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Curve and Area Computation Date: 02/04/2001. Circle through 3 points A B C O D E Angle BDO = Angle BEO = 90 Angle EDO = Angle DEO (isos. Tri.) Known:

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Presentation on theme: "Curve and Area Computation Date: 02/04/2001. Circle through 3 points A B C O D E Angle BDO = Angle BEO = 90 Angle EDO = Angle DEO (isos. Tri.) Known:"— Presentation transcript:

1 Curve and Area Computation Date: 02/04/2001

2 Circle through 3 points A B C O D E Angle BDO = Angle BEO = 90 Angle EDO = Angle DEO (isos. Tri.) Known: 3 angles, Line DE using SINE or COSINE rule to find OD/OE. Known Pt A, B, C Result: find Centre O, Radius OA, OB, OC.

3 Circle through 3 points n Point A, B, C are known coordinates. n Point D and E are mid-points on the chords. n N D = (N A + N B )/2; E D = (E A + E B )/2 n N E = (N B + N C )/2; E E = (E B + E C )/2 n Brg DO = Brg AB + 90; Brg EO= Brg BC + 90 n By brg-brg intersection, Coordinate of O –N O =[N D tan(Brg DO) - N E tan(Brg EO) - E D + E E ] / [tan(Brg DO) - tan(Brg EO)] –E O = E D + (N O -N D ) tan(Brg DO) –Radius =  [ (N O -N A ) 2 + (E O -E A ) 2 ]

4 Polygonal Area n By formula: n Area = 0.5 [(X 2 -X 1 )(Y 2 +Y 1 ) + (X 3 -X 2 )(Y 3 +Y 2 ) + …… + (X N -X N-1 )(Y N +Y N-1 ) +(X 1 -X N )(Y 1 +Y N )]

5 Curve Area n Sector of a circle –πr 2 (Included angle / 360); or –½r (arc length) where arc length = 2 πr(Included angle / 360) n Segment Area –(Sector Area - Triangle Area) where Δ Area = ½ab sinC


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