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Very steep rhombohedral calcite?
  
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Josele




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PostPosted: Oct 05, 2020 09:58    Post subject: Very steep rhombohedral calcite?  

This crystal was listed in Mindat as "scalenohedral and flat rhombohedral form". Somebody noticed that this description is wrong and it was changed to "steep and flat rhombohedral form".
It is correct?
The term "steep rhombohedron" is normally used for {401} form but I think this one has a much much more higher indice.



C_1.jpeg
 Locality:
Sweetwater Mine, Ellington, Viburnum Trend District, Reynolds County, Missouri, USA
 Dimensions: 6 x 4.3 x 4.1 cm
 Description:
foto © Rob Lavinsky
 Viewed:  916 Time(s)

C_1.jpeg



C_2.jpeg
 Mineral: Calcite
 Locality:
Sweetwater Mine, Ellington, Viburnum Trend District, Reynolds County, Missouri, USA
 Dimensions: 6 x 4.3 x 4.1 cm
 Description:
foto © Rob Lavinsky
 Viewed:  915 Time(s)

C_2.jpeg



C_smorf.jpg
 Description:
To find a form alike I must use a very high "h" Miller indice. ¿Is that possible?
 Viewed:  916 Time(s)

C_smorf.jpg


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James Catmur
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PostPosted: Oct 05, 2020 10:23    Post subject: Re: Very steep rhombohedral calcite?  

Looks like Mina Julia style
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PostPosted: Oct 05, 2020 10:32    Post subject: Re: Very steep rhombohedral calcite?  

The tall sloping faces are scalenohedron faces. The termination at the top is the flattened rhombohedron. Simple.
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Pete Richards
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PostPosted: Oct 05, 2020 10:52    Post subject: Re: Very steep rhombohedral calcite?  

First of all, the change is correct because the crystal shows no scalenohedra, but a shallow and a steep rhombohedron - one with faces relatively close to horizontal and one with faces close to vertical. While {401} is a common steep rhombohedron, I don't think "steep rhombohedron" should refer only to it. I would suggest that the boundary between steep and flat or shallow rhombohedra is the cleavage rhomb, {101}, with faces inclined slightly less than 45° from vertical (such angles are measured between the c-axis (vertical) and the normal to the face in question). That's just my opinion - I know of no official definition.

You have encountered one of the quirks of Miller indices! Because of their definition related to reciprocals of intercepts, at least in part, the Miller indices of forms increase rapidly as the form approaches a limit, such as a steep rhombohedron approaching a prism. Among other things, this means that measured crystals have increasing error in the calculation of their indices as the steepness of the form increases. It is much easier to measure the difference between {301} and {401} (4.45°) than it is between {30 0 1} and {31 0 1} (0.062°), or even {30 0 1} and {40 0 1} (0.48°), for example.

Unless faces are highly perfect optically, it will be impossible to determine exact indices for steep faces, as a result. Furthermore, such faces (in calcite, anyway) are usually not optically smooth and not optically flat. This means that different spots on the face actually have different indices, but in most cases we can't determine them very accurately anyway.

Using a form like {30 0 1} is perfectly acceptable in order to capture the overall shape of a crystal like this one. It is probably wise to mention that the indices were determined by experimentation to depict the shape of the crystal, but not actually measured.

An interesting consequence of the rhombohedral structure of calcite is that common rhombohedra tend to follow a particular sequence of alternating positive and negative rhombohedra, one in which, in order of increasing steepness, each form is beveled by the previous one (i.e. cuts its edges to form a face with parallel edges). The common forms follow the sequence:
(101), (021), (401), (081), (16 0 1)...
The shallow end of the sequence is
...(081), (401), (012), (101)...

The complimentary forms ...(102), (011), (201), (041)... are rarely seen, if ever.

These facts may be useful in choosing indices in the absence of measurements.

This ordered pattern of succession, which has its equivalents in scalenohedra and dipyramids, is part of what gives calcite its aesthetic crystal habits.



calc.jpeg
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 Viewed:  856 Time(s)

calc.jpeg



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Josele




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PostPosted: Oct 06, 2020 05:10    Post subject: Re: Very steep rhombohedral calcite?  

