I am trying to make an approximation of the length of day from sunrise to sunset, and the length of night from sunset to sunrise. My present approximation is crude (it assumes yesterday and tomorrow have equivalent values to today), but for now I am not specifically concerned with pinpointing yesterday sunset, today sunrise, today sunset, and tomorrow sunrise (yet). My goal is a calculation based on twelve equal hours per night (twelve equal to each other, not equal to a standard hour or daytime hour), and twelve equal hours per day.
What I am concerned with is that in my iOS app, the calculation is way off; a minute flies by in 5-6 (standard) seconds' time. When I use unmodified time, in other code from here, the clock moves at a standard pace, but when I try to get this code to feed the clock code, something is out of bounds.
The code I've been working on, as an approximation, is:
NSDate *now = [[NSDate alloc] init];
NSDate *factory = [[NSDate alloc] init];
NSDate *summerSolstice2013 = [factory initWithTimeIntervalSinceReferenceDate:_referenceSummerSolstice];
double distanceAlong = [now timeIntervalSinceDate:summerSolstice2013];
double angleAlong = M_PI * 2 * distanceAlong / (2 * (_referenceWinterSolstice - _referenceSummerSolstice));
double currentHeight = cos(angleAlong) * _latitudeAngle + _tiltAngle;
...
if (_secondsAreNatural)
{
_secondsAreShadowed = FALSE;
double dayDuration = 12 * 60 * 60 + 12 * 60 * 60 * sin(currentHeight);
double midday = fmod(24 * 60 * 60 * _longitudeAngle / (2 * M_PI) + 12 * 60 * 60, 24 * 60 * 60);
double sunrise = midday - dayDuration / 2;
double sunset = midday + dayDuration / 2;
double seconds = fmod([now timeIntervalSinceReferenceDate], 24 * 60 * 60);
double proportionAlong = 0;
if (seconds < sunrise)
{
_naturalSeconds = (seconds - sunset - 24 * 60 * 60) / (sunrise - sunset - 24 * 60 * 60);
}
else if (seconds > sunset)
{
_naturalSeconds = 12 * 60 * 60 * (seconds - sunset) / (sunrise + 24 * 60 * 60 - sunset) + 18 * 60 * 60;
}
else
{
_naturalSeconds = 12 * 60 * 60 * (seconds - sunrise) / (sunset - sunrise) + 6 * 60 * 60;
}
}
Are there any problems (given that this approximation can probably be refined to any extent) you can pinpoint in this code?
Thanks,
--EDIT--
The code I wrote above was fairly demanding in terms of the loose ends presented to someone reading it. I tried to take another pass, and rewrite it in simpler terms and with a purer mathematical model. I wrote, comments added:
NSDate *now = [[NSDate alloc] init];
NSDate *summerSolstice2013 = [[NSDate alloc] initWithTimeIntervalSinceReferenceDate:_referenceSummerSolstice];
double distanceAlong = [now timeIntervalSinceDate:summerSolstice2013];
// How far along are we, in seconds, since the reference date?
double angleAlong = M_PI * 2 * distanceAlong / (2 * (_referenceWinterSolstice - _referenceSummerSolstice));
// What's the angle if 2 π radians corresponds to a whole year?
double currentHeight = cos(angleAlong) * _latitudeAngle + _tiltAngle;
// _latitudeAngle is the angle represented by our latitude; _tiltAngle is the angle of the earth's tilt.
NSInteger day = 24 * 60 * 60;
// 'day' could have been called secondsInADay, but it was mean to reduce the number of multiplicands represented in the code.
// If we are in the endless day or endless night around the poles, leave the user with standard clock hours.
if (currentHeight > M_PI / 2)
{
_secondsAreShadowed = TRUE;
}
else if (currentHeight < - M_PI / 2)
{
_secondsAreShadowed = TRUE;
}
// Otherwise, calculate the time this routine is meant to calculate. (This is the main intended use case.)
else if (_secondsAreNatural)
{
_secondsAreShadowed = FALSE;
// closestDay is intended to be the nearest midnight (or, in another hemisphere, midday), not exactly in hours offset from UTC, but in longitude offset from Greenwich.
