Example 8.6 For the dimensioned pin assembly shown below, specify the allowable perpendicularity tolerance for each of the actual pin sizes specified. ger diameter of the stepped pin is f9.97 0.010 mm. Concentricity can be held to within 0.002 mm for both the stepped pin and the counterbored hole. Assume that concentricity is the only important feature other than the tolerances on the diameters (e.g. negligible straightness, perpendicularity, etc.). For each case below, determine the diameter and tolerance of the smaller diameter of the stepped pin that will permit the parts to mate as specified. Use bilateral, equally divided tolerances on the dimensions. If the required dimensions and tolerances cannot be found, clearly explain why... backward forward pause audioOn paused thisWav playing buttonClick buttonClick 4thisWav, audioOn, vol mmstatus clip @ = "playing" mmPause h = "paused" mmvolume Ethiswav = mmPlay notify repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, audioon, vol syserrornumber = 0 mmvolume clip mmplay 0 notify f<> 0 audioerror Repeat 1. At an actual size of 20.07, the pin is at its maximum material condition. The allowable perpendicularity is thus specified as 0.02. 2. At an actual size of 20.05, the pin is 0.02 less than its maximum material condition. Thus this amount may be added to the allowable perpendicularity. The perpendicularity allowance is now 0.04. 3. At an actual size of 20.03, the pin is 0.04 less than its maximum material condition. Thus this amount may be added to the allowable perpendicularity. The perpendicularity allowance is now 0.06. producing a line-to-line fit. Thus (5.00 - 0.02) = 4.98. 4. The "0.06 MAX" on the drawing specifies that the maximum perpendicularity allowance can be no greater than 0.06. Thus for all actual sizes less than 20.03, the perpendicularity allowance is always 0.06. table Actual Size Perpendicularity Allowance 20.07 0.02 20.05 0.04 20.03 0.06 20.01 0.06 19.09 0.06 19.07 0.06 19.05 0.06 19.03 0.06 n and bores to show the above dimensions, tolerances, and control features so the parts will fit with the desired clearance and interference. cover2 audioon false currframe notifybefore 4currframe, audioon = 2 ( > 2 cover1 audioon false currframe notifybefore 4currframe, audioon = 1 ( > 1 cover3 audioon false currframe notifybefore 4currframe, audioon = 3 ( > 3 cover5 audioon false currframe notifybefore 4currframe, audioon > 3 cover4 audioon false currframe notifybefore 4currframe, audioon = 4 ( > 4 cover6 audioon false currframe notifybefore 4currframe, audioon > 3 cover7 audioon false currframe notifybefore 4currframe, audioon > 3 cover8 audioon false currframe notifybefore 4currframe, audioon > 3 thisanim lastanim thiswav numframes lastwav enterpage reset leavepage step0 forward paused currframe thiswav audioerror playing numframes update reset 4numframes 4thiswav, thisanim, lastanim, lastwav reset 4currframe, |vol, audioon syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait r<> 0 audioerror ("txt" & i) " & i) "step0" update step1 step2 step3 step4 step5 step6 step0 Example 8.7 Assume that the 8 pin assembly shown below is an existing part. The pins each have a size f8.000 0.010 mm and are located as shown. You are to design the mating part with the same number holes as pins. The holes have a size f8.200 0.020 mm. The parts must fit together, with clearance, in any rotational orientation. Specify the locations of the holes on the mating part, in such a way that the maximum allowable tolerances are produced, which would allow the two parts to fit together. Assume that only size and location tolerances need to be considered. 1. For clearance problems, consider the largest pin, f8.010, and the smallest hole, f8.180. Not yet taking into consideration the location tolerances, the minimum diametrical clearance between the holes and the pins is 0.170. The minimum radial clearance, which is 1/2 the minimum diametrical clearance, is 0.085. 2. Now consider the location tolerances of the pins and holes. To achieve the largest tolerances and still have the parts fit, true position tolerancing and the maximum material conditions will be used. 3. The pin assembly was toleranced using conventional tolerancing. The location of each pin may vary from the nominal location by 0.025. This target area is represented by a 0.050 square. backward forward pause audioOn paused thisWav playing buttonClick buttonClick 4thisWav, audioOn mmstatus clip ; = "playing" mmPause c = "paused" mmPlay } notify repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, audioon, vol syserrornumber = 0 mmvolume clip mmplay 0 notify f<> 0 audioerror Repeat 4. The greatest distance the pin center can be from the nominal location is at any corner of the square target area. This displacement from the nominal location is 0.035, which is along the diagonal of the square. 5. The maximum radial pin displacement is subtracted from the minimum radial clearance between the pins and the holes. The remainder is the location tolerance of the holes. Thus the location tolerance for the holes about their true positions is (0.085) - (0.035) = 0.050. 6. The appropriate symbols are added to the drawing.m 7. The maximum material condition symbol is added to the hole location tolerance. With this condition, as the actual size of the hole gets larger, the hole location tolerance correspondingly increases. 8. Datums are added to ensure that measurements can be made accurately and consistently.k step7 step8 thisanim lastanim thiswav numframes lastwav enterpage reset leavepage answer forward paused currframe thiswav audioerror playing numframes update reset 4numframes 4thiswav, thisanim, lastanim, lastwav reset 4currframe, |vol, audioon syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait r<> 0 audioerror -- For minus one) ("txt" & i) "answer" update answer Example 8.3 The size of the pin in the part shown is 14.75 0.005 mm. Determine the size of the hole such that the diametric interference between the two parts is between 0.010 and 0.030 mm. Assume that only the sizes of the parts need to considered. If the parts were to be assembled by thermal conditioning, to what temperature difference would the hole need to be heated, or the pin cooled, to ensure that the parts always assemble with clearance?? 1. Consider the case where the diametric interference is the largest. This occurs when the hole is the smallest, and the pin is the largest. Thus (largest pin diameter) - (smallest hole diameter) = (largest diametric interference). Rewritten, the smallest hole diameter is (largest pin diameter) - (largest diametric interference) = (14.755) - (0.030) = (14.725). 2. Next, consider the case where the diametric interference is the smallest. This occurs when the hole is the largest and the pin is the smallest. Thus (smallest pin diameter) - (largest hole diameter) = (smallest diametric interference). Rewritten, the largest hole diameter is (smallest pin diameter) - (smallest diametric interference) = (14.745) - (0.010) = (14.735). 3. The limit sizes of the hole are thus 14.725 to 14.735 mm. This can be rewritten with bilateral and equally distributed tolerances as 14.73 0.005. backward audioon answer forward paused currframe thiswav audioerror numframes playing update buttonclick buttonclick 4currframe, numframes 4thiswav, vol, audioon / > 0 "answer" ("txt" & p - 1 z > 0 syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait = "8-3-" & <> 0 mmvolume mmplay audioerror update forward pause audioOn paused thisWav playing buttonClick buttonClick 4thisWav, audioOn mmstatus clip ; = "playing" mmPause c = "paused" mmPlay } notify repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, audioon, vol syserrornumber = 0 mmvolume clip mmplay 0 notify f<> 0 audioerror Repeat answer thisanim lastanim thiswav numframes lastwav enterpage reset leavepage answer forward paused currframe thiswav audioerror playing numframes update reset 4numframes 4thiswav, thisanim, lastanim, lastwav reset 4currframe, |audioon, vol syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait r<> 0 audioerror -- For minus one) ("txt" & i) "answer" update Example 8.2 The size of the hole in the part shown is 23.50 0.010 mm. Determine the size of the pin such that the diametric clearance between the two parts is between 0.015 and 0.045 mm. Assume that only the sizes of the parts need to considered. 1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475) 2. Next, consider the case where the diametric clearance is the largest. This occurs when the hole is the largest and the pin is the smallest. Thus (largest hole diameter) - (smallest pin diameter) = (largest diametric clearance). Rewritten, the smallest pin diameter is (largest hole diameter) - (largest diametric clearance) = (23.510) - (0.045) = (23.465) 3. The limit sizes of the pin are thus 23.465 to 23.475 mm. This can be rewritten with bilateral and equally distributed tolerances as 23.47 0.005.der the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)0.005 backward audioon answer forward paused currframe thiswav audioerror numframes playing update buttonclick buttonclick 4currframe, numframes 4thiswav, audioon, vol 9 > 0 "answer" ("txt" & z - 1 syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait = "8-2-" & <> 0 mmvolume mmplay audioerror update forward pause audioOn paused thisWav playing buttonClick buttonClick 4thisWav, audioOn mmstatus clip ; = "playing" mmPause c = "paused" mmPlay } notify repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, audioon, vol syserrornumber = 0 mmvolume clip mmplay 0 notify f<> 0 audioerror Repeat answer thisanim lastanim thiswav numframes lastwav enterpage reset leavepage ,%H.