--------------------_Geometrical Contructions 4.39
o
ParabolaFig. 4.56 Construction of a Parabolasuch that OV = Scm. Now OV is the abscissa.
- Construct the rectangle PQRS such that PS is the double ordinate and PQ = RS =
VO(abscissa). - Divide PQ and RS into any number of (say 8) equal parts as 1, 2, ... 8 and 11 21 .... 81
respectively, starting from P on PQ and S on SR. Join 1,2, ... 8 and 11,21 .... 81 with V. - Divide PO and OS into 8 equal parts as 1121 ...... 81 and 1'12\ ..... 8'1 respectively, starting
from P on PO and from S on SO. - From 11 erect vertical to meet the line VI at PI'
- Similarly from 21 , ... 8 1 erect verticals to meet the lines V2, .... V8 at P 2 .... Pg respectively.
- Also erect verticals from III 2\ ..... 8\ to meet the lines VI' .... V21 ..... V8^1 at PII .... P 21
'" .. P8^1 respectively. - Join P, PI' P 2 , ••••••• PI 7 .VI ..... Pll and S to represent the path ofthe ball which is a parabola.
Problem: Draw a parabolic arc with a span of 1000 nun and a rise of 800 mm. U.se rectangular
method. Draw a tangent and norm~l at any point P on the curve.
Solution: (Fig.4.58)
- Draw an enclosing rectangle ABCD with base AB = 1000 nun and height BC = 800 mm
using a suitable scale. - Mark the axis VH of the parabola, where V is the vertex and mid point ofline CD. Dividf
DV and AD into the same number of equal parts (say 4). - Draw a vertical line through the point II lying on the line DV. Join V with 1 tymg on the line