Interpolation of a 3D curve in space I'm doing a curious exercise. I have to find a method to interpolate the following ballistic trajectory in 3D space:
$$\left\{\begin{matrix}
x(t) = \frac{v\cos(\phi)\cos(\beta)}{k}(1-e^{-kt})+d_{0}\cos(\alpha)\\ y(t) = \frac{v\sin(\phi)\cos(\beta)}{k}(1-e^{-kt})+d_{0}\sin(\alpha)
\\ 
z(t) = \left ( \frac{v\sin(\beta)}{k}+\frac{g}{k^2}\right)(1-e^{-kt})-\frac{g}{k}t 
\end{matrix}\right.$$
provided a set of data $(x_n,y_n,z_n)$.
Therefore, to do this I think that I should eliminate the dependence on the parameter 't' and write the curve in the form of two different functions of one variable: $z(x)$ and $z(y)$:
$$z(x)= \left( v\sin(\beta)+\frac{g}{k}\right )\frac{x-d_0\cos(\alpha)}{v\cos(\phi)\cos(\beta)}+\frac{g}{k^2}\log\left(1-k\frac{x-d_0\cos(\alpha)}{v\cos(\phi)\cos(\beta)}\right) $$
$$z(y)= \left( v\sin(\beta)+\frac{g}{k}\right )\frac{y-d_0\sin(\alpha)}{v\sin(\phi)\cos(\beta)}+\frac{g}{k^2}\log\left(1-k\frac{y-d_0\sin(\alpha)}{v\sin(\phi)\cos(\beta)}\right) $$
Is this actually the best way to set the problem before interpolating?
 A: This is just a preliminary note, following the comment above.
Start with a simple case.
Suppose the model is
$$
\left\{ \matrix{
  x(t) = a\,t + b \hfill \cr 
  y(t) = c\,t + d \hfill \cr}  \right.
$$
and that you collected a sample $\{(x_n, \, y_n\}$ of data, supposedly synchronuously , at the same time $t_n$, but you do
not know the values of the $t_n$'s.
So you are going to interpolate  with a straight line.
But which linear regression model are you going to choose ?
Do you choose $x(y)$ ( $x$ dependent variable, $y$ explanatory variable) , or   $y(x)$ ?
Depending on the choice you'll get in general two different results.
Which model to choose depend on the amount (variance) of the "error" expected on $x$ and $y$, and on many other assumptions
you can make about its distribution (scedasticity in primis).
Refer to the Assumptions paragraph in the cited Wikipedia article.
If you can assume $y$ to be quite more "precise" than $x$, then you take $x(y)$, and viceversa.
But if they are comparable, then you shall pass to Total Least Squares Regression.
Refer for instance to this related post and various other on the subject.
