Symmetry vs. Position - Symmetry uses the same theory as Position, right?

Jim Wynne

Staff member
Admin
#11
NetScorpium said:
Hello JSW05,
Thank you for your answer. I had improve the file in order to explain better my doubt. Please look again.
I'm not sure what you're asking. It appears that your method of calculating the position callout in your attachment is correct; did you also need information about the concentricity callout?
 
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N

NetScorpium

#12
Hello JSW05,
The main target of my post is ensure that my measurement method for calculating the position callout and concentricity callout is the most correct. Please, analyse my method to measure the concentricity too, and tell me your opinion.
 

Jim Wynne

Staff member
Admin
#13
NetScorpium said:
Hello JSW05,
The main target of my post is ensure that my measurement method for calculating the position callout and concentricity callout is the most correct. Please, analyse my method to measure the concentricity too, and tell me your opinion.
Because of the difficulties involved in actually measuring concentricity error, I'm not sure where your data is coming from. In my experience, in almost every instance I've ever seen where concentricity was specified it was really control of position or runout that was needed. The input values for your formula must be the result of measurement for the median points of diametrically opposed elements of the feature being controlled. In practical terms, this involves using two indicators positioned 180 degrees apart on the controlled feature, and determing the net difference in the readings at different points along the feature, and then identifying the worst case, which is the greatest net difference between any two readings. Thus it appears that perhaps your formula may be as simple as finding the difference between two (perhaps differently signed) values.

If this isn't clear, or if I haven't answered your question, please post back and I'll give it another go.
 
B

Bill Ryan - 2007

#14
JSW05 has addressed your question on concentricity. Now for position.....

The calculation for position is useage of the Pythagorean Theorem (and multiply by 2). From the values I see you have reported (0.21 from basic and .05 from basic), I get a position of .432. Does that coincide with what you get?
 
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Jim Wynne

Staff member
Admin
#15
Bill Ryan said:
JSW05 has addressed your question on concentricity. Now for position.....

The calculation for position is useage of the Pythagorean Theorem (and multiply by 2). From the values I see you have reported (0.21 from basic and .05 from basic), I get a position of .432. Does that coincide with what you get?
Hi Bill

I get the same answer as you, so I was in error in my other response to the OP, when I apparently read 0.21 as .021 :bonk: . Interestingly, that error results in the same number as in the example spreadsheet (0.11), so it does look like there's a problem.
 
N

NetScorpium

#16
Bill Ryan said:
JSW05 has addressed your question on concentricity. Now for position.....

The calculation for position is useage of the Pythagorean Theorem (and multiply by 2). From the values I see you have reported (0.21 from basic and .05 from basic), I get a position of .432. Does that coincide with what you get?
Humm... well if I understand your calculation formula for position, the 0.432 is the result of:

Formula: *** h^2=[(C1)^2 + (C2)^2]
*** FinalResult=(h^2) * 2

Applying formula:
h^2=(50.00-49.79)^2 + (50.00-49.95)^2
h^2=0.0441 + 0.0025
h=0.216

FinalResult=0.216 * 2 = 0.432

OK, this I understand. And if I have only one plan
Example: |PositionSymbol|Ø 0.1|A|

or if I have tree plans
Example |PositionSymbol|Ø 0.1|A|B|C|

The Pythagorean Theorem (and multiply by 2) is the same to calculate these kind of positions? How? Where can I find more information about position and concentricity calculations?
 
N

NetScorpium

#17
JSW05 said:
Because of the difficulties involved in actually measuring concentricity error, I'm not sure where your data is coming from. In my experience, in almost every instance I've ever seen where concentricity was specified it was really control of position or runout that was needed. The input values for your formula must be the result of measurement for the median points of diametrically opposed elements of the feature being controlled. In practical terms, this involves using two indicators positioned 180 degrees apart on the controlled feature, and determing the net difference in the readings at different points along the feature, and then identifying the worst case, which is the greatest net difference between any two readings. Thus it appears that perhaps your formula may be as simple as finding the difference between two (perhaps differently signed) values.

If this isn't clear, or if I haven't answered your question, please post back and I'll give it another go.
Yes, I understand your explanation. To measure concentricity, I use a Coordinate Measuring Machine. For example, I measure a diameter and the median points of diametrically are placed to 0,0 (X=0,Y=0). Than I measure the other diameter and I calculate the distance between the two median points of diametrically. The result is my concentricity value.
 
B

Bill Ryan - 2007

#18
NetScorpium said:
The Pythagorean Theorem (and multiply by 2) is the same to calculate these kind of positions? How? Where can I find more information about position and concentricity calculations?
Yes, the calculation works for one, two, or three variables.

Just about any book on Geometric Dimensioning and Tolerancing will go into detail about Concentricity and Position, as well as all other symbols. A few I'll mention are "ANSI Y14.5", "Geometrics" by Lowell Foster, "Geometric Dimensioning and Tolerancing" by James Meadows, "Geometric Dimensioning and Tolerancing" by Al Neumann. You also might try "Googling" the term GD&T (or spelled out) and see what pops up.
 

Jim Wynne

Staff member
Admin
#19
NetScorpium said:
Yes, I understand your explanation. To measure concentricity, I use a Coordinate Measuring Machine. For example, I measure a diameter and the median points of diametrically are placed to 0,0 (X=0,Y=0). Than I measure the other diameter and I calculate the distance between the two median points of diametrically. The result is my concentricity value.
It sounds like you're measuring runout or position, not concentricity. In order to accurately measure concentricity you have to be able to rotate the part about the datum axis using diametrically opposed indicators as I advised before. It could very well be that the result your're getting will satisfy design intent, however. As I also said earlier, it's been my experience that when concentricity is specified it's usually by someone who understands the everyday definition of the word (two things that are coaxial) but isn't aware that it isn't that simple in the GD&T sense.
 
P

prabsreloaded

#20
i have heard that applying pythagoras theroem u get position and then multiplying it by 2 u get concentricity.... my doubt is why do we multiply by 2
 
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