how to calculate time difference between two longitudes?
87306
The time difference between each longitude (each degree) is 4 minutes. So if it is 12 noon at Greenwich (0 degree), it would be 12:04 pm at 1 degree meridian and so on. In India, the standard meridian is 82-and-half degree.
414376
The eastern most longitude is 97degree25 east in arunachal pradesh. ... There is a difference of 4 mins between 2 consecutive longitudes. So difference between 30 longitudes is 30 multiplied by 4 =120 min or 2 hrs. As a result there is a difference of 2 hours between local time of gujarat and arunachal pradesh.
435885
To get the time difference in a single time unit (hours ,minutes or seconds), you can perform the following calculations. Total minutes between two times: To calculate the minutes between two times, multiply the time difference by 1440, which is the number of minutes in one day (24 hours * 60 minutes = 1440).
1201243
['Difference in hours, ignoring minutes and seconds: =HOUR(B2-A2)', 'Difference in minutes, ignoring hours and seconds: =MINUTE(B2-A2)', 'Difference in seconds, ignoring hours and minutes: =SECOND(B2-A2)']
1020409
3 Answers. Use timeIntervalSince(_ anotherDate: Date) function to get difference between two dates. Call the function with two times to find the difference.
1201045
['function distance($lat1, $lon1, $lat2, $lon2) {', '$pi80 = M_PI / 180;', '$lat1 *= $pi80;', '$lon1 *= $pi80;', '$lat2 *= $pi80;', '$lon2 *= $pi80;', '$r = 6372.797; // mean radius of Earth in km.', '$dlat = $lat2 - $lat1;']
510702
['datetime1 = datetime. datetime(2000, 1, 1)', 'datetime2 = datetime. datetime(2010, 1, 1)', 'time_difference = relativedelta(datetime2, datetime1)', 'difference_in_years = time_difference. years.', 'print(difference_in_years, "years")']
953664
['Type a start time and end time. ... ', 'Set the h:mm AM/PM format. ... ', 'Subtract the two times. ... ', 'Set the h:mm format.']
359090
['Calculating the time difference between two TIMESTAMP datatypes is much easier than the old DATE datatype. ... ', 'SELECT SUBSTR(time1,1,30) "Time1", ... ', 'As you can see, the results are much easier to recognize, 8752 days, 10 hours, 34 minutes, and 24 seconds. ... ', 'SELECT SUBSTR(time1,1,30) "Time1",']
1110835
It takes an average of 24 hours from the time the sun is directly over a certain longitude to the next time the sun is directly over the same longitude. There are 360° of longitude around the whole planet. If you divide 24 hours by 360 degrees of longitude, you get 4 minutes per degree of longitude.
1437167
To find the time zone in hours of a particular location, you can take the longitude -- in degrees -- and divide it by 15. So, for example, 75° E would be 75/15 which equals 5. That translates to the time zone being 5 hours ahead of UTC or GMT time, which can also be labeled as UTC+5.
658499
['Add the hours.', 'Add the minutes.', 'If the minutes are 60 or more, subtract 60 from the minutes and add 1 to hours.']
283292
The earth has 360 imaginary lines called longitudes or meridians running vertically between the poles. Each of these longitudes is called a degree. ... The time difference between each longitude (each degree) is 4 minutes. So if it is 12 noon at Greenwich (0 degree), it would be 12:04 pm at 1 degree meridian and so on.
138018
var t = +(new Date()) - +(new Date("2013-02-20T12:01:04.753Z")); See discussion here. From there you can make a formula to convert the difference to minutes or hours .
526851
Method 1: Using strtotime() function $timestamp1 = strtotime ( $date1 ); $timestamp2 = strtotime ( $date2 ); echo "Difference between two dates is " . $hour = abs ( $timestamp2 - $timestamp1 )/(60*60) .
207686
To calculate the minutes difference between two dates, we should divide the diffInMilliSeconds with 60 seconds) and then get the remainder by dividing it by 60 (60 minutes per hour). After that, subtract the diffInMilliSeconds with the number of minutes in milliseconds (minutes * 60) .
496121
You can use distanceFromLocation to find the distance between two coordinates. Code Snippets: CLLocation *loc1 = [[CLLocation alloc] initWithLatitude:lat1 longitude:lng1]; CLLocation *loc2 = [[CLLocation alloc] initWithLatitude:lat2 longitude:lng2]; CLLocationDistance distance = [loc1 distanceFromLocation:loc2];