answered question

answers (3)

mysterygir...
1
Vote
BEST ANSWER  chosen by asker   |  mysterygirl89  |  June 23, 2009 01:48 AM
If it is broken as in there is no life in it at all then.
Well there are 24 hours in one day, there are 12 hours on the clock. So, say your broken clock reads 6:45. It will be correct at 6:45PM the afternoon and at 6:45AM the morning. Because the clock is broken it will not move

Same with the Month and day, just wait one year from that Year and *poof* they are the same again, minus the year... working on it if it was just running slow, cant seem to think of how to make an equation though. I will post the other answer in a comment on here if I figure it out
Asker's rating:  
mysterygirl89 was the first to answer and I was wrong - her answer IS correct! My question only makes sense if the clock is still working. A broken clock is STILL only right twice a day just not at the exact same moment every day.

Sorry for the refuted fact - I was dead wrong!

And I thought I was being so clever finding the .002 difference that I got lost in the idea and failed to realize my mistake!

Sorry mysterygirl89, I hope the tip makes up for my ignorance. :)

P.S. Thank you to every who helped clarify the error in my question.

voted helpful: verthorizon

Comment
mysterygir...
mysterygirl89  |  June 23, 2009 02:02 AM
Lets say the clock that is slow takes 2 hours to reach and hour… Sooo
Lets start at Noon with both clocks. 12=12 or 24, 1=13, 2=14, 3=15, 4=16, and so on, military time.
So basically it would take 24 hours.
X/2=Y This is the only one that worked, 24/2=12 Yeay!
silverhamm...
silverhammer  |  June 23, 2009 03:16 AM - Fact Refuted
The usual answer is "twice a day" but as discovered in eurekapizza's original question actual time is offset by "leap seconds" which would make clocks which you could see the seconds off by a specific amount per day.

How long would it take to completely cycle through HH:MM:SS of real time (taking the offset mentioned in the link) before it would match up again?

Then if your clock showed month and day (not year), how long before it would line up again (taking leap years and leap seconds into account)?

http://cr.yp.to/proto/utctai.html

Thanks for trying, do a little more research and try again. :)
verthorizo...
verthorizon  |  June 23, 2009 04:47 AM
Just to clarify:

We have a clock that is broken, and for simplicity sake, we'll say that its a 24 hour clock. It breaks at 12 o'clock. After 24 hours has passed, the clock is at noon but the 'real' time by rotation of the earth is actually noon plus 2 milliseconds. After another 24 hours has passed, the clock is once again in its disrepaired state, forever showing 12 o'clock, but time has progressed another 2 milliseconds so it is really noon and 4 milliseconds.

If this is the case, its simply dividing our perceived 24 hours by the difference between the real and perceived 24 hour day which is 2 milliseconds:

24 hours * (3600 seconds/ 1 hour) = 86,400 seconds in a perceived day
86, 400 seconds in a day / .002 seconds = 43,200,000 days
43,200,000 days * (1 year / 365.25 days) = 118 275.154 years

So about 118,275 years later, the minuscule extra milliseconds in a day finally add up to 24 hours.

Though you could argue that since UTC has a leap second every year or two it'll probably right in that time span.

But I still think mysterygirl has the right answer. The clock in my case is right once a day, whether or not you take into account the millisecond. At the point when it is 11:59:988 would be the correct time on the clock.

EDIT - For clocks that include both month and day, you just keep adding the above value, and checking to see if it lands on the correct day and month. I'm sure there's a mathematical way to solve, but a creating program would be easier.
mysterygir...
mysterygirl89  |  June 23, 2009 06:26 AM
That is if a clock is DEAD, BROKEN. So it is true that it will be correct 2 times because it never moves. So my fact is not bad. I even explained how. I did research, hence the military time.
mysterygir...
mysterygirl89  |  June 23, 2009 06:29 AM
@Verthorizon Thank you. It is correct that if a clock is DEAD or "Broken", meaning that non of the hand on the clock are moving, or like I said in a comment about if the clock is a little slow, as you can read in my other comment. Thanks
mysterygir...
mysterygirl89  |  June 23, 2009 06:35 AM
@silverhammer Read my fist comment, Starting "Lets say...." (^_^)V
mysterygir...
mysterygirl89  |  June 26, 2009 04:29 PM
It's ok, I knew you where just being stubborn like me, you were most likely meaning the question in a different way and were wanting your answer, I am the same way. Thanks though. I spent a lot of time on the question. I had papers all over the floor in my living room writing down equations. LOL. (^_^)V
albanian
0
Votes
albanian  |  June 23, 2009 05:00 AM
Yes the time varies but that would not stop a clock from being right twice a day. The time still passes through the same twenty four hours. The clock would just be correct slightly further than twelve hours apart. A 24 hr clock would be correct almost once a day. In other words, 1 pm would still occur once a day, so a clock stuck at 1 pm would be right at that point in time.

It is a clock which has not broken which will rarely be correct to the degree that you are speaking of.
Comment
badaspie
0
Votes
badaspie  |  June 23, 2009 07:52 AM
Leap seconds are added (or subtracted) empirically, by comparing the observed rotation of the earth with the atomic standard (Coordinated Universal Time or UTC). The adjustment is made to UTC to bring it into agreement with the time as determined by the earth's rotation. The earth's rotation rate varies on several different time scales in addition to the long-term slowing caused by the moon's tidal effects and is too complex to be mathematically modeled and predicted. Therefore, it is impossible to write a mathematical formula for calculating the exact times when a broken clock will be accurate.
Comment
140

ask any question

Top of Page
Buy Mahalo Dollars
WITH CREDIT CARD OR PAYPAL

Please log in to use this function.