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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//to find thelength of path of contact and the length of arc of contact
+clc
+//let module be 1
+m=1
+t1=28
+t2=45
+r=t1*m/2
+R=t2*m/2
+ra=r+m
+Ra=R+m
+phi1=14.5*%pi/180
+//10.8 => AB =(ra^2-r^2*cos(phi)^2)^(1/2)+(Ra^2-R^2*cos(phi)^2)^(1/2)-(r+R)*sin(phi)
+//AB=A+B-C
+A=m*(ra^2-r^2*cos(phi1)^2)^(1/2)
+B=m*(Ra^2-R^2*cos(phi1)^2)^(1/2)
+C=m*(r+R)*sin(phi1)
+AB=A+B-C
+p=%pi*m
+ABp=AB/%pi
+arc1=ABp/cos(phi1)//length of arc of contact
+phi2=20*%pi/180
+//10.8 => AB =(ra^2-r^2*cos(phi)^2)^(1/2)+(Ra^2-R^2*cos(phi)^2)^(1/2)-(r+R)*sin(phi)
+a=m*(ra^2-r^2*cos(phi2)^2)^(1/2)
+b=m*(Ra^2-R^2*cos(phi2)^2)^(1/2)
+c=m*(r+R)*sin(phi2)
+ab=a+b-c
+abp=ab/%pi
+arc2=abp/cos(phi2)//length of arc of contact
+printf("\nLength of path of contact\nWhen phi = 14.5 degrees = %.3fm\nWhen phi = 20 degrees = %.2fm\nLength of arc of contact\nWhen phi = 14.5 degrees = %.2fp\nWhen phi = 20 degrees = %.3fp\n",AB,ab,arc1,arc2)