% cover forward paused currframe thiswav audioerror playing update reset 4numframes 4thiswav, thisanim, lastanim, lastwav reset 4currframe, syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait r<> 0 audioerror " & i) ("cover" & i) update Example 8.1 The part drawing below has been removed from its header. Still there at least five errors or items that are considered poor practice in the dimensioning of the object. Find these items and note, in your mind, how these items should be corrected. Portions of the drawing may be enlarged by clicking on that area. Tolerance is larger than the feature size on the 5 mm thickness dimension. Tolerance on radius dimension is missing. Datum C is not defined for the position of the larger hole. A location tolerance appears on the larger hole, even though conventional tolerancing is already used. A location tolerance does not appear on the smaller holes, where true-position tolerancing is used. 9060,4140 buttonclick buttonclick "9060,4140" 4185,2685 buttonclick buttonclick "4185,2685" 4125,4140 buttonclick buttonclick "4125,4140" 4065,4125 buttonclick buttonclick "4065,4125" 6585,2460 buttonclick buttonclick "6585,2460" cover1 cover2 cover3 cover4 cover5 step1 step2 step4 step5 step3 backward audioon cover forward paused currframe thiswav audioerror playing update buttonclick buttonclick 4currframe, numframes 4thiswav, audioon, vol 9 > 0 ("cover" & m - 1 w <> 0 syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait = "8-1-" & <> 0 mmvolume mmplay audioerror update forward pause audioOn paused thisWav playing buttonClick buttonClick 4thisWav, audioOn mmstatus clip ; = "playing" mmPause c = "paused" mmPlay } notify repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, audioon, vol syserrornumber = 0 mmvolume clip mmplay 0 notify f<> 0 audioerror Repeat Portions of the drawing may be enlarged by clicking on that area.. table H%] LG* forward thisAnim lastanim currframe thisWav numframes lastWav update enterpage play0 forward currframe update leavepage animdone audioon thiswav audioerror tol5a audiodone audioon tol5b successful thiswav audioerror tol5a mmnotify paused step0 thiswav audioerror playing tol5a play0 paused step1 step0 thiswav audioerror playing tol5a play0on step1 play0off audioon tol5c paused step1 step0 thiswav audioerror playing play1 audioon tol5c step2 paused step1 thiswav audioerror playing play1on step1 play1off audioon tol5d step2 paused step1 thiswav audioerror playing play2 audioon tol5d step2 paused thiswav audioerror playing step3 play2on step2 play2off tol5e audioon step2 paused thiswav audioerror playing step3 play3 tol5e audioon step4 paused thiswav audioerror playing step3 play3on step3 play3off tol28 mmsystem.dll forward thisAnim lastAnim currframe thisWav timegettime numframes lastWav update enterpage n;dequeue mmsystem.dll forward currframe update leavepage audiodone animationstage thisanim hole3 animerror animdone cover1 tol28 audioon false step2 step4 thisanim step1 noanim step6 step5 thiswav audioerror yesplay cover2 animerror step3 Animationstage queue cover1 cover5 step2 cover7 cover6 step4 cover3 paused step1 cover4 step5 thiswav audioerror step6 playing cover2 step3 dequeue animationstage starttime cover5 }gyieldapp cover7 cover6 thisanim cover3 successful timegettime cover4 mmnotify (minimum hole size - shaft size - straightness tolerance) = minimum clearance 5.30 - 4.80 - 0.04 = 0.46.46 Conventional Here we show how a shaft is toleranced by conventional methods to produce a clearance fit with a hole of arbitrary size. The shaft has a straightness tolerance which is independent of its diameter.. Notice that as the diameter of the shaft decreases, the clearance, shown in yellow, increases. animationstage rptanim animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref, x, y syserrornumber = 0 x = 0 y = mmendpoint clip thisanim mmPlay stage "animationstage" hold animerror Repeat Animation step1 The diameter of the shaft ranges from 4.80 to 5.20. step2 The tightest fit occurs when the shaft is at its largest allowable diameter. The straightness tolerance of 0.04 is applied to the shaft to ensure a clearance of 0.06..f 0.06. step3 With conventional tolerancing, the straightness tolerance is independent of the diameter of the shaft. When the shaft diameter is decreased, the clearance increases. (minimum hole size - largest shaft - straightness tolerance) = minimum clearance 5.30 - 5.20 - 0.04 = 0.06 nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk correct misc.tbk incorrect false misc.tbk incorrect inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.6 A flatness specification on a surface does not require a reference datum surface. answer True. reference datums nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk incorrect misc.tbk incorrect false misc.tbk correct inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.5 If three different datum planes exist on a part, then they are also perpendicular to each other. answer Inconclusive. The plane may or may not be perpendicular to each other, depending on the desired feature orientation.. reference 3datums nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk correct misc.tbk incorrect false misc.tbk incorrect inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.4 The pin shown below cannot be dimensioned and toleranced to achieve a 0.005 to 0.030 clearance fit in the hole shown. answer True. The total dimensional tolerance of the hole is greater than the total tolerance of the fit. reference nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk incorrect misc.tbk incorrect false misc.tbk correct inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.3 The parts shown, as toleranced, will produce a clearance fit. answer Inconclusive. The fit is a transition fit. The parts will sometimes produce a clearance fit and sometimes an interference fit, depending on the actual size of the parts............ reference nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk incorrect misc.tbk correct false misc.tbk incorrect inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.2 The parts shown, as toleranced, will produce a clearance fit. answer False. The parts will produce an interference fit. reference 3datums Receptacle Pin Mount This is an example of non-Maximum Material Condition tolerancing on a pair of mating blocks. The task is to show that the parts will fit together with a clearance fit based on the information given in the engineering drawing. Clearances in the animation are shown as colored bands..). Clearances in teh animation are shown as colored bands. animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref, x, y syserrornumber = 0 -- x = 0 -- y = mmendpoint clip thisanim mmPlay stage "animationstage" hold animerror Repeat Animation step1 According to the given specifications, the diameters of the holes are allowed to be between 5.06 and 5.10, while the diameters of the pins are within the range 4.98 to 5.02. The longitudinal separation of the pins and the holes is between 19.98 and 20.02.m step2 To determine if the parts will always fit, we check to see if they fit under the most extreme conditions. With regard to size, we assume the holes have the smallest allowable diameter. (nominal hole size - lower tolerance) = minimum diameter 5.80 - 0.02 = 5.06 = 0.066.20 - 0.04 = 0.0666 step3 For the pins, we assume that they have the largest allowable diameter. (nominal pin size - upper tolerance) = maximum diameter 5.00 + 0.02 = 5.026 = 0.066.20 - 0.04 = 0.0666 step4 Position also factors into the worst-case scenario. In this case, we assume that the pins are separated by the greatest allowable distance. (nominal distance - upper tolerance) = maximum separation 20.00 + 0.02 = 20.02 0.066.20 - 0.04 = 0.0666 step5 We also assume that the holes are separated by the smallest allowable distance. (nominal distance - lower tolerance) = minimum separation 20.00 - 0.02 = 19.98 0.066.20 - 0.04 = 0.0666 step6 Finally, we check to see whether or not the points on the holes and pins separated by the longitudinal distance interfere. The parts will just fit, even under worst-case conditions. Assuming the holes are as far apart as possible and that the pins are as close together as possible gives similar results.. (separation of centers + 2 x radius) = total separation Holes: 19.98 + 5.06 = 25.04 Pins: 20.02 + 5.02 = 25.04 hole extents - pin extents = minimum clearance 25.04 - 25.04 = 0.00000 step7 Notice that with conventional tolerancing, the tolerances are absolute. If, for example, the holes have the largest (instead of the smallest) possible diameter, the position tolerances remain the same, while clearance increases. table ~ G"A G"` : table Actual Diameter Position Tolerance 5.06 0.02 5.07 0.03 5.08 0.04 5.09 0.05 5.10 0.06 5.10 + 0.14 = 5.24 5.00 + 0.24 = 5.24 4.90 + 0.34 = 5.24 4.80 + 0.44 = 5.24 cover5 ~ G"A cover4 cover3 cover2 cover1 animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref, x, y syserrornumber = 0 x >= 150 -- x = 150 -- y = x + 30 mmPlay clip stage "animationstage" hold notify animerror Repeat Animation The receptacle from the blocks on the preceding page is now toleranced. Now the problem is to determine what position tolerances are allowed for all possible hole diameters in increments of 0.01, filling in the table below. The solution will show that as the diameter of the holes increases, there is a corresponding increase in the position tolerance in order to keep the minimum clearance constant. The increased tolerance facilitates manufacturing and reduces costs.................. manufacturing and reduces costs... step1 The worst case conditions for the pin mount still correspond to the largest allowable pin diameter and the greatest possible pin separation. step2 The size tolerance of the holes remains the same. Maximum diameter: 5.08 + 0.02 = 5.10 Minimum diameter: 5.08 - 0.02 = 5.0666 step3 The position tolerance of 0.02 is applied at the MMC, where the holes are at the smallest possible diameter. (nominal pin size - lower tolerance) = minimum diameter 5.08 - 0.02 = 5.066 = 0.066.20 - 0.04 = 0.0666 (nominal distance - MMC tolerance + (2 x MMC radius)) = extents of holes 20.00 + 0.02 = 20.02 (extents of holes - extents of pins) = clearance 25.04 - 25.04 = 0.00 (MMC tolerance + hole size variation) = new position tolerance 0.02 + 0.01 = 0.03980.02 = 19.984 = 0.0666 step7 Completing the table is now straightforward. If the actual diameter of the holes is increased by 0.01, the position tolerance of the holes is also increased by 0.01. (nominal hole size + upper tolerance) = maximum diameter 5.00 + 0.02 = 5.02 (nominal distance + upper tolerance) = maximum separation 20.00 + 0.02 = 20.02 Extents of pins: 20.02 + 5.02 = 25.04k step4 At the worst-case condition, the holes are at the least possible separation. The parts now fit together with zero clearance, as before. step5 However, if the actual diameter of the holes is 5.07 (the MMC + 0.01), then the position tolerance on the holes is increased by a corresponding amount. step6 At worst case condition, where the holes are at the least allowed separation, the parts now fit together with zero clearance. (nominal separation - new tolerance + 2 x hole radius) = clearance Extents of Holes: 20.00 - 0.03 + 5.07 = 25.04 extents of holes - extents of pins = clearance 25.04 - 25.04 = 0.00 = 0.00 0.000 - 25.04 = 0.00 table table Actual Diameter Position Tolerance 5.04 0.00 5.05 0.01 5.06 0.02 5.07 0.03 5.08 0.04 5.09 0.05 5.10 0.06= 5.24 4.90 + 0.34 = 5.24 4.80 + 0.44 = 5.24 cover5 cover4 cover3 cover2 cover1 cover6 cover7 The size tolerance on the holes may be increased by setting the lower limit on the position tolerance to zero. This example shows how to determine what position tolerances are allowed for all possible hole diameters, in increments of 0.01, for the redesigned part. inst Datum A, then Datum B, and finally Datum C. This procedure is done to prevent any part distortion from causing the part to rock, and possibly altering the measurements. animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref, x, y syserrornumber = 0 x >= 100 -- x = 100 -- y = x + 30 mmPlay clip stage "animationstage" hold notify animerror Repeat Animation step1 The worst case conditions for the pin mount correspond to having the pins at the largest allowable diameter, and at the greatest possible separation. and the greatest possible pin separation. step2 The receptacle has been redesigned so that the diameter of the holes ranges from 5.04 to 5.10. step3 At the MMC, the holes are at the least allowable diameter. (nominal hole size - lower size tolerance) = MMC diameter 5.07 - 0.03 = 5.0466660.066.20 - 0.04 = 0.0666 (nominal distance - MMC tolerance + (2 x MMC radius)) = extents of holes 20.00 - 0.00 + 5.04 = 25.04 (extents of holes - extents of pins) = clearance 25.04 - 25.04 = 0.00 (nominal separation - new tolerance + 2 x radius) = extents of holes 20.00 - 0.01 + 5.05 = 25.04 (extents of holes - extents of pins) = clearance 25.04 - 25.04 = 0.00 (2 x pin radius + separation) = extents of pins Extents of Pins: 5.02 + 20.02 = 25.04 e + upper tolerance) = maximum separation 20.00 + 0.02 = 20.02 Extents of pins: 20.02 + 5.02 = 25.045.04.04 step4 The parts fit together with zero clearance. The holes cannot be moved, and the position tolerance is zero. step5 If the actual diameter of the holes is 5.05, then the position tolerance of the holes is increased to 0.01. The parts now fit together with zero clearance as before. step6 The pattern is now clear. If the actual diameter of the holes increases by 0.01, then the MMC designation on the position tolerance allows this tolerance to be increased by 0.01. The rest of the table is filled in accordingly. a) A clearance fit is required on the larger diameter of the stepped pin. The smaller diameter of the stepped pin must have 0.000 to 0.015 mm diametric clearance under any condition. Example 8.5 Consider the mating parts below. The counterbored hole has a hole diameter of f5 0.010 mm and a counterbore diameter of f10 0.010. The larger diameter of the stepped pin is f9.97 0.010 mm. Concentricity can be held to within 0.002 mm for both the stepped pin and the counterbored hole. Assume that concentricity is the only important feature other than the tolerances on the diameters (e.g. negligible straightness, perpendicularity, etc.). For each case below, determine the diameter and tolerance of the smaller diameter of the stepped pin that will permit the parts to mate as specified. Use bilateral, equally divided tolerances on the dimensions. If the required dimensions and tolerances cannot be found, clearly explain why... suball a) A clearance fit is required on the larger diameter of the stepped pin. The smaller diameter of the stepped pin must have 0.000 to 0.015 mm diametric clearance under any condition. b) A clearance fit is required on the larger diameter of the stepped pin. The smaller diameter of the stepped pin must have 0.005 to 0.030 mm diametric interference under any condition. c) Properly label, on the figure below, both the pin and bores to show the above dimensions, tolerances, and control features so the parts will fit with the desired clearance and interference. backward forward pause audioOn paused thisWav playing buttonClick buttonClick 4thisWav, audioOn, vol mmstatus clip @ = "playing" mmPause h = "paused" mmvolume Ethiswav = mmPlay notify repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, audioon, vol syserrornumber = 0 mmvolume clip mmplay 0 notify f<> 0 audioerror Repeat 1. Notice that the tolerance of the hole is 0.010, or a total tolerance of 0.020. The total tolerance on the clearance fit is only 0.015. The tolerance of the fit is tighter than the tolerance of one of the parts. Thus, no matter how precise the mating part is, the desired fit cannot be achieved. There is no solution for this part of the problem..he problem. 2. For part b), the problem is solved by first temporarily ignoring the concentricity tolerances. They will be added later. of the larger pin is f10.02 0.01. A = 10.03 3. For the least interference at the hole, the smallest pin must be larger than the largest hole by 0.005. Thus the smallest pin size is (5.000 + 0.010) + 0.005 = 5.015 ame as the smallest hole, producing a line-to-line fit. Thus (5.00 - 0.02) = 4.98. 4. For the most interference at the hole, the largest pin must be larger than the smallest hole by 0.030. Thus the largest pin size is (5.000 - 0.010) + 0.030 = 5.020 5. Using bilateral, equally distributed tolerances, the size of the pin at the hole is f5.0175 0.0025 6. Finally, the effect of concentricity must be checked. At the counterbore, the diametric clearance is calculated by (smallest hole) - (largest pin) = (10.00 - 0.01) - (9.97 + 0.01) = 0.01. The radial clearance is 1/2 the diametric clearance. Thus the radial clearance is 0.005. The concentricites tolerances of the pins and the bore are each 0.002, for a total radial displacement of only 0.004. Thus, even with concentricity tolerances, the part will still fit as desired. b) A clearance fit is required on the larger diameter of the stepped pin. The smaller diameter of the stepped pin must have 0.005 to 0.030 mm diametric interference under any condition. "Arial Narrow TITLE SCALE +/- TOLERANCES UNLESS INDICATED DRAWN BY APPROVED BY DATE HIGHER ASM surface quiz1 datumrel nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive quiz1 misc.tbk correct misc.tbk incorrect false misc.tbk incorrect inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.1 For small metal parts, dimensional tolerances of 0.100 mm are easily achievable with standard milling operations. answer True. reference machine nopix nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk correct misc.tbk incorrect false misc.tbk incorrect inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.10 For the part shown, the concentricity specification on the pin will be contained within a diameter of 0.025 when the actual diameter of the pin is 4.93. answer True. reference mmclast nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk incorrect misc.tbk correct false misc.tbk incorrect inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.9 The maximum material condition of a hole occurs when the actual size of the hole is at its largest acceptable dimension. answer False. MMC for a hole occurs when the hole is at is smallest acceptable dimension. reference nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk incorrect misc.tbk correct false misc.tbk incorrect inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.8 For parts fabricated to the tolerances shown, the concentricity specification will not permit all parts to be assembled with clearance on all surfaces on all occasions. answer False. Even when the concentricities are at their limit, the clearances are such that the parts will still fit together with clearance.. reference datumrel nopix buttonclick buttonclick false false buttonclick buttonclick False inconclusive L&inconclusive buttonclick buttonclick inconclusive Inconclusive misc.tbk incorrect misc.tbk correct false misc.tbk incorrect inconclusive thisanim lastanim thiswav lastwav enterpage audiodone animdone -- These scripts are where you the answer 's quiz question ? buttons (on /) will always Dsame message. It's -- just a matter creating u"correct" "incorrect" -- setting inconclusive display F one, since that be chosen. 8"misc.tbk" 4thisanim, lastanim, thiswav, lastwav audiodone animdone Quiz 8.7 A Total Indicated Runout (TIR) specification does not require a datum surface. answer False. The measured surface must be rotated about a cylindrical datum axis. reference datumrel STEEL buttonclick 4base1, base2, low1, low2, high1, high2 4thisAnim, ref "clearance" "interference" "transition" Zreal clear1 v * .1) clear2 = * .1) Cinter1 Ointer2 * .1) * .1) syserrornumber = 0 mmPlay clip stage "animationstage" hold animerror buttonclick loc item1 = item2 = item3 = item4 = loc1 = loc2 = x = ( y = ( "blowup" 8"ch8prob.tbk" currentpage "ex8-1" big1 = picture "ex81" big2 = big3 = big4 = size1 = size2 = targetx = x * ( targety = y * ( /2*15)), (- /2*15)) -- For a child /15 - /15 - a popup - the 35 a kludge! "view" ? + 35 = cursor "mag" = default buttonclick 4currframe, numframes 4thiswav, audioon, vol D <> 0 ("cover" & syserrornumber = 0 <> 1 (mmstatus clip = "playing" = "paused") mmstop wait = "8-1-" & mmvolume mmplay audioerror " & i) normalGraphic = icon "repeat" forwardsml" update buttonclick 4currframe, numframes 4thiswav, vol, audioon D <> 0 ("txt" & "answer" syserrornumber = 0 <> 1 (mmstatus clip = "playing" = "paused") mmstop wait = "8-2-" & mmvolume mmplay audioerror reset update normalGraphic = icon "repeat" forwardsml" buttonclick 4currframe, numframes 4thiswav, vol, audioon J = 0 "suball" q = 2 "subb" "subc" "answer" <> 0 ("txt" & syserrornumber = 0 <> 1 (mmstatus clip = "playing" = "paused") mmstop wait = "8-4-" & mmvolume mmplay audioerror reset update normalGraphic = icon "repeat" forwardsml" buttonclick 4currframe, numframes 4thiswav, audioon, vol D <> 0 ("txt" & "answer" syserrornumber = 0 <> 1 (mmstatus clip = "playing" = "paused") mmstop wait = "8-3-" & mmvolume mmplay audioerror reset update normalGraphic = icon "repeat" forwardsml" 4numframes 4thiswav, thisanim, lastanim, lastwav reset 4currframe, |vol, audioon syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait r<> 0 audioerror ("txt" & i) "suba" "subb" "subc" "suball" "answer" update 4numframes 4thiswav, thisanim, lastanim, lastwav reset 4currframe, |vol, audioon syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait r<> 0 audioerror ("txt" & i) "suba" "subb" "subc" "suball" "answer" update buttonclick 4currframe, numframes 4thiswav, audioon, vol > < 3 I <> 0 ("txt" & = 3 >= 5 "step3" "step4" "txt5" "step5" reset syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait <> 0 = "8-7-" & <> 0 mmvolume mmplay audioerror update normalGraphic = icon "repeat" forwardsml" buttonclick 4currframe, numframes 4thiswav, vol, audioon J = 0 "suball" q = 1 "subb" -- "subc" "answer" <> 0 ("txt" & syserrornumber = 0 <> 1 (mmstatus clip = "playing" = "paused") mmstop wait = "8-5-" & mmvolume mmplay audioerror reset update normalGraphic = icon "repeat" forwardsml" 4thisAnim, lastAnim, thisWav, lastWav audiodone 4thiswav, vol, audioon = "tol2a" syserrornumber = 0 mmvolume clip mmplay 0 notify [<> 0 audioerror animdone mmnotify cref, ccommand, cresult * = "successful" = "tol2b" = "tol2c" = vol 4currframe, numframes, thisAnim, lastanim, thisWav, lastWav lastAnim = thisanim 8 = "tol7a" g = 3 z = 0 update ("play" & & "off") play0 animdone audiodone 4thiswav, vol, audioon syserrornumber = 0 (mmstatus clip M = "playing" c = "paused") mmstop } wait }<> 0 audioerror "step0" play0on "step1" play0off play1 = "tol7b" mmvolume mmplay play1on "step2" play1off play2 = "tol7c" = vol play2on "step3" play2off play3 = "tol7d" = vol play3on -- should called "step4" play3off 4currframe, numframes, thisAnim, lastanim, thisWav, lastWav