To design shell structures and architectural compositions (Civil Engineering), gear teeth, Todesid other machine components (Mechanical Engineering), a few curves, such as parabola, ellipse, hyperbola, spiral, involute, and cycloids are called for. Parabola, ellipse, and hyperbola are called conic sections they are obtained by cutting a right circular cone by a plane in different positions as shown
PRACTICAL APPLICATION
1. The shape of the reflectors designed to focus light to a point, will be an ellipse, when a circular rod or a circular tube is cut by a plane which is not perpendicular to the axis, the cut shape will be an ellipse. If the connecting rod of an engine is assumed to be infinietly long as compared with the crank, the path of a point on the connecting rod will be an ellipse. Civil engineering constructions such as concrete arches, spanning a water channel are in the shape of an ellipse.
2. Reflectors for parallel beams such as search lights are in the shape of a parabola. The path traced out by a projectile and the path of a jet of water issuing from a vertical orifice are parabolic. In civil engineering constructions such as suspension bridges, the upper ends of the panels when joined will be of parabolic shape.
3. When pressure multiplied by volume is constant, the expansion curve of a gas or steam is represented by the rectangular hyperbola.
PARABOLA
When a cone is cut by a plane parallel to one of the generators of the cone (l.e., the elevation of the cone) and inclined to the axis, the section obtained is called “Parabola”
Q.Construct a parabola when its focus is 60 mm from its directrix.
Solution:
1. Draw a line AB vertically which indicates directrix and mark a convenient point “O” through which the axis OC is drawn horizontally.
2. Mark the focus point “F” at a distance of 60 mm from “O” on the axis.
3. Bisect OF and the middle point is named as vertex V. From V take some points 1, 2, 3…..and draw perpendicular lines through them.
4. 0-1 as a radius and “F” as a centre, cut the perpendicular drawn through point 1 on both the sides of the axis,
5. Similarly 0-2, 0-3…..etc., as a radia and F as centre intersect the corresponding perpendiculars at point 2, 3…….etc., respectively.
6. Join all these intersecting points passing through V by a smooth curve.
ELLIPSE
When a cone is cut across by a plane passing through the axis at an angle and cutting the slant surface of the cone on the opposite sides, the section is called an “Ellipse” along with cutting line 2.2 in both the
Q.Construct an ellipse when the distance of the focus from the directrix is 60 mm and eccentricity is 2/3.
Solution:.
Step 1: Draw a vertical line AB which is known as directrix.
Step 2: Take a convenient point “O” and draw a horizontal axis and mark a point F_{1} 60 mm from O.
Step 3: Divide O*F_{1} into five equal parts (The equal parts must be the sum of numerator and denominator of the eccenstricity.)
Step 4: V₁ F_{1} as a radius V_{1} as centre draw an arc which intersect the perpendicular drawn at V_{1} and mark it as C.
Step 5: Join a line from O to C and extend it further till it meets with a line drawn at 45° from F_{1} . Mark this point as “D”.
Step 6: From D draw a perpendicular on the axis which meets at vertex V_{2}
Step 7: Mark some points 1, 2, 3….between V_{1} and V_{2} and draw vertical lines through them. Name 1′, 2′, 3′ etc., the intersections of OC extension with those perpendiculars 1 – 1′ as a radius Fas centre, cut the arcs on both the sides of perpendicular drawn at 1. Similarly 2 – 2′ , 3 – 3′ …….etc., as radii and F_{1} as centre intersect the arcs on both the sides of corresponding perpendiculars.
Step 8: Join all these points by a smooth curve passing through V_{1} and V_{2} . Name it as an “Ellipse”.
HYPERBOLA
The section obtained when the plane passing through 4-4 makes an angle less than the angle made by the generator with the axis of a right circular cone.
RECTANGULAR HYPERBOLA :When cutting plane parallel to the axis cuts the base and generator as in 5-5 of Fig. 5.20 (a) the section obtained is rectangular hyperbola.
Q.Draw a rectangular hyperbola given a point Pon it at a distance of 20 mm and 15 mm from the two asymptotes.
Solution:
Draw OX and OY, the two asymptotes. Mark the point P at 20, 15 mm. Through P draw AB and CD parallel to OY and OX respectively. Draw a number of radial lines through o to cut PA at 1, 2, 3 etc., and PD at 1′, 2, 3′ etc., Draw a vertical line through 1′ and a horizontal line through 1, both will intersect at the point P1. Similarly for each of the other radial lines, obtain the points P2, P3 etc., Connect P1, P2, P3 etc., by a smooth curve.
Now draw radial lines from O to cut PC at 4′, 5′, etc., and PB at 4, 5 etc., Draw a horizontal through 4 and a vertical through 4′ to get P4. Similarly obtain the other points Ps. P6 etc. Connect P1, P4, P5, etc., by a smooth curve as shown in Fig. 5.30. Application of Rectangular Hyperbola to The Boyle’s Gas Law: The rectangular hyperbola is importance to an engineer since the curve PV = constant, the Boyle’s Law, for varying pressure and volume of a gas can be represented by a rectangular hyperbola
