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The Tolerancing Engineer Newsletter - November, 2012
...by our client company personnel and James D. Meadows using our ‘GD&T HOTLINE’

Subject: How Not to Conduct a Training Class

Jim,

I just attended a workshop on GD&T taught by a team of our in-house “experts”. They would read a passage out of the Y14.5 standard, and then ask for our opinions on what it meant. Since most of the people in the class were new to this stuff, we might as well have had cardboard question marks hanging over our heads. After we displayed a lot of shoulder shrugging and head shaking, they would say, “Ok then, let’s move on to the next section.

After a while, some of us got the gist of what was going on and decided to ask them what the passage meant. They said, “It’s open to interpretation.” We said, “Okay, what’s your interpretation?” Then they huddled up (a team of five “instructors”) and when the huddle broke, they said, “You mean you don’t know?” And I was the one who said, “Of course we don’t know. We’re here to learn.” And then they huddled up again. When they came out of their mini-conference, one said, “Well, I guess we hired the wrong people.” And I said, “Someone definitely hired the wrong people to teach this workshop.” I was told, in chorus, to “Just shut up.”

Many of the people in the class were so intimidated after that, that they refused to speak for the rest of the day. I was asked not to return for the next day of “training” and a letter rebuking my conduct was added to my personnel file.

Some of those who attended the second day said, the “instructor/experts” changed their tact. When they read a passage from the standard and asked the “students” what they thought it meant, if no one knew or asked a question, they said, “What’s wrong with you that don’t know?” Or, “What’s wrong with you that you’d ask a question like that? Are you stupid?” When one guy responded that he “must be stupid for showing up to a second day of this class”, he was asked to leave.

By the third day, they were down to one guy and he showed up just so he wouldn’t have to go back to work. It was Friday and he said he just wanted to sleep through the day and get to the weekend.

I don’t have a question for you. I just wanted to “whine and gripe” (a quote from my letter of misconduct).

John (The Whiner with a Gripe)

John,

What can I say? You obviously signed up for the deluxe training package.

Jim
 

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Figure 25-11 from my newest GD&T textbook


Subject: Question on Book-Minimum Wall Thickness Calculations Using Tolerance Stack-Up Analysis

Jim,

I’ve been working my way through your book Geometric Dimensioning and Tolerancing and had a question regarding one of your examples. In Step 1 of the solution for Figure 25-11 on page 496, I understand all of the calculations except that I would also expect there to be some inclusion of the datum shift allowed for Maximum Material Boundary on datum B for the central hole. Am I correct that this would add another .040 to the diameter of the Outer Boundary of the holes?

I’ve really enjoyed and learned a lot from the book so far.

Best regards,

Michael

From Page 497
Step 1:

.375 = LMC Hole
+ .035 = Geo. Tol. at LMC
Ř.410 = Outer Boundary Hole
+ .400 = Datum Feature Shift 
Ř.810 = Outer Boundary Hole with Pattern Shift
+ .400 = Position Tol. of D
Ř1.210 = Outer Boundary Holes with all Facets
R.605

Step 2:
4.990 = LMC Outside Diameter
- .040 = Geo. Tol. at LMC Ř 4.950 = R2.475
Ř4.950 = Inner Boundary of Outside Diameter 2

Step 3:
1.600 Basic Distance between Holes
x  2 (1.4142)
Ř 2.263 Bolt Circle Diameter
R1.1315

FIGURE 25-12 [Circuit Graphed]



Michael,

The tolerance on datum feature B can either be used as perpendicularity tolerance on the O.D., or it can be used as datum feature shift for datum feature D. Since it was used as perpendicularity tolerance on B, it would negate any shift of datum feature D from datum B being referenced at MMB. Notice step 2 where the perpendicularity tolerance is listed as .040. If it’s there, it locks up the B feature. One way to understand this is to picture the functional gage that inspects datum feature D. In this gage, datum feature B would be simulated by a gage hole that is a diameter of 5.030. If you look at step 2 and add the .040 to the 4.990 (instead of subtracting it as the step does), it shows that the O.D. would (at 5.030) lock itself up inside the gage hole which is that size.

