Sections
- Introduction
- Adding Time
- Subtracting Time
- Adding and Subtracting Angles
- A Note About Fractions of a Second and Fractions of an Arcsecond
- Quiz Yourself
Introduction
Adding and subtrating time units (hours, minutes, seconds) or the common angular units (degrees, arcminutes, arcseconds) is not quite the same as adding and subtracting numbers in our usual base-\(10\) system. The reason for this is that these quantities do not simply “roll over” to the next decimal value after reaching \(9\) as standard base-\(10\) does.
To see this think about simply counting the number of students in your class. If you have a particularly large class held in an auditorium, you may have \(250\) members in your class. Counting them is a simple matter of counting the first person as \(1\), the next as \(2\), the ninth as \(9\), the tenth as \(10\), the eleventh as \(11\), the ninety-ninth as \(99\), the one hundreth as \(100\), and so on, until you count the last student as \(250\).
Adding Time
As you know, the same form of counting doesn’t work when you keep track of time. Suppose your class starts at \(\mbox{9:30 AM}\) and ends at \(\mbox{10:20 AM}\). When the clock reaches \(\mbox{9:59}\) the next minute isn’t written at \(\mbox{9:60}\) and the one after that as \(\mbox{9:61}\); instead those times are written as \(\mbox{10:00}\) and \(\mbox{10:01}\), respectively. Although the minutes keep counting from \(0\) to \(59\), they recycle back to \(0\) the following minute while the hour increases by \(1\). Of course this isn’t completely true when the time reaches \(\mbox{23:59}\) (assuming you are counting up to \(24\), “military time”). The next minute becomes \(\mbox{0:00}\) and another day begins.
The situation is even more complicated when using the more common AM/PM notation. Starting at midnight, the clock doesn’t reset to \(\mbox{0:00}\) but instead goes from \(\mbox{11:59 PM}\) to \(\mbox{12:00 AM}\). Over the next hour the clock counts up to \(\mbox{12:59 AM}\) and then resets, but not to \(\mbox{0:00}\)! Instead, it resets to \(\mbox{1:00 AM}\). The same behavior happens \(12\) hours later going from \(\mbox{11:59 AM}\) to \(\mbox{12:00 PM}\) and from \(\mbox{12:59 PM}\) to \(\mbox{1:00 PM}\). This is fairly bizzare when you think about it.
Given how convoluted the counting scheme is when using two \(12\)-hour cycles that run from \(\mbox{1:00}\) to \(\mbox{12:59}\) before repeating, it is much easier to use the method of counting from \(\mbox{0:00}\) to \(\mbox{23:59}\) and then resetting to \(\mbox{0:00}\) again. You can then convert to the typical AM/PM clock time from there. Of course, if you are also keeping track of seconds, the last second of one day is given as \(\mbox{23:59:59}\) and the beginning of the next day is \(\mbox{0:00:00}\).
