The internal storage of a DATETIME value
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            by Peter Larsson
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        Published on Fri, 05 Nov 2010 14:10:11 GMT
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            2010/12/06
            16:58 UTC
        
        
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        SELECT  [Now],
        BinaryFormat,
        SUBSTRING(BinaryFormat, 1, 4) AS DayPart,
        SUBSTRING(BinaryFormat, 5, 4) AS TimePart,
        CAST(SUBSTRING(BinaryFormat, 1, 4) AS INT) AS [Days],
        DATEADD(DAY, CAST(SUBSTRING(BinaryFormat, 1, 4) AS INT), 0) AS [Today],
        CAST(SUBSTRING(BinaryFormat, 5, 4) AS INT) AS [Ticks],
        DATEADD(MILLISECOND, 1000.E / 300.E * CAST(SUBSTRING(BinaryFormat, 5, 4) AS INT), 0) AS Peso
FROM    (
            SELECT  GETDATE() AS [Now],
                    CAST(GETDATE() AS BINARY(8)) AS BinaryFormat
        ) AS d
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