Hope this helps.

Jim
 



Subject: Mistake on Page 42 and 47 of your Workbook? Variables Data Analysis

Hello, Mr. Meadows,

I purchased your book "Workbook and Answers for GDT" and going over problems. I realize that space is limited and sometimes detailed answers cannot be given. I so far cannot explain responses to some questions:

Question 16 on Page 42;
0.04 straightness at RFS of surface elements. The part has a taper as manufactured ranging from 100.1 to 99.9. If I were to look at surface elements, draw two parallel lines (.04 apart) to axis of part from 100.1, straightness is violated. 0.1 is greater than 0.04. Am I missing something?

16. Circularity, Straightness & Total Runout



Regarding the last part of Question 21 on Page 47

I don't understand the question and answer regarding the .003 out-of-straightness. There is no mention of straightness on the drawing and why would an inspector look at derived median. If the part is made within the limits of size.....I don't understand?

Paul

The Problem as stated for page 47:
Part 12135 – Bushing: Using the following variables data, assess the part on the following page and make assessments pertaining to its compliance to the tolerances depicted.
Available collected measurement data:
•Outside diameter measures .525 at its largest and .519 at its smallest.
•Part outside diameter is round to within .0002.
•FIM of I.D. while centered on and rotating about datum axis D is .002 at each cross section.
•Inside diameter is .2550 at its largest and .2549 at its smallest.
•Part thickness (shown in the left side view) varies between .196 and .195. It is rigid.
•Part thickness (shown in the left side view) has a slight uniform bow. Its derived median plane is out of straightness .003.



Paul,

Regarding page 42, in order for surface elements to be out of straightness, the surface would have to have pits, bumps or curves. None are shown. Taper has nothing to do with straightness.

On page 47, Rule #1 in ASME Y14.5 states that, unless otherwise specified, rigid parts must have perfect form at MMC. In gaging terms, the part must be able to fit the GO gage (which checks the MMC) to meet size requirements. That part’s MMC in that view is .197. The statement of the produced part’s condition is as follows:
•Part thickness (shown in the left side view) varies between .196 and .195. It is rigid.
•Part thickness (shown in the left side view) has a slight uniform bow. Its derived median plane is out of straightness .003.

When you add .003 to either .195 or .196, one can see that the part would not fit between two parallel planes that are .197 (the MMC envelope-simulated by a GO gage) and would fail to comply with Rule #1. In other words, it fails the size requirements.

Hope this helps.

James Meadows

 



Subject: Question on Composite Profile

Mr. Meadows,

I am hoping you can help me to understand the profile feature call out on this drawing. I explain my thoughts in RED next to the RED highlighted feature control frame.

Thanks in advance,

Mark



Mark,

Yes that’s correct. You must hold angle and location to the datum structure to within 4mm and shape/form to within 3mm. The 3mm shape control only applies to size if the features being controlled are closed loop features (such as oddly configured holes and shafts). In other words, it only applies if the controlled features have a size. I can’t tell if they do from the illustration you sent, but the note 3 DUCT OPENINGS would seem to indicate that they do.

Profile isn’t required to reference datums. Composite profile controls often don’t reference datums in the lower level (tighter toleranced) control. The tightest tolerance controls form and if it is an all-around or all-over control, it also controls size. If datums are referenced in the lower level of a composite profile (or position) control, they don’t control location, but can only control orientation (angle).

Jim Meadows
 



Subject: Basic Dimension Question

Hi Jim,

A question came up that we would appreciate your opinion on: can you have a “0.000” Basic dimension. The example would be when a datum is in the same plane as the holes that are being dimensioned

David,

A zero basic dimension applies where axes, center planes or surfaces are shown coincident (in the same place) on a drawing and geometric tolerances establish the relationship between the features. [Although this concept is not new to previous Y14.5 standards, this statement explicitly states the concept for the Y14.5-2009 standard.]
In other words, it is implied, so there’s no reason to put it on the drawing.

Jim
 



Subject: Positional Boundary

Jim,

I am a prior student of yours. Would you confirm or correct following interpretation to attached drawing requirement stating ASME Y14.100 applies.