lastAnim = thisanim 8 = "tol6a" g = 3 z = 0 update ("play" & & "off") play0 animdone audiodone 4thiswav, vol, audioon syserrornumber = 0 (mmstatus clip M = "playing" c = "paused") mmstop } wait }<> 0 audioerror "step0" play0on "step1" play0off play1 = "tol6b" mmvolume mmplay play1on "step2" play1off play2 = "tol6c" = vol play2on "step3" play2off play3 = "tol6d" = vol play3on -- should called "step4" play3off 4currframe, numframes, thisAnim, lastanim, thisWav, lastWav lastAnim = thisanim b = 3 u = 0 update ("play" & & "off") play0 animdone audiodone 4thiswav, audioon, vol = "tol5a" syserrornumber = 0 mmvolume clip mmplay 0 notify [<> 0 audioerror mmnotify cref, ccommand, cresult " = "successful" = "tol5b" vol, (mmstatus = "playing" = "paused") mmstop wait "step0" play0on "step1" play0off play1 = "tol5c" = vol play1on "step2" play1off play2 = "tol5d" = vol play2on "step3" play2off play3 = "tol5e" = vol play3on -- should called = vol "step4" play3off 4thisAnim, thiswav, lastAnim, lastWav, currframe 4numframes ) = thisWav = "tol9a" linkDLL "mmsystem.dll" timegettime () l = 7 update unlinkDLL " (play & & off) "highlight2a" "step0" animdone audiodone -- Following are the scripts animation table # runs hides Lleftovers #on shows results frewind play0on vol, audioon syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait mmvolume mmplay audioerror play1off , bounds1 = "tol9b" "step1txt" "highlight1" starttime = () < + 400 H0, 400 = default play1on , bounds2 = vol "step2txt" play2 = "tol9c" = vol H195, 0 () < + 1000 -- similar , but allows sounds + 1164 " = \ " + 435 play2off play2on = " = " = " = vol H0, 1200 H1950, 0 + 5820 " = + 2175 play3off = "tol9d" = vol U <> 0 "step3txt" "highlight3" play3on play4off = "tol9e" = vol "step4txt" "highlight4" play4on play5off = "tol9f" = vol "highlight5" "step5txt" play5on play6off = "tol9g" = vol "step6txt" "highlight6" play6on "step7txt" "highlight7" play7 = "tol9h" = vol play7off play7on -- shouldn't be called "Should Check buttonclick 4currframe, numframes 4thiswav, audioon, vol > <= 3 H > 0 ("txt" & } - 1 <> 0 = 4 "step5" "step3" "step4" syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait = "8-7-" & <> 0 <> 0 mmvolume mmplay audioerror update buttonclick 4currframe, numframes 4thiswav, audioon, vol 9 > 0 C = 4 "cover5" "cover6" "cover7" "cover8" E" & ("txt" & <> 0 -- PLAY AUDIO syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait = "8-6-" & <> 0 mmvolume mmplay audioerror update buttonclick 4currframe, numframes 4thiswav, vol, audioon 9 > 0 I = 1 "suball" p = 2 "subb" -- "subc" "answer" ("txt" & <> 0 syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait = "8-5-" & <> 0 mmvolume mmplay audioerror update buttonclick 4currframe, numframes 4thiswav, vol, audioon 9 > 0 I = 1 "suball" p = 3 "subb" "subc" "answer" ("txt" & <> 0 syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait = "8-4-" & <> 0 mmvolume mmplay audioerror update 4thisAnim, thisWav, lastWav, lastAnim 6 = "tol16" picture ( thisanim "instruct" animdone audiodone queue movie , thiswav, lastwav, audioon, vol Q & " syserrornumber = 0 & "pic") mmClose clip notify mmopen ~<> 0 animerror (mmstatus = "playing" = "paused") mmstop wait mmvolume mmplay audioerror mmnotify cref, ccommand, cresult 3 = " 9 = "successful" failed - t: " & stage "animationstage" hold : " & 4thisAnim, thisWav, lastWav, lastAnim 6 = "tol17" picture ( thisanim "instruct" animdone audiodone queue movie , thiswav, lastwav, audioon, vol Q & " syserrornumber = 0 & "pic") mmClose clip notify mmopen ~<> 0 animerror (mmstatus = "playing" = "paused") mmstop wait mmvolume mmplay audioerror mmnotify cref, ccommand, cresult 3 = " 9 = "successful" failed - t: " & stage "animationstage" hold : " & buttonclick 4currframe, numframes 4thiswav, vol, audioon D <> 0 ("txt" & l + 1 -- play f clip WAIT syserrornumber = 0 <> 1 (mmstatus = "playing" = "paused") mmstop wait = "8-6-" & mmvolume mmplay ("cover" & starttime = timegettime() !+ 500 yieldApp() = default audioerror reset update normalGraphic = icon "repeat" forwardsml" 4thisAnim, lastAnim, thisWav, lastWav Zcurrpic & "txt") "instruct" picture "picinstruct" audiodone animdone queue movie lastwav, thiswav, audioon, vol, V & " syserrornumber = 0 mmClose clip notify mmopen |<> 0 animerror (mmstatus = "playing" = "paused") mmstop wait mmvolume mmplay audioerror <> "pgrind" <> "cast" <> "extrude" & "pic" mmnotify cref, ccommand, cresult / = " 5 = "successful" -- failed - t: " & stage "animationstage" hold -- : " & 4numframes, currframe 4thiswav, thisanim, lastanim, lastwav b = 0 reset vol, audioon syserrornumber = 0 (mmstatus clip = "playing" = "paused") mmstop wait r<> 0 audioerror ("cover" & i) -- 1 through 4 "(4 + i)) -- 5 ("txt" & i) update 4thisAnim, thisWav, lastWav, lastAnim, currframe, numframes L = "tol25" X"mmsystem.dll" timegettime() k = 0 h = 3 update dequeue audiodone animdone 4thisanim, x, y, ref x = 1 y = 50 # = "mmc" syserrornumber = 0 mmopen clip mmplay stage "animationstage" hold [<> 0 animerror -- Show 4x, y, thiswav, vol, audioon, x = 1 y = 50 = = " yesplay = -- No fields }need be shown num = 1 x = 90 y = 150 = "shmmc1" "step1" "eq1" -- "table" "cover1" x = 150 y = 180 = "shmmc2" "step2" "eq2" -- "cover2" x = 180 y = 210 = "shmmc3" "step3" -- " & num) -- ("eq" & -- Start the sound mmvolume audioerror -- Always Animationstage" notify -- Remove -- -- -- "cover3" "cover4" "cover5" -- ("eq" & (mmstatus = "playing" = "paused") mmstop wait mmnotify cref, ccommand, cresult 4x, y, = "successful" x = 240 x = 210 x = 240 y = 270 starttime = + 1000 yieldapp() x = 180 x = 210 y = 240 + 1000 4thisAnim, thisWav, lastWav, lastAnim, currframe, numframes L = "tol27" X"mmsystem.dll" timegettime() k = 0 h = 7 update dequeue audiodone animdone 4thisanim, x, y, ref x = 1 y = 20 # = "hole2" syserrornumber = 0 mmopen clip mmplay stage "animationstage" hold [<> 0 animerror -- Show 4x, y, thiswav, vol, audioon, x = 1 y = 20 = = " yesplay = -- No fields }need be shown num = 1 x = 210 y = 230 = "mba1" "step1" "eq1" x = 230 y = 260 = "mba2" "step2" "eq2" x = 260 y = 280 = "mba3" "step3" "eq3" x = 280 y = 290 = "mba4" "step4" "eq4" -- "table" "cover1" x = 290 y = 310 = "mba5" "step5" "eq5" -- x = 310 y = 320 = "mba6" "step6" "eq6" -- "cover2" x = 410 y = 440 = "mba7" "step7" -- " & num) -- ("eq" & -- Start the sound mmvolume audioerror -- Always Animationstage" notify -- Remove -- -- -- -- "cover3" "cover4" "cover5" -- ("eq" & (mmstatus = "playing" = "paused") mmstop wait mmnotify cref, ccommand, cresult 4x, y, = "successful" x = 470 x = 440 x = 470 y = 500 starttime = + 1000 yieldapp() ge" x = 410 x = 440 y = 470 + 1000 4thisAnim, thisWav, lastWav, lastAnim, x, y, numframes, currframe R = "tol26" x = y = C = 0 V = 7 update dequeue animdone 4thisanim, x, y, ref x = 1 y = 20 # = "hole1" syserrornumber = 0 mmopen clip mmplay stage "animationstage" hold [<> 0 animerror audiodone -- Show 4x, y, thiswav, vol, audioon, x = 1 y = 20 = = " yesplay = -- No fields }need be shown num = 1 x = 20 y = 150 = "mb1" x = 150 y = 160 = "mb2" x = 160 y = 170 = "mb3" x = 170 y = 180 = "mb4" x = 180 y = 190 = "mb5" -- x = 190 -- y = 200 = "mb6" noanim = x = 190 y = 200 = "mb7" " & num) ("eq" & -- Start the sound mmvolume audioerror -- Always Animationstage" -- Remove ("eq" & (mmstatus = "playing" = "paused") mmstop wait 4thisAnim, thisWav, lastWav, lastAnim, x, y, numframes, currframe R = "tol24" x = y = C = 0 V = 3 update dequeue animdone 4thisanim, x, y, ref x = 1 y = 50 # = "mmcconvent" syserrornumber = 0 mmopen clip mmplay stage "animationstage" hold [<> 0 animerror audiodone -- Show 4x, y, thiswav, vol, audioon, x = 1 y = 50 = = " yesplay = -- No fields }need be shown num = 1 x = 50 y = 90 = "shole1" x = 90 y = 150 = "shole2" x = 150 y = 160 = "shole3" ("eq" & -- Start the sound mmvolume audioerror -- Always Animationstage" -- Remove (num) ("eq" & (mmstatus = "playing" = "paused") mmstop wait 4thisAnim, thisWav, lastWav, lastAnim, currframe, numframes L = "tol28" X"mmsystem.dll" timegettime() k = 0 h = 6 update dequeue audiodone animdone 4thisanim, x, y, ref x = 240 y = 260 & = "hole3" syserrornumber = 0 mmopen clip mmplay stage "animationstage" hold [<> 0 animerror -- Show 4x, y, thiswav, vol, audioon, x = 240 y = 260 @ = " yesplay = -- No fields }need be shown num = 1 x = 260 y = 280 = "mbb1" "step1" "eq1" x = 280 y = 310 = "mbb2" "step2" x = 310 y = 320 = "mbb3" "step3" "eq3" -- x = 100 -- y = 110 = "mbb4" "step4" "eq4" -- "table" "cover1" noanim = x = 320 y = 350 = "mbb5" "step5" "eq5" -- "cover2" x = 350 y = 380 = "mbb6" "step6" -- " & num) -- ("eq" & -- Start the sound mmvolume audioerror -- Always Animationstage" notify -- Remove -- -- -- "cover3" "cover4" "cover5" "cover6" "cover7" -- ("eq" & (mmstatus = "playing" = "paused") mmstop wait mmnotify cref, ccommand, cresult 4x, y, = "successful" x = 470 x = 440 x = 470 y = 500 starttime = + 1000 yieldapp() ge" x = 410 x = 440 y = 470 + 1000 ge" x = 380 x = 410 y = 440 + 1000 x = 350 x = 380 y = 410 + 1000 4base1, base2, low1, high1, low2, high2 4thisAnim, lastAnim, thisWav, lastWav, ref 1 = "tol8" "mmsystem.dll" timegettime() -- Variable key: - shaft (-) - hole (-) ?basic B"9" |" = = 10 B"10" }" = update "low1val" "high1val" "low2val" "high2val" "slider1" reset "transition" "clearance" "interference" B"9" = 10 B"10" audiodone animdone stage "animationstage" = "tryitout" syserrornumber = 0 mmOpen clip mmSeek mmShow Q<> 0 animerror .&, ; .&, ; .&, ; animationstage clearance thisAnim high2 base2 interference base1 high1 animerror transition buttonclick 1,!FX 1,!Fv 1,!Fl blowup ch8prob.tbk ex8-1 buttonclick mouseenter default mouseleave ,%H.