## 3 The Proposed Method
### Extracting the Parameters of Basic Footprints
#### 3.1.2 Describing the Basic Footprints' Outer Frames Using Bezier Curves  
As Bezier curves can be used to link discrete points by a smooth line, this approach can be used to describe the outer frame of each basic footprint. If the Bezier curve lies on an XY-plane, the parametric equation for the Bezier curves may be formulated as  
\[B(t)|_{t=[0,1]}=\sum_{i=0}^{n}\binom{n}{k}t^{i}(1-t)^{n-i}\begin{bmatrix}x_{i}\\ y_{i}\end{bmatrix} \tag{2}\]  
where \(\begin{bmatrix}x_{i}&y_{i}\end{bmatrix}^{T}\) are the control points.  
The outer frame of each basic footprint was divided into four Bezier curves. The four sampling points (see Figure 7) are the tip of the footprint (\(P_{t}=(0,0)\)), front-most end of the footprint (\(P_{f}=(x_{max},\)\(y_{b})\)), highest point of the footprint (\(P_{max}=(x_{b},\)\(y_{max})\)), and lowest point of the footprint (\(P_{min}=(x_{c},\)\(y_{min})\)). To allow a smooth curve to pass through these four points, \(P_{t}\) to \(P_{max}\) and \(P_{t}\) to \(P_{min}\) were linked by two quadratic Bezier curves, whereas \(P_{max}\) to \(P_{t}\) and \(P_{min}\) to \(P_{f}\) were linked by two cubic Bezier curves. In addition, a few more control points (\(P_{1}\)-\(P_{6}\)) were added to smoothen the Bezier curves, as shown in Figure 7. The control points for the outer frame of a basic footprint are then \(P_{t}\), \(P_{1}\), \(P_{max}\), \(P_{max}\), \(P_{2}\), \(P_{3}\), \(P_{f}\), \(P_{f}\), \(P_{4}\), \(P_{5}\), \(P_{min}\), and \(P_{min}\), \(P_{6}\), \(P_{t}\). The computation of the Bezier curves is described below, and we have used the \(P_{t}\)-to-\(P_{max}\) and \(P_{max}\)-to-\(P_{f}\) curves shown in Figure 8 as examples to illustrate this process.  
When \(n=2\) in Equation (2), the quadratic Bezier curve for \(P_{t}\)-to-\(P_{max}\) may be expanded as follows  
\[B(t)|_{t=[0,1]} = (1-t)^{2}P_{t}+2t(1-t)P_{1}+t^{2}P_{max}\] \[\frac{dB}{dt} = (2t-2)P_{t}+(2-4t)P_{1}+2 tP_{max} \tag{3}\]  
Figure 8: Control points of the Bézier curve.  
Figure 7: Additional control points.  
When \(n=3\) in Equation (2), the quadratic Bezier curve for \(P_{max}\)-to-\(P_{f}\) may be expanded as follows  
\[B(t)|_{t=[0,1]} = (1-t)^{3}P_{max}+3t(1-t)^{2}P_{2}\] \[+3t^{2}(1-t)P_{3}+t^{3}P_{f}\] \[\frac{dB}{dt} = (-3+6t-3t^{2})P_{max}+(3-12t+9t^{2})P_{2}\] \[+(6t-9t^{2})P_{3}+3t^{2}P_{f}\]  
If \(t=0\) and \(t=1\) are the endpoints of these curves, the gradients of the quadratic Bezier curve between these endpoints are  
\[\frac{dB}{dt}|_{t=0} = (-2)P_{t}+2P_{1}\] \[\frac{dB}{dt}|_{t=1} = (-2)P_{1}+2P_{max}\] \[\frac{dB}{dx}|_{t=0} = \frac{y_{1}-y_{t}}{x_{1}-x_{t}}\] \[\frac{dB}{dx}|_{t=1} = \frac{y_{max}-y_{1}}{x_{max}-x_{1}} \tag{5}\]  
Likewise, the gradients of the cubic Bezier curve between these endpoints are  
\[\frac{dB}{dt}|_{t=0} = (-3)P_{max}+3P_{2}\] \[\frac{dB}{dt}|_{t=1} = (-3)P_{3}+3P_{f}\] \[\frac{dB}{dx}|_{t=0} = \frac{y_{2}-y_{max}}{x_{2}-x_{max}}\] \[\frac{dB}{dx}|_{t=1} = \frac{y_{f}-y_{3}}{x_{f}-x_{3}} \tag{6}\]  
Given that the quadratic and cubic Bezier curves share an endpoint (\(P_{max}\), as shown in Figure 8), the gradient of Curves 1 and 2 at \(P_{max}\) are  
\[\frac{dy}{dx}|_{t=1} = \frac{y_{max}-y_{1}}{x_{max}-x_{1}}\] \[\frac{dy}{dx}|_{t=0} = \frac{y_{2}-y_{max}}{x_{2}-x_{max}} \tag{7}\]  
If Curves 1 and 2 are to be connected to each other at \(P_{max}\), the curve at \(P_{max}\) can only be differentiable if  
\[\frac{y_{max}-y_{1}}{x_{max}-x_{1}}=\frac{y_{2}-y_{max}}{x_{2}-x_{max}}. \tag{8}\]  
Therefore, \(P_{1}\), \(P_{max}\), and \(P_{2}\) must lie on the same line.