Section II.—Geometrical Problems.
To bisect a given Straight Line.
—Let A B ([Fig. 1]) be the given line. From A and B, with any radius greater than 1⁄2 A B, draw arcs cutting each other in C and D; then the line joining C D will bisect the line A B as at E.
Fig. 1.
Fig. 2.
Fig. 1.
Fig. 2.
To erect a Perpendicular to a given Straight Line.
—Let it be required to erect a line perpendicular to the point B ([Fig. 2]) in the line A B. From any point C above the line, with radius B C, describe an arc as A B D; join A C and produce the line until it cuts the arc in D, and join D B; then will D B be perpendicular to A B.
Fig. 3.
To divide a Line into any number of equal parts.
—Let it be required to divide the line A B ([Fig. 3]) into five equal parts. From B, at any angle, draw B C, and on the line B C lay off five equal parts, 1, 2, 3, 4, 5; then take a set square E, and make one of the sides containing the right angle coincide with the points A and 5, and to the other side apply a straight-edge D; then by passing the set square along the edge of the straight-edge and drawing lines from the points 4, 3, 2, 1, through the line A B, we shall have the line A B divided into five equal parts through the points 1′, 2′, 3′, 4′.
To draw a Line making, with another line, a given Angle.
—Let it be required to draw a line making with the line A B ([Fig. 4]) an angle of 35°. From a scale of chords, which will be found on most scales supplied with a set of instruments, take off 60°; from the point A, with this distance for radius, describe an arc C D; lay off on this arc the distance of 35° taken from the same scale of chords; from A draw a line through this point. Then will the line A E make with the line A B an angle of 35°. The same result may be more readily arrived at by means of a protractor. If the centre point of the protractor be placed on the point A and its base made to coincide with the line A B, we can from its circumference prick off the distance of 35°, and a line drawn from A through the point thus found will make, with the line A B, the required angle of 35°.
Fig. 4.
Fig. 5.
Fig. 4.
Fig. 5.
To bisect an Angle.
—Let B A C ([Fig. 5]) be the angle which it is required to bisect. From A, with any radius, describe an arc cutting the lines A B and A C in D and E; from D and E, with the same or any other radius, describe arcs cutting each other in F, and from A draw a line through F; this line will bisect the angle as required.
Fig. 6.
Fig. 7.
Fig. 6.
Fig. 7.
To construct an Equilateral Triangle on a given base.
—Let A B ([Fig. 6]) be the given base. From A and B, with radius A B, describe arcs cutting each other in C; join A C and C B, which will complete the required triangle.
To construct a Triangle, the lengths of the Sides being given.
—Let it be required to construct a triangle whose sides shall be equal respectively to 6, 5, and 4. Lay down the base A B ([Fig. 7]), making it equal to 6 divisions of the scale; from A with radius equal to 4 divisions, and from B with radius of 5 divisions of the scale describe arcs cutting each other in C; join A C and C B, which will complete the required triangle.
Fig. 8.
To construct an Angle equal to a given angle.
—It is required to draw a line making with the line D E ([Fig. 8]) an angle equal to that contained by the lines B A C. From A, with any radius, draw an arc F G, and from D, with the same radius, draw the arc H I, and make H I equal F G; then a line drawn from D through I will make, with the line D E, an angle equal to the angle B A C.
Fig. 9.
To construct a Triangle, the length of the base and the angles at the base being given.
—It is required to construct a triangle whose base shall equal 1 inch, and the angles at the base be 30° and 45° respectively. Having made the base A B ([Fig. 9]) of the required length, make the angles at A and B of the required magnitude in the manner already described (see [Fig. 4]), and the continuation of these lines meeting in the point G will complete the construction of the required triangle.
Fig. 10.
To describe a Circle which shall pass through three given points.
—Let A B C ([Fig. 10]) be the points through which it is required to draw the circle. From each of these points, with any radius, describe arcs cutting each other in D and E; join the points D and E, and the point where these lines intersect will be the centre from which to describe the circle which will pass through the points A B C as required.
Fig. 11.
To draw a Tangent to a circle.
