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我正在尝试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。我目前的近似值是粗略的（它假设昨天和明天具有与今天相同的值），但现在我并不特别关心确定昨天的日落、今天的日出、今天的日落和明天的日出（还）。我的目标是根据每晚 12 小时（彼此相等的 12 小时，不等于标准小时或白天的小时）和每天 12 小时相等的时间进行计算。

我担心的是，在我的 iOS 应用程序中，计算很差；一分钟在 5-6（标准）秒的时间内飞逝。当我在此处的其他代码中使用未修改的时间时，时钟以标准速度移动，但是当我尝试让此代码提供时钟代码时，有些东西超出了范围。

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;
    }
}

是否有任何问题（考虑到这个近似值可能会在任何程度上改进）你可以在这段代码中查明吗？

谢谢，

- 编辑 -

我上面写的代码在呈现给阅读它的人的松散末端方面要求相当高。我试图再通过一次，用更简单的术语和更纯粹的数学模型重写它。我写了，评论补充说：

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相似的行为：时钟指针以大约 10 倍的自然时间旋转，而它们应该在0.8 到 1.2 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预期的内容或错误之处？

谢谢，

4

1 回答 1

1

您的代码很难遵循，但我会尽力为您提供一些提示：

  • 现有的库可以计算给定日期的太阳角/方位角和日出/日落。使用 google 作为帮助，这里有一些相关资源：http ://www.esrl.noaa.gov/gmd/grad/solcalc/如果您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。

  • 对于您的代码，您说时间过得很快。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。我认为你在那里混合了很多情况。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); 参考时间 = previousDay + day * 3 / 4;

比例应该从 (previousDay + day * 3 / 4) 到 (closestDay + day * 3 / 4)，或者，换一种说法，从最接近一天的黄昏到黎明。但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）和对应的插值。

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。所以为了让时间流逝，你不需要知道太阳在哪里。

于 2013-10-27T18:09:30.790 回答