(Reduce the figure to an equal triangle, vertex at the top corner, then divide base of triangle into seven equal parts, and join to vertex, transfer the points to the figure that falls outside it, by the principle of similar triangles.)

Miscellaneous Problems.-The following problems are inserted owing to their frequent occurrence in engineering drawing, and the value of the methods employed in admitting of extension to similar questions. It is not at all uncommon to require the meeting points of lines which meet off the paper, or to draw an arc of a circle when the centre is inaccessible.

PROBLEM XXVIII. (Fig. 36)-To draw an arc of a circle through three given points, when the centre is inaccessible.

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Let A B C be the three given points. Join A C, A B, and B C. With A and C as cen

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3

1+

tres, and radius A C, draw arcs as shown, and produce A B and CB to meet the arcs in D and E respectively. Mark a number of small distances, E 1, 12, 23, &c., along the arc, below E, and equal distances, D 1', 1' 2', 2′ 3′, &c., above D. cutting at the point F. Continue the construc

Join A to D 2', and C to E2 by lines Then F is a point in the required arc. tion for finding other points between A and B, and for points between B and C in a similar way by taking equal distances below D and above E.

(This method depends upon the fact that all angles in the same segment of a circle are equal (Euclid iii., 21). Since E2 and D 2' are equal arcs of equal circles, the angles D A 2' and EC2 are equal (Euclid iii., 27), also the angles A G F and C G B are equal, being opposite angles (Euclid i., 15), and, therefore, the third angle in each triangle, the angles AFC and A B C are equal.)

EXAMPLES.

EX. 34.-Draw a tangent to a circle from a point in the circumference without using the centre of the circle.

EX. 35.-Three points, A B C, are on the circumference of a circle, and are thus situated, A B=3·25′′, BC=2·75′′, A C=5·5′′. Draw the arc of the circle passing through the points without using the centre. (Woods and Forest, 1886).

EX. 36.-Draw an arc of a circle of large radius, and then find points in continuation of the arc in one direction without

using the centre of the circle. In the same circle, equal angles are contained by equal segments.

(On the given arc draw any chord, A B, and draw an angle, ABC, in the segment. At any other point, D, in the arc, set off the length AC to E, and then set off the angle ACB at E, and the angle CAB at D, these two lines will meet at F, which will be a point in the arc, for the triangle DEF will be equal to triangle C A B.)

PROBLEM XXIX.-To bisect the angle between two straight lines without using their meeting points.

Draw a parallel to each line inside the angle at an equal distance from each. These two parallels will meet at a point if drawn at a sufficient distance from the given lines, and will make an angle equal to that between the given lines. Bisect this angle in the usual way, and the line will bisect the given angle.

EXAMPLES.

EX. 37.-Draw any two lines inclined to each other but not meeting, and bisect the angle between them, assuming the meeting point to be inaccessible.

EX. 38.-Draw any two lines at an angle to each other but not meeting, and find a point between them, such that its distance from the two lines shall be as 3: 5, also a point outside the lines so that its distance shall be as 1 : 4.

PROBLEM XXX. (Fig. 37).-To draw a line through a given point which shall meet at the intersection of two given inclined lines, when that intersection point is inaccessible.

R

These conditions frequently occur in graphical solutions of roof stresses, when an inclined wind pressure is taken into account. In such a case the inclined resultant, R, of the downward forces is known, and the reaction at P which is vertical, that end being on rollers, also the point, Q, through which the other reaction must pass. Then, knowing that the directions of the three forces, PRQ, must pass through the same point, it is required to find the direction of the reaction Q.

Fig. 37.

B

C

Let Q be the given point, and P B and RA the given lines. Draw any triangle, PA Q, starting from the given point Q, and having angular points at A and Q

Then start from any other point, B, on the line P B, and draw a similar triangle, BC D, so that BD is parallel to PQ, BC to PA, and CD to AQ; thus, obtaining the point D, then the point D is in the line joining Q to the intersection of P and R. and the line QD can therefore be drawn.

EXAMPLES.

EX. 39.-Draw any two lines inclined to each other but not meeting, and mark a point, P, outside and a point, Q, inside the lines. Then draw lines through P and Q to meet in the intersection of the given lines, without using that intersection.