Thank you for answering. Special thanks to Pete for his so well exposed reasoning. Is a pleasure read your comments and follow your approach to the matter. I always learn much more than a couple of things when Mr. Richards appears in a conversation.

Sorry if I say foolishness but these so high indice forms like {30 0 1} seems to me very extreme forms. A step every 30 or more unit cells! Crystallization thermodynamic conditions should be almost ideal for prismatic habit with a small touch of rhombohedral. Looks like very difficult that the narrow range of conditions in the border of different habits can be maintained long enough for a crystal of this size to grow. Wonders of nature...

Apart its bumpy texture, this crystal appears to have quite straight edges and flat faces. If it is not a true high indice form close to {30 0 1}, could be a prism face with rhythmical micro steps of a more common rhombohedral form, maybe {401}, {021}, {101} or even {012} with a rhythm that macroscopically equates something like {30 0 1} form.
Although conceptually are different things, are the same to our eyes.

Interesting the fact that all common rhombohedral forms are beveled by the lower form in the sequence. This implies, besides positive and negative alternation, that rhombohedral edges form exactly the same angle with c axis that the face of the lower form in the sequence. There must be a wonderful geometric reason for that, I have to look into it...



R_sequence.jpg
 Description:
{012} {101}{021}{401}{081}{16 0 1}{0 32 1}

A progressive sequence of positive and negative rhombohedra will conform lovely pseudo-bent faces with curved edges.
 Viewed:  775 Time(s)

R_sequence.jpg



{30 0 1}{012}.jpg
 Description:
{0 30 1}{012}

Crystal like that one at photos elongated in “c” axis direction. Pure trigonal!
 Viewed:  775 Time(s)

{30 0 1}{012}.jpg


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Bob Morgan




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PostPosted: Oct 06, 2020 14:10    Post subject: Re: Very steep rhombohedral calcite?  

Three steeper rhomb faces are below flatter ones on top. The other three steeper rhomb faces are above the three shallower ones on the bottom. Perhaps that makes it easier to visualize that both forms are rhombohedral.

Maybe the rough texture of the steep rhomb faces and their deviation from being prismatic is explainable as due to natural etching. The same might explain the other steeper rhomb faces you'd noticed. In that case it would be etching on an exposed edge.

In both cases such etching would be quite slow to produce flatter appearing surfaces rather than curved and more disturbed surfaces.
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PostPosted: Oct 06, 2020 15:06    Post subject: Re: Very steep rhombohedral calcite?  

It is indeed hard to understand how a crystal can grow such steep sides, as Josele says, requiring 16 steps down for every one out. It is almost as hard for me to understand how crystal growth produces flat faces of any form that has this kind of stepped structure (so-called S- or step-faces).

I don't think we can assume that the form of the crystal is an equilibrium form in the thermodynamic sense. Crystals do not grow in equilibrium. If they were in equilibrium, they would not grow. To assume a crystal is an equilibrium form is to assume that it had time to sit without growing, but with time to adjust its shape by moving atoms from one part of the crystal to another, until this process leads to no further changes. Snowflakes do this in snow packs as they become firn, but I'm not so sure about calcite.

The same issues about growing faces that are near-prisms also pertain to etching them to create the forms we see. Usually, etching rounds corners and edges. Unless "etching" is more like removing ions one place on the crystal and depositing them on another, which is more like an adjustment toward equilibrium, I don't think it offers a solution to this dilemma.

I acknowledge the existence of so-called vicinal faces and their possible relevance here, but I think that is another topic!

Final comment. Josele, it's great to see you drawing crystals! Your more recent drawing shows two negative rhombohedra. Your first drawing was more representative of the photo, with a shallow negative rhombohedron (012) and a steep positive rhombohedron (30 0 1). The crystal's form in the photo requires one of each kind, positive and negative. This second drawing does show a purely trigonal crystal, in the sense that every cross section is a triangle. But I don't think that can be done with a combination of negative and positive rhombohedra.

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PostPosted: Oct 06, 2020 18:34    Post subject: Re: Very steep rhombohedral calcite?  

