Determination of the best model for predicting soil available phosphorus based on soil organic carbon

Sanaa Hama Gharib Faraj, Kamal Hama Karim, Razan Omar Ali
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Abstract

There are several methods for determining the determination of soil available phosphorous. These methods require many chemicals and different devices, and determining soil available phosphorous in the laboratory is difficult, time-consuming, and costly. Consequently, pedotransfer functions more suitable and economical were used to predict the soil available phosphorous(AP) from soil organic carbon (OC). To predict soil available phosphorous (AP) from soil organic carbon (OC), five models were utilized, including exponential, linear, logarithmic, polynomial, and power models. Soil AP was predicted as a function of soil OC. The soil AP determined by laboratory tests was compared with the predicted soil AP based on the AP-soil OC model using the Bland-Altman method. The 95% limits of agreement for comparison of soil AP were determined with laboratory tests, and the soil AP pedotransfer function was computed at -1.707 and 1.542 mg kg -1. The mean soil AP difference between the two methods was - 0.083 mg kg -1. The polynomial model (AP = 38.173 - 4.462 x OC+ 0.146 x OC2) is given the best fit to predict soil available phosphorus, due to its high R2 (0.868) and low RMSE (0.846).
根据土壤有机碳确定预测土壤可利用磷的最佳模型
测定土壤可利用磷的方法有多种。这些方法需要许多化学试剂和不同的设备,在实验室测定土壤可利用磷既困难又耗时费钱。因此,我们采用了更合适、更经济的土壤转移函数来预测土壤有机碳(OC)中的土壤可利用磷(AP)。为了预测土壤有机碳(OC)中的土壤可利用磷(AP),我们使用了五种模型,包括指数模型、线性模型、对数模型、多项式模型和幂模型。土壤可利用磷是土壤有机碳的函数。使用布兰德-阿尔特曼法,将实验室测试确定的土壤 AP 与根据 AP-土壤 OC 模型预测的土壤 AP 进行比较。通过实验室测试确定了土壤 AP 比较的 95% 一致限,并计算出土壤 AP pedotransfer 函数为-1.707 和 1.542 mg kg-1。两种方法的平均土壤 AP 差值为 - 0.083 毫克/千克-1。多项式模型(AP = 38.173 - 4.462 x OC+ 0.146 x OC2)具有较高的 R2(0.868)和较低的 RMSE(0.846),因此是预测土壤可利用磷的最佳拟合模型。
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