% audioon cover forwardsml repeat paused currframe thiswav audioerror numframes playing buttonclick repeat forwardsml currframe numframes update audioon answer forward reset paused currframe thiswav audioerror numframes playing update buttonclick repeat forwardsml currframe numframes update audioon answer forward reset suball paused currframe thiswav audioerror numframes playing update buttonclick repeat forwardsml currframe numframes update audioon answer forward reset paused currframe thiswav audioerror numframes playing update buttonclick repeat forwardsml currframe numframes update thisanim lastanim thiswav numframes lastwav enterpage reset leavepage answer forward suball paused currframe thiswav audioerror playing numframes update reset thisanim lastanim thiswav numframes lastwav enterpage reset leavepage forward suball paused currframe thiswav audioerror playing numframes update reset step5 audioon forward step4 reset paused currframe thiswav audioerror numframes playing step3 update buttonclick repeat forwardsml currframe numframes update audioon forward reset suball paused currframe thiswav audioerror numframes playing update buttonclick repeat forwardsml currframe numframes update thisAnim lastAnim thisWav lastWav enterpage leavepage audioon tol2a thiswav audioerror audiodone animdone audioon successful tol2a thiswav audioerror tol2b tol2c mmnotify tol7a forward thisAnim lastanim currframe thisWav numframes lastWav update enterpage play0 forward currframe update leavepage animdone audiodone tol7a paused step0 thiswav audioerror playing play0 tol7a paused step1 step0 thiswav audioerror playing play0on step1 play0off audioon tol7b paused step1 step0 thiswav audioerror playing play1 audioon step2 tol7b paused step1 thiswav audioerror playing play1on step1 play1off audioon step2 tol7c paused step1 thiswav audioerror playing play2 audioon step2 tol7c paused thiswav audioerror playing step3 play2on step2 play2off audioon step2 tol7d paused thiswav audioerror playing step3 play3 audioon tol7d paused step4 thiswav audioerror playing step3 play3on step3 play3off tol6a forward thisAnim lastanim currframe thisWav numframes lastWav update enterpage play0 forward currframe update leavepage animdone audiodone tol6a paused step0 thiswav audioerror playing play0 tol6a paused step1 step0 thiswav audioerror playing play0on step1 play0off audioon tol6b paused step1 step0 thiswav audioerror playing play1 audioon step2 tol6b paused step1 thiswav audioerror playing play1on step1 play1off audioon step2 tol6c paused step1 thiswav audioerror playing play2 audioon step2 tol6c paused thiswav audioerror playing step3 play2on step2 play2off audioon step2 tol6d paused thiswav audioerror playing step3 play3 audioon step4 tol6d paused thiswav audioerror playing step3 play3on step3 play3off H%l F mmsystem.dll forward thisAnim lastAnim tol9a currframe thiswav timegettime numframes lastWav update enterpage .&, " mmsystem.dll forward step0 currframe highlight2a update leavepage animdone audiodone audioon paused tol9a step0 thiswav audioerror highlight2a playing LGplay1off play0on paused tol9a step0 thiswav audioerror highlight2a playing play0 .&+ +E starttime audioon step1txt paused bounds1 highlight1 step0 thisWav timegettime audioerror tol9b playing default play1 step1txt bounds1 highlight1 play1off audioon step2txt highlight2 step1txt paused bounds1 highlight1 thiswav audioerror tol9b playing bounds2 play1on .&+ +E starttime audioon step2txt highlight2 step1txt paused highlight1 thisWav timegettime audioerror playing default tol9c bounds2 play2 step2txt highlight2 bounds1 highlight1 bounds2 play2off audioon step2txt highlight2 paused bounds1 highlight1 Hplay3off thiswav audioerror highlight2a playing tol9c bounds2 play2on audioon step2txt paused thiswav audioerror highlight2a playing step3txt tol9d Gplay2off step3 highlight3 play3 step3txt step3 highlight3 play3off play3 {Hplay4off play3on audioon step4txt step4 highlight4 paused Hplay3off tol9e thiswav audioerror playing play4 step4txt step4 highlight4 play4off Hplay5off play4 play4on audioon tol9f paused step5txt step5 highlight5 thiswav audioerror playing {Hplay4off play5 step5txt step5 highlight5 play5off play5 OIplay6off play5on audioon Hplay5off paused thiswav audioerror step6txt tol9g step6 playing highlight6 play6 step6txt step6 highlight6 play6off step6 step7txt step6txt highlight7 highlight6 play6on tol9h audioon step6 paused thiswav step7txt audioerror step6txt highlight7 playing highlight6 play7 step7txt highlight7 play7off Should not be called. Check the scripts. play7on step5 audioon forward step4 paused currframe thiswav audioerror playing step3 update buttonclick cover audioon cover8 cover5 forward cover7 cover6 paused currframe thiswav audioerror playing update buttonclick audioon forward suball paused currframe thiswav audioerror playing update buttonclick audioon answer forward suball paused currframe thiswav audioerror playing update buttonclick thisAnim lastAnim tol16 thisWav lastWav enterpage thisAnim instruct leavepage animdone audiodone audioon thisAnim lastAnim paused instruct thiswav audioerror playing lastwav animerror queue animationstage mmopen mmClose thisAnim mmopen failed - result: successful mmClose failed - result: animerror mmnotify thisAnim lastAnim tol17 thisWav lastWav enterpage thisAnim instruct leavepage animdone audiodone audioon thisAnim lastAnim paused instruct thiswav audioerror playing lastwav animerror queue animationstage mmopen mmClose thisAnim mmopen failed - result: successful mmClose failed - result: animerror mmnotify .&+ +E ,%H.% starttime audioon cover }gyieldApp forward reset paused currframe thiswav timegettime audioerror numframes playing default update buttonclick repeat forwardsml currframe numframes update thisAnim lastAnim thisWav lastWav enterpage picinstruct thisAnim picture instruct currpic leavepage audiodone animdone picinstruct audioon thisAnim pgrind picture lastAnim paused instruct thiswav audioerror extrude playing lastwav animerror currpic queue animationstage mmopen mmClose thisAnim successful animerror mmnotify thisanim lastanim currframe thiswav numframes lastwav enterpage reset leavepage V, #> cover forward paused currframe thiswav audioerror playing numframes update reset tol25 mmsystem.dll forward thisAnim lastAnim currframe thisWav timegettime numframes lastWav update enterpage n;dequeue mmsystem.dll forward currframe update leavepage audiodone animationstage thisanim animerror animdone cover1 audioon tol25 false step2 shmmc2 step1 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animationstage hole1 thisanim animerror animdone audiodone Animationstage audioon false tol26 thisanim noanim thiswav audioerror yesplay animerror queue paused thiswav audioerror playing dequeue forward thisAnim lastAnim currframe thisWav numframes lastWav tol24 update enterpage n;dequeue currframe leavepage animationstage thisanim mmcconvent animerror animdone audiodone Animationstage audioon false thisanim shole1 shole2 thiswav audioerror yesplay shole3 tol24 animerror queue paused thiswav audioerror playing dequeue high2val slider1 low2val mmsystem.dll high1val thisAnim lastAnim low1val update high2 shaft thisWav timegettime base1 high1 lastWav high1 high2 base2 enterpage high1 high2val clearance low2val high1val high2 low1val update base2 shaft interference base1 high1 slider1 high2 transition reset mmsystem.dll reset leavepage audiodone animationstage tryitout thisAnim animerror animdone drill extrude grind lathe tryitout datum1 datum2 datum3 datum4 cylindricity flatness perpendicularity runout straightness angularity roundness parallelism concentricity convent mmcconvent hole1 hole2 hole3 sawwav drillwav lathewav millwav grindwav edmwav extrudewav castwav 8-1-1 8-1-2 8-1-3 8-1-4 8-1-5 8-2-1 8-2-2 8-2-3 8-3-1 8-3-2 8-3-3 8-4-1 8-4-2 8-4-3 8-4-4 8-4-5 8-4-6 8-4-7 8-4-8 8-5-1 8-5-2 8-5-3 8-5-4 8-5-5 8-5-6 8-6-1 8-6-2 8-6-3 8-6-4 8-7-1 8-7-2 8-7-3 8-7-4 8-7-5 8-7-6 8-7-7 8-7-8 pgrind tol2a tol2b tol2c pgrindwav tol5a tol5b tol5c tol5d tol5e tol6a tol6b tol6c tol6d tol7a tol7b tol7c tol7d tol9a tol9b tol9c tol9d tol9e tol9f tol9g tol9h tol10 tol11 tol12 tol13 tol14 tol15 tol16 tol17 tol23 tol24 tol25 tol26 tol27 tol28 straightnesswav flatnesswav roundnesswav cylindricitywav angularitywav perpendicularitywav parallelismwav concentricitywav runoutwav tirwav objectives shole1 shole2 shole3 shmmc1 shmmc2 shmmc3 4numf anim\cast.flc anim\drill.flc anim\edm.flc anim\extrude.flc anim\grind.flc anim\mill.flc anim\lathe.flc anim\saw.flc anim\tryitout.flc 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8-5-4 audio\8-5-5.wav 8-5-5 audio\8-5-6.wav 8-5-6' audio\8-6-1.wav 8-6-1{ audio\8-6-2.wav 8-6-2 audio\8-6-3.wav 8-6-3# audio\8-6-4.wav 8-6-4w audio\8-7-1.wav 8-7-1 audio\8-7-2.wav 8-7-2 audio\8-7-3.wav 8-7-3s audio\8-7-4.wav 8-7-4 audio\8-7-5.wav 8-7-5 audio\8-7-6.wav 8-7-6o audio\8-7-7.wav 8-7-7 audio\8-7-8.wav 8-7-8 anim\fgrind.flc rind.k audio\page2a.wav audio\page2b.wav audio\page2c.wav audio\page3a.wav audio\fgrind.wav audio\page5a.wav audio\page5b.wav audio\page5c.wav audio\page5d.wav audio\page5e_a.wav audio\page6a.wav audio\page6b.wav audio\page6c.wav audio\page6d_a.wav audio\page7a.wav audio\page7b.wav audio\page7c.wav audio\page7d_a.wav audio\page8a.wav audio\page9a.wav audio\page9b.wav audio\page9c.wav audio\page9d.wav audio\page9e.wav audio\page9f.wav audio\page9g.wav audio\page9h.wav audio\page10.wav audio\page11.wav audio\page12.wav audio\page13.wav audio\page14.wav audio\page15.wav audio\page16.wav audio\page17.wav audio\page23.wav audio\page24.wav audio\page25.wav audio\page26.wav audio\page27.wav audio\page28.wav audio\sf1.wav \sf1.wa3(* audio\sf2.wav audio\sf3.wav audio\sf4.wav audio\sf5.wav audio\sf6.wav audio\sf7.wav audio\sf8.wav audio\sf9.wav audio\sf10.wav audio\chapter8.wav audio\shole1.wav audio\shole2.wav audio\shole3.wav audio\shmmc1.wav audio\shmmc2.wav audio\shmmc3.wav audio\mb1.wav audio\mb2.wav audio\mb3.wav audio\mb4.wav audio\mb5.wav audio\mb6.wav audio\mb7.wav audio\mba1.wav audio\mba2.wav audio\mba3.wav audio\mba4.wav audio\mba5.wav audio\mba6.wav audio\mba7.wav audio\mbb1.wav audio\mbb2.wav audio\mbb3.wav audio\mbb4.wav audio\mbb5.wav audio\mbb6.wav mmsystem.dll thisAnim lastAnim thisWav timegettime lastWav enterpage animdone audiodone 4thisAnim, lastAnim, thisWav, lastWav, ref < = "tpt" 4 = "tolb" "mmsystem.dll" timegettime() animdone audiodone True position tolerancing has an allowable error which is independent of direction. Notice that the shape outlined by the allowable location for the center of the hole is a circle, whose perimeter points are equidistant from the center. The circle circumscribes the square and is 57% larger than the square on the previous page. This technique maximizes the allowable error within a given tolerance and still allows the mating parts to fit. The increased tolerance facilitates manufacturing, and reduces costs.osts. animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation datums tol12 datum1 thisAnim lastAnim thisWav lastWav enterpage animdone audiodone 4thisAnim, thisWav, lastWav, lastAnim 7 = "datum1" < = "tol12" animdone audiodone Datums Datums are reference surfaces from which measurements are made. They are identified by capital letters between dashes and framed by a rectangle. A datum is usually defined in a view where it appears in edge view. In this case, the datum is the best plane placed on the face of the box. animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation thisAnim lastAnim tol13 thisWav lastWav datum2 enterpage animdone audiodone 4thisAnim, thisWav, lastWav, lastAnim 7 = "datum2" < = "tol13" animdone audiodone This datum is the best cylinder that just fits over the shaft. animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation tol14 thisAnim lastAnim datum3 thisWav lastWav enterpage animdone audiodone 4thisAnim, thisWav, lastWav, lastAnim 7 = "datum3" < = "tol14" animdone audiodone A hole is the inverse of a shaft, and has a corresponding cylindrical datum that just fits inside the hole. animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation 3datums datum4 thisAnim tol15 lastAnim thisWav lastWav enterpage animdone audiodone 