, (i = 0, 1, . . . , m), i,n i,0 Qi,1 = Pi,n n-1,n-1 (n
, (i = 0, 1, . . . , m), i,n i,0 Qi,1 = Pi,n n-1,n-1 (n ) (Pi,n – Pi,n-1 ), (i = 0, 1, . . . , m), ^ 0,n-1 (1 ) ) ( 1 ) ) – ( – ( Qi,2 = – n-2,n-2 (n-1 n-2,n^ 1)2n-1 Pi,n-2 n-2,n-2 (n^-) n-2,n^ 1)2n-1 0,n-2 ^2 ) 1n-1 ( two 0,n-2 two 1,n-1 ( two n-2,n-2 (n )n-1,n-1 (n ) n-1,n-1 (n )0,n-2 ( 1 ) n-1,n-1 (n ) – – ^ ^ ^ ^ ^ ^ two 0,n-2 (two )1,n-1 (2 )2 0,n-2 ( 2 ) 1,n-1 ( two ) 0,n-2 ( two ) 1,n-1 ( 2 ) n-1,n-1 (n ) ^ n-1,n-1 (n )0,n-2 (1 ) n-2,n-2 (n )n-1n-1 (n ) Pi,n-1 – ^ ^ ^ ^ ^ 2 0,n-2 ( 2 ) 1,n-1 ( two ) 0,n-1 ( 1 ) 0,n-2 ( two ) 1,n-1 ( two ) n-1,n-1 (n ) n – -1,n-1^ )n 1 Pi,n 0,n-1 ( 1 0,n-2 (^2 )1,n-1 (^2 )two (i = 0, 1, . . . , m) hold, where is any positive real quantity. Proof. When two adjacent C-B ier surfaces satisfy the G2 smooth continuity situation, the G0 and G1 continuity circumstances has to be happy at the joint first. In brief, two C-B ier(17)Mathematics 2021, 9,12 ofsurfaces will have to have a common boundary and Scaffold Library Description frequent tangent plane. For the G0 continuity situation of C-B ier surfaces, we’ve ^^ ^ ^^ ^ R1 (s, 1; 1 , . . . , n , 1 , . . . , m ) = R2 (s, 0; 1 , . . . , n , 1 , . . . , m ). Right after simplifying the above situations, the following boundary points Qi,0 = Pi,n are obtained. Now, for the smooth continuity condition of G1 , any two adjacent C-B ier surfaces have a prevalent tangent plane at a joint point with the frequent boundary, i.e., there’s a continuous tangential derivative in the boundary point, as well as the following situations need to be happy: R (s, 1; 1 , . . . , n , 1 , . . . , m ) R1 (s, 1; 1 , . . . , n , 1 , . . . , m ) t 1 s ^^ ^^ ^ ^^ ^ ^ ^^ ^ = R2 (s, 0; 1 , . . . , n , 1 , . . . , m ) R2 (s, 0; 1 , . . . , n , 1 , . . . , m ). t s(18)By additional simplification based on [21], Equation (18) can be Benidipine custom synthesis simplified as follows: ^^ ^ ^^ ^ R (s, 1; 1 , . . . , n , 1 , . . . , m ) = R2 (s, 0; 1 , . . . , n , 1 , . . . , m ) t 1 t exactly where is any real constant. By calculating the above values, we’ve Qi,1 = Pi,n n-1,n-1 (n ) (P – Pi,n-1 ), (i = 0, 1, . . . , m), ^ 0,n-1 (1 ) i,n (19)(20)which are the values necessary by the continuous condition of smooth G1 within the s path. Similarly, for the continuity of G2 within the s path, the two surfaces need to possess the exact same standard curvature at typical boundary [22,23] and satisfy2 two ^^ ^ ^^ ^ R1 (s, 1; 1 , . . . , n , 1 , . . . , m ) = 2 2 R2 (s, 0; 1 , . . . , n , 1 , . . . , m ) t2 t 2 2 ^^ ^^ ^ ^^ ^ ^ ^^ ^ 2 f (s) R2 (s, 0; 1 , . . . , n , 1 , …, m ) f two (s) 2 R2 (s, 0; 1 , . . . , n , 1 , . . . , m ) ts s ^^ ^^ ^ ^^ ^ ^ ^^ ^ R2 (s, 0; 1 , . . . , n , 1 , . . . , m ) h(s) R2 (s, 0; 1 , . . . , n , 1 , . . . , m ) t s(21)where and are any genuine constants, and h(s) and f (s) are linear functions. For our comfort and helpful calculation, we take into consideration h(s) = f (s) = = 0. Therefore, Equation (21) might be simplified as follows: 2 2 ^^ ^ ^^ ^ R (s, 1; 1 , . . . , n , 1 , . . . , m ) = 2 2 R2 (s, 0; 1 , . . . , n , 1 , . . . , m ). two 1 t t By calculating the above values, we haveQi,two = – n-2,n-2 (n-1 )n-2,n-1 (n-1 ) n-2,n-2 (n-1 )n-2,n-1 (n-1 ) Pi,n-2 ^ ^ ^ ^ 0,n-2 (two )1n-1 (two )two 0,n-2 (two )1,n-1 (two )two n-1,n-1 (n ) (n )n-1,n-1 (n ) n-1,n-1 (n )0,n-2 ( 1 ) – – n-2,n-2 ^ ^ ^ ^ ^ ^ 0,n-2 (2 )1,n-1 (2 ) 0,n-2 (2 )1,n-1 (two )two 0,n-2 (two )1,n-1 (two )2 ^ n-1,n-1 (n ) n-2,n-2 (n )n-1n-1 (n ) n-1,n-1 (n )0,n-2 (1 ) Pi,n-1 – ^ ^ ^ ^ ^ 0,n-1 (1 ) 0,n-2 (two )1,n-1 (2 ) 0,n-2 (2 )1,n-1 (two )two n-1,n-1 (n ) (n ) – n-1,n-1 1 Pi,n ^1 ) ^2 )1,n-1 (2 )two ^ 0,n-1 ( 0,n-2 ((22)(23)(i = 0, 1, . . . , m)which satisfie.