double closestDay;
if (fmod(distanceAlong, day) < .5 * day)
{
closestDay = distanceAlong - fmod(distanceAlong, day);
}
else
{
closestDay = day + distanceAlong - fmod(distanceAlong, day);
}
// As we go through the calculations, for the most part we keep up information on the previous and next days, which will to some degree be consulted at the end.
double previousDay = closestDay - day;
double nextDay = closestDay + day;
// For the three days, what proportion of the way along are they from the solstices?
double closestDayAngleAlong = M_PI * 2 * closestDay / (2 * (_referenceWinterSolstice - _referenceSummerSolstice));
double previousDayAngleAlong = M_PI * 2 * previousDay / (2 * (_referenceWinterSolstice - _referenceSummerSolstice));
double nextDayAngleAlong = M_PI * 2 * nextDay / (2 * (_referenceSummerSolstice - _referenceSummerSolstice));
// What angle are we placed by on the year's cycle, between _latitudeAngle + _tiltAngle and -latitudeAngle + _tiltAngle?
double closestDayHeight = cos(closestDayAngleAlong) * _latitudeAngle + _tiltAngle;
double previousDayHeight = cos(previousDayAngleAlong) * _latitudeAngle + _tiltAngle;
double nextDayHeight = cos(nextDayAngleAlong) * _latitudeAngle + _tiltAngle;
// Based on that, what are the daylight durations for the three twenty-four hour days?
double closestDayDuration = day / 2 + (day / 2) * sin(closestDayHeight);
double previousDayDuration = day / 2 + (day / 2) * sin(previousDayHeight);
double nextDayDuration = day / 2 + (day / 2) * sin(nextDayHeight);
// Here we use both morning and evening for the closest day, and the previous day's morning and the next day's evening.
double closestDayMorning = closestDay + (day / 2) - (closestDayDuration / 2);
double closestDayEvening = closestDay + (day / 2) + (closestDayDuration / 2);
double previousDayEvening = previousDay + (day / 2) + (previousDayDuration / 2);
double nextDayMorning = nextDay + (day / 2) + (nextDayDuration / 2);
// We calculate the proportion along the day that we are between evening and morning (or morning and evening), along with the sooner endpoint of that interval.
double proportion;
double referenceTime;
if (distanceAlong < closestDayMorning)
{
proportion = (distanceAlong - previousDayEvening) / (closestDayMorning - previousDayEvening);
referenceTime = previousDay + day * 3 / 4;
}
else if (distanceAlong > closestDayEvening)
{
proportion = (distanceAlong - closestDayEvening) / (nextDayMorning - closestDayEvening);
referenceTime = closestDay + day * 3 / 4;
}
else
{
proportion = (distanceAlong - closestDayMorning) / (closestDayEvening - closestDayMorning);
referenceTime = closestDay + day * 1 / 4;
}
// Lastly, we take both that endpoint and the proportion of it, and we get the number of seconds according to the daylight / nighttime calculation intended.
_naturalSeconds = referenceTime + proportion * day / 2;
I was hoping to make the code clearer and easier to grasp, and I think I did that, but it is displaying similar behavior to my previous attempt: the clock hands spin by at about ten times natural time when they should be within a factor of .8 to 1.2 of standard hours/minutes/seconds.
Any advice? Has my edited code been any clearer either about what is intended or what is wrong?
Thanks,
Your code is hard to follow, but I'll try to get you some tips:
There are existing libraries out there that compute solar angle/azimuth and sunrise/sunset for a given date. Use google as a help, here's some relevant resources: http://www.esrl.noaa.gov/gmd/grad/solcalc/ If you don't find any useful source code, I could post some.
Do not use double to calculate with dates and times. That's confusing and results in errors. Use a data type that is intended to store dates.
For your code, you say that the time is running to fast. Since referenceTime and day in the last line are constant (at least for half a day), the error must be in proportion. I think you're mixing to many cases there. The interpolation should go from the start of the range to the end, so in the case
proportion = (distanceAlong - previousDayEvening) / (closestDayMorning - previousDayEvening);
referenceTime = previousDay + day * 3 / 4;
proportion should run from (previousDay + day * 3 / 4) to (closestDay + day * 3 / 4), or, described differently, from the dusk to dawn of closestDay. But it's completely unclear how this interpolation should work.
Try to draw a diagram of the different cases (I believe there should only be two, one for day and one for night) and the corresponding interpolation.
