Why Should You Divide by x-1 for Evenly Spaced Events Throughout the Day?

[Square1]

Lets say you want to do something x times per day. Importantly, you want to start doing this thing at the begining of the day and end at the end of the day (start of period end of period). The first thing I think about when I want to start doing something like this is divide the day's hours by x. But I find you have to divide the day x-1, to land the activity at wanted times.

Ex: I want to water a plant three times a day form 9 to 12. Intuitively I want to go 15hrs/3 = water every 5 hrs. But starting at 9, I would end up watering at the following times: 9am, 2pm, 7pm as supposed to wanting to do it at 9am, 12am, and in the middle of the day. 15hrs/3-1 however gives me 7.5. I would get what I want and start at 9, 4:30, and midnight.

So the algorithm is to do the division by x-1 when wanting to place the intervals at the start. Why is that? Can someone elaborate on this? Throw some terminology around? Thanks.

 

[Mark44]

Yes, if you want to water a plant three times a day, from 9AM to 12 midnight, break the time interval (15 hrs) into two pieces, each of 7.5 hours.

If memory serves, the idea here is called fencepost counting. If you build a fence with two rail sections, you will need three (2 + 1) fenceposts. The times that you water are analogous to the fenceposts, and the times between are like the rail sections.

 

[HallsofIvy]

Lets say you want to do something x times per day. Importantly, you want to start doing this thing at the begining of the day and end at the end of the day (start of period end of period).

This is a bit confusing. If you "want to do something x times a day", doing "at the beginning of the day" and at the beginning of the day" then the end of one day is the beginning of the next day. That is, if x= 3, you would do it at midnight, then two more times during the day with the next midnight being the first time of the next day. So there will be 24/3= 8 hours between times: do it at midnight, 8:00 AM, and 4:00 PM.

The first thing I think about when I want to start doing something like this is divide the day's hours by x. But I find you have to divide the day x-1, to land the activity at wanted times.

Ex: I want to water a plant three times a day form 9 to 12.

So when you say "day", you don't mean a 24 hour day but starting at 9:00 AM and ending at 12:00 midnight? So you do it at 9:00 AM and midnight which leaves one time in the middle: 15/2= 7.5 hours so 9:00+ 7:30= 4:30 PM.
(Assuming you want the times equally spaced which you never actually say.)

Intuitively I want to go 15hrs/3 = water every 5 hrs. But starting at 9, I would end up watering at the following times: 9am, 2pm, 7pm as supposed to wanting to do it at 9am, 12am, and in the middle of the day. 15hrs/3-1 however gives me 7.5. I would get what I want and start at 9, 4:30, and midnight.

So the algorithm is to do the division by x-1 when wanting to place the intervals at the start. Why is that? Can someone elaborate on this? Throw some terminology around? Thanks.

 

[Square1]

Thanks all. And thanks Mark for the fenceposting word. I learn math better it seems when I have words to use and remember instead of just numbers, algorithms etc...

 

[bahamagreen]

Never heard it called fenceposting, but that is good; it shows up in time problems a lot.

So does the x-1 thing...

If you ask, "How many times in a day does the hour hand point to 12?" you run into this problem right away.

If you think "times in a day" means times in just one day, the answer is 3, because the endpoints are closed, so you count starting midnight, noon, and ending midnight.

If you think of "times in a day" means a rate, as in times per day over some indefinite number of days, then you have to use the x-1 thing and make each day have one open and one closed end to its interval (each day only has one midnight and a noon, so as to not count one day's ending midnight it shares with the next day's starting midnight twice).
For a finite number of days, either the first or last day will have 12 pointed 3 times, all the rest 2 times.
For an infinite munber of days, each day points 12 only 2 times.

Then the answer converges to 2 as the number of days increases.

for 2 days, 12 is pointed 5 times so 5/2
for 20 days, 12 is pointed 41 times so 41/20
for 200000 days 12 is pointed 400001/200000

so for x days, 12 is pointed (2X+1)/x