The snub anti-prisms are convex for n-gonal bases up to an n of 4.744567331.. The 19/4 snub anti prism (n=4.75) is very close to this limit. This is difficult to see on the full shape as it involves four revolutions of the snub faces around the base 19/4-gon. Click here for a cut-away showing only one revolution of the snub faces. Interestingly, despite the departure from coplanarity being only 0.1 degrees, this is visible on the VRML model.

The proof follows:

Position the prism with **O** at the origin and with
a prismatic base in the XY plane.

**R** is the centre of revolution and is placed on
the X-axis.

**S**, **O** and **S'** are vertices of the prismatic
base.

**L**,** M** and **N** are snub vertices.

Co-ordinates of **R** are *( r, 0, 0)*

Co-ordinates of **S** are *( c, s, 0)*

As **OR** = **SR** this gives *r = 1/2c*

Let co-ordinates of **L** be *( -x ,y, -z)*

Co-ordinates of **N** are *( 2x, 0, -2z)*

As triangles **LNO** and **MNO** are coplanar then

or

Distance **LO** = 1 so

Distance **LS** = 1 so

Solving the above gives

As **L** and **M **are identical snub vertices then
they are the same distance from the axis of revolution

So **LR** = **NR** (ignoring the z component)

i.e. the distance *( -x, y)* to *( r, 0)* is
equal to the distance *( -2x, 0)* to *( r, 0)*

Equating the two gives the quadratic

Which solves to give

The vertex and the central angle of a polygon are related by the formula:

where theta is half of the central angle. (Proof is here)

Turning back to the prismatic face, the general n-gon has the sum of its central angles given by

So in full

Which numerically is **n
= 4.74456733125464..**