A Chapter on the Construction of Some Figures and Curves in Practical Plane Geometry Useful in Ornament
Definitions and names of figures from 1 to 13.
An Equilateral triangle is a triangle which has three equal sides. Fig. 1.)
An Isosceles triangle is that which has only two sides equal. Fig. 2.)
A Scalene triangle is that which has three unequal sides. Fig. 3.)
A Right-angled triangle is that which has a right angle. Fig. 4.)
An Acute-angled triangle is that which has three acute angles. (Fig. 5.)
A Parallelogram is a four-sided figure which has its opposite sides parallel. Fig. 6.)
A Rhombus is a four-sided figure which has all its sides equal, but its angles are not right angles. Fig. 7.)
A Lozenge is a square set angle-wise. Fig. 8.)
Note.—A square, an oblong, a rhombus, and a rhomboid are all species of parallelograms.
A Diamond is composed of two equilateral triangles set back to back. Fig. 9.)
All other four-sided figures are called Trapeziums. If one opposite pair of sides be parallel, and the other pair not, the figure is called a Trapezoid. Fig. 10.)
Polygons.—A Polygon is a plane rectilineal figure contained by more than four straight lines.
A Regular Polygon is that which has its sides equal, and its angles also are equal.
An Irregular Polygon may have unequal sides and unequal angles, or unequal sides and equal angles, or equal sides and unequal angles. In this chapter regular polygons are only treated of.
Polygons are named according to the number of sides or angles they may have. A polygon having
| 5 | sides is | a | Pentagon. |
| 6 | “ | a | Hexagon. |
| 7 | “ | a | Heptagon. |
| 8 | “ | an | Octagon. |
| 9 | “ | a | Nonagon. |
| 10 | “ | a | Decagon. |
| 11 | “ | a | Undecagon. |
| 12 | “ | a | Dodecagon. |
| 13 | “ | a | Tridecagon. |
| 14 | “ | a | Tetradecagon. |
| 15 | “ | a | Pentadecagon. |
| 16 | “ | a | Hexadecagon. |
| 17 | “ | a | Heptadecagon. |
| 18 | “ | an | Octadecagon. |
| 19 | “ | a | Nonodecagon. |
| 20 | “ | a | Bisdecagon. |
Figs. 11, 12, and 13 are self-explanatory.
Fig. 14 From a given point D without to draw Tangents to a given circle A B C.
Join E the centre of the circle D.
Bisect D E in F. With F as centre and F E radius describe the circle D B E cutting the given circle in A and B. Draw the required tangents from D to touch the given circle at A and B. N.B.—A tangent to a circle or arc is always at right angles to a radius drawn to the point of contact.
Fig. 15 To draw an Exterior Tangent to two given circles A B and C D K.
Join the centres E and F cutting the circumference of the larger circle at K. Bisect E F in G. From K in the line K F cut off a part K P equal to the radius of the smaller circle E B.
With centre G and radius K F describe a semicircle; with F as centre and radius F P describe a circle. The semicircle cuts this circle at H. Join F H, and produce it to C. At E draw E A parallel to F C. Join A C, which is the exterior tangent required.
Fig. 16 To draw an Interior Tangent to two given circles B E and F D.
Join the centres E and F. Bisect E F in G, and describe a semicircle on E F. From K on the larger circle mark off K J and E F equal to the radius of the smaller circle, and with F as centre and F J as radius describe an arc passing through semicircle at H. Join F H cutting the larger circle at C, and draw E A parallel to F H. The points of contact are A and C, through which the interior tangent is drawn.
Fig. 17 Within a given circle to describe any Regular Polygon—say a Pentagon.
Draw the diameter A F and divide it into the same number of parts as the required polygon is to have sides—in this case it will be five parts. To divide the diameter into the number of equal parts, draw a line A X any angle to A F. Set off any convenient measurement five times on this line. Join point 5 to F, and draw the lines 4, 4´, 3, 3´, &c., parallel to 5 F to meet the diameter. With A and F as centre and A F as radius describe arcs intersecting at L. From
L draw a line through the Second division on A F at point 2´ cutting the circumference at B. Join A B. This is the length of the side of the required polygon. Set off the length of the side A B around the circumference at C, D, and E. Join the points A, B, C, D, E to complete the required pentagon.
