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Cases of clip a line segment-pq
Case 1: As we determine a new value of tE that is value of parameter t for any potentially entering (PE) point we select tmax as: tmax = max {tmax, tE}. The initial value of tmax = 0 is considered.
Case 2: As we find a new value of tL that is value of parameter t for any potentially leaving (PL) point then tmax value is updated by tmin = min {tmin, tL}. The initial value of tmin = 1 is considered.
At last when value of t for all edges are determined and if tmax < tmin then line is visible else not. As well as line is visible from
P + tmax (Q - P) to P + tmin (Q - P) that is
P + t (Q - P) hence tmax ≤ t ≤ tmin
tmax < tmin (draw line)
tmax < tmin (reject line)
i need help in my Engineering graphic assignment
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