fide of the river, and measured back 170 yards, when I observed I was higher than the object. I took the angle of depression of the mark by the river side 42° 78', of the H
bottom of the object 72° 8', and of its top 78° 20'. Required the height and distance of the object.
Here, because the angle ABC is 42° 18' the angle BAC is 47° 42;
consequently, its supplement, the angle BAD will be 132° 18. And since all the angles of a triangle are equal to two right angles, and that the angle DBA is 29° 50', the remaining angle BDA will be 17° 52' Again, the angle CDE is a right angle, of which the angle BDC is a part; therefore, the angle BDE is 72° 8', and the angle at E 101° 40'; also the angle DBE will be 6° 12'.
To find the dift. of the object. .
To find BD. As fine ADB 17° 52' 9.48686 | As sine BDA 17° 52' 9.48686 is to AB 170 2.23045
is to AB 170
2.23045 So is fine ABD 29° 50' 9.69677 | fois fineBAD=132°18' 9.86902
Being on a horizontal plane, I took the angle of elevation of the summit of a hill, and of the top of a tower built upon it, and found them to be 48° 20' and 61°25'. I then measured back 150 yards, and found the angle subtended by the height of the tower above the plane to be 38° 19'. Required the height of the tower.
The exterior angle CBD, is equal to both the interior and opposite angles, CAB, ACB; but CAB is 38° 19'; therefore, ACB will be 23° 6': and since all the angles of a triangle are equal to two right angles, angle ABC will be 118° 35'. Or it is the supplement of the angle CBD; also angle BCD is 28° 35', and CEB will be 138° 20'.
To find BC.
To find the tower's height. As fine an. ACB 23° 6' 9.59366 As fine CEB 138° 20' 9.82269 is to AB 150 2.17609
is to BC 237 2.37475 So is finean. A 38°19' 9.79240
Sois fine CBE 13° 5' 9-35481
To BC 237
11.72956 2.37483 To the height of the
tower 80.7 190687
EXAMPLE IV. Plate 5. fig.2
From a window on a level with the bottom of a steeple, I took the angle of elvation of the top of the steeple 50° ; from another window, 20 feet perpendicular above the former, I took another angle of the top of the steeple 45° 15' Required the height and distance of the steeple.
Because the angle ACD is a right angle, of which the angle SCD=50° is a part, the angle SCA will be 40°, consequently, the alternate angle CSD will also be 40°. And fince the angle SAB is 45° 15', and the angle BAD a right angle : therefore, the whole angle SAC 135° 15', and the angle ASC 4° 45'.
To find CS.
To find the height of the steeplé. As sine ASC 4° 45' 8.91807 | Assec. ang. SCD 50° 10.19193 is to AC 20 1.30103
is to SC 170 2.23045 So is sineSAC 135° 15'9.84758 So is tan. SCD 50° 10.07619 11.14861
12.30664 To CS 170 2.23054 | To the height SD