Ryaboshtan O.F., Milenin A.M., PhD

Kharkiv Petro Vasylenko National Technical University of Agriculture

Description of the portion of the surface of the second order of smoothness

Let the edges of the spatial four curvilinear polygon (portion of the surface) so that the opposite two of its sides depend on the parameter u (uÎ[0,1]), the remaining two - from v (vÎ[0,1]). The equation of any line portion of the surface for the i- th coordinate xi (i=1,2,3) will be denoted by (u, v). Hand portion of the surface described by the functions (u, 0)³, (u, 1)³, (0, v) ³, (1, v) ³     i=1,2,3.

The equation of the surface portions of the two-dimensional contours for the   i-th coordinate in the form (index coordinates omitted):

,    (1)

where L(u) – row matrix.

   ,                                     (2)

where li – transients features that ensure compliance with the conditions specified differential;

L'(v) – a column matrix, the transpose of L(u), its function  li(v) depend on v.

The structure functions (u, 0)v and (u, 1)v defines the direction of the transverse shear along the edges (u, 0) and (u, 1), that is tangent to the lines connecting corresponding points of these edges. For u = 0 their values must match the values of the derivative along the edge (0, v) respectively, for v = 0 and v = 1. Therefore, they can take the following:

,                                  (3)

The function l1(u) has l1(0) = 1, l1(1) = 0 and zero values of their derivatives up to and including second at u = 0 and u = 1, ensures the fulfillment of the requirements of the harmonization of the function at the nodes of portions of the surface. After a similar reasoning, we have:

                                    (4)

Here , , ,  – values of the derivatives of  u functions (u, 0), (u, 1) in the relevant portions of the surface nodes; , , ,  – values of the derivatives v functions (0, v), (1, v) in the same units.

Similarly we define the function:

    ,                           (5)

where , , ,  – the value of the second derivatives of (0, v), (1, v) in the nodes of portions of the surface;

, , ,  – the values of second derivatives of u the functions (u,(u,0),1) in the same units.

Differentiating (3) to u, we have:

,                                               (6)

that gives . Similarly, we have the vanishing of the mixed derivatives , , , , , ,  a portion of the surface sites. In general, the function of the transverse first and second derivatives must be chosen with an obligatory condition of compatibility of their values at the nodes, namely:

,                 

         ,                                   (7)

,               

Functions (3), (4) and (5) allow a docking surface portions of the second order with respect smoothness at the boundaries of portions at zero values of the mixed derivatives in the nodes. Consider, how is the boundary conditions. We write equation (1) schematically as follows:

     (8)

notation means that the equation is the sum of the terms of the matrix, multiplied by appropriate factors l³(u) (left) è l³(v) (bottom). We set u = 0. Then l1(0)=1, the remaining l³(0)=0, ³ = 2¸6, which highlights in the second row of the matrix, all of whose members, except (0, v), respectively members of the first row (for u = 0) with opposite sign, that is, in summation, they cancel out, regardless of the values of l³(v), ³ = 1¸6. We have (0, v) = (0, v). The value of u = 1 singles out the third row of the matrix, whose members are reduced in the summation of the members of the first line, which leads to (1, v) = (1, v).

The mutual reduction of members of the rows and columns has been possible because of the above mentioned structure functions "transverse" first and second derivatives at the boundaries of portions of the surface.

Differentiating (8) for u. Function l³(u) for u are replaced by their derivatives l³`(u), functions in the first row of the matrix will be replaced by their derivatives: ,, …, .

The equality u = 0 selects the fourth row of the members of which cancel each other with members of the first line so that we have =. Similarly u = 1 yields  = . For v = 0 is allocated the second column, whose members, except for , being reduced to the members of the first column, so =. For v = 1 we have=.

We define second mixed derivative, obtained by differentiating the above equation by v. The members of the line l³(v) are replaced by their derivatives l³`(v), the function will be the first column , , …, .

If u = 0       l3`(0) = 1, the remaining l³`(0) = 0, What distinguishes the fourth row of the members of which cancel each other with members of the first row. We have =. When v = 0           l3`(0) = 1, l³`(0) = 0. In a joint review of the first and fourth columns have =. Similarly, the remaining boundary conditions satisfied by the second mixed derivative.

The mixed derivative  portion of the surface in the space of its nodes, in general, will have zero values. In this equation (8) unit (4´4) members of the matrix in the lower right part (the value of the mixed derivatives , …) can be replaced by zeros. If the terms of the problem requires that the mixed derivatives of the parameters (to determine the coordinates õ³) have zero values at the nodes of the surface portion, then in equations (4) and (5) must befunction la replaced by another function, ensuring fulfillment of the given values of the mixed derivatives.

The above method of describing the portions of the surface of the second order of smoothness allows one to successfully describe interscapulum gas turbines, as a description of the boundary surfaces of portions of space.