Summary: The introduction can be calculated accurately one kind the new algorithm that involute spline broach pours horny tine measured value, the program of C language computation that wrote this algorithm. Graph the mathematical calculation of 2 RA of graph of tooth form of 1 involute spline the raises my factory production involute spline gear of 1 problem exports a product to poured horn to raise strict requirement to gear, when the involute spline broach that because this designs treatment,this gear uses, must pour horny measured value to undertake mathematical calculation to broach. To the graph a shown involute spline tooth form, the parameter such as B of A of Rb of radius of S of transverse tooth thickness of Ra of A of Z of number of the modulus M of foregone spline, tine, pressure angle, trail, reference circle, basic circle, length pouring wine cup, angle pouring wine cup, pour horny measured value to broach H(pursues 1 medium EO is worth) move of primary consideration method is as follows: ① Ch=acotb; ② RA=ra+ch; ③ reference circle is alveolar wide W=pm-s; ④ pours horn to stop be in =2rA(w/mz+inva-invaA) of ︵ of alveolar wide WA=AB, the AA=arccos(rb/rA) in type; AB=2rAsin(180wA/2prA) of ⑤ WA corresponding chord; ⑥ CD=AB+2a; ⑦ D=arcsin(CD/2ra) ; ⑧ Dx=90 ° - B-d; ⑨ H=EO=racosdx. Analyse afore-mentioned computation measure is knowable, (3) ～ (9) is theoretic all correct. But by the graph 1 knowable, RA is not equal to Ra+a, be less than Ra+a however. Accordingly, the RA that by RA=ra+a computation gives is not exact value, however approximation, this is the insufficient place of this calculation method. Accordingly, must use undertake be begginged accurately solving to RA and H via improvement new algorithm. Of 2 RA begging solution accurately to if the graph shows coordinate 2 times,be built is, set bit of C(x0, y0) , a(xA, YA) . By the graph 2 knowable: XA=x0+a, YA=y0+acotb, reason A nods coordinate to be A(x0+a, y0+acotb) , because because C dot is on spline trail,this has RA=[(x0+a)2+(y0+acotb)2]½(1) , because A dot is on involute,reason has X02+y02=ra2(2) , reason has Q=tanaA-aA(3) by the graph vertical of couplet of 2 knowable AA=arccos{rb/[(x0+a)2+(y0+acotb)2]½}(4) (3) , type (4) , have Q=tan(arccos{rb/[(x0+a)2+(y0+acotb)2]½})-arccos{rb/[(x0+a)2+(y0+acotb)2]½}(5) by the graph ︵ of 2 knowable Q=1EF - the D=-arctan(x0+a)/(y0+acotb) in D2rb(6) type because EF ︵ is basic circle essentially alveolar wide Wb, and the computational formula of Wb is Wb=wcosa+mzcosainva(7) accordingly, type (7) generation enters type (6) can get Q=1wcosa+mzcosainva+arctanx0+a2rby0+acotb(8) your form (5)= type (8) , can get Q=tan(arccos{rb/[(x0+a)2+(y0+acotb)2]½})-arccos{rb/[(x0+a)2+(y0+acotb)2]½}=1wcosa+mzcosainva+arctanx0+a2rby0+acotb(9) couplet vertical (2) and type (9) , can get equation group X02+y02=ra2q=tan(arccos{rb/[(x0+a)2+(y0+acotb)2]½})-arccos{rb/[(x0+a)2+(y0+acotb)2]½}=1wcosa+mzcosainva+arctanx0+a2rby0+acotb(10) to be in type (in 10) , divide X0, outside Y0, other parameter is worth all foregone may beg piece, because this is theoretic,can beg solution X0 through this equation group, y0 (actually X0, the computation seeking solution of Y0 is very trival, accordingly must computer of have the aid of undertakes) of process designing computation, next according to type (the accurate solution that 1) seeks a RA. the accurate solution of RA era joins the computational move of the first introduction, can get accurate broach to pour H of horny measured value. Graph 3 laws trying a value beg flow chart of solution equation group the law trying a value of 3 equation group begs solution to be in type (in 10) , x0, y0 already below consist in radical sign, consist in trigonometric function, in returning trigonometric function, because this uses groovy algorithm very unintelligible piece. It is this to be able to use law trying a value to seek solution. type (the X02+y02=ra2 in 10) is out of shape for Y0=(rA2-x02)½ . To each given X0 value all has corresponding Y0 value, because this can decide C nods coordinate C(x0, y0) , and dot C(x0, y0) is located in over round X02+y02=ra2. Because A nods coordinate to be A(x0+a, x0+acotb) , because this A is nodded,also can decide subsequently. Next simply differentiate X0, whether does Y0 accord with type (10)(namely whether is A dot on involute) , if accord with, criterion X0, y0 is the solution of equation group namely. As a result of type (9) essence is type (5)= type (8) , because this can be out of shape its,be type (5)- type (8)=0. As a result of type (5) , type (the value of 8) all is real number, want to make both compares difficulty equally absolutely, because need to differ both cost control to be able to think inside some limits two type are equal only here, if become - 0. 00001 ≤ type (5)- type (8) ≤ 0.
00001 when can think two type are equal, namely X0, y0 accords with type (10) , for the solution of equation group. The need when seeking solution chooses the initial value of X0. If adopt computation of increase by degrees to X0, x0 initial value must be taken left is nodded in C; If X0 initial value is taken,right side is chosen in C, should decrease successively to X0 computation. Initial value of the X0 when computation is chosen automatically by the computer. Use law trying a value to if the graph is shown 3 times,beg the order flow that sees equation group. 4 computation program and example beg the computational efficiency that sees equation group to rise, used C language to write program of the following computation. Although use method trying a value (try a value to X0) seek solution, but do not need initial value of manual input X0 (choose) automatically by the computer. Need to input involute spline parameter only, can achieve computational result. # Include "stdio.
H" # Include "math.
H" # Define PI 3.
141562652 Double Inv(double Num) (return Tan(num)-num;) Double X0, y0, z, o, c, da, db, a, b1, b2, s, m, p1, AA, b, dc, ax, m, dk, w, ac, wc, b, fx, h, aa, sd, sx, xs; Main() {printf("c=M =d K="); Printf("%1f%1f%1f" , &c, &M, &dk); Printf("m= Z= O= Da= A="); Scanf("%1f%1f%1f%1f%1f" , &m, &z, &o, &da, &A)); Db=m*z*cos(A*PI/180); If(fmod(z, 2)==1) Ax=acos(db*cos(PI/(2*z))/(M+dk)); Else Ax=acos(db/(M+dk)); S=m*z*(PI/z-dk/db+inv(A*PI/180)-inv(Ax)); W=PI*m-s; Aa=acos(db/da); Sd=da*w/(m*z)-da*(inv(Aa)-inv(A*PI/180)); Sx=da*sin(sd/da); Xs=-(sx/2+c); X0=xs; A100:x0=x0+0.
0001; Y0=sqrt(da*da/4-x0*x0); AA=acos((db/2)/sqrt((x0+c)*(x0+c)+(y0+c/tan(o*PI/180))*(y0+c/tan(O*PI/180)));B1=inv(AA); B2=(w*cos(A*PI/180)+m*z*cos(A*PI/180)*inv(A*PI/180))/db+atan((x0+c)/(y0+c/tan(o*PI/180))); B=B1-B2; (if B>0.
0001 | | B<- 0.
0001)goto A100; Dc=sqrt((x0+c)*(x0+c)+(y0+c/tan(o*PI/180))*(y0+c/tan(o*PI/180)))*2; Ac=acos(db/dc); Wc=dc*(w/(m*z)+inv(A*PI/180)-inv(Ac)); B=sin(wc/dc)*dc; Fx=90-o-180*asin((b+2*c)/(m*z))/PI; H=da*cos(Fx*PI/180)/2; Printf("x0=%1f Y0=%1f/n" , x0, y0); Printf("db=%1f AA=%1f/n" , db, AA); Printf("B=%1f S=%1f/n" , b, s); Printf("dc=%1f H=%1f/n" , dc, example of computation of H); Getcha(); } : Parameter of broach of foregone involute spline: M=2.
5mm, z=18, a=30 ° , s=3.
76mm, ra=21.
335mm, pour wine cup for 0.
45 ° of 5 × . Use what primary consideration method gets to pour H=16 of horny measured value.
958mm, and the mathematical calculation method that uses article introduction gets H=16.
882mm. Visible, the computation of method of two kinds of calculation needs a value for 0.
076mm, and the computational value of mathematical calculation method wants a few smaller. Accordingly, pouring horn to ask not severe case falls to gear, use former calculation the method is simpler; And pouring horn to ask severe case falls to gear, should use mathematical calculation method, and program of computer of have the aid of assists computation. CNC Milling