—I. Let B ([Fig. 11]) be the point from which it is required to draw the tangent. Draw the radius O B, and at B erect a perpendicular (see [Fig. 2]); then will the line B D be a tangent to the circle. II. It is required to draw a tangent from the point E in the same circle. Draw the radius O E extending beyond the circumference to F, and make E G equal to E F. From F and G, with any radius, describe arcs cutting each other in H and I; then a line drawn through these points will be a tangent to the circumference at E.
Fig. 12.
To find the Centre of a circle.
—From any point in the circumference, as B, ([Fig. 12]), describe an arc cutting the circumference in A and C, and from A and C, with the same radius, describe arcs cutting the first arc in two points; through the points of intersection draw lines to the interior of the circle, and the point O where these lines intersect will be the centre of the circle.
Fig. 13.
To draw lines which shall be Radii of a circle, the centre being inaccessible.
—Having laid off on the circumference of the arc, the distances apart of the radii, as A, B, C, &c. ([Fig. 13]), from each of these points, with radius greater than a division, describe arcs cutting each other as at a, b, c, &c., join A a, B b, C c, &c., and the lines so drawn will be radii of the circle as required.
Fig. 14.
To construct an Oval, the width being given.
—Draw the line A B ([Fig. 14]) equal to the width, and bisect A B by C D (see [Fig. 1]). From the point of intersection E, with radius E A or E B, describe the circle A C B F, and from A and B through F, draw the lines A G and B H. From A, with radius A B, describe the arc B G, and from B, with the same radius, describe the arc A H; also from F, with radius F G or F H, describe the arc G D H, which will complete the required oval.
Fig. 15.
To construct a Square on a given line.
—Let A B ([Fig. 15]) be the given line. At A erect a perpendicular (see [Fig. 2]), and from A, with radius A B, describe an arc cutting the perpendicular in C; also from B and C, with the same radius, describe arcs cutting each other in D; join C D and B D, which will complete the required square.
Fig. 16.
To construct a square that shall be a Multiple of any given square.
—Let A B C D ([Fig. 16]) be the given square, and let it be required to construct a square that shall contain 2, 3, 4, &c., times its surface. Draw the diagonal B C, then the square described on B C will be double the square A B C D. Lay off D E, equal to B C, and draw C E; then the square described on C E will be three times the square A B C D. In the same manner lay off D F, equal to C E, and the square described on C F will be four times the square A B C D; and so for any multiple of the square A B C D.
Fig. 17.
To construct a square that shall be equal to 1⁄2, 1⁄4, &c., of any given square.
—Let A B C D ([Fig. 17]) be the given square. On A B, as a diameter, describe the semicircle A G B, and erect at the centre E the perpendicular E G. Draw G B, which will be the side of a square equal to one-half of A B C D. Lay off B F, equal to one-fourth of A B, and erect the perpendicular F H; then the square described upon H B will be equal to one-fourth of A B C D. In the same manner a square may be constructed equal to any part of A B C D.
Fig. 18.
To construct a square that shall be in any Proportion to a given square.
—Let A B C D ([Fig. 18]) be the given square. It is required to construct a square which shall be to A B C D as 2 is to 5. Upon the side A B as a diameter describe the semicircle A F B, and divide the line A B into five equal parts. At the second point of division erect the perpendicular E F and join A F; the square described upon A F will be to the given square A B C D as 2 is to 5.
Fig. 19.
To construct, upon a given base, a Rectangle, which shall be similar to a given rectangle.
—Let A E F G ([Fig. 19]) be the given rectangle. It is required to construct upon the base A B, one that shall be similar to A E F G. Produce A E and lay off the given base from A to B; draw the diagonal A G and produce it indefinitely. Erect a perpendicular to A B at B, and from the point D where it intersects the diagonal produced, draw D C perpendicular to A F produced. Then A B C D will be similar to A E F G. All rectangles having their diagonals in the same line are similar.
Fig. 20.
To describe a regular Pentagon on a given line.
