# Influence of Vegetation Filter Strip on Slope Runoff, Sediment Yield and Nutrient Loss
## 4 Discussion
### Relationships between Erosion Indexes and Influencing Factors
Vegetation canopy and litter form a continuous cover to reduce soil erosion [48]. In addition, the effect of different vegetation settings on erosion was also inconsistent. There was no significant difference in the runoff and sediment yield of two vegetation-covered slopes (\(p>0.05\)). Previous studies have shown that more than 70% of soil loss variability is attributable to rainfall intensity and slope [49]. Rainfall intensity and slope jointly affected the effect of vegetation to prevent runoff and sediment production, and they played different roles in terms of nutrient loss. According to regression equation coefficients, VFS2 was hardly affected by the slope factor in reducing the runoff TP loss. When the slope was covered by vegetation, rainfall intensity and slope had no obvious effect on the TP content of sediments. The study indicated that root system may be the reason for high content of soil organic carbon and TN in the grassland. The presence of vegetation prolongs the initial runoff time, thereby increasing the soil infiltration rate, and the TN and TP in runoff and sediment will infiltrate. Moreover, it is strongly affected by slope factor when vegetation reduced sediment TN loss, and there was a negative correlation between slope and sediment TN loss. As the slope gradient becomes steeper, higher velocity and volumes of runoff takes away more soil particles [50]. [PERSON] [51] found that the loss of TN and TP was closely related to soil loss. 57% of TN and 75% of TP were lost with sediment, reflecting that sediment was the main carrier of slope nutrients. Consequently in this research, the reduction of nutrient loss by vegetation on slopes was mainly limited by the slope. \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline
**Erosion Index** & \begin{tabular}{c} **Vegetation** \\ **Coverage Type** \\ \end{tabular}  & **Mean** & \begin{tabular}{c} **Regression Equation** \\ **Coefficient**\(R^{2}\) \\ \end{tabular}  & \begin{tabular}{c} **Determination** \\ **\(p\) Value** \\ \end{tabular}  & **N** \\ \hline \multirow{4}{*}{Sediment (g)} & L & 171.232  & \begin{tabular}{c} \(Y=-416.856+7.100\times R+8.3323\times S\) \\ 87.252  & \begin{tabular}{c} \(Y=-193.807\times 3.399\times R+3.938\times S\) \\ 84.736  \\ \end{tabular}  & 0.682 & 0.000 & 18 \\  & Z & 87.252  & \begin{tabular}{c} \(Y=-193.807\times 3.399\times R+3.938\times S\) \\ 84.736  & \(Y=-180.704\times 3.255\times R+3.661\times S\) \\ \hline \multirow{4}{*}{Runoff (L)} & L & 23.882  & \
\begin{tabular}{c} \(Y=-11.675\times 0.504\times R+0.306\times S\) \\ 84.737  & 0.893 & 0.000 & 18 \\  & Z & 16.425  &
\begin{tabular}{c} \(Y=-11.157\times 0.347\times R+0.377\times S\) \\ 85.85  & 0.858 & 0.000 & 18 \\  & Z & 16.460  & \(Y=-0.040\times 0.010\times R+0.006\times S\) & 0.865 & 0.000 & 18 \\  & Z & 0.469  & \(Y=-0.085+0.007\times R+0.009\times S\) & 0.878 & 0.000 & 18 \\  & Z’ & 0.425  & \(Y=-0.067+0.008\times R+0.001\times S\) & 0.942 & 0.000 & 18 \\ \hline \multirow{4}{*}{TP in sediment (g/kg)} & L & 0.701  & \(Y=0.583+0.001\times R+0.005\times S\) & 0.626 & 0.001 & 18 \\  & Z & 0.603  & \(Y=0.599-0.001\times R+0.004\times S\) & 0.284 & 0.081 & 18 \\  & Z’ & 0.583  & \(Y=0.572+0.000\times R+0.002\times S\) & 0.212 & 0.168 & 18 \\ \hline \multirow{4}{*}{TN in runoff (mg/L)} & L & 16.754  & \(Y=18.224-0.017\times R+0.003\times S\) & 0.216 & 0.161 & 18 \\  & Z & 14.112  & \(Y=17.334-0.042\times R-0.019\times S\) & 0.910 & 0.000 & 18 \\  & Z