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I propose just to insert a legend to clearly identify the pylon.
Perhaps a frame with z could help to understand the view.

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same remark here

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Binary file added docs/user_guide/assets/angle_line_side.drawio.png
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53 changes: 53 additions & 0 deletions docs/user_guide/ug_cable_model.md
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Expand Up @@ -35,6 +35,59 @@ $$

Another way to see the cable plane is to rotate the cable plane.

### Line angles and arm lengths

![Image not available](./assets/angle_line_direct.drawio.png "Cable plane image")

We want to compute $a_{cable}$ and $b_{cable}$ depending on $a$, the line angles, the arm lengths and the chain deplacements

$A_i$ and $A_{i+1}$ (**attachments points**) are the points where cable is attached at the end of the suspension chains, respectively on left and right support. These are the points we want at the end.

$E_i$ and $E_{i+1}$ (**edge_arm points**) are the points of the end of the arms, respectively on left and right support

$C_i$ and $C_{i+1}$ (**center_arm points**) are the points on the support where the arms are attached

$G_i$ and $G_{i+1}$ (**support ground points**) are the points on the ground where the support is set. More precisely, this point is the barycenter of the legs of the support.


![Image not available](./assets/angle_line_points.drawio.png "Cable plane image up")

![Image not available](./assets/angle_line_side.drawio.png "Cable plane image side")


$a_{cable} = \sqrt{(x_{A_i}-x_{A_{i+1}})^2 + (y_{A_i}-y_{A_{i+1}})^2}$

$b_{cable} = |z_{A_i}z_{A_{i+1}}|$


We want the coordinates of $A_i$ and $A_{i+1}$ in the span frame of the support $i$



#### Computing support ground $G_i$ coordinates

Support ground coordinates $G_i$ are built recursively. Starting from $G_i$, we can build $\overrightarrow{G_{i}G_{i+1}}$, then get coordinates of $G_{i+1}$

$\overrightarrow{G_{i}G_{i+1}}$ length is $a$. Its direction can be determined using the line angles $\gamma_i$

Then center arm coordinates $C_i$ can be easily computed by adding the altitude of the arm.


#### Computing edge arms $E_i$ coordinates

In order to get $E_i$, we use point $E_i'$, which is the equivalent point if there wasn't any line angle.\
Coordinates of $E_i'$ are $(x_{C_i}, L_l, z_{C_i})$. Same x and z coordinates than center arm points, but with an arm length along the y axis.\
Then we can get $E_i$ by applying a rotation of $-\frac{\gamma_{l}}{2}$ around the z axis.

Same thing for $E_{i+1}$, we use point $E_{i+1}'$.\
Coordinates of $E_{i+1}'$ are $(x_{C_{i+1}}, L_l, z_{C_{i+1}})$.\
Then we can get $E_{i+1}$ by applying a rotation of $+\frac{\gamma_{l}}{2}$ around the z axis.

#### Computing attachement $A_i$ coordinates

From coordinates of $E_i$ and $E_{i+1}$, you can get coordinates of $A_i$ and $A_{i+1}$, by adding $dep_x$ and $dep_y$ to their coordinates.

$dep_z$ can be calculated using the two other coordinates and the insulator length (TODO).

### Catenary model

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