We construct a spiral starting at point R that consists of circular arcs with decreasing radius.
Let d be the planned thickness of the neck.
Construct D with RD = d and angle RDS = 90° + ε/2.
Let F be the intersection of TZ and RD.
Let G be the point on RD with angle FGZ = 90° + ε/2
Let k be the circle with center Z and radius ZG.
M₁, M₂, ... are elements of k with GM₁ = M₁M₂ = M₂M₃ = ... = d.
Note that the angles M_i+1ZM_i are equal to ε.
Define A₀ := R. Recurrence definition of A_i for positive integers.
Define A_i as intersection of ray M_iM_i+1 and circle through A_i-1 centered at M_i.
Define A_n as the first point of the sequence A_i that lays above TZ. For ε = 15°:  n=7.
If the arc A_n-1A_n crosses line segment TZ then choose a smaller value for d. For ε = 15° the spiral passes through T for d ≈ 0.60446659260427 units. (RS = 2 units)
Finally choose a point H on the spiral and draw a path from H to T (for example a circular arc or a straight line segment).

A V-Swan is called perfect if the spiral starting at R passes exactly through the point T.
The perfect thickness d depends on the angle ε. So far we do not know a straight forward construction, therefore we have to determine the optimal d by iteration in advance.
d is optimal if T is an element of the circle through A_n centered at M_n.
Example: For ε = 18°:  d ≈ 0.50412062747023 is optimal.
For ε = 18°:  n=6 and A₅(-4.000733, -0.001690) is very close to T(-4, 0).

Tiling patches will be added later.

Appendix 1

Maximal angle ε for which a perfect V-Swan exists

For d = 0 the lines TZ, RZ, and SZ have a length of 4 units. And as the length of RS is 2 units: sin ε = 1/4 =0.25.
Thus ε < 2 arc sin 0.25 = 28.955024371860° .