When you draw a Fibonacci spiral in a quantized way , or more clearly from a golden rectangle geometrical construction based on Fibonacci numbers, and select one of arc nodes extremity to be the center of a rotation, three nodes away from it the spiral create a boundary as shown on the picture , is there a trigonometric formula which explain that boundary condition ? "phi-spiral"
Third and fourth arc in your construction have centers which are aligned with the "first" point: as a consequence, a circle with center at the first point and passing through the fourth point is tangent to both arcs, giving the illusion of what you call "a defined boundary".
As a matter of fact, all arc endpoints in the spiral (blue points in diagram below) lie between exactly two arc centers (red points).
EDIT.
For a true golden spiral (red curve in diagram below), i.e. a logarithmic spiral with polar equation $r=a e^{b\theta}$ and $b=\ln\phi/(\pi/2)$, this feature is lost.
The evolute (locus of centers of curvature) of a logarithmic spiral, is another logarithmic spiral with polar equation $r=ab e^{b(\pi/2+\theta)}$ (black curve below). The evolute is rotated by $\pi/2$ with respect to the original spiral: the center of curvature of a point $A$ on the red spiral is the point $C$ on the black spiral and $\angle AOC=\pi/2$, where $O$ is the coordinate center. The blue circle (center $C$ and radius $AC$) is then the osculating circle at $A$.
For a golden spiral, however, point $C$ is not on the spiral itself: point $A'$ on the red spiral is near to $C$ but the circle with center $A'$ and radius $A'C$ (dashed in the diagram) is quite different from the osculating circle.
Notice that for different values of $b$ the evolute can be the the same as the spiral itself (see here for details). In that case you have exactly the feature you are looking for: every point on the spiral is the center of the osculating circle at a point which is an angle $2n\pi-\pi/2$ after it, where $n$ is an integer positive number.