T values

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  • T values

    Please, who can help me??
    Now , all of a sudden, engineers and managers want to know the exact meaning of the T on the measuring report.

    I always thought that it was the shortest distance from the measured point to the actual point. Now there is a whole debate what it means.

    In the help function from PCDmis, the description is not very clear.

    Has anyone got a better description.

    Cheers
    sigpic If only all problems were so simple

  • #2
    Here is a old post.

    T Value


    T represents deviation along a vector. It is quite simply a distance with a nominal value of zero. It is found by projecting a measured point onto a surface normal vector (the “nominal vector”) of the associated nominal point. The distance from this projected point to the nominal point is the T value.

    Graphically, this can be represented as follows:


    If you want to step through a T value by hand, this is what you need to do:

    1) Collect your given values.
    You will need the X, Y, and Z (xn yn zn) of the nominal point, the I, J, K (in jn kn)of the nominal vector, and either the I, J, and K (ia ja ka)of the vector defined by the actual point or the X, Y, and Z (xa ya za) of the actual point.

    2) Find the vector defined by the actual point.
    You can compute this vector as follows:














    3) Unitize the normal vector (a.k.a. shorten its length to 1.0)
    PC-DMIS for Windows is a kind and forgiving software package. If you acquired the normal vector from PC-DMIS, then it is already unitized and you need only do a name change...(in jn kn) becomes (iu ju ku). You then can skip to step 4.
    a) find the length of the normal vector
    Add up the squares of the vector components and take the square root of that value:



    b) use the length to unitize the normal vector
    Divide each vector component by the vector length to unitize the vector:




    4) Project the vector defined by the actual point onto the unitized normal vector.
    This is accomplished by computing a dot product of these two vectors. The final answer is a single number which is the T value. A dot product is obtained by multiplying the two i components, the two j components, and the two k components of the two vectors then adding them all together:



    Assuming that the surface normal vector is pointing away from the surface on which the nominal point lies, a positive value of T means that the material is too “high” while a negative value of T means that the material is too “low”.

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    • #3
      looks like pics didn't come thru

      here it is again

      Image2.jpg

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      • #4
        What Tested is giving you is the B&S explanation and it is rather convoluted. I've attached the document so you might can read it better. I wish it was as easy as the square root of the deviation. That was much easier to explain in the MM4 days. Good luck.
        Attached Files
        When the going gets weird, the weird turn pro. Hunter S. Thompson

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        • #5
          IF you use SNAP points, it IS sqrt(xdev^2+ydev^2+zdev^2)
          sigpic
          Originally posted by AndersI
          I've got one from September 2006 (bug ticket) which has finally been fixed in 2013.

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          • #6
            Originally posted by jmgreen View Post
            What Tested is giving you is the B&S explanation and it is rather convoluted. I've attached the document so you might can read it better. I wish it was as easy as the square root of the deviation. That was much easier to explain in the MM4 days. Good luck.
            Thank you for the document on T values. Good ammo for when the knotheads come calling.
            sigpic

            James Mannes

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