—Let A B ([Fig. 20]) be the given line. Bisect A B at C, draw C F perpendicular to A B, and make C D equal to A B. Draw A D and produce it indefinitely; make D E equal to half A B. From A as a centre, with A E as a radius, describe an arc cutting the perpendicular C D in F; and from A F and B as centres, with radius A B, describe arcs cutting each other in G and H; join A G, B H, F G and F H; then A G F H B will be the pentagon required.
Fig. 21.
To describe a regular Hexagon.
—With a radius equal to the length of one side of the required hexagon, describe a circle ([Fig. 21]), and set off the same radius round the circumference of the circle, which will be thus divided into six equal parts. Join the points thus found, and the required hexagon will be completed as A B C D E F.
Fig. 22.
To draw a Parabola, the base and height being given.
—Let C A ([Fig. 22]) equal half the base, and C D the height. From the point D draw D E parallel and equal to A C, and from the point A draw A E parallel and equal to C D. Divide D E and A E similarly, making the end E of A E correspond to the end D of E D. Through 1, 2, &c., in DE draw 1, 1; 2, 2, &c., parallel to D C. Join D to the several points 1′, 2′, &c., in A E. The parabola will pass through the points of intersections of these lines with the verticals drawn from D E to C A.
Fig. 23.
Fig. 24.
Fig. 25 a.
Fig. 25 b.
Fig. 23.
Fig. 24.
Fig. 25 a.
Fig. 25 b.
To draw an Ellipse.
—I. By means of a piece of string and pins. Place the diameters A B and C D ([Fig. 23]) at right angles to each other, and set off from C half the major axis at E and F; then will E and F be the two foci in the ellipse. Fix a pin at E and another at F; take an endless string equal in length to the three sides of the triangle E F C and pass it round the pins, stretch the string with a pencil G, which will then describe the required ellipse. II. From the centre O ([Fig. 24]) describe a circle of the diameter of the minor axis of the required ellipse. From the same centre, describe another circle with a diameter equal to its major axis. Divide the inner circle into any number of equal parts as 1, 2, &c., and through these points draw radii cutting the outer circle in 4, 3, &c. From 1, 2, &c., draw horizontals, and from 3, 4, &c., draw perpendiculars cutting each other in E F, &c.; the curve traced from C through the points C E F A, &c., will complete the curve of the required ellipse. III. Let A B ([Fig. 25 a]) be the major and C D the minor axis of the required ellipse. On any convenient part of the paper draw two lines F G, F H ([Fig. 25 b]) at any angle with each other. From F with the distance E C or E D, the semi-axis minor, describe an arc cutting the lines F G, F H, in I and K; and from F with the distance E A or E B, the semi-axis major, describe the arc L M. Join I M, and from L and K draw lines parallel to I M, cutting F G, F H, in N and O. From A and B ([Fig. 25 a]) set off the distance F N ([Fig. 25 b]) in points N′, and from these points as centres, with F N as radius, describe an arc of about 15° on each side of the major axis. From C and D ([Fig. 25 a]) set off on the minor axial line the distance FO ([Fig. 25 b]) in points O′, and from these points as centres, with radius FO, describe arcs of about 15° on each side of the axis C D. To obtain any number of intermediate points take a slip of paper ([Fig. 25 a]) and mark upon one edge, with a sharp-pointed pencil, 1, 3, equal to the semi-axis major, and 2, 3, equal to the semi-axis minor. If the slip of paper be now applied to the figure and moved over it in such a manner that the point 2 is always in contact with the major axis, and the point 1 in contact with the minor axis, the outer point 3 will describe a perfect ellipse, any number of points in which can be marked off as the “trammel” is moved into successive positions.
For this last method, which in practice is by far the best, we are indebted to Binns’ ‘Orthographic Projection.’
Fig. 26.
To construct a Semi-Elliptical Arch.
—The span A B ([Fig. 26]) and rise C D being given, divide C A and C B into any number of equal parts. Through the point D, draw E F parallel to A B, and from the points A and B erect the perpendiculars A E and B F. Divide A E and B F similarly to C A and C B. Produce C D and make C G equal C D. From D draw lines to the points 1, 2, 3, &c., in the lines A E and B F; also from G draw lines through the points 1, 2, 3, &c., in the line A B, and produce these lines until they cut those drawn from D to the corresponding numbers in A E and B F. Through the points thus obtained draw the curve of the ellipse.