But: What are you trying to achieve after all? The resulting time is just a forward running time, it is actually independent of latitude or longitude or time of day. So to make the time run, you don't need to know where the sun is.
Related
Modified Question.
My fitness app will calculate the number of calories burned based on a calculated value for each second. I have a timer that will allow the app to pick back up should it. I can't get the running sum to continue calculating when the app goes into the background. I tried to place the running sum inside of a DispatchQueue but not getting the sum as expected. Any guidance is greatly appreciated.
Here's the code I have placed inside the function that updates the timer.
//MARK: - Update Timer Label
func updateTimerLabel() {
interval = -Int(timerStartDate.timeIntervalSinceNow)
time = interval
let hours = interval / 3600
let minutes = interval / 60 % 60
let seconds = interval % 60
print("Current interval = \(interval)")
timerLabel.text = String(format:"%02i:%02i:%02i", hours, minutes, seconds)
DispatchQueue.global(qos: .background).async {
if self.activityArray[self.currentArrayRow].2 <= 4.5 {
self.cps = self.activityArray[self.currentArrayRow].2 * Double(self.user.userWeightInKilo) / 3600
self.runningCPS = self.runningCPS + self.cps
print("MET \(self.activityArray[self.currentArrayRow].2) <= 4.5 * KG (\(Double(self.user.userWeightInKilo))) * HR (\(Double(self.user.userHeartRate))) / MaxHR (\(Double(self.user.maxHeartRate)) * interval \(Double(self.interval)) / 3600. Gives a cps 0f \(self.cps) and a runningCPS of \(self.runningCPS) ")
} else {
self.cps = self.activityArray[self.currentArrayRow].2 * Double(self.user.userWeightInKilo) * Double(self.user.userHeartRate) / Double(self.user.maxHeartRate) / 3600
self.runningCPS = self.runningCPS + self.cps
print("MET \(self.activityArray[self.currentArrayRow].2) > 4.5 * KG (\(Double(self.user.userWeightInKilo))) * HR (\(Double(self.user.userHeartRate))) / MaxHR (\(Double(self.user.maxHeartRate)) * interval \(Double(self.interval)) / 3600. Gives a cps 0f \(self.cps) and a runningCPS of \(self.runningCPS) ")
}
}
activeLabel.text = String(format: "%0.1f", runningCPS) + " Calories Burned"
}
This is the follow up of a previous question of mine.
In a nutshell, I am trying to follow this tutorial step-by-step: https://jtauber.github.io/mars-clock/ to get to Coordinated Mars Time, but I got stuck right before the end. My code works fine up until the end (some values are more accurate than in the tutorial because I went back to the source from NASA: https://www.giss.nasa.gov/tools/mars24/help/algorithm.html ):
double millis = ( [[NSDate date] timeIntervalSince1970] * 1000 );
NSLog(#"millis: %f", millis);
double JDUT = ( 2440587.5 + (millis / 86400000) );
NSLog(#"JDUT: %f", JDUT);
double JDTT = ( JDUT + (37 +32.184) / 86400);
NSLog(#"JDTT: %f", JDTT);
double J2000Epoch = ( JDTT - 2451545.0 );
NSLog(#"J2000Epoch: %f", J2000Epoch);
double MSD = ( (( J2000Epoch - 4.5 ) / 1.0274912517) + 44796.0 - 0.0009626 );
NSLog(#"MSD: %f", MSD);
The only step remaining is actually calculating Coordinated Mars Time, using this equation:
MTC = mod24 { 24 h × MSD }
The problem is that I have no idea how. I tried to use modf( (double), (double *) ) but no idea how it actually works. I tried it the way below, but it gave me an incorrect answer (obviously as I have really no idea what I am doing). :(
double MSD24 = (24 * MSD);
double MCT = modf(24, &MSD24);
NSLog(#"MCT: %f", MCT); // Result: 0.000000
Any help would be much appreciated. Thank you very much!
p.s.: Notice that I use Objective-C; I do not understand swift unfortunately! :(
Carrying on from the code you gave, I tried:
CGFloat MTC = fmod(24 * MSD, 24);
and got
// 19.798515
which was right according to the web page you cited at the moment I tried it.
