日照角度和时间纬度有什么关系?

2025-03-15 21:31:48
推荐回答(1个)
回答1:

正午太阳高度角=90°-该地与太阳直射点纬度差

时间问题,地球是球体立体几何要学好
H表示正午太阳高度角。

对于北半球而言,H=90°-(φ-δ);

对于南半球而方,H=90°-(δ-φ)。

假设春分日(秋分日也可,太阳直射点在赤道)

某时刻太阳直射(0°,120°e)这一点,120°e经线上各点都是正午

这点离太阳直射点的纬度距离是0度
此时,(0°,120°e)的太阳高度角就是90°

另外一个观测点,(1°n,120°e)与太阳直射点的纬度差为1度

此时,这一点的太阳高度角为89°(涉及立体几何计算,我就不详细推导了)

聪明的你肯定知道,(1°s,120°e)与太阳直射点的纬度差也是1度

因此,当地的太阳高度角也是89°

同一时刻,下列各观测点,报告的太阳高度角度数如下:

南北纬2度(与太阳直射点相距2纬度):88°(=90°-2°)

南北纬3度(与太阳直射点相距3纬度):87°(=90°-3°)

南北纬10度(与太阳直射点相距10纬度):80°(=90°-10°)

南北纬30度(与太阳直射点相距30纬度):60°(=90°-30°)

南北纬80度(与太阳直射点相距80纬度):10°(=90°-80°)

南北纬90度(与太阳直射点相距90纬度):0°(=90°-90°)

但是,这个“纬度差”的计算可是有讲究的:

设太阳直射点纬度为θ°,观测点纬度δ°

如果θ与δ在同一半球,则“纬度差”为|θ-δ|(θ减δ差的绝对值)

如果θ与δ在异半球,则“纬度差”为θ+δ

说起来好像很麻烦,其实只要脑袋里有个地球的模型就简单了

比如太阳直射点是北纬10°,观测点是北纬30°,纬度差当然是20°

如果太阳直射点是南纬10°,观测点是北纬30°,纬度差当然是40°

事实上,计算“正午太阳高度角”,根本就不要考虑“正午”这个因素

只要用90°减去观测点与太阳直射点的纬度差,得出的就是正午太阳高度角。

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