Fig. 27.
To draw the Gothic Equilateral Arch.
—From the points A and B ([Fig. 27]), with radius A B equal to the span, describe the arcs B C and A C. By joining C to A and B we obtain an equilateral triangle from which this arch derives its name.
Fig. 28.
To draw the Gothic Lancet Arch.
—In this arch, the centres E and D ([Fig. 28]) from which the arcs are struck, are situate outside of and in a line with the points of springing A and B; thus it is constructed on an acute-angled triangle, as will be seen by joining C to A and B.
Fig. 29.
To draw the Gothic Obtuse Arch.
—This arch, called sometimes the Drop-Arch, is constructed on an obtuse-angled triangle; the centres E and D ([Fig. 29]) being situate below and within the points of springing A and B.
Fig. 30.
To draw the Gothic Tudor Arch.
—On the line of springing A B ([Fig. 30]), take any two points as F and G, so that A F is equal to G B. Draw F E and G D cutting each other on the bisecting line through C; from F and G, with radius F A or G B, describe the short arcs, and from E and D, with radius E C or D C, describe the arcs meeting in C.
Fig. 31.
To draw the Moorish Horse-Shoe Arch.
—The centres E and D ([Fig. 31]) from which the arcs forming this arch are struck, are situate above and within the points of springing A and B. One of the most graceful forms of this arch is obtained when the height of the points E and D above the line of springing and their distance from the bisecting line through C are equal to one-third of the span A B.
Fig. 32.
To draw the Gothic Ogee Arch.
—The most pleasing form of this arch is that constructed on an equilateral triangle, in the following manner. Having drawn the equilateral triangle A B C ([Fig. 32]), draw F G parallel to A B. Bisect the sides A C and C B and produce the bisecting lines to F G and H, which will complete the triangle F G H similar and equal to the triangle A B C. From H, with radius H A or H B, describe the arcs A E and B D, and from F and G, with the same radius, describe the arcs E C and C D.
Fig. 33.
Fig. 34.
Fig. 33.
Fig. 34.
To draw the Roman Cyma Recta and Cyma Reversa.
—Join A B ([Fig. 33]) and bisect A B in C. From the points C and B, with the distance B C, describe arcs cutting each other in E; and from A and C, with the same radius, describe arcs cutting each other in D; from D, with the same radius, describe the arc A C, and from E describe the arc C B. The projection of the upper end of the curve over the under, as F B, is generally equal to the height, A F, of the moulding. The same description applies to the Cyma Reversa ([Fig. 34]) letter for letter.
Fig. 35.
To draw the Gothic Trefoil.
—Having drawn the equilateral triangle A B C ([Fig. 35]), bisect the angles and produce the bisecting lines D E F which will bisect the sides of the triangle in G H I. From A B and C as centres, with radius A H or A I, equal to half the side of the triangle, describe the arcs K L M, and those concentric with them, and from the centre O of the triangle describe the outer circles and concentric arcs, which will complete the figure.
Fig. 36.
To draw the Gothic Quatrefoil.
—Draw the square A B C D ([Fig. 36]); bisect the sides of the square at I K L M and produce the bisecting lines to E F G H. From the angles A B C D of the square as centres, with radius A I or A M equal to half the side of the square, describe the arcs P N R S, and draw the outer concentric arcs. The circles, completing the figure, are drawn from the centre O of the square.
Fig. 37.
To construct the Gothic Cinquefoil.
—Having drawn the regular pentagon A B C D E ([Fig. 37]), bisect the angles and produce the bisecting lines to F G H I K, which will cut the sides of the pentagon in a, b, c, d, e. From A B C D and E as centres, with radius A a or A b, equal to one-half of the side of the pentagon, describe the arcs L M N P R, and draw the outer concentric arcs and those concentric with them. The circles are drawn from the centre O of the pentagon, as in the preceding example.