Diametrical Positioning for a Plane? Need help interpreting drawing

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  • Diametrical Positioning for a Plane? Need help interpreting drawing

    Hello All,

    I'm confused by a drawing we got from a customer. I got involved in this project at the 11th hour so I don't want to go to the customer yet explaining I don't understand their drawing until I'm sure I cannot interpret the drawing.

    I've attached a drawing I made of the part, so as not to use the actual part print. On the drawing I wrote my explanation of how I interpret the drawing and my questions.

    Thanks for any help.

    Attached Files

  • #2
    DAT C= 1 square pocket

    treat the square pocket like you would any circular tertiary datum and clock to it.
    keep it simple.
    Last edited by BIGWIG7; 12-29-2018, 01:54 PM.
    Che Guevara is a communist scumbag.


    • #3
      You certainly could have a square pocket on a diametrical tolerance zone, but if it's not called out as a square pocket (if it's as shown, then it's not) then then a diametrical zone is not right.
      "This is my word... and as such is beyond contestation."


      • #4
        Clock to it as mentioned already, it is your datum. Is the center specified? You'll probably need it to rotate to print coordinates. The callout really only controls distance from center (got CAD nominals?). I'm not defending the print just giving them a number to fulfill their callout. They specify the location of the target and the size of the bullseye and you tell them how close they are with a radilal (doubled) and the linears from center. Don't over complicate it.

        Just my .02 at my age

        sigpicHave a homebrew


        • #5
          Thanks for the replies.

          I added some dimensions to the drawing.

          I'm a rookie at GD & T, so I'm trying to figure out what physically has to be within the 0.005" diametrical true position (the "target"). Is it supposed to be the center of the lug? One axis is defined by Datum C. Then would the other be to factor in the 7.985" length and the 6.30" radius to the center? (dimensions were added on the updated drawing). If not that, then what has to be within the 0.005" diametrical tolerance zone? The 7.985" length is ± .005" and the 6.30" radius call out is ± .010". It seems to me neither would work as part of the equation because the tolerances are too large to stay within the 0.005" diametrical true position tolerance zone, if the target is the center of the lug.

          Then there is that .875 basic dimension. Does that mean that the other half of the lug is supposed to be .8745, because the lug width is 1.7495" ± .0005"? Or ... ?

          Thanks again for any help on this.




          • JacobCheverie
            JacobCheverie commented
            Editing a comment
            I don't think your updates saved.

        • #6
          As is, you have a "Width" being used as Datum C. It cannot be a diametrical TP.

          For a square slot/lug, the centroid/axis is what would have to be within your tolerance zone, just like a cylinder.
          "This is my word... and as such is beyond contestation."


          • #7
            I've added the attachment I was referring to in my previous post, if anyone wants to comment I would appreciate any help with this.

            Attached Files


            • #8
              [TP][Ø 0.005][A][B] diametrical tol zone is a width, without a tertiary datum is contradictory.
              --Let's chock off the diameter symbol as a typo. Speak with your supervisor and have them confirm with customer, that this TP is a pair of planar tolerance zones as pictured below/attached.

              The basic 0.875" offset is linear, to a perceived ray line from B. This basic offset "Clocks" the sides of the 1.75" boss to be parallel to a ray line from center of B datum out to the midline of the boss.
              Because your Radius of 6.3 and your 7.985" dimensions are NOT BASIC (square box around feature of size) the true position callout does not control the second axis of position. Also the DIM is tied only to the width of the boss, the TP has not control of how close or far it is away from B.
              If you intersect the A datum to the B cylinder, and the mid-plane of the two measured planes to get your 1.75" width, you would have your tertiary C datum point to use for future dimensional controls.
              Last edited by louisd; 01-08-2019, 10:50 AM.


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