The high indices as shown in the crystal drawing program are also a result of using the old a:c ratio of 1:0.854.
The new value that actually reflects the crystal structure according to new measurements is 1:3.419.

Edit: so the unit cell of calcite is 4 times longer along the c-axis than people thought before the refined measurements.

Maslen, E.N., Streltsov, V.R., and Streltsova, N.R. (1993) X-ray study of the electron density in calcite, CaCO3. Acta Crystallographica B: 49: 636-641.

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PostPosted: Oct 07, 2020 11:01    Post subject: Re: Very steep rhombohedral calcite?  

Bob, thanks for your comment.

Amir, thank you for note the a:c relation in calcite. Smorf must upgrade this unit cell data.

Pete, thanks for notice the positive - negative combination. My last drawing was mistaken. Then combinations positive - positive and negative - negative are crystallographically forbidden. That means edges of both forms never are aligned, all own edges of every form meet a face in the border between forms. Another matter that must have an elegant geometric explanation.

Link to a calcite with very steep rhombohedral form: https://www.mindat.org/photo-833692.html



C_2.jpg
 Description:
Combinations of two forms with equant development.

Certainly combinations of opposite sign forms have better stylish aspect compared to repeated sign combinations. We have been lucky!
 Viewed:  598 Time(s)

C_2.jpg



C_7.jpg
 Description:
The weird but possible combination: elongated very steep positive rhombohedron with a negative flat one at terminations.

Pure trigonal with a perfect hexagonal section at center.
 Viewed:  598 Time(s)

C_7.jpg



C_oldcell.jpg
 Description:
{101} form with old and new unit cell data.

{101} form with new data corresponds exactly to old {401} form.

Miller indices of cleaved rhombohedron with new data will be {104}
 Viewed:  600 Time(s)

C_oldcell.jpg


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PostPosted: Oct 15, 2020 06:16    Post subject: Very steep rhombohedral calcite? - Polarity  

Rummaging through the drawers I found a Chinese calcite with likely the same flat and steep rhombohedra combination. Crystals up to 2 cm allowed to measure some angles that point to steep rhombohedral form with indices close to {16 0 1}.

The polarity of this rhombohedral combination is confusing me. If usually rhombohedral combinations alternate positive and negative forms, this case seems an exception with two forms of the same sign.



C_rr.jpg
 Mineral: Calcite
 Locality:
Guangxi Zhuang Autonomous Region, China
 Dimensions: 9 x 6 x 5 cm
 Description:
All crystals have same combination of {0 16 1} and {012} or {16 0 1} and {102} forms, the same than crystal at the beginning of this thread.
 Viewed:  402 Time(s)

C_rr.jpg



P1230023.jpg
 Locality:
Guangxi Zhuang Autonomous Region, China
 Dimensions: FoV: 2.5 cm
 Description:
I checked a dozen of crystals, all with their long edges forming an angle of about 14º (+/- 0.5º)
 Viewed:  402 Time(s)

P1230023.jpg



P1230022.jpg
 Locality:
Guangxi Zhuang Autonomous Region, China
 Dimensions: FoV: 2.5 cm
 Description:
Relative position of faces indicates that both rhombohedrons have the same polarity (positive + positive or negative + negative)
 Viewed:  401 Time(s)

P1230022.jpg



RRR.jpg
 Locality:
Sweetwater Mine, Ellington, Viburnum Trend District, Reynolds County, Missouri, USA
 Dimensions: 6 x 4.3 x 4.1 cm
 Description:
Scheme at left fits the real crystal.

Edges of termination meet steep rhombohedral face in his narrow side.

foto © Rob Lavinsky
 Viewed:  404 Time(s)

RRR.jpg


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PostPosted: Oct 15, 2020 11:20    Post subject: Re: Very steep rhombohedral calcite?  

The drawing models the outline of crystal surfaces well for a steep rhombohedral form.

The faces, however. are stepped, and I suspect the reflections indicate a steeper rhomb or perhaps even the prism form.

With a reflecting goniometer, angles between faces would discern which faces are involved.

This can be done using a laser pointer aimed outward perpendicular from a wall and getting reflections from two adjacent faces with a little trigonometry.

