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01. When direct indexing: |
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the index crank is not used |
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02. The formula for direct indexing eight holes in the circumference of a piece of barstock is: |
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03. When milling a square and using the 36 hole plate circle on the direct indexing plate, how many holes are required for the milling of each side? |
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04. In order to revolve the spindle one complete revolution on most dividing or indexing heads, the indexing crank must be revolved: |
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05. The calculated indexing for cutting 156 teeth on a gear blank is: |
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index=40/N=40/156=10/39=sp/hp |
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06. The calculated simple indexing for cutting 25 teeth on a gear blank is: |
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index=40/N=40/25=1 3/5=1 45/75 |
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07. To drill 2 holes 105 degrees apart using direct indexing and the 24 hole plate circle, the number of spaces to move is: |
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360/24=15 degrees 105/15=7 spaces |
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08. When angular indexing, 3 1/2 turns of the index crank will index the work through an angle of: |
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index=D/9 D=index*9=3.5*9=31.5 degrees |
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09. When differential indexing, for negative rotation using compound gearing you should use one idler (T/F) |
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simple differential: + --> 1 idler - --> 2 idlers Compound differential + --> 2 idlers - --> 1 idler |
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10. Using differential indexing, which gear train would be used to index for 175 divisions? Approximate = 160 |
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Indexing=40/A Gearing=(A-N)40/A=(160-175)40/160=-3.75 =72*40/(32*24), 1 idler |
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11. Before differential indexing can be accomplished: |
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disengage the lock or stop pin |
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12. The milling machine is set to cut a right hand helix, to cut a left hand helix: |
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swivel the table clockwise and add an idler |
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13. Using a milling machine with a standard pitch leadscrew, (4tpi) which of the following gear trains would be used to cut a 28-inch lead? (Gears available 24, 28, 32, 40, 44, 48, 64, 72, 86, 100) |
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Gearing=Lmc/Lsc=(1/4*40)/28=.357=DR/dn =24*40/(48*56)=LS/SP |
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14. During the linear graduating process, the table and the index head are connected with gears, two turns of the index crank will move the table: |
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40turns --> 1rev --> 1/4=.25 2 turns --> 2/40=.05 rev --> .25*.05=.0125" |
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15. The calculated indexing required for a linear graduation of .03125 inch is: |
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40 --> 1 --> .25 5 <--- .125 <--- .03125 |
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16. When linear graduating the table is moved by: |
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17. The circular/ radial graduating method would produce ... |
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Vernier protractor scales or radial lines about a circle |
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22. If driving gear has 40 teeth and revolves at 300 rpm, and is in mesh with a driven gear having 15 teeth, the rpm of the driven gear would be: |
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23. What is the pitch diameter of each gear whose ratio is 2.5:1 and their center distance is 7 inch? |
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PDg=Ng/(Ng+Np)*2*CD
4 inch and 10 inch |
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24. Three calculations required to mill a helical gear that are not required to produce a spur gears are: |
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helix angle, change gears, lead of helix |
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25. To ensure that a helical groove or spline has the same contour as the form cutter: |
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swing the table to the helix angle |
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26. Helical gears that connect parallel shafts must have ... hands of helix |
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27. Involute milling cutters for cutting helical gears on a milling machine are selected according to Normal Diametral Pitch (T/F) |
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28. Helical gears cannot connect: |
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29. Helical gears require more horse power to turn than a similar spur gear (T/F) |
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30. If the circular pitch of a helical gear with a 25 degrees helix angle is .221" the normal circular pitch will be: |
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Cos(helix angle) = DP/NDP=NCP/CP, NCP=.2 |
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31. Which of the following ratios is an example of hunting tooth ratio: |
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32. To cut helical gears with crossing shafts on a horizontal milling machine, swing: |
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table clockwise for both gears |
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33. When cutting a helical gear: |
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disengage the index plate locking device |
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34. To ensure that a worm and worm gear run correctly, the: |
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circular pitch must equal the linear pitch |
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35. To find the centre distance of a worm and worm gear: |
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add pitch diameters and divide by two |
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36. A worm gear has 50 teeth and a pitch diameter of 3. inch. The linear pitch of the worm: |
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Cir = pi*PD = CP*N LP=CP=pi*PD/N = 3.14*3/50 = .188" |
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37. A 4 start worm with a linear pitch of .25" drives a 40 tooth worm gear. If center distance is 2.2". What is lead angle of the worm? |
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Tan (lead angle) = Lw/Cir Cir = pi*PDw = CP*Np CD=1/2*(PDw+PDg)
PDg=Cir/pi=CP*Ng/pi=3.183 PDw=2CD-PDg=1.217 lead angle = actan(Lw/(pi*PDw)=14.66 degrees |
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38. One manufacturing method used to produce worm gears is: |
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39. The pressure angle of a 29 degrees worm is |
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40. One application for a coarse pitch worm and worm gear is a: |
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41. A worm gear has 75 teeth and is driven by a 5 start worm. If the center distance is 4. What is the pitch diameter of the worm? |
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See the formular specify by saying |
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42. One application for a fine pitch worm and worm gear is: |
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43. If a spur and worm gear have the same DP and number of teeth: |
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their pitch diameters will be equal (PD=N/DP) |
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44. When cutting bevel gear teeth, the index head is swung to the: |
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45. When blanking out a bevel gear, the most important angle to consider would be: |
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46. The chordal thickness of a bevel gear would be measured at the: |
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47. Which tooth form is not found on a bevel gear? |
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48. Bevel gears with curved teeth can be produced in a: |
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49. Bevel gear cutters are: |
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thinner than spur gear cutters |
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