The sort of thing his page actually shows, e.g. "19:49:38" or whatever (at the time I tried it), is merely a string representation of that number, treating it as a number of hours and just dividing it up into minutes and seconds in the usual way. Which, I suppose, brings us to the second part of your question, i.e. how to convert a number of hours into an hours-minutes-seconds representation? But that is a simple matter, dealt with many times here. See NSNumber of seconds to Hours, minutes, seconds for example.
So, carrying on once again, I tried this:
CGFloat secs = MTC*3600;
NSDate* d = [NSDate dateWithTimeIntervalSince1970:secs];
NSDateFormatter* df = [NSDateFormatter new];
df.dateFormat = #"HH:mm:ss";
df.timeZone = [NSTimeZone timeZoneWithAbbreviation:#"GMT"];
NSString* result = [df stringFromDate:d];
NSLog(#"%#", result); // 20:10:20
...which is exactly the same as his web page was showing at that moment.
And here's a Swift version for those who would like to know what the "mean time" is on Mars right now:
let millis = Date().timeIntervalSince1970 * 1000
let JDUT = 2440587.5 + (millis / 86400000)
let JDTT = JDUT + (37 + 32.184) / 86400
let J2000Epoch = ( JDTT - 2451545 )
let MSD = (( J2000Epoch - 4.5 ) / 1.0274912517) + 44796.0 - 0.0009626
let MTC = (24 * MSD).truncatingRemainder(dividingBy: 24)
let d = Date(timeIntervalSince1970: MTC*3600)
let df = DateFormatter()
df.dateFormat = "HH:mm:ss"
df.timeZone = TimeZone(abbreviation: "GMT")!
df.string(from:d)
How to show a countdown time duration until the next alarm
Code:
TimeOfDay _nextSalah(List<SalahModel> salahs) {
DateTime now = DateTime.now();
List<TimeOfDay> times = [];
int currentSalah;
salahs.forEach((s) => times.add(s.time));
times.add(TimeOfDay(hour: now.hour, minute: now.minute));
times.sort((a, b) => a.hour.compareTo(b.hour));
currentSalah = times.indexWhere((time) => time.hour == now.hour);
return TimeOfDay(hour: times[currentSalah].hour, minute: times[currentSalah].minute);
}
But the time difference is wrong and it doesn't animate. Also how to make sure the time difference works when it's the same day and time of the next day i.e. now is Dec 1 2:30 PM and I want to get the difference on Dec 2 6:15 AM.
It does not work because TimeOfDay represents a time during the day, independent of the date that day might fall on or the time zone. The time is represented only by hour and minute.
If you want a countdown that spans multiple days a DateTime must be used and the time difference evaluation needs some math before formatting the result string, something like:
String nextTime(DateTime nextAlarmTime) {
List<int> ctime = [0, 0, 0, 0];
DateTime now = DateTime.now();
int diff = nextAlarmTime.difference(now).inSeconds;
ctime[0] = diff ~/ (24 * 60 * 60); // days
diff -= ctime[0] * 24 * 60 * 60;
ctime[1] = diff ~/ (60 * 60); // hours
diff -= ctime[1] * 60 * 60;
ctime[2] = diff ~/ 60; // minutes
ctime[3] = diff - ctime[2] * 60; // seconds
return ctime.map((val) => val.toString().padLeft(2, '0')).join(':');
}
I work with decimal times in Lua and make arithmetical operations on them.
For example 124500+5=124505 (12:45:05)
What formula can avoid 60 digits problem?
124459+5=124504 (not 124464)
How can I resolve it?
You are mixing formation with calculation. The best way is to transform your time "string" in a real number:
12:45:05 -> 12 * 60 * 60 + 45 * 60 + 05 = 45905
The function could look like this:
function time_to_number(t)
return (math.floor(t / 10000) * 60 * 60) + ((math.floor(t / 100) % 100) * 60) + (t % 100)
-- you can also use % 10000 if the hours are limited to two digits
end
Now you can calculate on the seconds.
To format the value back you can use this function
function time_split(t)
local hour = math.floor(t / 3600)
local min = math.floor((t % 3600) / 60)
local sec = (t % 3600) % 60
return hour, min, sec
end
I have used many brackets for readability, which are not all required.
This question already has answers here:
How to convert milliseconds into human readable form?
(22 answers)
Closed 6 years ago.
I need to convert an milliseconds into Days, Hours, Minutes Second.
ex: 5 Days, 4 hours, 13 minutes, 1 second.
