#根據緯度，水壓計算水深

public static double getData(double latitude, double pWater){ double x = Math.Pow(Math.Sin(latitude / 57.29578), 2);//Console.WriteLine(x);double g = 9.780318 * (1.0 + (5.2788 * Math.Pow(10, -3) + 2.36 * Math.Pow(10, -5) * x) * x) + 1.092 * Math.Pow(10, -6) * pWater;double depth = ((((-1.82 * Math.Pow(10, -15) * pWater + 2.279 * Math.Pow(10, -10)) * pWater - 2.2512 * Math.Pow(10, -5)) * pWater + 9.72659) * pWater) / g;return depth;}

#根據溫度和電導率計算出鹽度
其中傳入的溫度是90年溫標下得出的溫度，而我需要用68年的溫標。不需要轉換請自行去掉第二行代碼

public static double getSalinity(double temperature, double conductivity){double R = conductivity / 42.914;//90年溫標轉為68年溫標double tempt68 = temperature * 1.00024;double rt = 0.6766097 + 0.0200564 * tempt68 + 0.0001104259 * tempt68 * tempt68 - 0.00000069698 * tempt68 * tempt68 * tempt68 + 0.0000000010031 * tempt68 * tempt68 * tempt68 * tempt68;double RT = R / rt;double RT2 = RT * RT;double SQRTRT = Math.Sqrt(RT);double sal12 = 0.008 - 0.1692 * SQRTRT + 25.3851 * RT + 14.0941 * SQRTRT * RT - 7.0261 * RT2 + 2.7081 * RT2 * SQRTRT;double a = (tempt68 - 15) / (1 + 0.0162 * (tempt68 - 15));double sal2 = (0.0005 - 0.0056 * SQRTRT - 0.0066 * RT - 0.0375 * SQRTRT * RT + 0.0636 * RT2 - 0.0144 * RT2 * SQRTRT) * a;double sail68 = sal12 + sal2;return sail68;}

總結

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