4thisAnim, thisWav, lastWav, lastAnim 7 = "datum4" < = "tol15" animdone audiodone Multiple datums may be used on a single object. In this three plane datum system, the planes are at right angles to each other. The order of the planes is important, since the object contacts the first plane at three points, the second at two, and the third at only one point. This procedure is done to prevent any rocking due to part distortion, which could possibly alter the measurements...g the measurements.es are not required to be perpendicular to each other..... Datum surfaces are not required to be perpendicular to each other..... animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation surface cylindricitypic roundnesspic flatnesspic 0Modern o9_(o9 $o9v% o91&o9 o9H#o9 instruct Surface features may be dimensioned and toleranced independently or relative to datums. Examples of each type of feature are shown above with the standard shorthand symbols used to represent them. The limits are denoted by the green surfaces. Surface Features Independent Features animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation straightness queue buttonclick mouseenter default mouseleave buttonclick queue "straightness" = cursor = default Straightness queue cylindricity buttonclick mouseenter default mouseleave buttonclick queue "cylindricity" = cursor = default Cylindricity roundness queue buttonclick mouseenter default mouseleave buttonclick queue "roundness" = cursor = default Roundness flatness queue buttonclick mouseenter default mouseleave buttonclick queue "flatness" = cursor = default Flatness straightnesspic *o#U- 0Modern 5'G#5'O( 3G#5'G# ;,G#;,O( Click one of the above shorthand symbols.for an explanation. straightness buttonclick buttonclick The designated feature elements must lie within a region bounded by parallel lines that are separated by the given tolerance. For instance, for a straightness tolerance of f0.20 on the diameter of the circle, the axis of the cylinder must lie within a cylindrical region of diameter 0.20. - click to remove - F%*%C% - click to remove - flatness buttonclick buttonclick - click to remove - The specified surface must lie between two parallel planes separated by the given tolerance at all points on the surface. The parallel planes do not have to be perpenidcular or parallel to any feature of the given part, but they must be parallel to each other. roundness buttonclick buttonclick - click to remove - The boundary of every cross-sectional element of the indicated feature (e.g. a cylinder) must lie within two concentric circles whose radii differ by the specified tolerance. Each cross-section is considered independent of the others. cylindricity buttonclick buttonclick b-F-_- - click to remove - The specified cylindrical surface must lie between two concentric cylinders whose radii differ by the indicated tolerance. The axes of these cylinders do not have to be aligned with any element of the part. datumrel animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation concentricitypic perpendicularitypic angularitypic parallelismpic perpendicularity queue buttonclick mouseenter default mouseleave buttonclick queue "perpendicularity" = cursor = default Perpendicularity Related to Datums angularity queue buttonclick mouseenter default mouseleave buttonclick queue "angularity" = cursor = default Angularity parallelism queue buttonclick mouseenter default mouseleave buttonclick queue "parallelism" = cursor = default Parallelism concentricity queue buttonclick mouseenter default mouseleave buttonclick queue "concentricity" = cursor = default Concentricity runout queue buttonclick mouseenter default mouseleave buttonclick queue "runout" = cursor = default Runout queue buttonclick mouseenter default mouseleave buttonclick queue "tir" = cursor = default runoutpic tirpic Click one of the above shorthand symbols.for an explanation. instruct These are features which are toleranced relative to datums, which are shown in purple. The limits are denoted by the green surfaces. angularity buttonclick buttonclick The indicated surface must lie between two parallel planes that are separated by the given tolerance and form the desired angle with the referenced datum plane. lerance of f0.20 on the diameter of the circle, the axis of the cylinder must lie within a cylindrical region of diameter 0.20. - click to remove - H&,&E& - click to remove - perpendicularity buttonclick buttonclick For surfaces, the specified surface must lie between two parallel planes that are separated by the given tolerance and are perpendicular to the referenced datum plane. For cylinders, the nominal cylinder axis must lie within a cylindrical region that has a diameter given by the tolerance and is perpendicular to the referenced dtum plane. - click to remove - parallelism buttonclick buttonclick The indicated surface must lie between two parallel planes that are separated by the given tolerance and are parallel to the referenced datum plane. aightness tolerance of f0.20 on the diameter of the circle, the axis of the cylinder must lie within a cylindrical region of diameter 0.20. - click to remove - n-R-k- - click to remove - concentricity buttonclick buttonclick All points of the axis of the specified cylinder must lie within a cylindrical region whose diameter is given by the tolerance. The axis of the cylinder and the axis of the referenced cylindrical datum must coincide. of the cylinder must lie within a cylindrical region of diameter 0.20. - click to remove - - click to remove - runout buttonclick buttonclick A dial indicator is placed on the part normal to the true geometric shape of the specified surface, and the part is rotated 360 degrees about the axis of the referenced datum. The total indicator movement must be less than the given tolerance. This is done for each section of the surface. click to remove - l4P4i4 - click to remove - buttonclick buttonclick Runout is measured for each section of the surface. The difference between the maximum and minimum indicator readings for the entire procedure must be less than the tolerance. (These readings may occur at different points along the feature.) ithin a cylindrical region of diameter 0.20. - click to remove - - click to remove - thisAnim tol23 lastAnim thisWav lastWav enterpage stage2 stage1 animerror animdone audiodone animerror leavepage 4thisAnim, thisWav, lastWav, lastAnim 6 = "tol23" animdone syserrornumber = 0 mmopen clip "hmmc" smmc" mmplay stage "stage1" hold p<> 0 animerror audiodone mmisopen mmclose wait Maximum Material Condition (MMC)a stage1 stage1 animerror buttonclick buttonclick syserrornumber = 0 mmPlay clip "smmc" stage "stage1" hold G<> 0 animerror Repeat Animation The maximum material condition is the state in which the part, not the feature, is at its greatest volume. For a shaft, this occurs at its largest dimension, while for a hole, it is when the feature is at its smallest. stage2 stage2 animerror buttonclick buttonclick syserrornumber = 0 mmPlay clip "hmmc" stage "stage2" hold G<> 0 animerror Repeat Animation (minimum hole size - largest shaft - straightness tolerance) = minimum clearance 5.30 - 5.20 - 0.04 = 0.06 Using Maximum Material Condition= The shaft from the preceding page may be toleranced using the Maximum Material Condition to keep the clearance constant. The stated straightness tolerance is now applied only at the maximum material condition. As the shaft diameter decreases, the starightness tolerance is increased. The steps in the following example illustrate this by showing how to fill in the tolerance table show below. As the solution progresses, notice that the minimum clearance must remain constant.a animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref, x, y syserrornumber = 0 x >= 180 x = 180 y = x + 30 mmPlay clip stage "animationstage" hold notify animerror Repeat Animation step3 The rest of the table is completed by the same method. As the actual diameter is decreased by 0.10, the straightness tolerance is increased by 0.10. Using MMC tolerances allows for variable tolerancing. The parts still fit with the proper clearance, and the shaft is easier to manufacture to specifications. to specifications. (minimum hole size - shaft size - straightness tolerance) = minimum clearance 5.30 - 5.10 - 0.14 = 0.06.06 table table Diameter Straightness Tolerance Total 5.20 + 0.04 = 5.24 5.10 + 0.14 = 5.24 5.00 + 0.24 = 5.24 4.90 + 0.34 = 5.24 4.80 + 0.44 = 5.24 5555555 cover5 cover4 cover3 cover2 cover1 step2 If the diameter of the shaft is decreased by 0.10, the straightness is not held constant, but rather increased by an equivalent amount. Notice that the clearance returns to its original size. step1 The straightness of the shaft is toleranced to 0.04 at the MMC, when the shaft is at its maximum allowable diameter. The minimum clearance is 0.06, as before. c.wav a) Determine the minimum nominal size of diameter A, the large pin diameter, e.g. A 0.01 mm.. b) Determine the maximum nominal size of diameter B, the small pin diameter, e.g. B 0.01 mm.. c) Properly label, on the figure below, both the pin and bores to show the above dimensions, tolerances, and control features so the parts will fit with the desired clearance and interference.erence.. Example 8.4 The pin assembly shown below requires a clearance fit on the smaller diameter, and an interference fit on the larger diameter. The nominal diameter of the bore and counterbore are 5 mm and 10 mm, respectively. The tolerances are 0.02 mm on the bore diameters, and 0.01 mm on the pin diameters. Assume that concentricity is the only important feature other than the tolerances on the diameters (e.g. negligible straightness, perpendicularity, etc.). The tightest concentricity that can be held is 0.005 mm for the pins, and 0.01 mm for the bores. suball a) Determine the minimum nominal size of diameter A, the large pin diameter, e.g. A 0.01 mm. b) Determine the maximum nominal size of diameter B, the small pin diameter, e.g. B 0.01 mm. c) Properly label, on the figure below, both the pin and bores to show the above dimensions, tolerances, and control features so the parts will fit with the desired clearance and interference.erence...................... backward forward pause audioOn paused thisWav playing buttonClick buttonClick 4thisWav, audioOn, vol mmstatus clip @ = "playing" mmPause h = "paused" mmvolume Ethiswav = mmPlay notify repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, audioon, vol syserrornumber = 0 mmvolume clip mmplay 0 notify f<> 0 audioerror Repeat 1. The clearance fit on the smaller diameter does affect the interference fit on the larger diameter, so the calculation of the larger diameter should be done first. To guarantee an interference fit at A, its minimum size must be the same as the largest hole, producing a line-to-line fit. Thus (10.00 + 0.02) = 10.02......02 2. Since the tolerance of A is 0.01, the lower limit is added to the minimum size pin to get the nominal size. Thus the size of the larger pin is f10.02 0.01. A = 10.03 3. Now consider the clearance fit at B. Temporarily, ignore the concentricity tolerances. They will be added later. To guarantee clearance at B, its maximum size must be the same as the smallest hole, producing a line- to-line fit. Thus (5.00 - 0.02) = 4.98. 