It is likely that you are so accustomed to the weird behavior of clock arithmetic that you really haven’t given it much thought, but suppose you find yourself needing to add an interval of time to a specified time to arrive at another time. For example, suppose that the current time of day is \(\mbox{1:45 PM}\) and you text your friend, telling her that you will meet her in the library in \(45\) minutes. Your friend immediately knows that you will be meeting her at \(\mbox{2:30 PM}\), but how did you and your friend do that arithmetic? Without realizing it you did something similar to, but slightly different from adding \(46\) apples to \(145\) apples to arrive at \(191\) apples. When adding the apples you start by adding the right-most digits \(6 + 5\) to get \(11\). The right-most \(1\) in your initial step becomes the right-most digit in the final answer, but the left-most \(1\) in your initial step gets added to the next decimal place to the left (you “carry the \(1\)”) so that the next step is to add \(1 + 4 + 4\) to get \(9\) in the second digit from the right in the final answer. Since the result for the second digit didn’t exceed \(9\) you don’t need to “carry forward” another digit, and your final answer is \(191\) apples. In the familiar column form of the process you did the following:
| \(1\) (carried) | |||
| \(1\) | \(4\) | \(5\) | |
| \(+\) | \(4\) | \(6\) | |
| \(1\) | \(9\) \((1+4+4)\) | \(1\) \((10+1)\) |
Returning to the time example, the difference is in when/how you carry the digits forward based on cycles running from \(0\) to \(59\) rather than from \(0\) to \(9\). First, write \(\mbox{1:45 PM}\) in the \(24\) hour format as \(\mbox{13:45}\) and then add \(45\) minutes to it. It is generally easier to think about adding all of the minutes together rather than the right-most places and then the left-most places for minutes. In our example this would mean \(45 + 45 = 90\) minutes. But \(90\) minutes is \(60\) minutes plus \(30\) minutes, or \(1\) hour plus \(30\) minutes. Therefore, you need to carry a \(1\) over to the right-most hours column with an addition \(30\) remaining in the minutes columns. To see this in column notation:
| \(1\) (carried) | |||
| \(13\) | \(:\) | \(45\) | |
| \(+\) | \(45\) | ||
| \(14\) \((1+13)\) | \(:\) | \(30\) \((60+30)\) |
The \(\mbox{14:30}\) time in \(12\) hour AM/PM format is \(\mbox{2:30 PM}\).
As another example of adding times, suppose the current time is \(\mbox{11:30 AM}\), or \(\mbox{11:30}\) in the \(24\) hour format, and you are told that an important event will occur in \(2\) days, \(22\) hours, \(45\) minutes from now. How would you add that amount of time to the current time to find out when the event will occur? The added complication in this case is that days have \(24\) hours in them, while hours have \(60\) minutes.
Start by adding the \(45\) minutes to the current \(30\) minutes, giving \(75\) minutes. But \(75\) minutes is \(1\) hour plus \(15\) minutes. Carrying the extra \(1\) hour to the next step in the addition means adding \(1 + 11 + 22 = 34\) hours. But since \(34\) hours is \(1\) day plus \(10\) hours, \(10\) will end up in the hours column and an additional \(1\) day will added to the other \(2\) days to give \(3\) days. In column format this becomes
| Days | Hr | Min | ||
|---|---|---|---|---|
| \(1\) (carried) | \(1\) (carried) | |||
| \(11\) | \(:\) | \(30\) | ||
| \(+\) | \(2\) | \(22\) | \(:\) | \(45\) |
| \(3\) \((1+2)\) | \(10\) \((24+10)\) | \(:\) | \(15\) \((60+15)\) |
If today is Sunday, then the event will occur three days from now on Wednesday at \(\mbox{10:15 AM}\).
Subtracting Time
Subtracting one time from another is just the reverse of adding times. Rather than “carrying” additional time to the next largest time unit (adding \(1\) to the hours column when you found that \(30\) minutes plus \(45\) minutes gives \(75\) minutes, or \(1\) hour plus \(15\) minutes) you may need to “borrow” time from the next largest time unit. For example, suppose the current time in \(24\) hour format is \(\mbox{22:15}\) (equivalent to \(\mbox{10:15 PM}\) and you want to find out what time it was \(8\) hours and \(35\) minutes ago. Since subtracting \(35\) in the minutes column from \(15\) in the minutes column would give a negative time (also wierd), you will need to “borrow” \(1\) hour from the \(22\) hours, allowing you to add \(60\) minutes to the \(15\) minutes, then the subtraction can continue as usual. (Recall that in base-\(10\) arithmetic you would always “borrow” \(10\) from the next highest decimal place, reducing the value of the next highest place by \(1\).) In column format the process just described looks like:
| \(21\) (1 hr “borrowed” for mins) | \(75\) \((60 + 15)\) | ||
| \(22\) | \(:\) | \(15\) | |
| \(-\) | \(8\) | \(:\) | \(35\) |
| \(13\) \((21 – 8)\) | \(:\) | \(40\) (\(75-35)\) |
The \(24\) hour format for the desired time is \(\mbox{13:40}\) which is equivalent to \(\mbox{1:40 PM}\). Of course you should check your answer by making sure that when you add \(\mbox{13:40}\) to \(\mbox{8:35}\) you do in fact arrive at your starting time of \(\mbox{22:15}\).
