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A Programmer's Geometry by Adrian Bowyer

By Adrian Bowyer

Programming for special effects calls for loads of uncomplicated geometric operations. the obvious solution to software those is frequently inefficient or numerical volatile. This booklet describes the simplest methods to those simple approaches, delivering the programmer with geometric suggestions in a sort that may be without delay included into this system being written. it really is without delay appropriate to special effects, but additionally to different programming projects the place geometric operations are required

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Example text

W e will now show how to cull most of t h e s e trivial cases by direct comparisons of t h e endpoints involving only subtractions. 7. The 51 effectiveness of tests of this sort depends on t h e orientation of t h e line segments in t h e coordinate system. For the method to be described we can only guarantee to reject the possibility of intersection when circles drawn with each line s e g m e n t as a diameter a r e disjoint Consider the lines KL and M N as above. different lines M N . e. not vertical or horizontal) cases, depending on t h e orientation of KL: 7 8 4 i 1 /.

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.

5. Unfortunately, it is not t h e n particularly easy to decide w h e t h e r t h e intersection of t h e unbounded lines lies within t h e segments. Instead w e d e t e r m i n e t h e parametric equations of t h e lines in such a way that t h e line KL has a p a r a m e t e r s running from 0 at Κ to 1 at L, and MN has a p a r a m e t e r t running from 0 at M to 1 at N. T h e solution of t h e set of simultaneous equations then gives t h e intersection point as: (X Ν s = (χ Ν (χ t = L - χ Μ Μ - ν\ • V -ν- - χ )(y Κ Μ - χ (Χ Ky Ν )(y M L - y - y M M Χχ Xx - χ ; M L κ - X ) κ V V - y Xx - χ ) Κ M V" - y (y N κ M Xx L - χ ) κ If t h e values of both s and t a r e in t h e range 0 to 1, t h e n the intersection is within both segments, and t h e actual coordinates can be found from e i t h e r of t h e parameters.

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