## 3 Formulation and Flowchart
### Formulation  
In the original FDM, the equilibrium equations for the node \(i\) (\(x_{i}\), \(y_{i}\), \(z_{i}\)) can be written as the formulas below (Figure 5). \[N_{ij}\frac{x_{j}-x_{i}}{L_{ij}}+N_{ik}\frac{x_{k}-x_{i}}{L_{ik}}+N_{il}\frac{x _{l}-x_{i}}{L_{il}}+N_{im}\frac{x_{m}-x_{i}}{L_{im}}+F_{xi}=0 \tag{1}\]  
\[N_{ij}\frac{y_{j}-y_{i}}{L_{ij}}+N_{ik}\frac{y_{k}-y_{i}}{L_{ik}}+N_{il}\frac{y _{l}-y_{i}}{L_{il}}+N_{im}\frac{y_{m}-y_{i}}{L_{im}}+F_{yi}=0 \tag{2}\]  
\[N_{ij}\frac{z_{j}-z_{i}}{L_{ij}}+N_{ik}\frac{z_{k}-z_{i}}{L_{ik}}+N_{il}\frac{ z_{l}-z_{i}}{L_{il}}+N_{im}\frac{z_{m}-z_{i}}{L_{im}}+F_{zi}=0 \tag{3}\]  
where, for example, \(N_{ij}\) and \(L_{ij}\) are, respectively, the axial load and the length of the segment \(i\)-\(j\). The system of Equations (1)-(3) is non-linear since it depends on the nodes coordinates, which are the solutions for the system. The FDM transforms Equations (1)-(3) into a linear system of equations by forcing the ratio \(N/L\) at each segment (i.e., the ratio between the axial force and its length) to be equal to a previously given value \(q\) (the so-called force density). Thus, Equations (1)-(3) can be written as the equations below. \[q_{ij}\cdot(x_{j}-x_{i})+q_{ik}\cdot(x_{k}-x_{i})+q_{il}\cdot(x_{l}-x_{i})+q_{ im}\cdot(x_{m}-x_{i})+F_{xi}=0 \tag{4}\]  
\[q_{ij}\cdot(y_{j}-y_{i})+q_{ik}\cdot(y_{k}-y_{i})+q_{il}\cdot(y_{l}-y_{i})+q_{ im}\cdot(y_{m}-y_{i})+F_{yi}=0 \tag{5}\]  
\[q_{ij}\cdot(z_{j}-z_{i})+q_{ik}\cdot(z_{k}-z_{i})+q_{il}\cdot(z_{l}-z_{i})+q_ {im}\cdot(z_{m}-z_{i})+F_{zi}=0 \tag{6}\]  
This original formulation can be adapted to obtain three-dimensional anti-funcular arches for a planar arch bridge contained within the \(x\)-\(z\) plane (Figure 6). For the intermediate nodes (\(2\leq i\leq n-1\)) located at the given \(x_{i}\) abscissae, the vertical equilibrium of every node can be written as the equation below. \[\frac{N_{i-1}}{L_{i-1}}(z_{i-1}-z_{i})+\frac{N_{i}}{L_{i}}(z_{i+1}-z_{i})+F_{zi }=0 \tag{7}\]  
where \(N_{i}\) and \(L_{i}\) are the axial force and the length, respectively, of the segment \(i\). Note that, in Equation (7), the previous notation used in cable meshes has been simplified for only one cable/arch by using only one sub-index. The given \(z\)-coordinates of the \(1\) and \(n\) nodes to complete the system of \(n\) linear equations. \[z_{1}=z(x_{1}) \tag{8}\]  
\[z_{n}=z(x_{n}) \tag{9}\]  
Figure 5: FDM: Node of a cable mesh. In our case, the force density \(q\) can be expressed as follows. \[N_{ij}\frac{y_{j}-y_{i}}{L_{ij}}+N_{ik}\frac{y_{k}-y_{i}}{L_{ik}}+N_{il}\frac{y _{l}-y_{i}}{L_{il}}+N_{im}\frac{y_{m}-y_{i}}{L_{im}}+F_{yi}=0 \tag{2}\]0  
where \(N_{x,i}\) is the longitudinal component (along the \(x\)-axis) of the axial force of the segment \(i\). Since \(x_{i}\) are given, \(z_{i}\) can be obtained by simply defining \(N_{x,i}\). Therefore, Equation (7) transforms into the formula below. \[N_{ij}\frac{y_{j}-y_{i}}{L_{ij}}+N_{ik}\frac{y_{k}-y_{i}}{L_{ik}}+N_{il}\frac{y _{l}-y_{i}}{L_{il}}+N_{im}\frac{y_{m}-y_{i}}{L_{im}}+F_{yi}=0 \tag{2}\]1  
Similarly to Equations (8), (9), and (11), the equations of equilibrium for the nodes can be written within the \(x\)-\(y\) plane.