Question

对于一个新项目，我喜欢计算月相。到目前为止，我还没有看到任何代码可以做到这一点。我不想依赖在线服务。

我尝试了一些功能，但它们并非100％可靠。我尝试过的功能：

NSInteger r = iYear % 100;

    r %= 19;
    if (r>9){ r -= 19;}
    r = ((r * 11) % 30) + iMonth + iDay;
    if (iMonth<3){r += 2;}
    r -= ((iYear<2000) ? 4 : 8.3);
    r = floor(r+0.5);

另一个：

float n = floor(12.37 * (iYear -1900 + ((1.0 * iMonth - 0.5)/12.0)));
    float RAD = 3.14159265/180.0;
    float t = n / 1236.85;
    float t2 = t * t;
    float as = 359.2242 + 29.105356 * n;
    float am = 306.0253 + 385.816918 * n + 0.010730 * t2;
    float xtra = 0.75933 + 1.53058868 * n + ((1.178e-4) - (1.55e-7) * t) * t2;
    xtra = xtra + (0.1734 - 3.93e-4 * t) * sin(RAD * as) - 0.4068 * sin(RAD * am);
    float i = (xtra > 0.0 ? floor(xtra) :  ceil(xtra - 1.0));
    float j1 = [self julday:iYear iMonth:iMonth iDay:iDay];
    float jd = (2415020 + 28 * n) + i;
    jd = fmodf((j1-jd + 30), 30);

和最后一个

        NSInteger thisJD = [self julday:iYear iMonth:iMonth iDay:iDay];
    float degToRad = 3.14159265 / 180;
    float K0, T, T2, T3, J0, F0, M0, M1, B1, oldJ = 0.0;
    K0 = floor((iYear-1900)*12.3685);
    T = (iYear-1899.5) / 100;
    T2 = T*T; T3 = T*T*T;
    J0 = 2415020 + 29*K0;
    F0 = 0.0001178*T2 - 0.000000155*T3 + (0.75933 + 0.53058868*K0) - (0.000837*T + 0.000335*T2);
    M0 = 360*[self getFrac:((K0*0.08084821133)) + 359.2242 - 0.0000333*T2 - 0.00000347*T3];
    M1 = 360*[self getFrac:((K0*0.07171366128)) + 306.0253 + 0.0107306*T2 + 0.00001236*T3];
    B1 = 360*[self getFrac:((K0*0.08519585128)) + 21.2964 - (0.0016528*T2) - (0.00000239*T3)];
    NSInteger phase = 0;
    NSInteger jday = 0;
    while (jday < thisJD) {
        float F = F0 + 1.530588*phase;
        float M5 = (M0 + phase*29.10535608)*degToRad;
        float M6 = (M1 + phase*385.81691806)*degToRad;
        float B6 = (B1 + phase*390.67050646)*degToRad;
        F -= 0.4068*sin(M6) + (0.1734 - 0.000393*T)*sin(M5);
        F += 0.0161*sin(2*M6) + 0.0104*sin(2*B6);
        F -= 0.0074*sin(M5 - M6) - 0.0051*sin(M5 + M6);
        F += 0.0021*sin(2*M5) + 0.0010*sin(2*B6-M6);
        F += 0.5 / 1440; 
        oldJ=jday;
        jday = J0 + 28*phase + floor(F);
        phase++;
    }
    float jd = fmodf((thisJD-oldJ), 30);

所有人的工作越来越少，但没有人真正给出2017年和2018年的正确满月日期。

是否有人能够正确计算月相 - 同时也基于时区？

编辑：我只想要Moonphases的功能。 SwiftAA提供了更多功能，只在应用程序中产生了不需要的开销。

准确计算月相

0 个答案: