# i nclude<stdio.h> }r@yBUW
# i nclude<stdio.h> *,IK4F6>:
# i nclude<math.h> SN]g4}K-
#define PI 3.1415926 k>7bPR5Mw
void ydgl(); *")*w> R
float d,am,ro,e,h,p,dt0,d0,s,ds,r,al; oOC&w0
int f; 7mf&`.C
np
main() Z4rk$K'=1w
{ printf("Determaination the prime circle of cam\n"); *ra>Kl0
printf("----------------------------\n"); A^#\=ZBg1
printf( "input ro=");scanf("%f",&ro); 5+;Mc[V3-
printf("input h=");scanf("%f",&h); V)l:fUm2
printf("input e=");scanf("%f",&e); SOn)'!g
printf("input dt0=");scanf("%f",&dt0); 1U/RMN3`
am=0;p=PI/180; 6`20
printf("The initial data:ro=%f\n",ro); iy_Y!wZ{
printf("h=%f e=%f dt0=%f\n",h,e,dt0); zBu@a:E%H
do ar6+n^pi0]
{ro=ro+5; C#^y{q
for(f=0;f<=dt0;f=f+2) bX,#z,
{d=PI*f/dt0; wDGb h=
ydgl(); MkL2I+*
} MIn_?r
} mC$y*G
while(am>30*p); !|UX4
printf("The intermediate results: am=%f\n",am/p); /t`\b
[
printf("ro=%f\n",ro); ;{L[1OP%e
do ?|+e*{4k
{ro=ro-1; =y _KL
if(ro<e) BgQ/$,
break; oBo*<6
y_}vVHT,
for(f=0;f<=dt0;f=f+2) [P =P8-5
{d=PI*f/dt0; NjpWK;L
ydgl(); AU}kIm_+
} 2#sFY/@
} B^r?N-Z A
while((am>30*p)||(am<=29.5*p)); Q?1J<(oq9
printf("The final results:max alfa=%f\n",am/p); 6]~/`6Dub
printf(" min ro=%f\n",ro); "a(4])
getch(); E;{RNf|
} FHw%ynC
void ydgl() \X _}\_c,d
{ d0=dt0*p; \)r M C]
s=0.5*h*(1-cos(d)); z!O;s
ep?/
ds=0.5*h*PI*sin(d)/d0; <%Nf"p{K
r=sqrt(ro*ro-e*e); B=L!WGl<!
if((s+r)==0) d"06
gp
return; iD G&