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+clear;
+clc;
+printf("\t\t\tExample Number 8.5\n\n\n");
+// shape-factor algebra for cylindrical reflactor
+// Example 8.5 (page no.-397-398) 
+// solution
+
+d = 0.6;// [m] diameter of long half-circular cylinder
+L = 0.2;// [m] length of square rod
+// we have given figure example 8-5(page no.-397) for solution of this problem and take the nomenclature as shown, 
+// from symmetry we have 
+F21 = 0.5;
+F23 = F21;
+// in general, F11+F12+F13 = 1. to aid in the analysis we create the fictious surface 4 shown in figure example 8-5 as dashed line.
+// for this surface 
+F41 = 1.0;
+// now, all radiation leaving surface 1 will arrive either at 2 or at 3. likewise,this radiation will arrive at the imaginary surface 4, so that F41 = F12+F13 say eqn a
+// from reciprocity
+A1 = %pi*d/2;// [square meter]
+A4 = L+2*sqrt(0.1^(2)+L^(2));// [square meter]
+A2 = 4*L;// [square meter]
+// so that 
+F14 = A4*F41/A1;// say eqn b
+// we also have from reciprocity
+F12 = A2*F21/A1;// say eqn c
+// combining a,b,c, gives
+F13 = F14-F12;
+// finally
+F11 = 1-F12-F13;
+printf("value of F12 is %f ",F12);
+printf("\nvalue of F13 is %f ",F13);
+printf("\nvalue of F11 is %f ",F11);
+
+
+
+
+
+ 
+\ No newline at end of file