The same process can be followed when subtracting more than \(24\) hours from another date and time. If the current date is June \(12\) and the current time is \(\mbox{18:30}\), what was the date and time \(35\) days, \(11\) hours, \(50\) minutes ago?
You should immediately see a problem; the day you are looking for must be in May! One approach to this dilemma is to “borrow” the number of days in May (\(31\)) and add them to the \(12\) days in June, so that there would be \(43\) days to start with in the “days” column. Of course, if we were subtracting \(63\) days instead of \(35\) days, we would also need to “borrow” the \(30\) days from April as well! You can see how this can get complicated quickly given the varying number of days in each month, with added complication of leap years. (A point of trivia for the next time you appear on a game show: It is for this reason that astronomers use a time system that simply counts the number of days since an arbitrary start date/time which happens to be noon on January \(\mbox{1, 4713}\) BCE in Greenwich, England.)
Carrying out the calculation, including “borrowing” the days from May, is represented in column notation as:
| Borrowed | |||||
|---|---|---|---|---|---|
| Month | Day | Hr | Min | ||
| May | \(43\) | \(17\) | \(90\) | ||
| June | 12 | 18 | : | 30 | |
| \(-\) | May | \(35\) | \(11\) | \(:\) | \(50\) |
| \(8\) | \(6\) | \(:\) | \(40\) |
The resulting day and time is May \(8\) at \(\mbox{6:40 AM}\).
Adding and Subtracting Angles
Adding and subtracting angles measured in degrees, arcminutes, and arcseconds is fundamentally no different than adding and subtracting units of time with the obvious exception that there are \(360^\circ\) in a circle and only \(24\) hours in one day. This is because there are \(60\) minutes in one hour and \(60\) seconds in one minute, while there are \(60\) arcminutes (also represented as \(60^\prime\)) in one degree and \(60\) arcseconds (\(60^{\prime\prime}\)) in one arcminute. As a result, we can proceed just as we did in the examples in the two previous sections.
As an example, add \(53^\circ\ 38^\prime\ 56^{\prime\prime}\) and \(45^\circ\ 52^\prime\ 36^{\prime\prime}\). Since arcminutes and arcseconds run from \(0\) to \(59\) it is going to be necessary to carry quantities over to the next highest value. In column notation:
| \(1^\circ\) | \(1^\prime\) | ||
| \(53^\circ\) | \(38^\prime\) | \(56^{\prime\prime}\) | |
| \(+\) | \(45^\circ\) | \(52^\prime\) | \(36^{\prime\prime}\) |
| \(99^\circ\) | \(31^\prime\) | \(32^{\prime\prime}\) |
Now consider what happens when the same two angles are subtracted instead. Since \(56^{\prime\prime}\) is greater than \(36^{\prime\prime}\) the arcseconds column can be subtracted in a straight-forward way. However, in the arcminutes column \(38^\prime\) is less than \(52^\prime\) so in order to subtract the values, \(1^\circ = 60^\prime\) will need to be “borrowed” from the degrees column. As a result, the number of degrees is decreased by \(1^\circ\) while the number of arcminutes is increased by \(60^\prime\). As a result, the calculation becomes:
| \(52^\circ\) | \(98^\prime\) | ||
| 53\(^\circ\) | 38\(^\prime\) | \(56’^{\prime\prime}\) | |
| \(-\) | \(45^\circ\) | \(52^\prime\) | \(36^{\prime\prime}\) |
| \(7^\circ\) | \(46^\prime\) | \(20^{\prime\prime}\) |
A Note About Fractions of a Second and Fractions of an Arcsecond
It is often the case that observations or experiments result in measurements that are a small fraction of a second or a small fraction of an arcsecond. When this happens, rather than further subdividing time or angular units by continuing factors of \(60\), simple decimal notation is used instead. For example, suppose that a time measurement results in the full duration of an event being recorded as \(4\ \mbox{h}\), \(37\ \mbox{min}\), \(12.82\ \mbox{s}\), while an intermediate measurement occurs at \(2\ \mbox{h}\), \(55\ \mbox{min}\), \(43.45\ \mbox{s}\). How much time elapsed between the two measurements?
