If we replace the path TAR of the nonagon with a circular arc, only seven vertices remain. This allows us to describe the tile as a partially curved heptagon. The name seahorse was coined by Ortwin Schenker.
Various arcs are possible, but we seek an elegant one for which the thickness of neck (and belly) increases strictly monotonically from point U down to point W. To achieve this the points R, M₁, and M₂ (defined below) have to be colinear, where M₁ is the center of the circular arc TR and M₂ is the center of the circular arc US.
Start the construction with the 5 points Z, T, U, R, and S known from the nonagon.
g is the perpendicular bisector of ZR, h is perpendicular to TZ through R, and M₀ = g ∩ h.
f is the perpendicular bisector of TR, k is the circle through Z and R, centered at M₀, and M₁ = f ∩ k is the center of the circular arc TR.
M₂ is the rotation of M₁ around Z by ε, this point is the center of the circular arc US.

Let Q be the midpoint of T and Z. The path from Q to T can be replaced with a circular arc. Choose a suitable center M₅ for the arc along the perpendicular bisector of TQ.
Reflect this arc across Q to obtain the curve from Z to T.
Rotate the curve TZ around Z by ε to generate the curve UZ. Reflect the curve UZ across point S to obtain the curve VW.
All tiling properties of the Vorderberg Nonagons remain fully preserved in a completely curved octagon.




Tiling patches will be added later.

Now the circular arc starting at R shall end at a point A instead of point T.
Choose A on the left and below T. If the internal angle at T (∠ATZ) is equal to 180°-ε then the internal angle at V is equal to ε.
M₀ and k are constructed in the known way.
Let M₁ be the intersection of k with the perpendicular bisector e of AR.
Draw the circular arc from A to R centered at M₁ and the straight line segment AT.
Finally apply the known transformations to obtain a V-seahorse with wide head.