The axis measures the computational theory of umbriferous and basic theorem and method

Of Applied Engineering, nanchang Insititute Of Aeronautical Technology)Abstract:SOme Researches Concerning The Calculation Of Fundamental Theorem Of Axonometric Projection Have Done In This Paper.

The Method For Calculating The Location And Motion Parameter Of Tetrahedron Based On Its Projection As Well As The Method For Calculating The Location And Motion Parameter Of The Projection Figure According To The Tetrahedron Have Been Offered.

Based On Such Calculations And Methods, a New Fundamental Theorem Of Axonmetric Projection Containing Quantitative Relation Has Been Established.

Key Words:TEtrahedron, projective Transformations, spining introductive wave is overcome - making waltz theorem is an axis measure umbriferous basic theorem, no matter countershaft measures umbriferous theory, still be the application that countershaft measures umbriferous theory, this theorem has important sense. Wave gram - theorem is Xuhuaerci (abbreviation, wave - make theorematic ① ) : What any is not degenerate is complete at 4 o'clock form, can be being regarded is the tetrahedral axis that an appearance already decided is measured umbriferous. Alleged and complete at 4 o'clock form it is to point to by the dot of 4 general positions (apical) the figure that reachs 6 acme to make even line place. 4 acme are absent to go up point-blank together at 4 o'clock form call blame degenerate at 4 o'clock form. Xie Yuju write " advanced descriptive geometry learns " lieutenant general wave overcomes - make waltz theorem specified as (abbreviation wave - make theorematic ② ) : Any is tetrahedral OK the axis is measured umbriferous the blame degenerate that already decided for an appearance at 4 o'clock form. The distinction of two kinds of afore-mentioned views is, before one kind is will tetrahedral make similar shift, after that one kind is to will be not degenerate at 4 o'clock form make similar shift, document [1] what prove in fact is latter, alternate according to likeness again principle infer is former. Look from qualitative angle, two kinds of afore-mentioned views do not have substaintial distinction, but look from mensurable angle, two kinds of afore-mentioned views are abhorrent. In fact, wave gram - make waltz theorem involve qualitative one side only, did not involve ration this on one hand. Research axis measures umbriferous and main theorematic and mensurable problem namely the problem that its are calculated is article research theoretically. The computational theory that the axis measures umbriferous and basic theorem will understand an axis to measure projection drawing to provide the basiccest academic basis for the computer. The applied value with wait for a respect to have fundamental is made in animation of modelling of robot vision, geometry, computer. The main problem that the article studies the issue of 1 article research has two, one is wave - the calculation that makes theorematic ① is theoretical, another is wave - the calculation that makes theorematic ② is theoretical. Wave of geometrical model attempt - the computational issue that makes theorematic ① is: The graph on will tetrahedral ABCD() make commutation of motion of dimensional rigid body and similar shift, make its arrive at ′ of D of ′ of C of ′ of B of positional A ′ , make ′ of D of ′ of C of ′ of B of tetrahedral A ′ is in some axis below umbriferous direction is measured umbriferous for complete at 4 o'clock form Abcd. Specific computation problem is: Foregone tetrahedral ABCD and complete at 4 o'clock form Abcd, beg parameter of rigid body motion and similar commutation coefficient. Wave - the computational issue that makes theorematic ② is: Will complete form Abcd made commutation of motion of dimensional rigid body and similar shift at 4 o'clock, make its achieve ′ of D of ′ of C of ′ of B of positional A ′ , make the axis that in some umbriferous direction issues measures tetrahedral ABCD umbriferous for complete at 4 o'clock ′ of D of ′ of C of ′ of B of form A ′ . Specific computation problem is: Foregone tetrahedral ABCD and complete at 4 o'clock form Abcd, beg parameter of rigid body motion and similar commutation coefficient. Campaign of dimensional rigid body is general dissoluble move for whirligig and translation. Countershaft is measured umbriferous for, parameter of motion of rigid body translation can not decide completely, after add other is restrained, translation motion parameter is obtained more easily, study the computational issue of parameter of rigid body whirligig for this article key. The commutation matrix that sets rigid body whirligig is: If beg a matrix R, can get each to decompose the parameter of rigid body whirligig below means conveniently. Be based on afore-mentioned considerations, might as well the A(0 that set 　 , 0, 0) , b(X1, 0, 0) , c(X2, y2, 0) , d(X3, y3, a(0 of Z3) 　 　 , 0, 0) , b(x1, 0, 0) , c(x2, y2, 0) , d(x3, y3, 0) 　 and similar commutation coefficient is K. 2 - allow theorematic computation theory and method 2.

