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GEOMETRICAL CONSTRUCTIONS
2.1 Lines and Angles
1. To bisect a straight line: o Draw a line AB of any convenient length o With A and B as centers and more than half of AB as radius draw arcs to intersect C and D, respectively. o Join CD, which is the perpendicular bisector of AB so that the line AB will be divided into two equal parts. 2. To bisect an angle: o Draw any convenient angle BAC. o With A as center and R1 as radius, draw an arc intersecting the sides of the angle BAC at D and E respectively. o With D and E as centers, and more than half of DE as radius (i.e. R 2 ), draw two arcs intersecting at F. o Join AF, which is the bisector of angle BAC. 3. To divide a straight line into any number of equal parts: o Draw a line AB o At B draw a line BD perpendicular to AB. o Hold one end of the scale such that nine parts of unit length, coincides with the line BD at point C. o Join AC and mark the nine parts on it. o Draw from these measured points lines parallel to BD; then line AB will be divided into nine equal parts. 4. To trisect an angle: o Let you draw angle BAC. o On one of the sides (AB) of the angle, measure AD, which is equal to unit length. o At D, erect a perpendicular DE to AC and draw DH parallel to line AC. o Take a graduated straight edge and hold it in such a way that the distance between E and H must be equal to twice AD and at the
same time the straight edge must pass through the vertex of the angle. o Draw a line AH. o Angle HAC is 1/3 of angle BAC.
5. To trisect a line: o Draw a line AB that is going to be divided into three equal parts. o Using the 30/60 deg triangle, at A and B draw angles of 30 degrees so that the two lines meet at O.. o Hold the T-square parallel to the line AB, and set the set square such that the edge of the set square passes through the vertex O as shown. o Now reverse the setsquare and repeat the process. o The line AB is divided into three equal parts such that AC = CD = DB. 6. To divide the area of a triangle into five equal parts: o Draw a triangle ABC. o Bisect AC. o With O as center and OA as radius, describe a semi circle. o Divide AC into five equal parts. o At points 1, 2, 3 and 4 draw perpendiculars that will intersect the semi circle at 1’, 2’, 3’, and 4’. o With C as center and C-1’ as radius, draw an arc intersecting at 1’’ line parallel to AB thorough 1’’. o Repeat the process so that the triangle ABC is divided into five equal parts. 7. To construct an angle by tangent method: 1 o (To draw an angle of some degree without using a protractor) o Draw a line AB of any convenient length. o At B draw a perpendicular BC on it that measures CB = AB* tanq; Where AB is the initial length, and q is the given angle. 2.2 Construction of Regular Polygons 1. To construct a triangle given the three sides: An angle equal to a given angle o Draw the angle BAC o Draw the line A’C’ of any convenient length o With A’ as center and R as radius, draw an arc cutting A’C’ at D’ o With D’ as center and R’ as radius, cut the former arc at E’
o Draw the perpendicular bisector ED. o At B, draw a perpendicular. o With B as center and BA as radius draw an arc intersecting the perpendicular at C. o Join AC o The perpendicular bisector DE will intersect the chord AC at the point 4 and arc AC at point 6. o Bisect the distance and mark point 5. o Step off the distance 4-5 equal to 6-7, 7-8, 8-9, 9-10 etc. on the perpendicular bisector. o If a hexagon is required, use point 6 as center and 6-A as radius and draw a circle. o With AB as radius and A and B as centers, describe arcs to locate I and F. o With I and F as centers and AB as radius, describe arcs to locate H and G. o The hexagon ABFGHI is constructed.
