c++ - 在等角投影上绘制卫星覆盖区域

Question

我需要在等角投影上绘制卫星观测区的边界。我找到了这个公式 (1) 和图:

sin(fi) = cos(alpha) * sin(fiSat) – sin(alpha) * sin (Beta) * cos (fiSat);

sin(lambda) = (cos(alpha) * cos(fiSat) * sin(lambdaSat)) / cos(asin(sin(fi))) +
              (sin(alpha) * sin(Beta) * sin(fiSat) * sin(lambdaSat)) / cos(asin(sin(fi))) -
              (sin(alpha) * cos(Beta) * cos(lambdaSat))/cos(asin(sin(fi)));

cos(lambda) = (cos(alpha) * cos(fiSat) * cos(lambdaSat)) / cos(asin(sin(fi))) +
              (sin(alpha) * sin(Beta) * sin(fiSat) * cos(lambdaSat)) / cos(asin(sin(fi))) -
              (sin(alpha) * cos(Beta) * sin(lambdaSat)) / cos(asin(sin(fi)));

地球各个平面的横截面:

与方程组(2)同图:

if sin(lambda) > 0, cos(lambda) > 0 then lambda = asin(sin(lambda));
if sin(lambda) > 0, cos(lambda) < 0 then lambda = 180 - asin(sin(lambda));
if sin(lambda) < 0, cos(lambda) < 0 then lambda = 180 - asin(sin(lambda));
if sin(lambda) < 0, cos(lambda) > 0 then lambda = asin(sin(lambda));

地球经度的引用角方案:

Where:  alpha – polar angle;
    fiSat, lambdaSat – latitude, longitude of satellite;
    Beta – angle which change from 0 to 2*Pi and help to draw the observation zone;
    fi, lambda – latitude, longitude of point B on the border of observation zone;

我从 0 到 2*Pi 循环重复 (1) 和 (2) 两个公式来绘制观察区域的边界。但是在(2)方程组中我不太确定。

在 [-180;-90]、[-90;90]、[90;180] 区间内，区域绘制正确。

中心在 -35;45:

中心在 120;60:

中心在 -120;-25

但是在 -90 度和 90 度的边界上它变得凌乱:

中心在 -95;-50

中心在 95;30

你能帮我算一下公式(1)和(2)或者再写一个吗？

double deltaB = 1.0*M_PI/180;
observerZone.clear();
for (double Beta = 0.0; Beta <= (M_PI * 2) ; Beta += deltaB){
    double sinFi = cos(alpha) * sin(fiSat) - sin(alpha) * sin(Beta) * cos(fiSat);
    double sinLambda = (cos(alpha) * cos(fiSat) * sin(lambdaSat))/cos(asin(sinFi)) +
                             (sin(alpha) * sin(Beta) * sin(fiSat) * sin(lambdaSat))/cos(asin(sinFi)) -
                            (sin(alpha) * cos(Beta) * cos(lambdaSat))/cos(asin(sinFi));
    double cosLambda = (cos(alpha) * cos(fiSat) * cos(lambdaSat))/cos(asin(sinFi)) +
                            (sin(alpha) * sin(Beta) * sin(fiSat) * cos(lambdaSat))/cos(asin(sinFi)) -
                            (sin(alpha) * cos(Beta) * sin(lambdaSat))/cos(asin(sinFi));
    if (sinLambda > 0) {
        if (cosLambda > 0 ){
            sinLambda = asin(sinLambda);
            sinFi = asin(sinFi);
        }
        else {
            sinLambda = M_PI - asin(sinLambda);
            sinFi = asin(sinFi);
        }
    }
    else if (cosLambda > 0) {
        sinLambda = asin(sinLambda);
        sinFi = asin(sinFi);
    }
    else {
        sinLambda = -M_PI - asin(sinLambda);
        sinFi = asin(sinFi);
    }
    Point point;
    point.latitude = qRadiansToDegrees(sinFi);
    point.longitude = qRadiansToDegrees(sinLambda);
    observerZone.push_back(point);
}

Answer 1

我解决了我的问题。 (1)式中计算cosLambda时应为+而不是-。

double cosLambda = (cos(alpha) * cos(fiSat) * cos(lambdaSat))/cos(asin(sinFi)) +
                            (sin(alpha) * sin(Beta) * sin(fiSat) * cos(lambdaSat))/cos(asin(sinFi)) +
                            (sin(alpha) * cos(Beta) * sin(lambdaSat))/cos(asin(sinFi));

抱歉打扰了。