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Basic dimension

Shane-o
BASIC DIMENSIONS USE G D &T TOLERANCING, THE QUESTION IS I HAVE BASIC DIMENSIONS COMING FROM X & Y AXIS. AM I SUPPOSED TO "CUT" THE TOLERANCE IN HALF WHEN DIMENSIONING THE BASIC DIMENSION? example X AND Y ARE BOTH [2.077] THE TRUE POSITION CALLOUT IS ALLOWING UP TO .015 SHOULD I USE .0075 FOR X AND .0075 FOR MY Y TOLERANCE?
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  • +1,000,000

    There is a correct mathematical formula for calculating true position RFS. Simple and quick. Square deviation of X and Y, add them together, take the square root of that number and multiply that by 2. That would be the formula you give your operator.

    Or, as I tell the imagineers when they get uppity:

    Position is calculated as double the root of the sum of the squares of the axial deviations. Slight smile That usually shuts them up for a few minutes while they try to sort it out in their heads.

    You can also draw a circle bisected vertically and horizontally with a line drawn from the intersection of the bisectors to a point on the circle which is the point created at the intersection of a line drawn perpendicular to one of the bisectors, and show them that the triangle created can be solved by Pythagoras' theorem, and then point out that the hypotenuse is the radius, which is half the diameter, which is denoted by the number in the FCF (if preceded by the diameter symbol).
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  • +1,000,000

    There is a correct mathematical formula for calculating true position RFS. Simple and quick. Square deviation of X and Y, add them together, take the square root of that number and multiply that by 2. That would be the formula you give your operator.

    Or, as I tell the imagineers when they get uppity:

    Position is calculated as double the root of the sum of the squares of the axial deviations. Slight smile That usually shuts them up for a few minutes while they try to sort it out in their heads.

    You can also draw a circle bisected vertically and horizontally with a line drawn from the intersection of the bisectors to a point on the circle which is the point created at the intersection of a line drawn perpendicular to one of the bisectors, and show them that the triangle created can be solved by Pythagoras' theorem, and then point out that the hypotenuse is the radius, which is half the diameter, which is denoted by the number in the FCF (if preceded by the diameter symbol).
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