{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 10: Toothed Gearing" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1, Page 320" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "Teeth=48\n", "pitch=.75 #in\n", "\n", "#Calculations\n", "D=Teeth*pitch/math.pi\n", "\n", "#Result\n", "print \"The pitch diameter is %.3f in\"%D" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The pitch diameter is 11.459 in\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2, Page 320" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "T=48#teeth\n", "pd=4#diametral pitch\n", "\n", "#Calculations\n", "D=T/pd#pitch diameter\n", "p=math.pi/pd#the circular pitch\n", "\n", "#Results\n", "print \"The pitch diameter = %.f in\\nThe circular pitch = %.4f in\"%(D,p)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The pitch diameter = 12 in\n", "The circular pitch = 0.7854 in\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3, Page 321" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "T=48\n", "m=6#mm ; module\n", "\n", "#Calculations\n", "D=m*T\n", "p=math.pi*m\n", "dia=D/10#cm\n", "P=p*0.0393700787#inches\n", "\n", "#Results\n", "print \"Pitch diameter = %.1f cm\\nCircular pitch = %.4f in\"%(dia,P)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Pitch diameter = 28.0 cm\n", "Circular pitch = 0.7421 in\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4, Page 334" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "phi=20*math.pi/180\n", "\n", "#Calculations\n", "#Solution a)\n", "ar=1\n", "t1=2*ar/math.sin(phi)**2#from equation 10.7\n", "T1=math.ceil(t1)\n", "#Solution b)\n", "aw=1.\n", "t2=2*aw/((1+3*math.sin(phi)**2)**(1./2)-1)#from euation 10.6\n", "T2=math.ceil(t2)\n", "#solution c)\n", "t=1.\n", "T=3.\n", "A=(t/T)*(t/T+2)\n", "t3=2*aw*(t/T)/((1+A*math.sin(phi)**2)**(1./2)-1)#from 10.5\n", "T3=math.ceil(t3)\n", "\n", "#Results\n", "print \"Smallest number of teeth theoretically required in order to avoid interference on a pinion which is to gear with\"\\\n", " \"\\na) A rack , t= %.f\\nb) An equal pinion , t= %.f\\nc) A wheel to give a ratio of 3 to 1 , t= %.f\"%(T1,T2,T3)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Smallest number of teeth theoretically required in order to avoid interference on a pinion which is to gear with\n", "a) A rack , t= 18\n", "b) An equal pinion , t= 13\n", "c) A wheel to give a ratio of 3 to 1 , t= 15\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5, Page 338" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "t=25\n", "phi=20*math.pi/180\n", "\n", "#Calculations\n", "#let pitch be 1 \n", "R=t/(2*math.pi)#R=t*p/(2*math.pi)\n", "Larc=1.6#1.6*p\n", "#AB=Larc*math.cos(phi)\n", "AB=Larc*math.cos(phi)\n", "Ra=(4.47+13.97)**(1./2)#by simplifying AB+2{(Ra^2-R^2*cos(phi)^2)-R*sin(phi)} and using p =1\n", "Addendum=Ra-R\n", "#writing p in place of p=1\n", "\n", "#Result\n", "print \"Addendum required = %.2fp\"%Addendum" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Addendum required = 0.32p\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6, Page 338" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "#let module be 1\n", "m=1\n", "t1=28\n", "t2=45\n", "\n", "#Calculations\n", "r=t1*m/2\n", "R=t2*m/2\n", "ra=r+m\n", "Ra=R+m\n", "phi1=14.5*math.pi/180\n", "#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)\n", "#AB=A+B-C\n", "A=m*(ra**2-r**2*math.cos(phi1)**2)**(1./2)\n", "B=m*(Ra**2-R**2*math.cos(phi1)**2)**(1./2)\n", "C=m*(r+R)*math.sin(phi1)\n", "AB=A+B-C\n", "p=math.pi*m\n", "ABp=AB/math.pi\n", "arc1=ABp/math.cos(phi1)#length of arc of contact\n", "phi2=20*math.pi/180\n", "#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)\n", "a=m*(ra**2-r**2*math.cos(phi2)**2)**(1./2)\n", "b=m*(Ra**2-R**2*math.cos(phi2)**2)**(1./2)\n", "c=m*(r+R)*math.sin(phi2)\n", "ab=a+b-c\n", "abp=ab/math.pi\n", "arc2=abp/math.cos(phi2)#length of arc of contact\n", "\n", "#Results\n", "print \"Length of path of contact\\nWhen phi = 14.5 degrees = %.3fm\\nWhen phi = 20 degrees = %.2fm\"\\\n", " \"\\n\\nLength of arc of contact\\nWhen phi = 14.5 degrees = %.2fp\\nWhen phi = 20 degrees = %.3fp\"%(AB,ab,arc1,arc2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Length of path of contact\n", "When phi = 14.5 degrees = 6.092m\n", "When phi = 20 degrees = 4.97m\n", "\n", "Length of arc of contact\n", "When phi = 14.5 degrees = 2.00p\n", "When phi = 20 degrees = 1.685p\n" ] } ], "prompt_number": 7 } ], "metadata": {} } ] }