# i nclude<stdio.h> yAOC<d9 E
# i nclude<stdio.h> y$WS;#
# i nclude<math.h> Rd2[xk
#define PI 3.1415926 39"'Fz?1
void ydgl(); bpkn[K"(
float d,am,ro,e,h,p,dt0,d0,s,ds,r,al; J [1GP_
int f; B?jF1F!9
main() QO8/?^d
{ printf("Determaination the prime circle of cam\n"); &7 ,wdG
printf("----------------------------\n"); aUnm9ur
printf( "input ro=");scanf("%f",&ro); \vI_%su1N
printf("input h=");scanf("%f",&h); A)xI.Q6
printf("input e=");scanf("%f",&e); =xgW$c/yB
printf("input dt0=");scanf("%f",&dt0); }~7>S5
am=0;p=PI/180; }V 1sY^C
printf("The initial data:ro=%f\n",ro); #\}hN~@F
printf("h=%f e=%f dt0=%f\n",h,e,dt0); PSRGlxdO
do -$7Jc=:>
{ro=ro+5; ?F-,4Ox{/
for(f=0;f<=dt0;f=f+2) <,m}TTq
{d=PI*f/dt0; ,GK>|gNsb
ydgl(); X**wRF
} ^C(AMT
} DT*/2TH*l
while(am>30*p); Ct B>
s7
printf("The intermediate results: am=%f\n",am/p); !o`al` q'
printf("ro=%f\n",ro); [\VzI\vb
do E( TY%wO
{ro=ro-1; eA!aUu
if(ro<e) As"'KR
break; Qwb@3{
z
>pq<}R6
for(f=0;f<=dt0;f=f+2) TqWvHZX
{d=PI*f/dt0; {vaq,2_w
ydgl(); &?~> I[^~
} kr3ZqMfeI
} ^!A{ 4NV
while((am>30*p)||(am<=29.5*p)); +Y.As
printf("The final results:max alfa=%f\n",am/p); 8J)x>6
printf(" min ro=%f\n",ro); D,NjDIG8
getch(); C ZJW`c/
} zNBG;\W
void ydgl() j*CnnM#n
{ d0=dt0*p; o}[wu:>yk
s=0.5*h*(1-cos(d)); 6ds&n#n
ds=0.5*h*PI*sin(d)/d0; cM55
vVd
r=sqrt(ro*ro-e*e); .9`.\v6R
if((s+r)==0) Lg'z%pi
return; z8gp<5=
al=atan((ds-e)/(s+r)); g
>X!Q
if(al>am) EB)0 iQ
am=al; f5'+F-`N
}