Enter the start and end times. If the span crosses into another day, also enter the start and end dates (for times on the same day, you can leave the dates blank).
Enter the hour with AM or PM (5:30 PM is hour "5 PM", minute "30"). The 24-hour clock (hour "17") works too. The hours are required. Blank minutes and seconds count as 0. Fill in both dates or leave both blank. If the dates are blank and the end time is earlier than the start time, the start is treated as the day before (over midnight).
Result
Enter the start and end times (and the dates, if the span crosses into another day) in the fields on the left and press "Calculate". The result will appear here.
Table of Contents
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What you can do on this page
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What is this calculation used for?
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How to Use
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Formula
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Symbols and terms
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Good to know before you start
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How to calculate it in Excel
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How to calculate it in Google Sheets
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How to calculate it in Python
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How to write it in LaTeX and other math languages (copy and paste)
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How to have ChatGPT do the calculation
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DataChef Features
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Related Features
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NumberChef Calculators List
What you can do on this page
- Enter two times, such as 8:30 to 17:30, and you get the elapsed time between them in hours, minutes and seconds (here, 9 h) right away
- Add a start date and an end date, and spans that cross into other days, such as July 10 at 22:00 to July 12 at 6:30, are shown as "1 d 8 h 30 min" (leap years included)
- The result also shows the same span as total hours (a decimal, such as 32.5 h), total minutes and total seconds
- If the dates are blank and the end time is earlier than the start time, the start is treated as the day before, so spans over midnight work too
- A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page
Enter the hours with AM or PM, such as "8 AM" or "5 PM" (the 24-hour clock, such as "17", works too). To find hours worked minus a break, the related page "Work Hours Calculator" is more convenient.
What is this calculation used for?
Finding the travel time of an overnight bus or flight (travel)
An overnight bus that leaves at 10:00 PM and arrives at 6:30 the next morning takes \(23400 - 79200 + 86400 = 30600\) seconds = 8 hours 30 minutes, even though the date changes on the way.
Knowing the travel time makes a big difference when you estimate how long you can sleep on board or plan what to do after you arrive. Even when a schedule does not list the travel time, as with local lines or connections, you can work it out yourself from the departure and arrival times.
Tracking how long you sleep (health)
If you go to bed at 11:30 PM and get up at 6:15 AM, you slept 6 hours 45 minutes. Because the span goes over midnight, the calculation that treats the start as the day before (adding 86400 seconds) works as is.
Sleep time is one of the most basic health records. Write down when you go to bed and when you get up each day, and you can use this subtraction to find, for example, how many hours you slept on average this week.
Timing a marathon or race (sports)
A runner who starts a marathon at 9:00 and finishes at 13:57 has a time of 4 hours 57 minutes. Even if you forget your stopwatch, you can find the elapsed time by subtraction as long as you know the start and finish times.
Timing in track, swimming, cycling and other sports is exactly this: the difference between two times. Handling borrowing in base 60 for minutes and seconds correctly is the basis for comparing results accurately.
Logging study or project time (self-management)
If you study from 7:15 PM to 9:45 PM, that is 2 hours 30 minutes of study for the day. Write down the start and end times, and you can add up each day's total later with this same subtraction.
Study logs for an exam, time logs for remote work, and more: turning "from when to when" into "how many hours" is something you do almost every day.
Counting the time left until a deadline or event
"How many hours are there from 18:00 on July 10 until a 9:00 AM deadline on July 12?" With 2 days between the dates, \(86400 \times 2 + 32400 - 64800 = 140400\) seconds = 39 hours (1 day and 15 hours).
Thinking of the time left in hours makes a work plan much more realistic than a vague feeling of "I still have 2 days". Countdowns to a rocket launch or an exam are the same kind of time difference across dates.
