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+
+
+//example 12.9
+//calculate depth of downstream cut-off
+clc;funcprot(0);
+//given
+Q=1000;        //discharge of river
+L=256;         //crest length of diversion
+f=1.1;         //silt factor
+seg=1/6;       //safe exit gradient
+hfl=103;       //high flood level
+cf=100;        //reduced level of downstream concrete floor
+H=2.4;         //maximum static head of weir
+b=40;          //length of concrete floor
+
+q=Q/L;
+R=1.35*(q^2/f)^(1/3);
+rld=hfl-1.5*R;
+d=cf-rld;
+d=round(d*100)/100;
+mprintf("depth of downstream cut-off=%f m.",d);
+
+alpha=b/d;
+lambda=(1+(1+alpha^2)^0.5)/2;
+G=H/(d*%pi*(lambda)^0.5);
+//since G<seg
+mprintf("\n G=%f. <1/6./nfloor is safe against failure by piping.",G);
+