I don't think so. No acceleration above 0.61c is used or provides momentum. The 0.86c is the apparent VECTORED
V (in your example). But, I don't think that your example is correct since you seem to be using two-dimensional (graph-paper) trig here, while with excluding time.
In V calculations, a Y velocity at 0.61c and an X velocity at 0.61c takes time to reach the "end-point" from where you measure the Hypotenuse. The "actual V" of the vectored triangle cannot exceed the V of the greater of the XV or the YV if time is included. The "apparent
" V can though, as is seen in apparent superluminal expansion around some supernova remnants.