The process is exactly the same as subtracting two time units demonstrated in the Subtracting Time section above. The only change is that subtracting the seconds column requires consideration of the usual base-\(10\) process of subtracting numbers with values to the right of the decimal point. To see this:
| \(3\) | \(96\) | \(72.82\) | |||
| \(4\) | \(:\) | \(37\) | \(:\) | \(12.82\) | |
| \(-\) | \(2\) | \(:\) | \(55\) | \(:\) | \(43.45\) |
| \(1\) | \(:\) | \(41\) | \(:\) | \(29.37\) |
Analyizing how the calculation was performed, first note that \(12.82\ \mbox{s}\) is less than \(43.45\ \mbox{s}\) so \(1\) minute was “borrowed” from the minutes column, which made the value in that column \(36\) rather than the original \(37\). The value in the seconds column was also increased by \(1\ \mbox{min} = 60\ \mbox{s}\) to give \(12.82 + 60 = 72.82\ \mbox{s}\) before the subtraction occurred. Once the calculation in the seconds column was finished, it was time to work on the minutes column. However, again we had the top number (now \(36\)) less than the lower number (\(55\)), requiring that \(1\) hour (\(60\) minutes) be “borrowed” from the hours column. This reduced the top hours column value to \(3\) while raising the minutes column value to \(36 + 60 = 96\). Now it becomes possible to finish the calculation. Notice that once the seconds column was modified to have a larger number minus a smaller number (by “borrowing” from the adjacent minutes column), the subtraction is exactly like calculating the difference in two numbers in base-\(10\).
The process is no different when dealing with fractions of arcseconds.
Quiz Yourself
To make sure that you understand how to work with time and angular units, try the problems below.
- The clock time is \(\mbox{21:59:35}\) in \(24\) format. What time is it \(3\) minutes, \(45\) seconds later.
- An east-coast baseball game went into extra innings. The game started at \(\mbox{8:30 PM}\) and ended at \(\mbox{1:18 AM}\) the next day. How long was the game?
- The latitude of Minneapolis, Minnesota is \(44^\circ\ 58^\prime\ 33^{\prime\prime}\) N and the latitude of Ogden, Utah is \(41^\circ\ 13^\prime\ 22^{\prime\prime}\) N. What is the difference in latitude between the two cities?
- The longitude of Minneapolis, Minnesota is \(93^\circ\ 16^\prime\ 25^{\prime\prime}\) W. Ogden, Utah is \(18^\circ\ 41^\prime\ 49^{\prime\prime}\) west of Minneapolis. What is the longitude of Ogden?
- A certain star brightens, dims, and then brightens again with a well defined period of \(12\) days, \(4\) hours, \(32\) minutes. The last time it is was at its brightest was on August 17 at \(\mbox{9:42 PM}\). When do you expect the star to be at its brightest again (in \(24\) hour format)?
Answers
- \(\mbox{22:03:20}\)
- \(4\) hours, \(48\) minutes
- \(3^\circ\ 45^\prime\ 11^{\prime\prime}\)
- \(111^\circ\ 58^\prime\ 14^{\prime\prime}\) W
- August 30 at \(02:14\)