1 - make theorematic ① 　 be below this kind of circumstance, complete form Abcd was not moved at 4 o'clock, because this is umbriferous direction cannot the plane that parallel is in at Abcd, namely planar Z=0, reason can set umbriferous direction to be V(l, m, 1) . Tetrahedral ABCD alternates ′ of D of ′ of C of ′ of B of positional A ′ arrives below action in dimensional whirligig R and likeness. Nod B ′ right now, c ′ , d ′ part of one's job has not been located in dot B, c, d and parallel at umbriferous direction V go up point-blank, can get accordingly: 　 of K(X3r21+Y3r22+Z3r13)-y3=Km(X3r31+Y3r32+Z3r33) of 　 of 　 of K(X3r11+Y3r12+Z3r13)-x3=Kl(X3r31+Y3r32+Z3r33) of 　 of 　 of K(X2r21+Y2r22)-y2=Km(X2r31+Y2r32) of 　 of 　 of K(X2r11+Y2r12)-x2=Kl(X2r31+Y2r32) of 　 of R21=mr31 　 of KX1r11-x1=KX1lr31 　 　 arranges afore-mentioned posture to be able to get: K(r11-lr31)=a1 　 　 (3-1)r21-mr31=0 　 　 (3-2)K(r12-lr32)=a2 　 　 (3-3)K(r22-mr32)=a3 　 　 (3-4)K(r13-lr33)=a4 　 　 (3-5)K(r23-mr33)=a5 　 　 (3-6) uses formula among them (3-1)(3-3)(3-5) can get: K(1+l2)=a12+a22+a42 　 　 (3-8) uses formula (3-2)(3-4)(3-6) can get: K2(1+m2)=a32+a52 　 　 (3-9) uses formula (3-1 ～ 6) can get: K2lm=a2a3+a4a5 　 　 (3-10) from type (3-8 ～ 10) can get: (3-11) among them 　 of B1=a2a3+a4a5b2=a12+a22+a42b3=a32+a52 　 (3-12) can get thereby: (3-13) 　 gets K of similar commutation coefficient and umbriferous direction V(l, m, after 1) , use formula (3-1)(3-2) and R112+r212+r312=1 can be gotten rotate the element in matrix R (R11, r21, r31) , recycle (3-3 ～ 6) and can beg a torsion coming back blast the other element in R, can beg a ′ of C of ′ of B of tetrahedral A ′ thereby the position of D ′ . From on can see, k of similar commutation coefficient has two solution, its absolute value is equal, the symbol is contrary. Umbriferous direction V(l, m, 1) also has two solution, this two solution about umbriferous plane (namely the plane that Abcd is in) symmetrical, can see torsion coming back thereby blast R will have 8 solution. Namely the position of ′ of D of ′ of C of ′ of B of tetrahedral A ′ will have 8 places. Be based on afore-mentioned analysises and seek solution, from qualitative with ration two respects consider, wave - make theorematic ① should complement for: What any is not degenerate is complete at 4 o'clock form, can be being regarded is an appearance already decided tetrahedral umbriferous. Similar commutation coefficient has two solution, its absolute value is equal the symbol is contrary; Umbriferous direction also has two solution, this two solution are symmetrical at umbriferous plane; After commutation tetrahedral have 8 places. 2.