4. Since the tolerance of B is 0.01, the upper limit is subtracted from the maximum size to get the nominal size. Thus the size of the smaller pin, not considering concentricity tolerances, is f4.97 0.01. 5. Now the concentricity tolerances must be added. The centers of the pins can be non-concentric by up to 0.005 radially. To compensate, the diameter of the smaller pin must be reduced by 2 x 0.005 = 0.010 to guarantee a clearance fit. 6. The centers of the bores can also be non-concentric by up to 0.01 radially. To compensate, the diameter of the smaller pin must be reduced by an additional 2 x 0.01 = 0.02 to guarantee a clearance fit. 7. The combination of the concentricity tolerances of the pins and the bore reduce the smaller pin diameter by a total of 0.03. Thus the smaller pin size must be f4.94 0.01. B = 4.94.. 8. The drawing is labeled as shown. Note the use of datums for the concentricity specifications... answer thisAnim lastAnim objectives thisWav lastWav enterpage leavepage audiodone animdone 4thisAnim, lastAnim, thisWav, lastWav , = "objectives" audiodone animdone Chapter 8: Geometric Tolerancingg The objectives of this chapter are to: Define various methods for specifying the tolerances on object dimensions. Demonstrate proper tolerancing techniques to achieve desired fits on mating objects. Show examples of basic fabrication techniques to produce guidelines in practical tolerancing. buttonclick buttonclick Begin Lesson backward n;dequeue forward currframe queue update buttonClick buttonClick 4currframe, numframes dequeue 7 - 1 update buttonclick forward n;dequeue forward currframe numframes queue update buttonClick repeat forwardsml currframe numframes update buttonClick 4currframe, numframes dequeue 7 + 1 D = 0 update buttonclick normalgraphic = icon "repeat" forwardsml" pause audioOn paused thisWav audioerror playing buttonClick buttonClick 4thisWav, audioOn, vol syserrornumber = 0 mmstatus clip ^ = "playing" mmPause = "paused" mmvolume Ethiswav = mmPlay notify audioerror repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, currframe, audioon, vol syserrornumber = 0 mmvolume clip mmPlay thisWav `<> 0 audioerror Repeat tolerancing step0 machine datums step0 shaft In an interference fit, the shaft is larger than the hole. To put the two together may require force, or thermal treatment, where the hole is heated to make it expand, or the shaft is cooled to make it contract. The hole below is the same as on the preceding page, 10 0.2 mm. However, the shaft must now be sized to produce an interference fit with an allowance of 1.2 mm and a tolerance of 0.6 mm. Interference Fitt step0 Start by clicking the forward arrow below.ow. step1 The lower limit of the hole is found by subtracting the tolerance from the basic size. 10 - 0.2 = 9.8 mm >= 9.8 mm= 9.8 mm step3 The lower limit of the shaft is found by subtracting the tolerance from the upper limit of the shaft. 11.0 - 0.6 = 10.4 mm Therefore, the diameter of the shaft is described by a bilateral and equal tolerance of 10.7 mm 0.3 mm. 10.7 0.3 mm mmmmmm step2 The upper limit of the shaft is found by adding the allowance to the lower limit of the hole.e.hole. 9.8 + 1.2 = 11.0 mm <= 11.0 mm 9.8 mm 1.2 mm mm= 9.8 mm backward V, #> forward currframe update buttonClick buttonClick 4currframe, numframes (play & ( ) - 1) & on) C - 1 update buttonclick forward V, #> play0 forward currframe numframes update buttonClick repeat forwardsml currframe numframes update buttonClick 4currframe, numframes (play & ( * + 1)) ? + 1 E & off) play0 o = 0 update buttonclick normalgraphic = icon "repeat" forwardsml" pause audioOn paused thisWav audioerror playing buttonClick buttonClick 4thisWav, audioOn, vol syserrornumber = 0 mmstatus clip ^ = "playing" mmPause = "paused" mmvolume Ethiswav = mmPlay notify audioerror repeat audioon thiswav audioerror buttonclick buttonclick 4thiswav, audioon, vol syserrornumber = 0 mmvolume clip mmplay 0 notify f<> 0 audioerror Repeat shaft A fit describes the way two or more parts come together. There are three main types of fits. In a clearance fit, one part is always smaller than the other, so there is space between the two. In contrast, for an interference fit, the inner part is larger than the outer one. The minimum clearance or maximum interference between the two parts is called the allowance. When the allowance is equal to zero, the part sizes match exactly and a "line-to-line" fit is produced. In the example below, there is a hole with a basic size of 10 mm, 0.2 mm. The shaft must be sized with a tolerance of 0.4 mm to provide a clearance fit with an allowance of 0.1 mm. step0 step0 Start by clicking the forward arrow below.ow. step1 The lower limit of the hole is found by subtracting the tolerance from the basic size. 10 - 0.2 = 9.8 mm >= 9.8 mm= 9.8 mm step3 The lower limit of the shaft is found by subtracting the tolerance from the upper limit of the shaft. 9.7 - 0.4 = 9.3 mm Therefore, the diameter of the shaft is described by a bilateral and equal tolerance of 9.5 mm, 0.2 mm.................................. 0.2 mm.8 mm step2 The upper limit of the shaft is found by subtracting the allowance from the lower limit of the hole. 9.8 - 0.1 = 9.7 mm <= 9.7 mm= 9.8 mm 0.1 mm mm= 9.8 mm shaft A transition fit describes two parts which are sometimes in interference and sometimes in clearance. The hole below is the same one used on the previous examples. The shaft is to be sized to produce a transition fit with an allowance of 0.1 mm and a tolerance of 0.4 mm. Transition Fititt step0 step0 Start by clicking the forward arrow below.ow. step1 The lower limit of the hole is found by subtracting the tolerance from the basic size. 10 - 0.2 = 9.8 mm >= 9.8 mm= 9.8 mm step3 The lower limit of the shaft is found by subtracting the tolerance from the upper limit of the shaft. 9.9 - 0.4 = 9.5 mmmm Therefore, the diameter of the shaft is described by a bilateral and equal tolerance of 9.7 mm 0.2 mm...bilaterally, is 9.7 mm, 0.2 mm. 0.2 mmm mmmmmm step2 The upper limit of the shaft is found by adding the allowance to the lower limit of the hole.e.hole. 9.8 + 0.1 = 9.9 mmm <= 9.9 mmm 9.8 mm 0.1 mm0 mm 9.8 mm convent mmsystem.dll thisAnim lastAnim thisWav timegettime lastWav enterpage animdone audiodone 4thisAnim, lastAnim, thisWav, lastWav, ref < = "convent" 8 = "tola" "mmsystem.dll" timegettime() animdone audiodone In conventional tolerancing, the allowable error is dependent on its direction. The shape traced by the allowable location for the center of the hole is a rectangle. Its diagonals are longer than its length and breadth. A rectangular area is traced by the allowable locations of the center of the hole. Note that its diagonals are longer than either its length or breadth. animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation True Position Tolerancing Compute! animationstage Try it out! Adjust the basic sizes and upper and lower tolerances for a shaft and hole system to create clearance, interference, and transition fits. shaft high1 high1val update high1 update notifybefore update 4high1 = "."& ,!J " high1val slider1 high1 high1 update buttonclick ,!J " ,!J2Y starttime high1val slider1 timegettime high1 high1 update buttonstilldown buttonclick 4high1 update "slider1" "high1val" f < 9 t + 1 starttime = timegettime() + 50 4temp = rgbfill = 196, 0,0 high1val slider1 high1 high1 update buttonclick ,!J2Y starttime high1val slider1 timegettime high1 high1 update buttonstilldown buttonclick 4high1 update slider1 "high1val" d > 0 r - 1 starttime = timegettime() + 50 4temp = rgbfill = 196, 0,0 high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos high1 enterdrop high1 leavedrop high1 high1 objectdropped enterdrop obj = "high1" leavedrop B = " objectdropped p, pos slider1 high1 high1 update update 4high1 low1val update update notifybefore update 4low1 = "."& enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low1 xL>M\ slider1 update update 4low1 ,!J " low1val slider1 update buttonclick ,!J " ,!J2Y starttime low1val slider1 timegettime update buttonstilldown buttonclick 4low1 update slider1 "low1val" b < 9 p + 1 starttime = timegettime() + 50 4temp = rgbfill = 196, 0,0 low1val slider1 update buttonclick ,!J2Y starttime low1val slider1 timegettime update buttonstilldown buttonclick 4low1 update slider1 "low1val" b > 0 p - 1 starttime = timegettime() + 50 4temp = rgbfill = 196, 0,0 base1 base1 buttonclick buttonclick 4base1 base1 buttonclick buttonclick 4base1 = 10 Nominal Size Shaft +txtUY high2 high2val update high2 update notifybefore update 4high2 = "."& ,!J " high2 high2val high2 slider1 update buttonclick ,!J " ,!J2Y starttime high2 high2val high2 slider1 timegettime update buttonstilldown buttonclick 4high2 update "slider1" "high2val" f < 9 t + 1 starttime = timegettime() + 50 4temp = rgbfill = 196, 0,0 high2 high2val high2 slider1 update buttonclick ,!J2Y starttime high2 high2val high2 slider1 timegettime update buttonstilldown buttonclick 4high2 update slider1 "high2val" d > 0 r - 1 starttime = timegettime() + 50 4temp = rgbfill = 196, 0,0 high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos high2 enterdrop high2 leavedrop high2 high2 objectdropped enterdrop obj = "high2" leavedrop B = " objectdropped p, pos slider1 high2 high2 update update 4high2 low2val update update notifybefore update 4low2 = "."& enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 enterdrop leavedrop objectdropped enterdrop obj leavedrop objectdropped v, pos 4low2 slider1 update update 4low2 ,!J " low2val slider1 update buttonclick ,!J " ,!J2Y starttime low2val slider1 timegettime update buttonstilldown buttonclick 4low2 update slider1 "low2val" b < 9 p + 1 starttime = timegettime() + 50 4temp = rgbfill = 196, 0,0 low2val slider1 update buttonclick ,!J2Y starttime low2val slider1 timegettime update buttonstilldown buttonclick 4low2 update slider1 "low2val" Z > 0 h - 1 starttime = timegettime() + 50 4temp = rgbfill = 196, 0,0 base2 base2 buttonclick buttonclick 4base2 base2 buttonclick buttonclick 4base2 = 10 Nominal Size clearance Clearance interference Interference transition Transition table step0 Standard fits for cylindrical parts are recommended by ANSI. Sizes, allowances, and tolerances are given in tables. In this example, the table above is used to apply tolerances to a 1/2 inch diameter hole for a given shaft to obtain a class RC-2 (running and clearance) fit. ANSI has separate tables for each type of fit.s separate tables for each type of fit.... step3 0.5000 + 0.0004 = 0.5004 step4 0.5000 + 0.0000 = 0.5000 step5 0.50000 + -0.00025 = 0.49975000 step6 0.50000 + -0.00055 = 0.49945000 step1txt The hole has a basic diameter of inch, which is in the 0.40 - 0.71 inch range. step2txt The type of clearance fit desired is of class RC 2. The classes vary in the tightness of the fit. step3txt To find the upper limit of the hole, the upper tolerance is added to the basic size. Note that values shown in the table are in thousandths of an inch. step4txt The lower tolerance is added to the basic size to find the lower limit of the hole. The hole can now be described by a bilateral and equal tolerance of 0.5002 0.0002. step5txt The upper limit of the shaft is found by adding the upper tolerance to the basic size. step6txt The lower tolerance is added to the basic size to find the lower limit of the shaft. Tolerance Tables highlight1 highlight2 highlight4 highlight5 highlight6 highlight2a highlight3 highlight7 step7txt The shaft can now be described by a bilateral and equal tolerance of 0.4996 0.00015. Notice that the loosest and the tightest fits are both clearance, and match the maximum and minimum amounts of clearance given on the table. :HDMEDIAPATH pagesys.sbk =^addToSysBooks thisAnim Tolerancing thisWav statusBar menusys.sbk enterBook addToSysBooks 4thisWav, thisAnim addToSysBooks "pagesys.sbk" menusys. statusBar = "Tolerancing" nnewBook System Arial Arial Arial :CDMEDIAPATH Arial Arial Arial Arial TBKWidgets Arial New Roman Arial Arial New Roman Arial P E:\ Arial New Roman Arial New Roman Arial Arial Symbol ew Roman Symbol Times New Roman Times New Roman Arial New Roman Symbol ew Roman Arial New Roman Arial Graphics Interactive ~(z(z( machine instruct Click one of the above machining processes.r an explanation. animationstage animationstage thisAnim animerror buttonclick buttonclick 4thisAnim, ref syserrornumber = 0 mmPlay clip stage "animationstage" hold N<> 0 animerror Repeat Animation Machining Processes pgrind queue buttonclick buttonclick queue "pgrind" Precision Grinding Sawing queue buttonclick buttonclick queue "saw" Milling queue buttonclick buttonclick queue "mill" Extrusion extrude queue buttonclick buttonclick queue "extrude" Drilling drill queue buttonclick buttonclick queue "drill" Lathe lathe queue buttonclick buttonclick queue "lathe" Grinding queue grind buttonclick buttonclick queue "grind" Casting queue buttonclick buttonclick queue "cast" EDMSawing queue buttonclick buttonclick queue "edm" sawtxt In sawing, a moving blade with teeth removes material completely through the part along a line. The blade is usually a band or circular in shape. For hand sized parts, this process can easily achieve tolerances of 2 mm. Tolerances of 0.2 mm are considered challenging for this operation. drilltxt In drilling, a spinning cylindrical toolbit creates circular holes in a stationary part. For common hole sizes less than 25 mm in diameter, this process is can easily achieve tolerances of 0.1 mm. An additional reaming operation can achieve tolerances of 0.01 mm. Tolerances of 0.001 mm are considered challenging for this operation. lathetxt In a turning operation on a lathe or a boring mill, the object is rotated at while a stationary cutting tool is used to remove material. These machines can easily achieve precisions of 0.01 mm for hand sized parts. Precisions of 0.001 mm are considered challenging for this operation. milltxt In a milling operation a rotating cutting tool is moved across a surface to smooth and shape it. Milling machines can easily achieve precisions of 0.02 mm for hand sized parts. Precisions of 0.002 mm are considered challenging for this operation. grindtxt In rough grinding, the material on a part is removed by a rotating abrasive wheel. Objects can be easily rough ground to a tolerance of 1 mm for hand sized parts. Tolerances of 0.1 wheel. Objects can be easily rough ground to a tolerance of 1 mm for hand sized parts. Tolerances of 0.1 mm are considered challenging for this operation. edmtxt In Electrical Discharge Machining, or EDM, a current is passed between a thing conductive wire and a conductive part. The sparks produce charge buildup and thermal stress, causing particles to break off or vaporize. An EDM operation can easily achieve precisions of 0.05 mm for hand sized parts. Precisions of 0.005 mm are considered challenging for this operation. extrudetxt Extrusion processes are usually reserved for production quantities. Heated material under pressure is forced through a die. Extrusion can easily achieve precisions of 0.2 mm for hand sized parts. Precisions of 0.02 mm are considered challenging for this operation. casttxt Casting and Molding operations are usually reserved for production quantities. Material is melted and poured into a mold before cooling and hardening. As a rule of thumb, casting can easily achieve precisions of 0.5 mm for hand sized parts. Precisions of 0.05 mm are considered challenging for this operation. Actual tolerances would depend on the type of casting operation. pgrindtxt Extremely fine precisions and surface finishes are created with precision grinding. A rotating abrasive wheel removes small amounts of material with each pass of the workpiece. The workpiece can be moved linearly, as with a milling machine, or rotated, as with a lathe. With precision grinding, sub-micron flatness, roundness, and surface roughness can be achieved. picinstruct currpic buttonclick buttonclick 4currpic picture "picinstruct" sawpic drillpic edmpic grindpic lathepic millpic picinstruct Click the picture to make it disappear. picture picinstruct currpic buttonclick buttonclick 4currpic picture "picinstruct" Show Picture &File E&xit Alt+F4 Exit the program &Navigate navigate &First Page Ctrl+Home first &Next Page Page Down &Previous Page Page Up previous &Go to Page... Ctrl+G Introduction intro Go to Introduction chapter Objectives intro1 History intro2 Usefulness intro3 Sketching sketch Go to Sketching chapter Objectives Techniques Objects Cartooning Engineering Drawings formDraw Go to Formal Drawings chapter Objectives Format Working Drawings Othogonal Projection ortho Go to Orthogonal Projection chapter Objectives orth1 Theory orth2 Standard Views orth3 Auxiliary Views orth4 Common Practices orth5 orth6 Pictorials pictorials Go to Pictorials chapter Objectives pict1 Oblique View pict2 Isometric View pict3 Perspective View pict4 pict5 Sections sections Go to Sections chpater Objectives Full Section Half Section Offset Section Broken-Out Section Revolved Section Removed Section Common Practices Dimensioning dimension Go to Dimensioning chapter Objectives Definitions Guidelines Common Shorthand Tolerancing tolerance Go to Tolerancing chapter Objectives Definitions Practical Fabrication Tolerances True Position Datums Surface Features Descriptive Geometry descGeom Go to Descriptive Geometry chapter Objectives Basic Principles and Relationships Line Visibility Distance Between Lines Edge Views and True Shapes Dihedral Angles Intersection of a Line and a Plane Intersection of Two Planes Intersection of a Plane and a Solid Intersection of Solids Surface Developments Contours and Cut-and-Fill Shadows &Main Menu Ctrl+Alt+Home Go to the main menu &Options options &Audio Mute Ctrl+M OnOff Turns audio on or off Volume... setVolume Set the volume of audio &Page Controls controls Displays/Hides the Navigation Control Bar &Help Instructions F1 tutor How to use the program About the Authors authors Information about the authors Prof. Dennis K. Lieu Chris Casey Su Shien Pang Paul Krueger Allison Okamura Acknowledgments others Copyright Info copyright intro pause audioOn paused thisWav playing buttonClick buttonClick 4thisWav, audioOn mmStatus clip 4 = "playing" mmPause d = "paused" mmPlay notify repeat audioOn thisWav buttonClick buttonClick 4thisWav, audioOn mmPlay clip notify Repeat smpause smpausedis backwardsml backwardDis forwardsml forwardDis pauseup pauseDis blank uppress downpress repeat wwwwp wwwwp **""""""**" ********** ********* ******** "" intro pause repeat lastAnim thisAnim thisWav lastWav enterPage leavePage AnimDone AudioDone 4lastWav, thisWav, lastAnim, thisAnim 4audioOn, startTime B"repeat" /"intro" --switch the qones disabled enabled buttons here AnimDone AudioDone You have reached the end of Tolerancing. To review the material, go to the next page. To continue to Descriptive Geometry, click the button below.e button below. nextLesson XdescGeom buttonClick buttonClick descGeom &Go to next lesson! intro pause audioOn paused thisWav audioerror playing buttonClick buttonClick 4thisWav, audioOn, vol syserrornumber = 0 mmStatus clip W = "playing" mmPause = "paused" mmvolume Pthiswav = mmPlay notify audioerror repeat audioOn thisWav audioerror buttonClick buttonClick 4thisWav, audioOn, vol syserrornumber = 0 mmvolume clip thiswav = mmPlay notify \<> 0 audioerror Repeat tolerancing Tolerancing No specified dimension can be exact, since it is impossible to make a part with perfect precision. Instead, each dimension has an allowable error or tolerance. A smaller or tighter tolerance increases the cost of manufacturing the part. The largest possible tolerance which also ensures the fit and function of the device is usually used to minimize cost.. Limits - the maximum and minimum sizes allowable for the finished part. Tolerance - the difference between the upper and lower limits. thisAnim lastAnim thiswav lastWav enterpage leavepage audiodone animdone 4thisAnim, thiswav, lastAnim, lastWav = thisWav = "tol3" audiodone animdone Representing Tolerances There are several ways to present tolerances, any of which may be used on a given dimension. The following tolerances are equivalent. The basic size, shown in blue, is the chosen size from which tolerances are applied and is shown in decimal form. This differs from the nominal size, which is a whole or fractional number describing the general size of the dimension.. V = 5 reset 4currframe, syserrornumber =`@D liPe U F2T;6IE?C4B ?@>A@ U_G}P ?@A@h VjeEYU DPUUUj ,J]3 9DRG=%A@ AZ4K:E=-SD?N8H2B \SD0[d ?v$P2bG9l+V `ptop =S-G7B[$K3E;?09W 7`B*I< f>fFBP(3c FwG@REk lwwDD'roK E!`6Q>- KAY91;H' A+XG0=P;'E?b< ;LAa0HTO\,F yf R`z"w K(9Z/B >OFT6"+3^ ;S!F/@N6(DJ2Y=V4Sw] A/\8K>QE5H;W,@1D D08Gt:SwPS}t 0Modern 0.6 A %p(p% 'U%h'8% ${'a Arial "System Arial Class RC I Class RC 2 Class RC 3 Nominal Standard Standard Standard Size Range, Tolerance Limits Tolerance Limits Tolerance Limits Inches Clear- Hole Shaft Clear- Hole Shaft Clear- Hole Shaft Over To Values shown below are in thousandths of an inch 0-0.12 +0.25 -0.25 -0.55 0.12-0.24 -0.15 -0.15 -0.35 0.24-0.40 +0.25 -0.35 -0.45 0.40-0.71 -0.25 -0.25 -0.45 -0.55 0.71-1.19 -0.55 osoft Word Times New Roman "Arial Narrow Limit 2Modern Unilat eral (posi tive) Unilat eral (nega tive) Bilate ral (equal Bilate ral (unequ +0.1" +0.4" 0Moder 0Moder 0.25G \38(\3 \33#\3 0Modern Section A-A M1@7M1 A4@7A4 ):N9): ):Z6): 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