The right and bottom sides of rectangular cutout must meet position .005 callout with respect to datums B & C. The left and top sides of cutout must conform to size tolerancing. Also, would any of the fillet radii 4 places need to meet position .005 as well?

Regards,

Michael



Michael,


The widths must meet the size requirements of 1.500-1.510 and .500-.510, and the positional boundary control states that the surface of those widths may not violate a virtual condition boundary that measures MMC minus .005 positional tolerance if the widths are holes/slots or if they are shafts/tabs the inviolate boundary would be the MMC plus .005. These boundaries are located at the true position (perfect location and orientation) from the datums E and C.

The problem, as I see it, is that they located the surfaces of the widths (with basic dimensions), whereas they should have located the centerplane of the widths (with basic dimensions-one of which is implied to be zero). They think they are locating the individual surfaces, but should be locating the centerplanes. The position tolerance zones are in the middle of the widths (two parallel planes that are .005 apart at MMC). The local notes (BOUNDARY) just mean they want you to simulate the effect of the position tolerances on the surfaces of the width by simulating a gauge. These functional gauges make sure that no portion of the width violates its virtual condition boundary as calculated in the first paragraph of this response. If these widths are slots, then picture the gage pin as .495 X 1.495 and located right at true position (which is not currently defined correctly). The surface of the widths must fit over the gage pin to comply with the position control.

I don’t think that they know that position isn’t used to locate planar surfaces with basic dimensions. It is used to locate the middle of the feature (in this case the centerplanes). Usually the features can be measured to determine if either 1) the centerplanes have violated the tolerance zones which are two sets of parallel planes .005 apart at MMC or 2) the surfaces of the controlled features have violated the virtual condition boundary (as the gauge I described would do). However, the local note BOUNDARY says do the second thing.

I hope this helps.

Jim
 



Subject: Total Runout Question

Hi Jim,

You recently taught a class at NASA which my branch attended. We have been holding group drawing “reviews” within the branch to help sharpen what we learned in your class. The most recent one we did a question surfaced that I thought you might clarify.


Attached is a copy of a total runout scheme along a compound datum axis where both datums have total runout applied. This differs from the approach of initially applying cylindricity then using total runout and referencing the compound datum axis (I attached the example from your book). Are these the same situation? Are both schemes then correct?

Your input on this would be appreciated.

David

Illustration from Y14.5-2009:



David and All Interested Parties,

Yes, even though the illustration you sent from my textbook is the more linear progressive way to define the part, the use of Runout or Total Runout on all of the diameters, including the datum features that will construct the compound axis they are measured from is, in some ways, more logical. The premise is that since each diameter alone doesn’t generate enough surface area to stabilize the part to measure other diameters from, each doesn’t create enough stability to measure the other datum feature from.
Illustration from my textbook:



Since each diameter generates an axis that is different from the compound axis, you can say that each must be coaxial (using Runout, Total Runout, Concentricity or Position) to the common axis formed simultaneously from both together. I explain this when going over the example you sent from my textbook and I have a worksheet in the workbook on page 114 that analyzes the four controls when used that way. In the worksheet, each possibility is explained as an option, and only one would be selected.



Illustration from my workbook (Pick One of These Options-Position vs. Runout vs. Total Runout vs. Concentricity):



Some feel uncomfortable in saying each datum feature diameter is controlled to their compound axis, in that if we were to “chuck-up” on both, they would be obscured and not measurable. The way around that is to not “chuck-up” on both, but rather probe them with a CMM and analyze the collected data, or put both in Vee-blocks which leaves them exposed to measure, or center drill the part and put it between centers, then measure every diameter to see how far off they are from each other. If everything is measured from the same axis, whether that axis is the proper datum axis or not, then every diameter is related to each other to within the sum of their tolerances to that common axis. In other words, if the part measures within the tolerance, it is good. If not, another measurement tactic can be used.

At any rate, the result of using either the linear progressive method in my textbook, or the “one-fell-swoop” method on page 114 in my workbook (and in the example you sent me from another source) can reap the same or similar results.

I hope this helps.

Jim Meadows

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