2 make theorematic ② 　 set umbriferous direction to be V(l, m, n, ) . Below this kind of circumstance, complete the axis that 6 of ′ of D of ′ of C of ′ of B of form A ′ also are edge of tetrahedral ABCD each arris was measured at 4 o'clock umbriferous. Accordingly, the node that applies two letters to show two diagonally sides (if ′ of D of ′ of diagonally edge A is mixed,the node of ′ of B ′ C is E ′ (F ′ ) , the corresponding dot E of E ′ belongs to arris edge AD, the corresponding dot F of F ′ belongs to BC, as a result of 3 o'clock brief more umbriferous than be being measured in the axis rigid body moves and reaching a space is changeless below similar commutation, reason can get the dot E that gets photogenic correspondence of AD of edge of the arris on tetrahedral ABCD and BC from E(f) and F according to following condition. (′ of D of ′ of Aed)=a ′ E) ′ of C of ′ of =(AED)(bfc)=(b ′ F) =(BFC) sets E(X4, y4, z4) , f(X5, y5, z5) , v of umbriferous direction: (Basis of 4-1) 　 nods B ′ , c ′ , d ′ , part of one's job has not been located in dot B, c, d, and parallel at umbriferous direction V go up point-blank, can get Klx1r31-Knx1r11+nX1=0Kmx1r31-Knx1r21=0Klx2y31+Knx2r11-Kny2r12+nX2=0Kmx2r31+Kly2r32-Knx2r21-Kny2r22+nY2=0Klx3r31+Kly3r32-Knx3r11-Kny3r12+nX3-lZ3=0Kmx3r31+Kmy3r32-Knx3r21-Kny3r22+nY3-mZ3=0 via arranging: (4-2) among them 　 B3=nx3 of 　 of 　 of B2=nx2 　 of 　 of 　 of 　 of 　 of A10=my3b1=nx1 of 　 of 　 of 　 of A9=ly3 　 of 　 of 　 of 　 of A8=my2 　 of 　 of 　 of 　 of A7=ly2 　 of 　 of 　 of 　 of 　 of A5=lx3a6=mx3 of 　 of 　 of 　 of A4=mx2 　 of 　 of 　 of 　 of A3=lx2 　 of 　 of 　 of 　 of A2=mx1 　 of 　 of 　 of A1=lx1 　 　 , if Kr11, kr12, kr21, kr22, kr31, kr32 regards unknown number, use formula (4-2) can decide afore-mentioned unknown are worth exclusively. Use torsion coming back blast the orthogonal property of R, can get 　 (4-3) can be gotten thereby rotate the element in matrix R (R11, r21, r31, r12, r22, r32) . Recycle 　 (4-4) can be gotten rotate the other in matrix R 3 elements. From on can see, umbriferous direction V(l, m, n) is only one solution, and K of similar commutation coefficient has two solution, its absolute value is equal the symbol is contrary, rotate thereby matrix R also has two solution, namely complete ′ of D of ′ of C of ′ of B of form A ′ had two places at 4 o'clock. Be based on afore-mentioned analysises and seek solution, from qualitative with ration two respects consider, wave - make theorematic ② should complement for: Any is tetrahedral it is OK to have the axis is measured umbriferous the blame degenerate that already decided for an appearance at 4 o'clock form. Umbriferous way is only one solution, similar commutation coefficient has two solution, its absolute value is equal the symbol is contrary; After commutation complete at 4 o'clock form have two places. The article studied 3 last words the computational issue that the axis measures umbriferous and basic theorem, gave out wave - the computational theory that makes theorematic ① and theorematic ② and computation are formulary, built the axis that contains mensurable relationship to measure umbriferous and basic theorem accordingly. Point out look from qualitative angle, wave - the distinction that makes theorematic ① theorematic ② do not have essence, but look from mensurable angle, these two theorem are put in the difference with obvious move, namely wave - make theorematic ① have 8 Jie Erbo - make theorematic ② only two solution. The research of the article is the complement that countershaft measures umbriferous theory, also have fair applied value at the same time. The geometrical modelling that if apply at surveying a plan from the axis,the three-dimensional information of tectonic object will be an object calculates what motor-driven draws to make provide academic basis and means. CNC Milling