FMF - Friends of Minerals Forum, discussion and message board

[Pete Richards]

[Tobi]

[Jesse Fisher]

Another Naica fluorite, which according to Pete's analysis would be cube-dodechahedral habit. From the find circa 2005.

[Bob Morgan]

Pete,
Some of us call these 'Mayan Temple" habit fluorites. Yes, these are the result of smaller overgrowths on octahedral crystals. I have a specimen from Naica that has an octahedral phantom sprinkled with small galena crystals, that can be seen through clear cube faces at the terminations of the underlying octahedron. This particular specimen has overgrowths that are cubes with small dodecahedral faces and larger trapezohedral faces and the occasional small octahedral face.
The original octahedral stage is an indication of a higher temperature of formation, which is supported by the octahedral habit of the small galena crystals. Subsequent overgrowths then were lower temperature aligned deposits more characteristic of cube shapes.
'Myan Temple' habit fluorites come from other places. I've seen them from Dalnegorsk, China and even places in the US that I can't recall.
Larger overgrowths tend to be on the octahedral terminations with smaller ones down the octahedral edges and even smaller ones stepping all over underlying octahedral faces.

[Pete Richards]

[Bob Morgan]

The picture you feature under your name is cube-octahedral.

[Bob Morgan]

The picture I referenced is under Carles Millan's name.

[Johan Kjellman]

[Bob Morgan]

Yes.

[Pete Richards]

I am not sure how to interpret Jesse's fluorite. I cannot see any view that is unambiguous. It appears that the cube faces are clear, maybe slightly pebbled, and that the other faces are frosted.

In the bottom-most fluorite in his specimen, the lower left corner of the forward-facing cube face points to the edge between a pair of the other faces, beyond which is a face that is just barely visible because it is nearly perpendicular to the view. This face holds the answer. If it is a cube face, then the habit is cube-octahedral; if it is one of the other, frosted faces, the habit is cube-dodecahedral.

The crystal on the upper left also helps a little. It has a cube face oriented forward and slightly down. Above and to the right of this face are two of the other faces. To the top-right of these is another face, sloping backwards. It is hard to see clearly, but it appears to me to be frosted. If so, this crystal also indicates a cube-dodecahedral habit; if instead this is a cube face, then the habit is cube-octahedral.

In my view, and with considerable uncertainty, the habit of this one is, like the other examples, cube-dodecahedral - (100)(110). Perhaps Jesse will look at his specimen from all angles and let us know!

[Bob Morgan]

Pete,
Points well stated.
I looked at photos in Mindat of fluorite on sphalerite. Nearly all were cube- dodecahedral that could be discerned clearly.

[Johan Kjellman]

I agree again - and withdraw my claim.
And Pete, you point directly at what I now think so many of us, although experienced, keep on misinterpreting these combos. We are all fast and confident in identifying the first prominent cube face. Then a "second face" a bit too quickly identified as a cube and, voila, you have a cube-octahedron.
Maybe we all should slow down and look for a third alternative cube face. I (think I) see them in both crystals you described on Jesse's specimen.The thing is, and this may be part of the problem, these cube faces seem relatively smaller than the obvious large one.

cheers

[Pete Richards]

[Johan Kjellman]

I came back to this thread, having forgotten and identified the first image first as a cube-octahedron, but then, after counting the exposed faces of the most obvious crystal (I could see 8) I came to the conclusion that it must be a cube-dodecahedron. Because you can never see more faces than half of what the crystal have from one side. I am quite confident that this applies even to distorted crystals, even stepped crystals if you count repeatedly oriented faces as 1.

So, I propose the following "half-face" method, not bulletproof, but works in many instances.
cube + octahedron = 6 + 8 faces = 14, one should not be able to see more than 7 faces
cube + dodecahedron = 6 + 12 faces = 18, i.e. not more than 9

I attach a hastily prepared sketch. The top two crystals are different but oriented in a way that minimize the number of the exposed faces. This illustrates the general problem.
On the bottom row I have tilted both crystals, and you can now count the exposed faces.