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T O P I C    R E V I E W
Xaime94 Posted - 05/15/2018 : 07:40:30 AM
Origin Ver. and Service Release (Select Help-->About Origin): OriginPro2015
Operating System:Windows 7

I got a question with regards to date data types, let's say yyMMdd HH:mm:ss. How does Origin count the value of the integral of the function f(T), where T corresponds to date data? I mean, what's the unit of the integral of funtion f(T) over time? Let's suppose the values of function f(T) are in Watts, then the unit of integral of f(T) is: Watts * (what unit?)

4   L A T E S T    R E P L I E S    (Newest First)
cpyang Posted - 05/20/2018 : 8:17:50 PM
Hi Xaime94,

The date column is internally stored as Julian Days, so the actual numeric value unit is a Day or 24 hours, so 0.5 = 12 hours.

CP
Drbobshepherd Posted - 05/15/2018 : 5:21:42 PM
X,

Yes, if f(t)is power in watts, and t is time in days, then the integral
⌠T
│ f(t)dt = energy in watt*days.
⌡0

Multiply by 86400 to express answer in joules; or, multiply by 0.024 to convert answer to kW*hr. These are more common units of energy.

DrBob
Xaime94 Posted - 05/15/2018 : 10:08:36 AM
Drbobshepherd,

Thank you, so the unit of integrated curve, is [Watts * 24h]?

In other words, if I got the time interval for example: 2018-03-12 06:29:21 - 2018-03-12 06:35:47 (which is 386 second), then the calculated integral over f(t),where t is in data type will be equal:
(the value of integrated curve f(t), where t in seconds)/24*60*60?

Because 1 day=24h=24*60*60=86400s
Drbobshepherd Posted - 05/15/2018 : 09:24:47 AM
Xaime94,

The way time is displayed in Origin is just a format interpretation. The way time data is stored is the floating point value of a date on the Julian calendar. The unit of the Julian calendar is a day.

Please verify this fact using the help files. I am not an employee of OriginLab Inc. I am just a user like yourself.

DrBob


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