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Capability calculations for True Position

VinniUSMC
Has anyone come across this as a means of calculating Cpk of a true position?

http://documentation.statsoft.com/ST...tionCapability

I think it should be possible, but I'm not sure if this is the answer or not. I was thinking about how something along the lines of the complex plane (or vectors/trig) could be used to identify grouping of a positional tolerance based on its quadrant within a unit circle, and then could be used to calculate Cpk. It's not unheard of. The normal distribution of a multivariate system is calculable, so why wouldn't a statistic that is related to the normal distribution in 1 variable work for higher dimensions?

I'm thinking that there must be a way to calculate the variance of position based on the the quadrant it falls in. Whether it's a trigonometric function, or a complex function.

Anyone have any input as I dive into this blackhole?

Curiosity killed the cat, hopefully I'm not a cat.

  • As I understand the Volvo doc, here is my "translation" :


    Don't forget that I work essentialy on prototypes, so I'm not user of those tools Disappointed.

    I think 1 or 2 things might not be right (or I could be reading the Volvo standard wrong).

    When the points are projected onto the line, they should be projected perpendicular to X, not dependent on the angle of the projected line, shouldn't they? So, when the line is at 0 degrees, the farthest point in -X, and the farthest point in +X should have the same X coordinate no matter what angle the projection line is at. But, that would mean that when approaching 90 degrees, the Y coordinates would approach +/- infinity (which doesn't make sense, and is why I think I could also be wrong) and the distance to barycenter would also approach infinity. Which would also make the Cp never good.

    Also, I think the barycenter should not change while rotating. The projected line should rotate about the barycenter of the initial input. "By iteratively rotating the constructed plane through the point of balance in different directions" (page 28).
  • The points where the relevant plane intersects the tolerance circle are determined. This means that the distance from the tolerance circle to point of balance can be calculated, in two directions. By dividing these distances by three standard deviations, two Cpk values are obtained.

    Those distances are calculated in V14 and V16, and assigned as dist1 and dist2, then divided for each angle by 3 std dev. It gives cpk1 and cpk2.

    By means of the calculated standard deviation and the distance between the two points of intersection, which makes up the tolerance zone in the relevant direction, Cp can be calculated for the orientation given by the plane in question.
    dist is the sum of abs(dist1) and abs(dist2). (T20)
    Idivided it by 3 std dev, but it's not explained in the text, maybe t should be std dev ?
    I don't know, I just try to help without really understanding Disappointed... I'm french !


  • I think 1 or 2 things might not be right (or I could be reading the Volvo standard wrong).

    When the points are projected onto the line, they should be projected perpendicular to X, not dependent on the angle of the projected line, shouldn't they? So, when the line is at 0 degrees, the farthest point in -X, and the farthest point in +X should have the same X coordinate no matter what angle the projection line is at. But, that would mean that when approaching 90 degrees, the Y coordinates would approach +/- infinity (which doesn't make sense, and is why I think I could also be wrong) and the distance to barycenter would also approach infinity. Which would also make the Cp never good.

    Also, I think the barycenter should not change while rotating. The projected line should rotate about the barycenter of the initial input. "By iteratively rotating the constructed plane through the point of balance in different directions" (page 28).

    In my tests, the barycenter doesn't change if you change values of angle (E7 and F7). It change only if you change points.
    a and b (R4 and S4) are the coefficient of the line equation, they change with the angle.
    I understand that the points are projected perp to the line and not perp to X.

    It's not easy to write in english a user manual.... And the "after sale service" is bad !
  • in reply to JEFMAN
    I don't think this standard was originally written in English either. Your understanding of it is at least as good as mine.

    I do think your calculations for Cp and Cpk are right. As far as I understand. It's just how the data is projected onto the line that i am not sure about.

    Any Volvo people on here that can provide how Volvo performs the projection? Slight smile
  • in reply to JEFMAN
    Sorry, you're absolutely right about barycenter. I was misreading.

    Yes, it's not clear whether projection should be normal to X or to the projection line. I agree.
  • in reply to JEFMAN
    1-(page 26) All measured locations are projected in the constructed plane.
    2-(page 28)By iteratively rotating the constructed plane through the point of balance in different directions, repeating the
    projection of the points in the different directions and repeating the evaluation of Cp and Cpk in each iteration, the
    orientation for the largest dispersion of the amount of points can be produced as a basis for Cp and the orientation
    where the tolerance violation is at its greatest as a basis for Cpk.

    I agree about cp, because the calculation is not described, but cpk1 and cpk2 match to the description.
    Not sure how calculate cpk from them, I understand it's the min...

    to the rescue Wink
  • in reply to JEFMAN
    Yes, for Cp, projecting onto the projection line will always yield the largest dispersion at 0 or 90, but that's not the case in the pictures on page 28. That's why I think the projection needs to happen in the construction plane (0 degrees). But, as I already mentioned, that leads to a paradox that the answer will always be infinity, which is obviously not right either.
  • can we calculate the level of bromance here?
  • in reply to ruidoso
    Now that you're here, it's over 9000. Rolling eyes
  • Projection line parallel to X-axis (removing the Y-coordinate completely from the calculation).
    During the rotation of the plane/line intersecting the centre of gravity, the points X-coord never changes - only the Y-position of them (X-coord is constant).
    Find the rotation/angle where the dispersion of points (width) is at it's largest - this is the base for Cp calculation (perpendicular to the line/plane).
    Find the rotation/angle where the tolerance violation is at it's greatest - this is the base for Cpk calculation (perpendicular to the line/plane).

    This is over my head though, but you already knew that. I just looked at the pictures. Sunglasses
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