I cross-posted the question on the Origami-List, and no one had info on the Crawford model.
You CAN buy "A Constellation of Origami Polyhedra" by John Montroll from Amazon for about $10 - $12.
Best, Kim: Stellated Octahedron, OUSA Convention Book 1998, p. 34
This is 1 sheet.
But I did get a little more info:
[Anne LaVin]
http://mathworld.wolfram.com/StellaOctangula.html
This model appears to be an octahedron with somewhat squat
triangular-pyramidal points - each point's geometry is made from
collapsing the diagonal and only one of the "bookfold" creases of a
waterbomb base so you're left with a pyramid. Each face is a 1-1-rt(2)
triangle, in other words.
To the best of my understanding, neither Sam Ciulla's nor Kim Best's
models are, actually, a "Stellated Octahedron" - the only proper
stellation of the ocathedron is named the "Stella Octangula."
Which takes nothing away from the models, they just should be called
something else, mathematically-speaking. ("Octahedron with spike-ish
bits sticking out," however, fails to roll off the tongue.
But - mathworld again to the rescue - have just discovered that the
proper word is "cumulation". Thus, math-speaking. those two models are
"Cumulated Octahedrons." (
http://mathworld.wolfram.com/Cumulation.html,
where there's even an image of one.)
[Sy Chen]
John Montroll has both Stellated Octahedron (45-45-90 triangle) and
Stella Octangula (60-60-60 triangle) diagrammed in his book - A
Constellation of Origami Polyhedra. If you read the diagram carefully,
Both of them can be collapsed from regular cube. i.e. You can fold a
traditional waterbomb with extra pre-creases in the beginning to start
with. The final collapse can give you either of Stellated Octahedron
and Stella Octangula depending on your pre-crease pattern.
Actually Sam Ciulla designed a stellated Octahedron from regular
waterbomb. The approach is surprising simple and elegant. The diagram
can be found in Gay M. Gross' book - The Art of Origami.
Kim Best's model is a Stellated Octahedon. Although the shape is the
same. Ciulla's approach is my personal preference.
As for Crawford's Stella Octangula, I believe it is from a hexagon.
- Hank Simon