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diff --git a/278/CH23/EX23.19/ex_23_19.sce b/278/CH23/EX23.19/ex_23_19.sce
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+//find
+clc
+//solution
+//given
+W=5000//N
+dx=40//mm
+t1=850//N/mm^2
+t2=850//N/mm^2
+C=6
+G=80000//N/mm^2
+//ref fig 23.22
+//D1-D2=2*d1
+//D1=C*d1
+//D2=C*d2
+//d1/d2=1.5
+//W1/W2=2.25....eq1
+//W1+W2=W....eq2
+//from 1 and 2,we get
+W1=3492//N
+W2=1538//N
+K1=(4*C-1)/(4*C-4)+(0.615/C)
+K2=K1
+//d1=(K1*8*W1*C/(%pi*t1))^(0.5)
+printf("dia of spring wires is,%f mm\n",(K1*8*W1*C/(%pi*t1))^(0.5))
+printf("dia is ,say 10mm\n")
+printf("mean outer dia is,%f mm\n",6*d1)
+d1=10
+D1=6*d1
+printf("dia of spring wires is,%f mm\n",(K2*8*W2*C/(%pi*t2))^(0.5))
+printf("dia is ,say 6 mm\n")
+d2=6
+printf("mean outer dia is,%f mm\n",6*d2)
+D2=6*d2
+//n1=(8*W1*C^3)/(dx*G*d1)
+printf("number of turns are in outer coil,%f \n",1/[(8*W1*C^3)/(dx*G*d1)])
+printf("numbr of turns are say 6\n")
+n1=6
+n1b=n1+2
+Ls1=n1b*d1
+n2b=n1b*d1/d2
+n2=n2b-2
+printf("numbr of tuns in inner coil is,%f \n",n2)
+fL=Ls1+dx+0.15*dx
+printf("free length is,%f mm\n",fL)
+printf("outr dia of outr spring is,%f mm\n",D1+d1)
+printf("innr dia of outr spring is,%f mm\n",D1-d1)
+printf("outer dia of innr spring is,%f mm\n",D2+d2)
+printf("innr dia of innr spring is,%f mm\n",D2-d2)
+
+
+