the Meaning of ... [CLOSED [closed] - ios

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Closed 10 years ago.
This is stupid, but i have to know how ( % ) sign means, because i want to add days.
this is an example.
int seconds = 78120;
int forHours = (seconds1 / 3600),
remainder = (seconds1 % 3600),
forMinutes = remainder / 60,
forSeconds = remainder % 60;
NSString *Time = [NSString stringWithFormat:#"%02i:%02i:%02i",forHours,forMinutes,forSeconds];
Label.text = Time;
Result:
21:42:00
i want the result to be like ( 0 days, 21:42:00 ) like ( DD, HH:mm:ss )

It's called the modulo operation. It's whats left when you divide a number (and only take into account whole numbers).
Examples:
3 % 2 = 1
6 % 2 = 0
6 % 3 = 0
6 % 4 = 2

% (modulo) gives the remainder after division.
So you can add the separation for days at the start, then use modulo to get the number of seconds after removing those accounted for in days:
int seconds = 78120;
int days = seconds / 86400;
// Equivalent to: seconds = seconds - days * 86400 /*# seconds in a day*/;
seconds = seconds % 86400; // seconds remaining less than a day
int forHours = (seconds1 / 3600),
remainder = (seconds1 % 3600), // seconds remaining within an hour
forMinutes = remainder / 60,
forSeconds = remainder % 60; // seconds remaining less than a minute

The modulus (%) operator returns the remainder of the integer division.
a = 13 % 5;
Here, a will equal 3.
Try:
int fordays = seconds1 / 86400,
remainder = seconds1 % 86400,
forHours = remainder / 3600,
remainder = remainder % 3600,
forMinutes = remainder / 60,
forSeconds = remainder % 60;
1 day = 86400 seconds.

Related

math operations with hours,minutes and seconds

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.

Ruby Program time conversion

The task is to Write a method that will take in a number of minutes, and returns a string that formats the number into hours:minutes.
here's what I have so far:
def time_conversion(minutes)
minutes = (minutes / 60) % 60
hours = minutes / (60 * 60)
format(" %02d:%02d ", hours, minutes)
return format
end
it's not working out for me
Try this
def time_conversion(time)
minutes = time % 60
hours = time / 60
minutes = (minutes < 10)? "0" + minutes.to_s : minutes.to_s
return hours.to_s + ":" + minutes
end
Using division in Ruby returns a whole number, lowered to the previous number. Using modulus returns the remainder after division.
Ruby's Numeric#divmod is exactly what you want here. It returns both the quotient and remainder of a division operation, so e.g. 66.divmod(60) returns [ 1, 6 ]. Combined with sprintf (or String#%, it makes for an extremely simple solution:
def time_conversion(minutes)
"%02d:%02d" % minutes.divmod(60)
end
puts time_conversion(192)
# => 03:12
Well try
h = minutes/60
M = minutes%60

How to customize my timer display to show only minutes and seconds?

My Timer is displaying Minutes and Hours, but once it gets to 60 minutes it restarts from 0 Minute.
Should I get rid of the modulo ( % 60 ) for minutes.
I would like my timer to display for ex: 80:45 ( basically not stopping at 60 min once it reaches 1 hour)
var min = 0
var sec = 0
func stringFromTimeInterval(interval: NSTimeInterval) -> String {
let interval = Int(interval)
let seconds = interval % 60
let minutes = (interval / 60) % 60
//let hours = (interval / 3600)// I don't need the hours
return String(format: "%02d:%02d",minutes, seconds)
}
% 60 means that it will spit out a minutes value that is the remainder when divided by 60(minutes). This is most probably because for time in the form hh:mm, you want it to go from 5:59 to 6:00, not 5:60. So changing the following line will give you what you seek.
let minutes = (interval / 60) % 60 -> let minutes = interval / 60

Converting MPH to minute miles

I'm attempting to convert MPH into minute miles. I'm currently running code to do this by doing 60 / the miles per hour which gives me the result in minute miles.
For example 60/8mph = 7.5
However the answer I get I need to convert into minutes and seconds so that I would have 7 minutes 30 seconds. Is there a way I can get the numbers after the decimal point so I can multiply it by 60 to convert it to seconds, then add it back to the minutes.
You can use remainder,
double remainder = fmod(a_double, another_double);
should include <math.h>
Well, I don't know whether there is an existing class that handles this, but to answer your specific question, the fractional part of the decimal (mantissa?) would be:
((60 % 8) / 8.0f)
You can multiply that by 60.
Do it in seconds...
3600/8 = 450
450/60 = 7 remainder 30
= 7:30
It's pretty simple, you're on the right path actually.
What you need to do is:
Get Minutes
Get Seconds
Convert seconds from int to real time (0.5 to 30, etc..)
Add seconds to minutes
Get minutes by casting it to an Integer:
int minutes = 60/8;
Get seconds by using the remainder:
float seconds = 60%8;
Convert seconds to real time:
int realSeconds = seconds * 60;
Now get result back by adding both:
int totalSeconds = minuts + realSeconds;
Here's a little function that does it (typed directly to browser, probably won't compile)
#include <math.h>
int getMinuteMiles(float mph){
int minutes = 60/mph;
double seconds = fmod(60, mph);
int realSeconds = seconds * 60;
return minutes+realSeconds;
}

Where is the bug in these length-of-daylight/night approximations?

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.

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