If you want to try this, I have a powerpoint file to show the technique. It should work well on the reflective faces in your photos.
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PostPosted: Oct 15, 2020 14:35    Post subject: Re: Very steep rhombohedral calcite?  

Josele wrote:
Rummaging through the drawers I found a Chinese calcite with likely the same flat and steep rhombohedra combination. Crystals up to 2 cm allowed to measure some angles that point to steep rhombohedral form with indices close to {16 0 1}.

The polarity of this rhombohedral combination is confusing me. If usually rhombohedral combinations alternate positive and negative forms, this case seems an exception with two forms of the same sign.


I think your analysis of the Chinese crystals is correct, at least as far as having two forms of the same "sign", when the relationship I described in an earlier post would predict one positive and one negative. Some clarifications and further points:

Calcite provides a way to distinguish between two geometric possibilities such as you list above: {16 0 1} and {102} versus {0 16 1} and {012}, which look identical except that in the standard orientation one is rotated 60° relative to the other one. The distinction relies on the fact that by the definition of the geometry of calcite, the cleavage is always and must be {101}, not {011} - or {104} using the structural unit cell. The cleavage therefore serves as an absolute determinant of the location of the positive sector and thereby also the negative sector. I am confident that the cleavages would show that the second set of indices - {0 16 1} and {012} are correct in this case.

The "rule" that relates that cascade of rhombohedra I wrote about last time- the so-called Rhombohedral Rule - states that forms (hkil) are favored when their indices obey the relationship 2h + k + l = 3n, where n is any integer. All of those alternating rhombohedrons in my earlier comment (e.g. {401}) obey this rule, and none of the opposite ones (e.g.{041}) do. Most of the common scalenohedra and other forms of calcite also obey this rule, though there are many forms that do not. In my experience, it tends to be less reliable with higher indices, perhaps in part because these indices themselves are less reliable. But if we want to try to apply it in this unpredictable region, {0 17 1} conforms to the rule, and would be very close to the same steepness as {0 16 1}.

The point of my previous comment was to offer a set of very common forms of calcite that have a special relationship with each other. They are not the only common forms, even among the rhombohedra. Two common forms that are not in that series but conform to the rule are {045} and {054}. Both of these are fair approximations to {011}, which does not conform to the rule. Even so, it has been also reported, though quite uncommonly.

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PostPosted: Oct 17, 2020 13:57    Post subject: Re: Very steep rhombohedral calcite?  

Amir Akhavan wrote:
The high indices as shown in the crystal drawing program are also a result of using the old a:c ratio of 1:0.854.
The new value that actually reflects the crystal structure according to new measurements is 1:3.419. (snip)


Both of these statements are correct, and we should probably use the unit cell with c:a=1:3.419, based on the structure of calcite. The older cell was derived from the morphology in the days before x-rays, and was based on the assumption or postulate that the unit cell shape was revealed by the cleavage rhombohedron, which was assigned the indices {111}. Because many drawings based on the old morphological cell, with associated Miller indices, are still in circulation (in fact probably more than are based on the structural cell), it is important to specify which cell is being used to make a particular drawing. Too often this does not happen.

The crystal drawing program Smorf allows the default parameters for a drawing to be changed, so while it offers the value of 0.8543 for the c axis, one can easily change it to 3.419, or enter the true unit cell constants of a=b=4.9896, c=17.061.

The effect of switching from the morphological cell to the structural cell, four times as long along the c axis, is to quadruple the value of the last Miller index. Thus {101} becomes {104}, etc. This does tame the indices of steep faces - {16 0 1} becomes {16 0 4}, which reduces to {401}, for example, but has the opposite effect for faces that are close to perpendicular to the c-axis. The nice tame and common rhombohedron {012} now becomes {018}, and the less common rhombohedron {104} becomes {1 0 16}.

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PostPosted: Oct 19, 2020 12:39    Post subject: Re: Very steep rhombohedral calcite?  

What an informative thread! Thanks to Josele, Amir, James, John, Bob, and Pete for taking the time to prepare such thoughtful replies to such interesting questions.

Your efforts are widely appreciated. Thanks, again.
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