N.B.—A Regular Hexagon may be inscribed in a circle by setting off the length of its radius six times round the circumference, and joining the points.
Fig. 18 On a given line to construct any Regular Polygon,—say a Pentagon.
Produce the given line A B to R, and with B as centre and A B as radius describe a semicircle A C R. Divide the semicircle into as many parts as the polygon is to have sides—in this case five. Draw a line from point B to the second division point Q C. Bisect A B and B C to find P, which will be the centre of a circle passing through the points A B C. Mark off the points D and E, making the distances C D, D E, and E A each equal to A B. Join C D, D E, and E A to complete the required polygon.
Draw two diameters B F and H D at right angles to each other. Bisect angles H K B and B K D in the lines K A and K C. Produce the lines K A, K C, to meet the circumference at G and E. The eight points thus found on the circumference are joined to make the required octagon.
With each corner of the square as centres, and half the diagonal of the square as radius, describe arcs cutting the sides of the square at F, G, H, K, &c. Join these points to complete the required octagon.
Fig. 21 To describe a circle to touch two given straight lines A B and A C, one point of contact being given.
Bisect the angle B A C in A D. At C draw a perpendicular to A C, meeting A D at D. With D as centre and D C as radius describe the required circle.
Fig. 22 To inscribe a circle in a given triangle A B C.
Bisect any two of the angles as at B and C. The lines of bisection intersect at D. Produce B D to E. With centre D and distance D E inscribe the required circle.
Fig. 23 A square being given, to inscribe four equal circles each touching two others and two sides of the square.
Draw the diagonals and two lines parallel to the sides through the centre of the given square. Join the extremities of the latter lines to obtain the points 1, 2, 3, and 4. With these points as centres, and 1 E drawn perpendicular to C A as radius, inscribe the four required circles.
Fig. 24 A square being given, to inscribe four equal circles each touching two other and one side of the square.
Draw the diagonals and two lines through the centre parallel to the sides of the given square A B C D. Bisect any one of the angles made by a diagonal and one of the sides of the square, as at D. Produce the line of bisection until it meets the vertical centre line at point 1. With the central point O as centre and O 1 as radius, describe a circle to obtain the points 1, 2, 3, 4. These are the centres of the required circles.
N.B.—If the central portion made by the meeting of the four circles were removed, the remaining parts of the circles would form a figure known as the quatrefoil, a form common in architecture.
Fig. 25 To inscribe six equal circles in a given equilateral triangle A B C.
Bisect the angles of the given equilateral triangle as at E, and draw the bisection lines through to meet the centre of each side. Bisect the angle A B J to obtain the point D on C K. Through D draw G F parallel to A B, also F H and H G parallel to the sides of the triangle. With D as centre and D K as radius inscribe one of the required circles, and with the same radius and F, 2, H, 1, and G as centres inscribe the remaining circles.
Fig. 26 (1) Within a given circle to inscribe a hexagon. (2) Without the same circle to describe a hexagon. (3) Within the inner hexagon to inscribe three equal circles each touching each other and two sides of the hexagon.
(1) Mark off the length of the radius of the given circle B D F six times on the circumference as at D E F, &c. Draw the three diameters A D, B E, and G F, and produce them a little beyond these points. Join the points G, D, E, F, &c., by straight lines to produce the hexagon within the given circle. (2) Bisect the angle K O H, the line of bisection will cut the circle at point R. Through R draw H K parallel to B C. With O as centre and O H as radius describe a circle cutting the produced diameters at K, L, M, &c.
Join the latter points to produce the required hexagon without the given circle. (3) Join the points G, E, A. This will obtain the points 1, 2, 3 on the diameters. Draw 1, 4 perpendicular to G B. With 1, 4 as radius and 1 as centre describe one of the required circles. 3 and 2 are the centres of the other two required circles.