EX. 40.-Draw any two lines approaching each other but not meeting, and mark a point, P, between them. Then draw a circle to pass through P, and to touch the two given lines, without using their meeting point.

(First draw the line bisecting the angle, then draw any circle not passing through P, and touching the lines; next draw the line passing through P, to meet at the intersection of the lines cutting the circle in A. Join A to the centre of the circle, and through P draw a parallel to this line, meeting the bisecting line of the angle in a point which is the centre of the required circle.)

GENERAL PROBLEMS ON LOCI-PATHS OF POINTS IN LINKWORK-CONSTRUCTION OF CAMS.

THIS section deals with what is perhaps the most important part of practical plane geometry, when regarded as a part of engineering drawing. There are very few mechanical engineers who do not frequently require to trace out the paths of certain points in mechanisms, such as engine valve gears and straight line or parallel motions, or to determine the form of grooved or curved plates called "cams," such that a certain desired motion may be produced. The geometrical constructions employed consists of finding a number of points in the particular path, or in the curve of the cam, and then drawing as smooth a freehand curve as possible through the points, and the student should remember that the number of points found is entirely a matter for individual decision, and should be settled by a consideration of each particular case. As a general rule, there are certain important parts of the curves, such as where the directions

change, where it is advisable to find more points than at other parts. The first set of the following problems on lines and circles deals with conditions which cannot be met by the ordinary methods of constructive geometry, and should, therefore, be regarded as important.

It is customary to speak of the "locus" of a point rather than of the "path" of the point, locus being a mathematical term for the path in which any given point travels; as, for example, the locus of the centre of the crank pin of an ordinary engine is a circle, while the locus of a point on the piston is a straight line.

PROBLEM XXXI. (Fig. 38a).-To draw the locus of the centres of circles, touching a given circle, and passing through a given point—that is, to draw a curve, every point on which shall be equidistant from the circumference of the given circle and from the given point.

Let A be the given circle, and P the given point outside it. Join the point P to the centre of the circle by a line cutting the circle in B, and bisect the distance, BP, in C. Then C is evidently one point in the locus. Mark any point, 2, between C and B, and a point, 2', between C and P, so that C 2=C2'. With P as centre, and radius, P 2, draw an arc, and with A as centre and radius, A 2', draw a

second arc cutting the

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Fig. 38a.

first arc above and below the line AP at the points D and E. Then, as P2 B 2', the points D and E are evidently points in the required locus. Other points in the curve are found by marking other equal distances, as C3 and C3', on both sides of C, and proceeding as before. In drawing any circle from a point in the curve touching the given circle, and passing through the point P, the point should be joined to the centre of the circle and to P, to give the points of contact.

Notice that it is only necessary to draw the arcs at about where they intersect, also that it is better to take the first one or two points very near to C.

PROBLEM XXXII. (Fig. 386).-To find the centre of the circles touching a given line and two given circles—that is, to find a point which shall be equidistant from the line and circles

Let the circles have the centres A and B, and let CD be the given line. It is necessary first to find a curve equidistant from the circumference of the two circles, and then a second curve equidistant from one of the circles and the given line. These two curves will intersect in a point, which will be equidistant from the line and the circles, and will, therefore, be the centre of the required circle.

First find points in the curve equidistant from the two circles, exactly as in the last problem, by drawing arcs centres A and B through the points on A B as shown. Notice that when the two circles are of equal diameter, the curve is a straight line, and that when unequal, the curve bends towards the smaller circle. A point is a circle of indefinitely small radius.

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D

Fig. 386.

Next draw the curve equidistant from the circle B and the given line C D. Draw a perpendicular, B C, from the centre of the circle to the line, and bisect the distance CF in G. Then G is evidently one point in the curve. Mark equal distances on either side and proceed as before, noticing that the points required are the intersections of straight lines throgh 1', 2', 3', parallel to the given line, and circles through the points 1, 2, 3, drawn from the centre of the given circle.

EXAMPLES.

EX. 1.-Draw a straight line of indefinite length, and at any point, C, in it draw a perpendicular, CD, 13" long. Then draw a curve, such that all points on it shall be equidistant from the line and the point D.

(It will be seen later on that this curve is the mathematical curve known as a Parabola.)