{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ch:30 Chain drive"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 30-1 - Page 778"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "C is 300.00 mm \n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import sqrt, pi\n",
    "n1=17#\n",
    "n2=51#\n",
    "C=300#\n",
    "p=9.52#\n",
    "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n",
    "x=(Ln-((n2+n1)/(2)))**2#\n",
    "y=8*(((n2-n1)/(2*pi))**2)#\n",
    "z=Ln-((n1+n2)/2)#\n",
    "C=(p/4)*(z+(sqrt(x-y)))\n",
    "\n",
    "\n",
    "  # printing data in scilab o/p window\n",
    "print \"C is %0.2f mm \"%(C)#\n",
    "  "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 30-2 - Page 778"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Ln is 80  \n"
     ]
    }
   ],
   "source": [
    "from math import tan\n",
    "G=4#\n",
    "n1=17#\n",
    "n2=n1*G#\n",
    "N1=2300#\n",
    "Kc=1.2# #from table 30-2\n",
    "p=12.7# #fom table 30-1\n",
    "D1=p*n1#\n",
    "D2=p*n2#\n",
    "phi=2*10.6#\n",
    "x=tan(phi/2)# #phi/2 = 10.6deg, from table 30-3\n",
    "Da1=(p/x)+(0.6*p)#\n",
    "Da2=(p/x*4)+(0.6*p)#\n",
    "Cmin=Kc*((Da1+Da2)/2)#\n",
    "Ln1=(2*Cmin/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/Cmin))#\n",
    "Ln1=80#\n",
    "print \"Ln is %0.0f  \"%(Ln1)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 30-3 - Page 779"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " Pc is 2.56 KW  \n",
      "\n",
      " V is 3.332 m/s  \n",
      "\n",
      " T is 767.6 N  \n",
      "\n",
      " FOS is 13.94   \n",
      "\n",
      " L is 1254.01 mm  \n"
     ]
    }
   ],
   "source": [
    "from math import sqrt, pi\n",
    "N1=1000#\n",
    "N2=500#\n",
    "P=2.03*10**3# #from table 30-8\n",
    "K1=1.26#\n",
    "Ks=1#\n",
    "#let Pc be the power transmitting capacity of the chain\n",
    "Pc=P*K1/Ks#\n",
    "p=9.52#\n",
    "n1=21#\n",
    "n2=42#\n",
    "V=n1*p*N1/(60*10**3)#\n",
    "#Let the chain tension be T\n",
    "T=Pc/V#\n",
    "#Let the breaking load be BL\n",
    "BL=10700#\n",
    "FOS=BL/T#\n",
    "C=50*p#\n",
    "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n",
    "L=Ln*p#\n",
    "Pc=Pc*10**-3#\n",
    "print \" Pc is %0.2f KW  \"%(Pc)#\n",
    "print \"\\n V is %0.3f m/s  \"%(V)#\n",
    "print \"\\n T is %0.1f N  \"%(T)#\n",
    "print \"\\n FOS is %0.2f   \"%(FOS)#\n",
    "print \"\\n L is %0.2f mm  \"%(L)#\n",
    "\n",
    "#The difference in the value of L and T is due to rounding-off the values."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 30-5 - Page 780"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "L is 1678.25 mm  \n"
     ]
    }
   ],
   "source": [
    "from math import sqrt, pi\n",
    "G=2#\n",
    "P=5000#\n",
    "Ks=1.7#\n",
    "Pd=P*Ks#\n",
    "K2=1.7#\n",
    "p=15.88#\n",
    "n1=17#\n",
    "n2=n1*G#\n",
    "D1=n1*p#\n",
    "D2=n2*p#\n",
    "C=40*p#\n",
    "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n",
    "L=Ln*p#\n",
    "print \"L is %0.2f mm  \"%(L)#\n",
    "#The difference in the value of L is due to rounding-off the values."
   ]
  }
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