7. To construct a regular octagon given the distance across flats: 2 o Draw a circle with a given distance across flats as the diameter. o Draw the two tangents, one at the top and one at the bottom. o Draw vertical tangents using 45-degree setsquare and complete the regular octagon. 8. To construct any regular polygon given one side o Given the line LM, draw a semicircle with LM as radius o Divide the semicircle in to equal parts as the number of sides needed for the polygon o suppose the polygon is seven sided o Draw the radial line through the points1,2,3,4,5,and 6 o Point 2 ( the second division point) is always one of the vertices of the polygon and the line L2 is one side o Using point M as center and LM as radius, strike an arc across the radial line L6 to locate point N o Using the same radius, with N as center, strike another arc across L5 to establish O on L o
o Although the procedure may be continued with point O as the next center, more accurate results will be obtained if point R is used as the center for the arc to locate Q, and Q as the center for P 2.3 Construction of Curves used in Eng’g Drawing 2.3.1 Simple Curves and Arcs
1. To draw a circular arc of Radius R1 tangent to a given circular arc with R2 and a straight line o Given the lines and the circular arc with center O o Draw line CD parallel to AB at a distance R 1 o Using the center O of the given arc, and a radius equal to its radius plus or minus the required arc (R 2 +/-R 1 ), swing a parallel arc intersecting CD at P. P will be the center of the tangent circle. Mark the points of tangency T 1 and T 2 o Used in fillets and rounds on views of machine parts 2. To draw a circular arc of a given radius R 1 tangent to two given circular arcs o Given the circular arcs AB and CD with centers O and P, radius R 2 and R 3 respectively o Using O as center, R 2 +R 1 as radius, strike an arc parallel to AB. o Using P as center and R 3 +R 1 as radius, strike an intersecting arc parallel to CD o Mark point of tangency T 1 and T 2 that lie on the line of centers PS and OS 3. To draw circular arc of radius R tangent to two lines o Given the two lines AB and CD and radius R o Draw lines EF and GH parallel to the given lines at distance R o Mark the tangent points T 1 and T 2 that lie along the perpendiculars to the given lines through O o With O as center and radius R, draw the required arc T 1 T 2 o Used to draw fillets and rounds in machine parts 4. To draw a tangent through a point P on a circular arc having an inaccessible center o Given the arc PB, draw the chord PB
o Draw perpendicular lines from the points of divisions from the outer circle and horizontal lines from the divisions on the inner circle o The intersection of these points with A,B,C, and D will give the points on the ellipse o Using smooth curve join all these points to get the ellipse
2. Ellipse by four center method o Draw AB(major diameter) and CD(minor diameter) to intersect at O and join A and C o With OC as radius and center C, draw an arc to intersect AC at E o With AE as radius and center C, draw an arc to intersect AC at F o Bisect AF to intersect AO at G and CD produced at H. o Use G and H as centers for curves to form half of the ellipse o Make OJ equal to OH and OK equal to OG to give two more centers for the other half of the ellipse o Join JG, HK and JK o With centers H, J, G and K and radii HC, JD, GA and KB respectively, describe arcs to form the ellipse. o The tangent points of the four arcs are at points 1,2,3 and 4 3. Ellipse by intersecting lines method o Draw AB and CD (the major and minor axes, to bisect each other at point O) o Draw a rectangle EFGH of EH=AB and EF=CD o Divide AO and AE in to four equal parts o Join C to the points of divisions on AE o Join D to the point of division on OA and produce these lines to meet C 1 ,C 2 , and C 3 to give three points on the ellipse o Using horizontal and vertical ordinates from these points, obtain three points in each of the other three quadrants o Draw smooth curve through these points to get the required ellipse 4. Parabola inside a rectangle o Draw a given rectangle ABCD o Find the mid point and draw vv’ o Divide VB and BC in to four equal parts and number as show o Draw lines V-1, V-2, V- o Draw horizontal lines through 1’, 2’, 3’ o Intersection of V-1 with 1’ will give the point E o Similarly obtain points F and G
o Repeat the process for the upper half of the rectangle o Join all these points by smooth curve which will give you the parabola
