By Adrian Bowyer
Programming for special effects calls for a good number of easy geometric operations. the obvious solution to software those is usually inefficient or numerical risky. This publication describes the easiest methods to those straightforward systems, supplying the programmer with geometric recommendations in a sort that may be without delay integrated into this system being written. it really is without delay appropriate to special effects, but additionally to different programming initiatives the place geometric operations are required
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Additional resources for A Programmer's Geometry
1. 0)/TANT ELSE 56 Τ = 2 . -ACCY) TANT = THEN -XKJ/YKJ Τ = 3 . 0 + Τ = 3 . T2) THEN Neareat point ia on the Neareat point ia an arc ELSE endpoint ENDIF ENDIF If t h e nearest point is on t h e arc t h e n t h e distance from t h e point to t h e arc is simply. r = VC(x K - ν" +( y K " y ) j 2] " Γ ϋ A negative distance indicates that J is within t h e circle. 4 to find t h e χ and y coordinates corresponding to T1 and Τ2. The minimum of these two distances is taken as t h e answer. 2. a r c / s e g m e n t pair is quick This operation If t h e infinite line and t h e whole circle do not intersect, rejection of t h e If, however, t h e line s e g m e n t and arc a r e both short compared to t h e circle radius, many cases w h e r e t h e segment and t h e arc a r e quite distant from each other will remain υ η rejected until quite late in t h e computation.
ACCY) THEN A = —RK*XKJ*DENINV Β = -RK*YKJ*DENINV ELSE 30 J liea on circle ROOT = SQRT(ROOT) RKSIGN = RK Negate for RKSIGN other tangent A = (~YKJ*ROOT - RKSIGN*XKJ)*DENINV Β = (XKJ*ROOT - RKSIGN*YKJ)*DENINV ENDIF C = -(A*XJ + B * Y J ) ENDIF ENDIF If t h e coordinates of t h e tangent point a r e r e q u i r e d , they can be obtained from t h e a and b coefficients of t h e appropriate line: χ = χ y = y Κ Κ + ar + br Κ Κ 2 5 Tangents to a Circle Normal to a Line This problem always has two solution lines, one each side of t h e circle.
If t h e equation of t h e tangent line required is ax + by + c = 0 Then t h e coefficients a and b a r e obtained from: (y ( - ν - K - ν * *κ - ν w (χ T r ( y y } b = K K " j * Κ - χ ) 2 J " 2 + ( y K + (y κ - y ) (χ V 2 + (y K - 2] J (y κ - Γ /- κ 2 κ - V *<»K - * / * K (χ y y y / - r 2 ] K / and c can then be calculated from t h e fact that t h e tangent passes through J: 29 c = -ax J - by J T h e r e a r e normally two possible tangent lines, obtained by attaching a sign to r . Κ root is positive then the two tangents a r e obtained as follows: If the square If the point J lies on the circle the contents of t h e square root a r e zero, and t h e r e is only a single tangent If J is inside the circle t h e r e a r e no tangents, of course, and the root goes negative.
A Programmer's Geometry by Adrian Bowyer