Formula
Turning a time of day into seconds after midnight
Standard notation
(the usual math form)
\(T\)
\(=\)
\(3600\)
\(\times\)
\(h\)
\(+\)
\(60\)
\(\times\)
\(m\)
\(+\)
\(s\)
In words
(symbols replaced with words)
⑥
\(T\): time in seconds
\(=\)
②
\(3600\): seconds in 1 hour
\(\times\)
①
\(h\): hour
\(+\)
④
\(60\): seconds in 1 minute
\(\times\)
③
\(m\): minute
\(+\)
⑤
\(s\): second
The formula in words
①
Take the
\(h\): hour
②
and multiply it by
\(3600\): seconds in 1 hour
③
add the
\(m\): minute
④
multiplied by
\(60\): seconds in 1 minute
⑤
add the
\(s\): second
⑥
and you get the
\(T\): time in seconds
(how many seconds that time is after midnight)
Quick example
5:30 PM (17:30:00 on the 24-hour clock) in seconds is
time in seconds (63000)
\(=\)
seconds in 1 hour (3600)
\(\times\)
hour (17)
\(+\)
seconds in 1 minute (60)
\(\times\)
minute (30)
\(+\)
second (0)
\(T = 3600 \times 17 + 60 \times 30 + 0 = 61200 + 1800 = 63000\)
Key idea
If you try to work with times using three units at once (hours, minutes and seconds), it is easy to slip up when borrowing, because 60 seconds = 1 minute and 60 minutes = 1 hour. So this calculator first turns each time into one number: how many seconds after midnight it is. Once both times are in seconds, the rest is a plain subtraction.
For PM times, add 12 to the hour to get the 24-hour clock before you calculate (5:30 PM → 17:30). The one exception is 12 PM (noon), which stays 12.
Time difference (subtracting in seconds)
Standard notation
(the usual math form)
\(D\)
\(=\)
\(T_2\)
\(-\)
\(T_1\)
\(D\)
\(=\)
\(T_2\)
\(-\)
\(T_1\)
\(+\)
\(86400\)
In words
(symbols replaced with words)
③
\(D\): elapsed time
\(=\)
①
\(T_2\): end time in seconds
\(-\)
②
\(T_1\): start time in seconds
\(D\): elapsed time
\(=\)
\(T_2\): end time in seconds
\(-\)
\(T_1\): start time in seconds
\(+\)
④
\(86400\): seconds in 1 day
The formula in words
①
Take the
\(T_2\): end time in seconds
②
subtract the
\(T_1\): start time in seconds
③
and you get the
\(D\): elapsed time
(the number of seconds from start to end)
④
If you leave the dates blank and the end time is earlier than the start time (over midnight), the start is treated as the day before, so add the
\(86400\): seconds in 1 day
(the second formula)
Quick example
The elapsed time from 8:30:00 (30600 seconds) to 17:30:00 (63000 seconds) is
elapsed time (32400 s = 9 h)
\(=\)
end time (63000 s)
\(-\)
start time (30600 s)
\(D = 63000 - 30600 = 32400\)
\(32400 = 3600 \times 9\)
Key idea
The subtraction gives a negative number when the end time is the smaller number, as with a start at 23:45 and an end at 0:15 the next day. If you did not enter dates, this calculator assumes the start was on the day before (the span went over midnight) and adds one day, \(86400\) seconds (\(= 24 \times 3600\)). Example: from 23:45:30 to 0:15:10 the next day, \(910 - 85530 = -84620\), and \(-84620 + 86400 = 1780\) seconds = 29 min 40 s.
If the start and end times are the same, the result is treated as "0 s" (not as 24 hours later).
Subtracting in hours and minutes by hand (borrowing 60)
Standard notation
(the usual math form)
\(D\)
\(=\)
\(60\)
\(\times\)
\((h_2 - h_1)\)
\(+\)
\((m_2 - m_1)\)
In words
(symbols replaced with words)
④
\(D\): elapsed time in minutes
\(=\)
②
\(60\): minutes in 1 hour
\(\times\)
①
\((h_2 - h_1)\): difference in hours
\(+\)
③
\((m_2 - m_1)\): difference in minutes
The formula in words
①
Take the
\((h_2 - h_1)\): difference in hours
②
multiply it by
\(60\): minutes in 1 hour
③
add the
\((m_2 - m_1)\): difference in minutes
④
and you get the
\(D\): elapsed time in minutes
(even if the difference in minutes is negative, just add it as it is and the answer is right)
Quick example
The elapsed time from 9:58 to 13:57 is
elapsed time (239 min = 3 h 59 min)
\(=\)
minutes in 1 hour (60)
\(\times\)
difference in hours (13 − 9)
\(+\)
difference in minutes (57 − 58)
\(D = 60 \times (13 - 9) + (57 - 58) = 240 - 1 = 239\)
\(239 = 60 \times 3 + 59\)
Key idea
If you subtract "13:57 − 9:58" by hand, the minutes give "57 − 58", which does not work. Just as with regrouping in ordinary subtraction, borrow 1 hour (\(= 60\) minutes) from the 13 hours and read the problem as "12 hours 117 minutes − 9 hours 58 minutes". Then the minutes are \(117 - 58 = 59\) and the hours are \(12 - 9 = 3\), so the answer is 3 hours 59 minutes.