Thanks
if you don't want to do the calculation by yourself, you could go for such a solution.
I, however, know that is a kinda costly solution, so you need to be aware of potential performance issues in runtime – depending on how frequently you intend to invoke this.
NSTimeInterval _timeInSeconds = 123456789.123; // or any other interval...;
NSCalendar *_calendar = [NSCalendar calendarWithIdentifier:NSCalendarIdentifierGregorian];
NSCalendarUnit _units = NSCalendarUnitDay | NSCalendarUnitHour | NSCalendarUnitMinute | NSCalendarUnitSecond;
NSDateComponents *_components = [_calendar components:_units fromDate:[NSDate date] toDate:[NSDate dateWithTimeIntervalSinceNow:_timeInSeconds] options:kNilOptions];
NSLog(#"%ld Days, %ld Hours, %ld Minutes, %ld Seconds", _components.day, _components.hour, _components.minute, _components.second);
You can write your own function like this:
import UIKit
let miliseconds: Int = 24 * 3600 * 1000 + 3700 * 1000
// 1 day and 1 hour 1 minute 40 seconds
func convertTime(miliseconds: Int) -> String {
var seconds: Int = 0
var minutes: Int = 0
var hours: Int = 0
var days: Int = 0
var secondsTemp: Int = 0
var minutesTemp: Int = 0
var hoursTemp: Int = 0
if miliseconds < 1000 {
return ""
} else if miliseconds < 1000 * 60 {
seconds = miliseconds / 1000
return "\(seconds) seconds"
} else if miliseconds < 1000 * 60 * 60 {
secondsTemp = miliseconds / 1000
minutes = secondsTemp / 60
seconds = (miliseconds - minutes * 60 * 1000) / 1000
return "\(minutes) minutes, \(seconds) seconds"
} else if miliseconds < 1000 * 60 * 60 * 24 {
minutesTemp = miliseconds / 1000 / 60
hours = minutesTemp / 60
minutes = (miliseconds - hours * 60 * 60 * 1000) / 1000 / 60
seconds = (miliseconds - hours * 60 * 60 * 1000 - minutes * 60 * 1000) / 1000
return "\(hours) hours, \(minutes) minutes, \(seconds) seconds"
} else {
hoursTemp = miliseconds / 1000 / 60 / 60
days = hoursTemp / 24
hours = (miliseconds - days * 24 * 60 * 60 * 1000) / 1000 / 60 / 60
minutes = (miliseconds - days * 24 * 60 * 60 * 1000 - hours * 60 * 60 * 1000) / 1000 / 60
seconds = (miliseconds - days * 24 * 60 * 60 * 1000 - hours * 60 * 60 * 1000 - minutes * 60 * 1000) / 1000
return "\(days) days, \(hours) hours, \(minutes) minutes, \(seconds) seconds"
}
}
convertTime(miliseconds)
//result is "1 days, 1 hours, 1 minutes, 40 seconds"
try this
NSTimeInterval time = <timein ms>;
NSInteger days = time / (24 * 60 * 60);
NSInteger hours = (time / (60 * 60)) - (24 * days);
NSInteger minutes =(time / 60) - (24 * 60 * days) - (hours * 60);
NSInteger seconds = (lroundf(time) % 60);
I have write the easy code to do this in both Objective-c and Swift:
Swift
var milliseconds : double_t = 568569600;
milliseconds = floor(milliseconds/1000);
let seconds : double_t = fmod(milliseconds, 60);
let minutes : double_t = fmod((milliseconds / 60) , 60);
let hours : double_t = fmod((milliseconds / (60*60)), 60);
let days : double_t = fmod(milliseconds / ((60*60)*24), 24);
NSLog("seconds : %.f minutes : %.f hours : %.f days : %.f", seconds, minutes, hours, days);
Output - seconds : 9 minutes : 56 hours : 38 days : 7
Objective
double milliseconds = 568569600;
milliseconds = milliseconds/1000;
float seconds = fmod(milliseconds, 60);
float minutes = fmod((milliseconds / 60) , 60);
float hours = fmod((milliseconds / (60*60)), 60);
float days = fmod(milliseconds / ((60*60)*24), 24);
NSLog(#"seconds : %.f minutes : %.f hours : %.f days : %.f ", seconds, minutes, hours, days);
Output - seconds : 10 minutes : 56 hours : 38 days : 7