Christian Smith wrote:
> If you use the julianday representation, the integer component is the
> number of days since "noon in Greenwich on November 24, 4714 B.C", with
> the fractional part being the fraction of that day. Hence, the
> resolution is determined by the fractional component of the real number.
> Now, in the UK, I get the following:
> sqlite> select julianday('now');
> 2454295.1407767
>
> The integer component consumes probably 21 bits of the available 52 bits
> mantissa of an IEEE-754 64-bit real. That leaves 31 bits for the
> fractions of a day, giving a resolution of 1/24855 of a second:
> 2^31/(60*60*24) = 24855.134814814814814814814814815
>
> Plenty enough for milli-second resolution.
>
> Probably not very good for embedded applications if an FPU is not
> available.
I'm a little confused by the math... help me work this out.
sqlite> SELECT julianday('now');
2454295.20404931
That gives me days since the Julian epoch. If I multiply by 86400 I
should get seconds since the Julian epoch.
sqlite> SELECT julianday('now') * 86400;
212051105903.613
That leaves me three decimal points of precision for seconds. So
that's thousandths of a second? Where do you get 24000ths of a second?
--
Scott Baker - Canby Telcom
RHCE - System Administrator - 503.266.8253
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