The plat being drawn to a suitably large scale, the polygon may be reduced to a triangle, as mDn, and the area of the triangle as found from measurements on the plat may be employed to check the computation.

EXAMPLES.

129. Required contents and plats from the following notes:

3. 4 and 5. To be made by the student in the field.

130. Method of Radiation-It is sometimes a convenient method for area and plat to take the lengths and azimuths of lines from a single point, as A, to the corners of the field.

Thus, set the transit at A, turn on B, and note the reading; then on C. and note the reading, and so on around

to B again. Measure AB, AC, AD, etc. Compute the areas of the triangles, (Art. 75, Prob. 1), and add them.

QUERY.-How would the solution be affected by the point A being outside of the field? How would the triangles outside be rendered, relatively, negative?

EXAMPLES.

131. Required area and plat from the following notes:

3, 4 and 5. To be made by the student in the field.

STADIA AND GRADIENTER.

132. The Stadia, or Micrometer, is a compound crosswire ring or diaphragm, shown in Fig. 109, having three horizontal wires, of which the middle one is cemented to the ring as usual, while the others, bb and cc, are fastened to small slides, held apart by a slender brass spring hoop, and actuated

od

by independ

ent screws, dd, by which the

distance be

od

FIG. 109.

the two mov

able wires can

be adjusted to include a given space, as one foot on a rod one hundred feet distant. These wires will in the same

manner include two feet on a rod two hundred feet distant, or half a foot at a distance of fifty feet, and so on in the same proportion, thus furnishing a means of measuring distances, especially over broken ground, much more easily and even more accurately than with a tape or chain.

133. The principle of the stadia may be shown as follows:

a

Let LS be the object-glass of the telescope, AB = 1, portion of a rod held vertically in front of the object glass, and ab s, the image of AB as intercepted by the stadia wires bb and cc, (Fig. 109).

=

Let F be the focus of parallel rays, and C the center of the instrument, or the intersection of the plumb-line with the axis of the telescope.

Let f be the distance FO,f the distance do, and f, the distance OD.

and from similar triangles, f1: f2 :: s: r. (2)

Eliminating f, we obtain f2 =f-+f. (3)

S

That is, the distance of the rod from the object-glass equals the focal length of the lens multiplied by the ratio of the intercepted space on the rod to the space between the stadia wires, increased by the focal length of the lens.

To find the distance d from the center of the instrument to the rod, the distance CO = c must be added to ƒ1⁄2, giving

f2+ c = d = ƒ− +f+c. (4)

f

The quantities ƒ and c are constant for the same instrument. So also is s for any instrument carrying unadjustable wires.

Now, as the best avenue to the brain is by the way of the hands, the student will be best helped to understand the matter of the stadia by solving for himself the following problems:

134. Prob. 1 To find f.

SUGGESTIONS.-Focus the instrument upon some very distant object, for example the moon or a star, and measure the distance from the plane of the cross-wires to the plane of the object-glass.

135. Prob. 2. To find c.

SUGGESTIONS.-Focus the instrument upon an object at a distance of about 100 feet, and measure the distance from the center of the horizontal axis of the telescope to the object-glass.

136. Prob. 3. To find s.

SUGGESTIONS.-Set up and level the instrument over a point. Measure forward from the point a distance f+c, marking the point. From this point measure forward any convenient distance, as 400 or 500 feet, for a base. Level the telescope and direct it upon a leveling rod held vertically at the extremity of the base. Note carefully the space intercepted by the stadia wires.

Now, calling the base b, and the space on the rod r, we have, from formula (4),

Example. Given ƒ = 6 in., b = 400 ft., and r = 4.8 ft., to find s. Result, s 0.006 ft.

SCH, If the intercepted space on the rod should be required to be in a given ratio to the base, as 1 : 100, then must 8 0.01 f.

It is possible to set the wires so that any given reading of the rod shall correspond to any required length of base.

137. Prob. 4. To find the base corresponding to a given setting of the wires.

SUGGESTIONS.-Set up the instrument and measure from the plumb-line two distances, as 100 ft. and 1000 ft., as accurately as possible. Level the telescope and take carefully the wire interval at each distance. Do this four or five times, and take the mean of results as the value of r for that station. Substitute the values of d and the corresponding values of r in formula (4), forming, thus, two equations from which the values of the unknown f

quantities and (f+c) may be found.

The first of these quantities is the constant number by which any reading of the rod is to be multiplied to produce the corresponding base, and the second is the number to be added to the base in finding the corresponding distance.

=

Example. Given d 100, 1.25 and d 1000, 7 = 12.61, to find the base and distance for r = = 8.5.

Results, b = 673.20; d = 674.20.

138. Prob. 5. To make a stadia rod.

SUGGESTIONS.-Procure a rod 14 in. square at the end, or a narrow strip of board 12 or 14 feet in length, in the latter case stiffening the back by a piece fastened along the middle. Give the rod two good coats of white paint. When ready for use, set up the transit, measure from the plumb-line the distance f+c, and set a point. From this point measure forward 100 feet to a second point, at which have an assistant hold the rod. The rod should be