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+// sum 30-3
+clc;
+clear;
+N1=1000;
+N2=500;
+P=2.03*10^3; //from table 30-8
+K1=1.26;
+Ks=1;
+//let Pc be the power transmitting capacity of the chain
+Pc=P*K1/Ks;
+p=9.52;
+n1=21;
+n2=42;
+V=n1*p*N1/(60*10^3);
+//Let the chain tension be T
+T=Pc/V;
+//Let the breaking load be BL
+BL=10700;
+FOS=BL/T;
+C=50*p;
+Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*%pi))^2)*(p/C));
+L=Ln*p;
+Pc=Pc*10^-3;
+
+ // printing data in scilab o/p window
+ printf("Pc is %0.2f KW ",Pc);
+ printf("\n V is %0.3f m/s ",V);
+ printf("\n T is %0.1f N ",T);
+ printf("\n FOS is %0.2f ",FOS);
+ printf("\n L is %0.2f mm ",L);
+
+//The difference in the value of L and T is due to rounding-off the values.