# Re: 2017-03-06- UTC, TImeZone, DayLight Saving Shifts, Enconding

```> On Mar 19, 2017, at 4:14 AM, Walter H. <Walter.H@mathemainzel.info> wrote:
>
> Hello,
>
>> On 18.03.2017 16:18, Joe Touch wrote:
>>> ....
>>> a day begins at 00:00:00 and ends at 23:59:59
>>> (it is really a heavy thing when counting starts at zero ...)
>> Well, here in the US we start at 12:00am and never use zero values.
> then the end is 11:59 pm and never 12:00am (if you want it without seconds)
> or 11:59:59pm (if you want it with seconds)

Yes.

>>> and now look at e.g. the calendar of Microsoft Outlook and look for
>>> the beginning and ending of a celebration day ...
>> The begin and end of a workday are configurable parameters within
>> Outlook. All-day events start at the beginning of a day (12am for
>> 12-hour clocks, 00:00 for 24-hour). This is what should be expected.
> I  was not talking about a workday ..., I was talking about the general problem
> the last day of the year ends at 11:59 pm and not 12:00 am - as this is the time the next day starts;

Why is that a problem? Would it help to think of 12 being 0?

Then we'd have 0:00am (same as 00:00 in a 24-hour system) and 00:00pm (12:00 in a 24-hour system).

Except we couldn't make clock faces using Roman numerals, because they lack a zero. And it wouldn't match the system in current use.

> by the way 12:00 pm is not defined,

Then when did I have lunch yesterday?

Every day starts at 1200am, hits 1200pm at the start of the second half, and ends at 11:59:59.9999999....

Except in leap years when it ends at 11:59:60.9999999...

> as the day has only 24 hours = 1440 minutes = 86400 seconds

That depends on how you define seconds and whether a leap second is involved, but generally, yes.
When you count 0..59 there are 60 items for both minutes and seconds. Counting 12,1,2...9,10,11 yields 12.

You haven't shown a problem yet.

Joe
```

Received on Sunday, 19 March 2017 14:01:00 UTC