我试图估算从日出到日落的白天长度,以及从日落到日出的夜晚长度。我现在的近似是粗略的(它假设昨天和明天的值与今天相同),但现在我并不特别关注精确定位昨天的日落、今天的日出、今天的日落和明天的日出(还)。我的目标是基于每晚 12 个相等小时(12 个相等,不等于标准小时或白天小时)和每天 12 个相等小时进行计算。

我担心的是,在我的 iOS 应用程序中,计算是遥不可及的;一分钟在 5-6(标准)秒的时间内飞逝。当我使用未修改的时间时,在此处的其他代码中,时钟以标准速度移动,但是当我尝试使用此代码来提供时钟代码时,有些东西越界了。

作为近似值,我一直在处理的代码是:

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;
    }
}

您是否可以在此代码中查明任何问题(鉴于此近似值可能可以改进到任何程度)?

谢谢,

- 编辑 -

我在上面编写的代码在呈现给阅读它的人的松散末端方面要求很高。我尝试再次通过,并用更简单的术语和更纯粹的数学模型重写它。我写了,评论补充:
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 &pi; 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;

我希望使代码更清晰、更容易掌握,我想我做到了,但它显示出与我之前的尝试类似的行为:时钟指针以自然时间的大约十倍旋转,而当它们应该在一个因子内时.8 到 1.2 的标准小时/分钟/秒。

有什么建议吗?我编辑的代码是否更清楚地说明了意图或错误?

谢谢,

最佳答案

您的代码很难遵循,但我会尝试为您提供一些提示:

  • 现有的库可以计算给定日期的太阳角度/方位角和日出/日落。使用谷歌作为帮助,这里有一些相关资源: http://www.esrl.noaa.gov/gmd/grad/solcalc/ 如果你没有找到任何有用的源代码,我可以发布一些。
  • 不要使用 double 来计算日期和时间。这令人困惑并导致错误。使用旨在存储日期的数据类型。
  • 对于你的代码,你说时间过得很快。由于最后一行中的referenceTime 和day 是恒定的(至少半天),因此误差必须成比例。我认为你在那里混入了许多情况。插值应该从范围的开始到结束,所以在这种情况下

    比例 = (distanceAlong - previousDayEvening)/(closestDayMorning - previousDayEvening);
    引用时间 = 前一天 + 天 * 3/4;

  • 比例应该从 (previousDay + day * 3/4) 到 (closestDay + day * 3/4),或者用不同的方式描述,从closestDay 的黄昏到黎明。但是完全不清楚这种插值应该如何工作。

    试着画出不同情况的图(我相信应该只有两种,一种是白天,一种是晚上)和相应的插值。

    但是:你到底想达到什么目的?结果时间只是一个向前运行的时间,它实际上与纬度或经度或一天中的时间无关。所以为了让时间跑起来,你不需要知道太阳在哪里。

    关于ios - 这些日光/夜间长度近似值的错误在哪里?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/19437569/

    10-11 19:25