Error Bounded O set

1 (t) then the tangent vector ^ T ips direction as can be shown by its dot product with ^ T. Rewriting equation 13 as ^ T (t) = (x 0 (t); y 0 (t)) + (?y 00 (t); x 00 (t))d k ^ Tk + (y 0 (t); ?x 0 (t))k ^ Tk 0 d k ^ Tk 2 and substituting it into ^ T (t) ^ T(t) = " (x 0 (t); y 0 (t)) + (?y 00 (t); x 00 (t))d k ^ Tk + (y 0 (t); ?x 0 (t))k ^ Tk 0 d k ^ Tk 2 # (x 0 (t); y 0 (t)) = (x 0 (t) 2 + y 0 (t) 2) + (?y 00 (t)x 0 (t) + x 00 (t)y 0 (t))d k ^ Tk = (x 0 (t) 2 + y 0 (t) 2) ? d k ^ Tk since the last term of ^ T (t) is perpendicular to ^ T(t). Using equation 10: ^ T (t) ^ T(t) = (x 0 (t) 2 + y 0 (t) 2) ? d k ^ Tk = 8 > < > : (x 0 (t) 2 + y 0 (t) 2) ? (t)(x 0 (t) 2 +y 0 (t) 2) 3 2 d p x 0 (t) 2 +y 0 (t) 2 = (x 0 (t) 2 + y 0 (t) 2)(1 ? (t)d); > 0 (x 0 (t) 2 + y 0 (t) 2) + (t)(x 0 (t) 2 +y 0 (t) 2) 3 2 d p x 0 (t) 2 +y 0 (t) 2 = (x 0 (t) 2 + y 0 (t) 2)(1 + (t)d); < 0: Since C(t) is a regular curve, T(t) is never zero and so is (x 0 (t) 2 +y 0 (t) 2) which is positive everywhere. Therefore for cases were the mathematical normal, N(t), coincides with the ooset normal, N o (t) or > 0 we get sign(^ T (t) ^ T(t)) = sign((x 0 (t) 2 + y 0 (t) 2)(1 ? (t)d)) = sign(1 ? (t)d): (15) Now for small (t) or relatively straight curve (1?(t)d) is positive. When (t) reaches 1 d the expression becomes zero or ^ T (t) = 0 since ^ T(t) is never zero. if (t) is bigger than 1 d the expression becomes negative or ^ T (t) ipped its direction. If < 0 …