In ordinary subtraction you borrow 10, but time counts in groups of 60 (base 60), so you borrow 60. That is the only difference. The formula above does the same thing in one line by adding the difference in minutes even when it is negative.
Time difference across dates
Standard notation
(the usual math form)
\(D\)
\(=\)
\(86400\)
\(\times\)
\(n\)
\(+\)
\(T_2\)
\(-\)
\(T_1\)
In words
(symbols replaced with words)
⑤
\(D\): elapsed time
\(=\)
②
\(86400\): seconds in 1 day
\(\times\)
①
\(n\): days between the dates
\(+\)
③
\(T_2\): end time in seconds
\(-\)
④
\(T_1\): start time in seconds
The formula in words
①
Take the
\(n\): days between the dates
(counted on the calendar from the start date to the end date)
②
multiply it by
\(86400\): seconds in 1 day
③
add the
\(T_2\): end time in seconds
④
subtract the
\(T_1\): start time in seconds
⑤
and you get the
\(D\): elapsed time
Quick example
The elapsed time from 22:00 on July 10, 2026 to 6:30 on July 12, 2026 (2 days apart on the calendar) is
elapsed time (117000 s = 32 h 30 min)
\(=\)
seconds in 1 day (86400)
\(\times\)
days between the dates (2)
\(+\)
end time (23400 s)
\(-\)
start time (79200 s)
\(D = 86400 \times 2 + 23400 - 79200 = 117000\)
\(117000 = 86400 \times 1 + 3600 \times 8 + 60 \times 30\)
Key idea
The days between the dates, \(n\), counts only the dates on the calendar (July 10 → July 12 is 2 days, and February 29 in a leap year is counted correctly). The formula just adds "the seconds in \(n\) whole days" to "the difference between the two times of day", so even if the difference between the times (\(T_2 - T_1\)) is negative, adding and subtracting as written gives the right answer.
If the result is 1 day or more, you can also count every \(86400\) seconds as 1 day and write it as "1 d 8 h 30 min".
The basic method for a time difference is: turn each time into seconds after midnight, then subtract. If you enter no dates and the end is earlier than the start, add one day, 86400 seconds. If the span crosses dates, add the days between the dates × 86400 seconds.
Symbols and terms
Symbols
| \(h\), \(m\), \(s\) | h, m, s | The hour (\(h\)), minute (\(m\)) and second (\(s\)) of a time of day, on the 24-hour clock (example: for 5:30 PM, \(h = 17\), \(m = 30\) and \(s = 0\)). They stand for "hour", "minute" and "second". |
| \(T\) | T | A time of day turned into seconds after midnight, found with \(T = 3600h + 60m + s\). (Example - 17:30:00 → 63000 seconds) |
| \(T_1\), \(T_2\) | T sub 1, T sub 2 | The start time in seconds (\(T_1\)) and the end time in seconds (\(T_2\)). (Example - 8:30:00 → \(T_1 = 30600\), 17:30:00 → \(T_2 = 63000\)) |
| \(h_1, m_1\), \(h_2, m_2\) | h sub 1, m sub 1, h sub 2, m sub 2 | The hour and minute of the start time (subscript 1) and of the end time (subscript 2), used in the subtract-by-hand formula. |
| \(D\) | D | The elapsed time (how long it is from start to end). D stands for "duration". The calculator on this page finds it in seconds first, then shows it as "9 h" or "3 h 59 min". |
| \(n\) | n | The number of days from the start date to the end date, counting only the dates on the calendar. (Example - July 10 → July 12 gives \(n = 2\)) |
| \(3600\), \(86400\) | three thousand six hundred, eighty-six thousand four hundred | The seconds in 1 hour (\(60 \times 60 = 3600\)) and the seconds in 1 day (\(24 \times 3600 = 86400\)). |
Terms
| time of day | A point in time, such as 8:30. It is different from a length of time, such as 9 hours. This page finds the length of time between two times of day (the elapsed time). |
| length of time | How long something lasts, such as 9 hours. It is different from a time of day, such as 8:30. This page finds the length of time between two times of day (the elapsed time). |
| elapsed time | The length of time that passes from the start time to the end time. It is the answer this page finds. |
| 24-hour clock | A way of writing times of day with hours from 0 to 23, sometimes called military time in the US. Add 12 to PM hours (5:30 PM → 17:30, 10:00 PM → 22:00). Midnight is 0:00, and noon stays 12:00. |
| base 60 | Counting in groups of 60, where 60 of one unit make 1 of the next unit up. Sixty seconds make 1 minute and 60 minutes make 1 hour. It is easy to make mistakes when this is mixed with ordinary base 10, so this calculator first puts everything into seconds. |
| borrowing | In subtraction, taking 1 from the next place up when a place is too small to subtract from (also called regrouping). When you subtract times by hand, you borrow 60 instead of 10 (1 hour = 60 minutes, 1 minute = 60 seconds). |
| over midnight | When the date changes between the start and the end, as with an overnight bus or a night shift. If you calculate without dates, the end time is a smaller number than the start time, so the start is treated as the day before and one day, 86400 seconds, is added. (When you enter dates, the days between the dates are used instead.) |
| leap year | A year with a February 29 (about once every 4 years). When you count the days between two dates, a span that includes February 29 has one more day. This calculator counts it correctly, just as on the calendar. |
Good to know before you start
Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back over these topics is the fastest way forward.
