with this cutoff function you wrote, am I not retaining something like
5 s 304 mm
5.01 s 305 mm
when I really want the whole remainder x0>=L_F shifted to begin at the origin?
It depends on how you want to represent the final data. I think you need to think about this carefully, ie:
What does the final data representation look like?
But before considering this question, there is another issue:
The problem of choosing the leading profile is under-determined. We would probably want to use some sort of continuity / periodicity condition to choose the leading profile. Ideally we want periodicity - after all it is a machine that repeats things over and over.
could I not let my seeding y0 continue beyond L_V so as to avoid having to restart x0?
Yes you could and you would get correct results - in the sense that the job will still get done. But you would have to keep on "seeding" y0 arbitrarily to keep going. Since this is after all a machine, we expect some sort of periodicity and you won't get that by arbitrarily seeding y0.
Now that I would be getting a single profile (I suppose without having tried it that x0 and y1 are the same) that can be repeated by all hooks, ...
This is the key. In a sense we want all the profiles to merge smoothly into one another. Moreover the profile of one hook should look the same as the profile of any other hook - only shifted in time. So that should give us a condition to fix the leading profile shape, ie we want:
x(t+tau) = x(t)*cos(2*angle) + sqrt(L_F^2 - (x(t).^2)*sin(2*angle)^2) -- (1)
which can alternately be written in its original form (before we solved for the root):
x(t+tau)^2 + x(t)^2 - 2*x(t)*x(t+tau)*cos(2*angle) = L_F^2 ----- (1a)
What is the time tau? It is the time for which:
because that's when the next hook comes online (after the leading hook has moved L_F from the zero position). Note that we don't know this time yet because we don't yet know the profile. We have to simultaneously solve for both the time tau and the profile x(t) from the two equations (1) and (2) above.
How does one solve this? I'm not sure - I've never really come across a "delay" equation like the above, but it certainly looks like it should be solvable. Note that simple ansatz's like polynomials in t wont work since you will get more equations than unknowns. For example - 2nd degree polynomial (constant acceleration) would be of the form x(t) = at + bt^2. There is no constant term since x(0) = 0 (equivalent to x(tau) = L_F) . If we plug this in to eq (1a), we get 5 equations in three unknowns (a, b and tau).
Thus we need a more sophisticated ansatz, or we need to solve (1) numerically.
Can you think of an ansatz? How about x(t) = A(1 - e^(-bt)). What if we plug this into (1a) and try to equate coefficients (we will have e^(2bt) terms, e^(bt) terms and constant terms (so three equations) and three unknowns A, b and tau. Could you try it?
[edited for corrections]