Fig. 27 Within a given circle to inscribe any number of equal circles, each touching the circumference and two other circles.
Divide the circle in the same number of parts as the number of circles required—in this case five. Draw the five radii. Bisect the angles B D A and A D C. Draw E F perpendicular to D A. D E F is a triangle any two angles of which bisect as at D and E. From point 1 thus obtained on D A and radius 1 A inscribe a circle. From D as centre and D 1 as radius describe a circle cutting the five radii in points 1, 2, 3, 4, 5. With the latter points as centres and 1 A as radius describe the remaining required circles.
Fig. 28 This problem is worked in the same manner as Fig. 27, seven circles being inscribed instead of five in a given circle.
Fig. 29 To inscribe a trefoil, or three equal semicircles having adjacent diameters in a given circle.
Divide the given circle into six equal parts by marking off the length of the radius six times on the circumference. From the centre D to these six points draw radii. Bisect any of the six sectors as at E. Draw E C obtaining F on one of the radials. On either side of F draw lines from it to meet the alternate radials perpendicular to B D and D C, and join their extremities, thus making the equilateral triangle 1, 2, 3. On the sides of this triangle describe the three semicircles required by using points 1, 2, and 3 as centres, and 2 F as radius. The completed figure is the trefoil, and the inscribed three semicircles have their diameters adjacent.
Fig. 30 To describe an equilateral triangle within and without a given circle.
Draw six radii dividing the given circle into six equal parts. Join their alternate extremities as at L M N. This makes the required equilateral triangle within the circle. Draw tangents to the circle at L M and N, or lines at right angles to L O, M O, and N O. Produce the latter radii to meet the tangents at A B C. A B C is the equilateral triangle without the circle.
N.B.—It will be seen that the triangle B A C is made up of four similar triangles each equal to L M N. Also, if six of the smaller triangles, as A L M, were placed around points A B and C a hexagon would be formed. This figure is very useful in designing geometrical and other repeating all over patterns in ornament.
Conic Sections
The figures known as the Conic Sections are the Ellipse, the Parabola, and the Hyperbola.
The Cone may have other sections in addition to these, such as the section through any point below the apex, on the axis, and taken parallel to the base; this would be a circle, and a section through the apex perpendicular to the base would be an isosceles triangle.
The Ellipse is the curve of the section made by a plane passing obliquely through a cone from side to side.
The Parabola is the curve of the section made by a plane passing through a cone parallel to one of its sides.
The Hyperbola is the curve of a section made by a plane passing through a cone parallel to its axis, or inclined at a greater angle to its base than its side, but not through its apex.
Fig. 31 The elevation of a cone is shown at A B C. A section through point X at right angles to the axis of the cone is a Circle. A section passing through and across the cone from point X, but not at right angles to the axis, is an Ellipse, as at X 1. A section through X parallel to the opposite side A C is a Parabola, as at X 2. A section through X parallel to the axis, as at X 3, or a section through X at any other angle greater than the angle made by the side and base, as at X 4, is a Hyperbola.
Figs. 32, 33, and 34 show the actual shape of the sections X 1, X 2, and X 3 respectively.
Fig. 32 In this figure the major or transverse axis of the Ellipse is equal to X 1. To find the minor or conjugate axis bisect X 1 (Fig. 31) in H, draw through it F G parallel to A B, drop a perpendicular from F to f, and describe the semicircle f h g. From H drop a perpendicular to A B, and produce it to h to meet the semicircle, k h is then half the length of the minor axis of the Ellipse, as C D. Divide A E into any number of equal parts, and A G into the same number. Draw from C lines through the divisions as 1, 2, 3 &c., and from D lines to 1´ 2´ 3´ &c. The curve of the required Ellipse will pass through the intersections of these lines, as at 1´´ 3´´ 5´´ &c.
Fig. 33 In this figure, the Parabola, the line C D is equal to X 2 (Fig. 31), while A B is twice the length of D 2 (Fig. 31). Divide G B into any number of equal parts, and join the points of the divisions to C. Divide D B into the same number of equal parts, and draw lines from the points of division parallel to D C to meet the similar numbered lines drawn from B G; through these meeting points the curve of the Parabola will be drawn.