5. Parabola given the distance from the focus to the directrix o Draw the directrix AB o Through the mid point O, draw a perpendicular line and locate the focus F at a given distance from the directrix o The mid point of OF is the vertex of the parabola and call it V o Select the point C on the line OF o With F as center CO as radius draw an arc intersecting the perpendicular line drawn through C, both at top and at bottom o Likewise select points randomly on the line OF and obtain the points of the parabola o Join all points by using smooth curve 6. Hyperbola o Given the foci F1 and F2 and the transverse axis AB o Using F1 and F2 as centers, and any radius R1>F1B, strike arcs o With these same centers and a radius equal to R1-AB, strike arcs intersecting the first arcs o These intersecting arcs establish the position of four symmetrical located points (P1,P2, P3, and P4) using a single pair of radii o Additional sets of four points are obtained by assuming a different initial radius at each time o Repeat these procedure as outlined, until sufficient number of points have been located to determine a smooth curve 2.3.3 Cycloidal curves: Cycloid, epicycloids and Hypocycloid Definition: Cycloid: -is a curve traced by a point on the circumference of a circle which rolls with out slipping along a straight line Epicycloid:- is a curve traced by a point on the circumference of a circle, which rolls without slipping on the outside of the other circle Hypocycloid:- is a curve generated by a point on the circumference of a circle that rolls in a plane on the inside of another circle a. Cycloid
1’ and the center of the rolling circle moves to the new position C1. Since point 1 is in contact with the base circle at 1’, the generating point P moves in the clockwise direction and lies on the arc drawn through 1, and also at a distance of r from the new center C o With the new center C1 and radius r draw an arc intersecting the line drawn through point 1 o Similarly with P 2 as center and with the same radius r, cut an arc drawn through point 2 at P 2 o Likewise obtain points P 3 ,P 4 , etc and draw smooth curve passing through those all points
3. Hypocycloid o the method used to draw a hypocycloid is similar to the method to draw cycloid above 2.3.4 Involutes, Archimedean Spiral, and Helix Definition: Involute curve: - is the spiral curve traced by a point on a chord as it unwinds from around a circle or a polygon Archimedes’ spiral: - is a plane curve generated by a point moving uniformly around and away from a fixed point or it is generated by a point moving uniformly along a straight line while the line revolves with uniform angular velocity about a fixed point Cylindrical Helix:- is a space curve generated by a point moving uniformly on the surface of a cylinder. The point must travel parallel to the axis with uniform linear velocity while at the same time it is moving with uniform angular velocity around the axis. The curve can be thought of as being generated by a point moving uniformly along a straight line while the line is revolving with uniform angular velocity around the axis of the given cylinder 1. Involute of a circle o Draw the circumference in to a number of equal parts o Draw tangents through the division points o Along each tangent, lay off the rectified length of the corresponding circular arc, from starting point to the point of tangency o The involute of the circle is the smooth curve through these points
2. Involute of a polygon o Extend the sides of the polygon as shown o With the corners as centers, in order around the polygon, draw arcs terminating on the extended sides o The first radius is equal to the length of one side of the polygon o The radius of each successive arc is the distance from the center to the terminating point of the previous arc 3. Archimedean spiral o Divide the given circle in to a number of equal parts (say 12), and draw radial lines to the division points o Divide the radial line in to same number of equal parts as the circle o Number the division points on the circumference of the circle beginning with the radial line adjacent to the divided one o With the center of the circle as center, draw concentric arcs that in each case will start at a numbered division point on the divided radial line and will end at an intersection with the radial line that is numbered correspondingly o The arc starting at point 1 gives a point on the curve at its intersection with radial line 1, the arc starting at point 2 gives an intersecting point on radial line 2, etc o The spiral is a smooth curve drawn through these intersecting points 4. Helix o Lay out the two views of the cylinder o Divide the circular view of the cylinder in to a number of equal parts (say 12) o The lead should be measured along the contour element and divided in to the same number of parts o The necessary points are founded by projecting a numbered point on the circular view to the division line of the lead that is numbered similarly o A smooth curve joining these necessary points yields the required helix
