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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//To find number of teeth on each of the four wheels
+clc
+//given
+d=7//in; central distance
+k1=2*7*7//T1+t1/(2*7)=7
+k2=2*7*5//T2+t2/(2*5)=7
+G=9/1
+t1=(-(k1+k2)+((k1+k2)^2+4*(G-1)*(k1*k2))^(1/2))/(2*(G-1))
+a=ceil(t1)
+b=floor(t1)
+T1=k1-a
+T2=k2-a
+T3=k2-b
+G1=T1*T2/(a*a)
+G2=T1*T3/(a*b)
+dp=a/d
+//case b)
+tb1=23//let t1 = 23
+Tb1=k1-tb1
+Gb1=Tb1/tb1
+Gb2=G/Gb1
+tb2=k2/(Gb2+1)
+p=ceil(tb2)
+Tb2=k2-p
+l=Tb1-1
+m=tb1+1
+n=Tb2+1
+o=p-1
+Gb2=l*n/(m*o)
+printf("\na) No of teeth = %.f, %.f, %.f, %.f\nG = %.2f\n\nb) No of teeth = %.f, %.f, %.f, %.f\nG = %.2f\n\n",T1,T2,a,b,G2,l,m,n,o,Gb2)