Fig. 34 The only difference between the working of this figure—the Hyperbola—and the Parabola is that the lines which in the Parabola were drawn parallel to G B, are here drawn to a point E on C D produced, C D being equal to X 3 (Fig. 31). This point E is found by drawing the line from 7 on D B to E on C D produced, where C E equals twice X O (Fig. 31).
Fig. 35 To describe an Archimedean spiral of any number of revolutions—say three, the longest radius A B being given.
Divide the radius A B into three equal parts for the three revolutions. With B as centre and B A as radius describe a circle, and divide it into any number of equal parts—say eight, by drawing four diameters. Each of the three divisions on A B is divided into eight equal parts. With centre B and the point of each succeeding division as radius, describe arcs, meeting in following order the next nearest diameter as shown at arcs 1 1´´, 2 2´´, 3 3´´, &c. Through point 8 with radius B 8, the second division, describe a circle, and through point 16 with centre B describe a circle. In these two divisions arcs are drawn as described above for the division A 8, &c., to the next nearest diameter. The spiral is then drawn through the points thus formed on the diameters, which mark its path as at 1´, 2´, 3´, &c., until it ends in its centre at B.
Divide C F into 15 equal parts. With C as centre describe a circle A E B to form the eye of the volute, making the diameter 3⅓ of these parts. Bisect A C and C B in 1 and 4. On 1 4 draw a square, 1, 2, 3, 4. Produce the sides 1 2, 2 3, and 3 4 to G, H, and I respectively.
Divide 1 C into three equal parts. Draw lines parallel to 1 G through the points of division to P and L, which cut the line C 2 in the points 6 and 10. Through these points (6 and 10) draw lines to M and Q parallel to E H, cutting C 3 in 7 and 11. In the same way draw lines parallel to 3 I from 7 and 11 to N and R. The points 1, 2, 3, 4, 5, &c., will then form the centres of the series of quadrants which are to form the outer spiral that begins with the radius 1 F. To describe the inner spiral. A´ F´ in Fig. 36 (a) is equal to A F (Fig. 36). F´ S´ is made equal to the breadth of the fillet at the top F S. V´ F´ is drawn at right angles to F´ A´ and equal to C 1. By joining V´ A´ and drawing T´ S´ parallel to V´ F´, then T´ S´ is obtained which will be the length of half the side of the square for drawing the inner spiral. The method for obtaining the inner spiral is the same as for the outer.
Fig. 37 There is no geometric means of drawing a perfect catenary curve; at best we can only obtain it by an approximation in geometry. The curve is formed by suspending a chain from two points and pricking points along the curve of the chain. These points will mark the path of the catenary. In the accompanying figure three catenary curves are drawn from a chain suspended from points A and B.
Fig. 38—To draw a cycloid curve when the generating circle is given. In order to find the length of the line A B on which the circle rolls, and which must be the length of the circumference of the given circle, we must first find approximately that length by the following method. Draw the vertical diameter of the circle D C. Draw D M at right angles to D C, and make it three times the length of the radius of the circle; make an angle of 30° at E, and draw a line parallel to D M of any convenient length. The line E L making the angle of 30° cuts C B in L. Join M L. M L is the approximate length of half the circumference. Make A C and C B each equal to M L. Then A B is the length approximately of the circumference, drawn at right angles to C D on which the circle rolls. Divide now half the circle into eight equal parts, and draw a line from E S parallel to A B, and equal to M L. Divide E S into eight equal parts. From the points 1, 2, 3, &c., draw lines parallel to A C. With centres 1´, 2´, 3´, &c., and with radius E C, describe arcs cutting them at 1´´, 2´´, 3´´, &c. The curve A D, which must be drawn by free-hand, will then pass through these points. Complete the cycloid by drawing D B in a similar manner. The length A B can also be found approximately by dividing C D into seven equal parts, and taking A B = 22 of those parts.