| Time of day and elapsed time (Grades 2–3) |
|
| AM, PM and the 24-hour clock (Grades 2–3) |
|
| Subtraction with borrowing (Grades 2–3) |
|
| Division with remainders (Grades 4–5) |
|
| Decimals (Grades 4–5) |
|
How to calculate it in Excel
Copy the whole table below and paste it into cell A1 in Excel. It works as is.
Table to turn a time into seconds after midnight
| Hour (24-hour clock) | 17 |
| Minute | 30 |
| Second | 0 |
| Time in seconds | =B1*3600+B2*60+B3 |
Table for the time difference on the same day (works over midnight)
| Start: hour (24-hour clock) | 8 |
| Start: minute | 30 |
| Start: second | 0 |
| End: hour (24-hour clock) | 17 |
| End: minute | 30 |
| End: second | 0 |
| Start in seconds | =B1*3600+B2*60+B3 |
| End in seconds | =B4*3600+B5*60+B6 |
| Elapsed time (s) | =IF(B8<B7,B8-B7+86400,B8-B7) |
| Elapsed: hours | =QUOTIENT(B9,3600) |
| Elapsed: minutes | =QUOTIENT(MOD(B9,3600),60) |
| Elapsed: seconds | =MOD(B9,60) |
Table for a time difference across dates (serial values)
| Start date and time | =DATE(2026,7,10)+TIME(22,0,0) |
| End date and time | =DATE(2026,7,12)+TIME(6,30,0) |
| Elapsed (difference in serial values = days) | =B2-B1 |
| Elapsed time shown as h:mm:ss | =TEXT(B2-B1,"[h]:mm:ss") |
| Total hours (decimal) | =(B2-B1)*24 |
In the first table, enter the hour, minute and second of a time in B1 to B3, and B4 shows the seconds after midnight (63000 in this example).
In the second table, enter the start and end times in B1 to B6. B7 and B8 turn them into seconds (30600 and 63000), and B9 shows the elapsed time in seconds (32400). The IF function adds 86400 when the end is the smaller number, to handle spans over midnight. B10 to B12 split the seconds into hours, minutes and seconds with QUOTIENT (the whole-number part of a division) and MOD (the remainder). The answer is 9 hours 0 minutes 0 seconds.
The third table works across dates. When you build a date and time with the DATE and TIME functions, Excel stores it as a serial value (1 day = 1), so the subtraction in B3 alone gives the elapsed time as a decimal number of days (about 1.3542 days). The TEXT format "[h]:mm:ss" in B4 keeps counting hours past 24, so it shows "32:30:00", and B5 multiplies the serial value by 24 to get 32.5 total hours. Just replace the values in column B with your own.
In the second table, enter the start and end times in B1 to B6. B7 and B8 turn them into seconds (30600 and 63000), and B9 shows the elapsed time in seconds (32400). The IF function adds 86400 when the end is the smaller number, to handle spans over midnight. B10 to B12 split the seconds into hours, minutes and seconds with QUOTIENT (the whole-number part of a division) and MOD (the remainder). The answer is 9 hours 0 minutes 0 seconds.
