# i nclude<stdio.h> oz]3
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# i nclude<stdio.h> qG~6YCqii
# i nclude<math.h> \AoqOC2u
#define PI 3.1415926 rk;]7Wu
void ydgl(); o4yl3o
float d,am,ro,e,h,p,dt0,d0,s,ds,r,al; #k d9}
int f; A{)pzV25
main() )'7Qd(4WT
{ printf("Determaination the prime circle of cam\n"); eAP
8!
printf("----------------------------\n"); ='1hvv/
printf( "input ro=");scanf("%f",&ro); }Cfl|t<5f
printf("input h=");scanf("%f",&h); 2$t%2>1>@
printf("input e=");scanf("%f",&e); 6#jql
printf("input dt0=");scanf("%f",&dt0); E|RC|Sz=u
am=0;p=PI/180; aTH$+f1?Q
printf("The initial data:ro=%f\n",ro); d>MDC
.
j
printf("h=%f e=%f dt0=%f\n",h,e,dt0); e_!Z-#\J%
do \=|=(kt)
{ro=ro+5; 3PLA*n+%
for(f=0;f<=dt0;f=f+2) ?D9iCP~~
{d=PI*f/dt0; PX23M|$!
ydgl(); K(lVAKiP]
} CsT&}-C
} ;0 +Dx~
while(am>30*p); CHO_3QIz
printf("The intermediate results: am=%f\n",am/p); +mR^ I$9
printf("ro=%f\n",ro); p9\*n5{
do ([rSYKpi
{ro=ro-1; :#n>Q1}x
if(ro<e) `0_
Y| 4KB
break; _tjexS'
{(Mmv[y
for(f=0;f<=dt0;f=f+2) br k*;
{d=PI*f/dt0; ,(sE|B#s
ydgl(); ",Mrdxn7
} G^VOA4
} <u#
7K\:
while((am>30*p)||(am<=29.5*p)); #s>'IPc0
printf("The final results:max alfa=%f\n",am/p); ku}`PS0UGd
printf(" min ro=%f\n",ro); MwQt/Qv=
getch(); EASmB
} }[@Q**j(
void ydgl() b"trg {e
{ d0=dt0*p; P&:[pPG
s=0.5*h*(1-cos(d)); c(5XT[Tw
ds=0.5*h*PI*sin(d)/d0; 1#+|RL4o
r=sqrt(ro*ro-e*e); F)imeu
if((s+r)==0) vE#8&Zq
return; fshG ~L7S9
al=atan((ds-e)/(s+r)); Y8lZ]IB
if(al>am) ]Z=al`-
am=al; kv?DE4=;
}