Stroke end

The position modeling algorithm will lag behind the raw input by some distance. This algorithm iterates the previous dynamical system a few additional time using the raw input position as the anchor to allow a catch up of the stroke (though this prediction is only given by predict, so is not part of the results and becomes obsolete on the next input).

Algorithm :Stroke end
Input:

• Final anchor position $p\left\lbrack \text{end} \right\rbrack = \left( x\left\lbrack \text{end} \right\rbrack,y\left\lbrack \text{end} \right\rbrack \right)$ (From the original input stream)

• final tip state $q_{f}\left\lbrack \text{end} \right\rbrack = \left( p_{f}\left\lbrack \text{end} \right\rbrack = \left( x_{f}\left\lbrack \text{end} \right\rbrack,y_{f}\left\lbrack \text{end} \right\rbrack \right),v_{f}\left\lbrack \text{end} \right\rbrack,a_{f}\left\lbrack \text{end} \right\rbrack \right)$ (returned from the physical modeling from the last section, $\cdot_{f}$ signifies that we are looking at the filtered output)

• $K_{\text{max}}$ max number of iterations (sampling_end_of_stroke_max_iterations)

• $\Delta_{\text{target}}$ the target time delay between stroke (1/sampling_min_output_rate)

• $d_{\text{stop}}$ stopping distance (sampling_end_of_stroke_stopping_distance)

• $k_{\text{spring}}$ and $k_{\text{drag}}$ the modeling coefficients

initialize the vector $q_{o}$ with $q_{0}\lbrack 0\rbrack = \left. \left( \underset{p_{o}\lbrack 0\rbrack}{\underbrace{p_{f}\left\lbrack \text{end} \right\rbrack}},\underset{v_{o}\lbrack 0\rbrack}{\underbrace{v_{f}\left\lbrack \text{end} \right\rbrack}},\underset{a_{0}\lbrack 0\rbrack}{\underbrace{a_{f}\left\lbrack \text{end} \right\rbrack}} \right) \right.$
initialize $\Delta t = \Delta_{\text{target}}$
for $1 \leq k \leq K_{\text{max}}$
- calculate the next candidate $$\begin{aligned} a_{c} & = \frac{p\left\lbrack \text{end} \right\rbrack - p_{o}\left\lbrack \text{end} \right\rbrack}{k_{\text{spring}}} - k_{\text{drag }}v_{0}\left\lbrack \text{end} \right\rbrack \\ v_{c} & = v_{o}\lbrack 0\rbrack + \Delta ta_{c} \\ p_{c} & = p_{o}\lbrack 0\rbrack + \Delta v_{c} \end{aligned}$$

- if $\left. \parallel{p_{c} - p\left\lbrack \text{end} \right\rbrack} \right.\parallel < d_{\text{stop}}$ (further iterations won’t be able to catch up and won’t move closer to the anchor, we stop here), - return $q_{0}$
- endif

• if $\langle p_{c} - p_{o}\left\lbrack \text{end} \right\rbrack,p\left\lbrack \text{end} \right\rbrack - p_{o}\left\lbrack \text{end} \right\rbrack\rangle < \left. \parallel{p_{c} - p_{o}\left\lbrack \text{end} \right\rbrack} \right.\parallel$ (we’ve overshot the anchor, we retry with a smaller step)

  
  

  • $\Delta t \leftarrow \frac{\Delta t}{2}$,

  • continue (this candidate will be discarded, try again with a smaller time step instead),

- else
- $q_{0}\lbrack end\ +1\rbrack = \left( p_{c},v_{c},q_{c} \right)$ (We append the result to the end of the $q_{0}$ vector)
- endif

• if $\left. \parallel{p_{c} - p\left\lbrack \text{end} \right\rbrack} \right.\parallel < d_{\text{stop}}$,

  • return (We are within tolerance of the anchor, we stop iterating),

• endif,

Output : $\left\{ q_{o}\lbrack k\rbrack = \left( s_{o}\lbrack k\rbrack,v_{o}\lbrack k\rbrack,a_{o}\lbrack k\rbrack \right),0 \leq k \leq n\left( \leq K_{\text{max }} - 1 \right) \right\}$