The third table works across dates. When you build a date and time with the DATE and TIME functions, Excel stores it as a serial value (1 day = 1), so the subtraction in B3 alone gives the elapsed time as a decimal number of days (about 1.3542 days). The TEXT format "[h]:mm:ss" in B4 keeps counting hours past 24, so it shows "32:30:00", and B5 multiplies the serial value by 24 to get 32.5 total hours. Just replace the values in column B with your own.
How to calculate it in Google Sheets
Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.
Table to turn a time into seconds after midnight
| Hour (24-hour clock) | 17 |
| Minute | 30 |
| Second | 0 |
| Time in seconds | =B1*3600+B2*60+B3 |
Table for the time difference on the same day (works over midnight)
| Start: hour (24-hour clock) | 8 |
| Start: minute | 30 |
| Start: second | 0 |
| End: hour (24-hour clock) | 17 |
| End: minute | 30 |
| End: second | 0 |
| Start in seconds | =B1*3600+B2*60+B3 |
| End in seconds | =B4*3600+B5*60+B6 |
| Elapsed time (s) | =IF(B8<B7,B8-B7+86400,B8-B7) |
| Elapsed: hours | =QUOTIENT(B9,3600) |
| Elapsed: minutes | =QUOTIENT(MOD(B9,3600),60) |
| Elapsed: seconds | =MOD(B9,60) |
Table for a time difference across dates (serial values)
| Start date and time | =DATE(2026,7,10)+TIME(22,0,0) |
| End date and time | =DATE(2026,7,12)+TIME(6,30,0) |
| Elapsed (difference in serial values = days) | =B2-B1 |
| Elapsed time shown as h:mm:ss | =TEXT(B2-B1,"[h]:mm:ss") |
| Total hours (decimal) | =(B2-B1)*24 |
IF (a condition), QUOTIENT (the whole-number part of a division), MOD (the remainder), DATE, TIME and TEXT are written the same way in Google Sheets as in Excel. Copy the whole table, paste it into cell A1, and replace the values in column B with your own. The answers are the same as in Excel.
How to calculate it in Python
from datetime import datetime
# Start and end date and time (use the same date if the span does not cross into another day)
start = datetime(2026, 7, 10, 22, 0, 0) # July 10, 2026, 22:00:00
end = datetime(2026, 7, 12, 6, 30, 0) # July 12, 2026, 6:30:00
duration = end - start # elapsed time (a timedelta)
total_seconds = int(duration.total_seconds()) # total seconds
hours, remainder = divmod(total_seconds, 3600) # hours and the seconds left over
minutes, seconds = divmod(remainder, 60) # split the rest into minutes and seconds
days, day_remainder = divmod(total_seconds, 86400) # for the display with days
print(f"Elapsed time: {hours} h {minutes} min {seconds} s")
print(f"With days: {days} d {day_remainder // 3600} h {day_remainder % 3600 // 60} min")
print(f"Total hours: {total_seconds / 3600} h")
print(f"Total minutes: {total_seconds / 60} min")
print(f"Total seconds: {total_seconds} s")
Runs with the standard library only. Subtracting one datetime from another gives the elapsed time as a timedelta, and total_seconds() turns it into seconds. divmod returns the quotient and the remainder of a division at the same time: dividing the seconds by 3600 gives the hours and the rest, and dividing by 60 gives minutes and seconds.
Change the start and end date and time at the top (in the order year, month, day, hour, minute, second), then run the code.
