But the ratio O a: OA is invariable, for the terms are of constant length, and therefore the ratio Om: OM is invariable.

On this property the use of the instrument depends. Let the point O be fixed at M ; attach a tracing-point to follow the outline of the figure P M Q, and at M a pencil; while the point M follows the line P M Q, the pencil at m will trace a line pmq similar to it. By varying the distances Aa and A B, always preserving however the above conditions, we may make the ratio Om: OM any value we please less than unity.

There are several parts necessary to prepare the instrument for use. The centre O (fig. 162) is formed by the extremity of a clasp which will glide along the ruler O A, and may be fixed at any point by means of a screw. The support S of the clasp is fixed to the board by three pins. The four rulers are of constant length, but the tracing-point is borne by a clasp which may be fixed at any point of the ruler am by a screw. By this arrangement we can vary at will the ratios

B

Fig. 162.

of the distances a m and A M, and at the same time satisfy the condition that the three points, O, m, and M, shall be in the same straight line. The instrument is

supported on three small wheels or castors, r,r,r. The lengths of the rulers usually vary from eighteen inches to a yard, according to the size of the designs to be reduced.

To measure the distance of an inaccessible object. 191. By means of the foregoing principles of proportion we may measure the heights and distances of inaccessible objects.

1st Method. Let AI be the distance required, I being the inaccessible object (fig. 163). Take a point B in the line A I, and any other point C, and fix a staff at each. Measure AC and BC and take these distances on the lines produced, thus obtaining the points, A' and B'. Find now a point I' in the line A'B' and also in the line CI. Then it is easily demonstrated that the triangle A'CI=ACI and A'I AI.

2nd Method. Take any point B, measure A B, and find a point A' in the line I B, such that A' B A B. Fix a staff at A, B, and A', and also at the point C, the middle of A A'. Now find the point O at the intersection of the lines of A I and B ̊C, and the point I' at the intersection of A B and A'O. It is evident that the figure IOI' is symmetrical about the line BO and A I' A I (fig. 164).

3rd Method. Measure any convenient distance AB, in the line PA, either to or from P; and in any convenient direction measure a distance BCA B. Take two cords each equal to AB or BC, and let one end of one of the cords be held to the ground at A, and one end of the other at C. Now let a person take the other ends of the cords and bringing them together at D, keep the cords stretched; the figure A BC D will be a rhombus. Fix a staff at E, the point

Fig. 164.

of intersection of the lines PC and AD. Then DEDC=AE: AP. The first three lines may be measured and the fourth found.

To measure a vertical height, the foot of which is accessible.

192. 1st Method. Plant a staff AB higher than the level of the eye at a point A. Walk back with a smaller rod, and plant it so that on looking over the tops C and B, the top of the tower H is found to be in the same straight line. Measure the distances ED, D C, between the rods and tower, and the difference DB of the rods. The proportion

E

H

Fig. 165.

A

B

CD: DB DE: EH, gives the height EH (fig. 165.)

2nd Method. Make a right-angled isosceles triangle of cardboard (fig. 166). From two angles, A and B, draw lines to the middle points of the opposite sides, determining by their intersection the point D. Through D draw a line parallel to one of the equal sides containing the right-angle, and place a pin through a point O in this line. If Fig. 166. the triangle be supported by the pin, one line A C will be vertical, and another A B horizontal. To find the height of a tower, spire, or tree, stand in such a position that on looking along BC when the triangle is supported by the pin, the top of the object is seen in the same direction. The height of the tower EH (fig. 165) above the level of the eye, is equal to the line C E, the distance of the observer from the foot of the tower.

Questions for Examination.

1. Define proportion.

2. Prove that

Ist. If parallel lines cut two concurrent straight lines, and divide one of them into equal parts, they will also divide the other into equal parts.

2nd. When parallel lines cut two lines not parallel, the parts intercepted on the one are proportional to those intercepted on the other.

3rd. In the above case the intercepted segments of the parallels are proportional to the distances of their extremity from the point of intersection of the straight line. 4th. The segments of two parallels intercepted by several transversals through the same point are proportional. 3. Through a given point on land to draw a line parallel to a given straight line.

4. To divide a straight line into a given number of equal parts.

5. What is meant by a plan of a plot of ground.

6. Describe the diagonal scale.

7. To find a fourth proportional to three given straight lines. 8. To find a third proportional to two given straight lines. 9. To divide a line into two parts, which shall have to each other a given ratio.

10. To divide a line into two parts, such that a rectangle contained by the whole line and one of the parts shall be equal to the other part.

11. Show how to reduce a drawing-
Ist. By the method of parallels.
2nd. By an angle of reduction.
3rd. By the four-pointed compass.
4th. By the pantograph.

Theorems and Problems.

1. Find a point M in the line A B, produced such that A M2 = МВХАВ.

2. If d be the length of a line B C, M a point dividing it into A M a BM

two parts such that

=

a

b

b

and M', a point in the line pro

a + b

b d a-b

3. To trace a straight line upon the ground between two given points when an obstacle intervenes.

4. Through a given point A to trace on the ground a line parallel to a given straight line B C,

Ist. When BC is accessible.

2nd. When B C is inaccessible.

5. Produce a given straight line to such a distance that the square on the produced part may be double of the square on the given line.

6. Produce a straight line to such a distance that the square on the whole line may be double of the square on the given line.

7. Divide a straight line into two parts, such that the square on the whole may be double of the square on one of the parts.

8. Divide a straight line into two parts, so that the square on one part may be double of the square on the other.

9. From B, one of the angles of a triangle A B C, a perpendi