Race Hours

[Rebecca Smith]

I'm building log book to keep track of a racers stats during a long ultra
marathon race. Things such as time on the bike, time off, H20 intake,
electrolytes and the like.
When I say long I mean days! I'd like to have on the input form, the running
number of hours the race has been in progress. DateDiff only allows me "h"
or hours and this won't take me past 24. I've tried other combinations of
formats and formulas but am just spinning my wheels.
I'd sure appreciate some help on figuring out how many hours the race has
been in progress since the starting gun.
Thank you,
Becky

PS - Access 97

[WindAndWaves]

I always try to change time and date values into numbers rather than
dates/times. It simply works a lot simpler. Then, in the end, you convert
them back to hours/days or what have you.

0 = 30-dec-1899, each day from then adds one.

24-May-04 = 38131

to work out the difference between, for example today and three days ago
date()-(date()-3)

to work out the difference between, for example, today and a specific date:

int(date() - #12-12-03#)

hours are fractions (e.g. .5 = mid day).

hope this helps

I really found the date and time functions did not do it for me and gave me
all sorts of roblems.

"Rebecca Smith" <rp*****@pcez.com> wrote in message
news:10*************@corp.supernews.com...

I'm building log book to keep track of a racers stats during a long ultra
marathon race. Things such as time on the bike, time off, H20 intake,
electrolytes and the like.
When I say long I mean days! I'd like to have on the input form, the running number of hours the race has been in progress. DateDiff only allows me "h"
or hours and this won't take me past 24. I've tried other combinations of
formats and formulas but am just spinning my wheels.
I'd sure appreciate some help on figuring out how many hours the race has
been in progress since the starting gun.
Thank you,
Becky

PS - Access 97

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[Rebecca Smith]

WindAndWaves,
Thanks for the reply. Where the real problem seems to lie is when I try
format, let's say 49 hours. If I use Format(myTime, "h:m:s") only 1 hour
returns. If for instance I add in an extra 'h' all that does is to put a
zero in front of the 1. Any ideas?
Thanks,
B.

I always try to change time and date values into numbers rather than
dates/times. It simply works a lot simpler. Then, in the end, you convert them back to hours/days or what have you.

0 = 30-dec-1899, each day from then adds one.

24-May-04 = 38131

to work out the difference between, for example today and three days ago
date()-(date()-3)

to work out the difference between, for example, today and a specific date:
int(date() - #12-12-03#)

hours are fractions (e.g. .5 = mid day).

hope this helps

I really found the date and time functions did not do it for me and gave me all sorts of roblems.

[Bob Quintal]

"Rebecca Smith" <rp*****@pcez.com> wrote in
news:10*************@corp.supernews.com:

I'm building log book to keep track of a racers stats during a
long ultra marathon race. Things such as time on the bike,
time off, H20 intake, electrolytes and the like.
When I say long I mean days! I'd like to have on the input
form, the running number of hours the race has been in
progress. DateDiff only allows me "h" or hours and this won't
take me past 24. I've tried other combinations of formats and
formulas but am just spinning my wheels. I'd sure appreciate
some help on figuring out how many hours the race has been in
progress since the starting gun. Thank you,
Becky

PS - Access 97

the datediff() function will give you all you need.

store the starting datetime of the race, and calculate the
duration to now, I like to use seconds for my calcs
Dim ElapsedTime as double
ElapsedTime = Datediff("s",[race start],now())

To format it as hours and minutes, you need a custom format
routine. Put this in a module and call wherever needed.

public function sec2dur(duration as double) as string
dim hours as integer, mins as integer, secs as integer
hours = int(duration * 3600)
mins = int(duration*60) - hours*60
secs = duration - (hours*3600-mins*60)

sec2dur = format(hours,"000") & ":" & format(mins,"00") & ";" &
format(secs,"00")

end function

I store the start and end time for each lap in a table, and have
a query show the duration of each lap, I can get total times by
having the query select the ElapsedTime and summing that, and
formatting the result.

For your described use, just place the expression in a textbox on
the form, You will need to use the timer event to recalculate
every minute or so.
Bob Quintal

[WindAndWaves]

Bob got a good solution

here is my function for time:

Public Function HAS(DAT As Double) As String 'Dat is time in numbers (e.g.
1.5 = 36 hours)
'hours as string
'On Error GoTo err
Dim h As Long, m As Long, s As Long, Sec As Double
Sec = DAT
h = Int(Sec * 24)
m = Int(Sec * 1440) - (h * 60)
s = Round(Sec * 86400) - ((h * 3600) + (m * 60))
HAS = Format(h, "000") & ":" & Format(m, "00") & ":" & Format(s, "00")
xit:
Exit Function
err:
HAS = "00:00:00"
resume xit
End Function
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[Rebecca Smith]

Bob,
Thanks and a hat tip! It took me more than a few moments to get everything
up and running but once I did it worked perfectly. The real modification I
did to your code was to eliminate seconds and add the days. Once I saw the
solution you presented it was fairly easy to progress as required.
Thanks a lot.
B.

PS - WindAndWaves - I'll look over your solution to see how it differs and
then play with it a bit. Thanks!

the datediff() function will give you all you need.

store the starting datetime of the race, and calculate the
duration to now, I like to use seconds for my calcs
Dim ElapsedTime as double
ElapsedTime = Datediff("s",[race start],now())

To format it as hours and minutes, you need a custom format
routine. Put this in a module and call wherever needed.

public function sec2dur(duration as double) as string
dim hours as integer, mins as integer, secs as integer
hours = int(duration * 3600)
mins = int(duration*60) - hours*60
secs = duration - (hours*3600-mins*60)

sec2dur = format(hours,"000") & ":" & format(mins,"00") & ";" &
format(secs,"00")

end function
Bob Quintal