How to write it in LaTeX and other math languages (copy and paste)
Turning a time of day into seconds after midnight
T = 3600h + 60m + s
T = 3600h + 60m + s
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>T</mi><mo>=</mo>
<mn>3600</mn><mi>h</mi>
<mo>+</mo><mn>60</mn><mi>m</mi>
<mo>+</mo><mi>s</mi>
</mrow>
</math>
T = 3600h + 60m + s
totalSeconds = 3600*h + 60*m + s
T := 3600*h + 60*m + s;
T = 3600*h + 60*m + s;
T = 3600h + 60m + s
Time difference (subtracting in seconds)
D = T₂ − T₁ (T₂ ≥ T₁), D = T₂ − T₁ + 86400 (T₂ < T₁)
D = T_2 - T_1 \quad (T_2 \ge T_1), \qquad D = T_2 - T_1 + 86400 \quad (T_2 < T_1)
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>D</mi><mo>=</mo>
<msub><mi>T</mi><mn>2</mn></msub><mo>−</mo>
<msub><mi>T</mi><mn>1</mn></msub>
<mo>,</mo><mspace width="0.7em"/>
<mi>D</mi><mo>=</mo>
<msub><mi>T</mi><mn>2</mn></msub><mo>−</mo>
<msub><mi>T</mi><mn>1</mn></msub>
<mo>+</mo><mn>86400</mn>
</mrow>
</math>
D = T_2 - T_1, D = T_2 - T_1 + 86400
durationSeconds = If[t2 >= t1, t2 - t1, t2 - t1 + 86400]
dur := `if`(t2 >= t1, t2 - t1, t2 - t1 + 86400); # D is reserved in Maple (the differential operator), so the name dur is used
if t2 >= t1, D = t2 - t1; else, D = t2 - t1 + 86400; end
D = T_2 - T_1 (T_2 ≥ T_1), D = T_2 - T_1 + 86400 (T_2 < T_1)
Subtracting in hours and minutes by hand (borrowing 60)
D = 60(h₂ − h₁) + (m₂ − m₁)
D = 60(h_2 - h_1) + (m_2 - m_1)
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>D</mi><mo>=</mo>
<mn>60</mn>
<mo>(</mo><msub><mi>h</mi><mn>2</mn></msub><mo>−</mo><msub><mi>h</mi><mn>1</mn></msub><mo>)</mo>
<mo>+</mo>
<mo>(</mo><msub><mi>m</mi><mn>2</mn></msub><mo>−</mo><msub><mi>m</mi><mn>1</mn></msub><mo>)</mo>
</mrow>
</math>
D = 60(h_2 - h_1) + (m_2 - m_1)
durationMinutes = 60*(h2 - h1) + (m2 - m1)
dur := 60*(h2 - h1) + (m2 - m1); # D is reserved in Maple (the differential operator), so the name dur is used
D = 60*(h2 - h1) + (m2 - m1);
D = 60(h_2 - h_1) + (m_2 - m_1)
Time difference across dates
D = 86400n + T₂ − T₁
D = 86400n + T_2 - T_1
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>D</mi><mo>=</mo>
<mn>86400</mn><mi>n</mi>
<mo>+</mo><msub><mi>T</mi><mn>2</mn></msub>
<mo>−</mo><msub><mi>T</mi><mn>1</mn></msub>
</mrow>
</math>
D = 86400n + T_2 - T_1
durationSeconds = 86400*n + t2 - t1
dur := 86400*n + T2 - T1; # D is reserved in Maple (the differential operator), so the name dur is used
D = 86400*n + T2 - T1;
D = 86400n + T_2 - T_1
How to have ChatGPT do the calculation
You are a time calculation assistant. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing). The start is July 10, 2026 at 22:00:00 and the end is July 12, 2026 at 6:30:00 (both on the 24-hour clock). Find each of the following: 1. The elapsed time (in the form "X hours Y minutes Z seconds") 2. The elapsed time with days (in the form "X days Y hours Z minutes") 3. The total hours (as a decimal), total minutes and total seconds Find the elapsed time by subtracting Python datetime objects, and split it into hours, minutes and seconds with divmod. Show the formulas you used and the numbers from the execution result.
How to Use
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1Enter your numbersType the numbers you want to calculate with into the input fields
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2CalculatePress the "Calculate" button
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3Check the resultThe result appears on the spot. The same page also explains the idea behind the calculation and the formula
DataChef Features
Easy and Free
Unlimited conversions for free.
No technical knowledge required.
Intuitive and user-friendly operation.
No technical knowledge required.
Intuitive and user-friendly operation.
No Registration Required
Available immediately after access.
Can be used without registering personal information.
Can be used without registering personal information.
Safe and Secure
Fully SSL encrypted communication.
Automatic file deletion by clicking "download".
Automatic file deletion by clicking "download".
Fast
High-speed site access
and rapid file conversion.
and rapid file conversion.
No Watermark
No watermark.
No attribution required.
No attribution required.
Commercial Use Available
Free for commercial use.
No need to contact us for commercial use permission.
No need to contact us for commercial use permission.
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Adds and subtracts lengths of time such as "2 hours 30 minutes"
Date and Time Calculator (Add or Subtract Hours, Minutes and Seconds)
Adds days, hours, minutes and seconds to a start date and time, or subtracts them, and shows the day of the week
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