This is a good question.

Angles between lattice planes are investigated in

O. Karpenkov, Geometry of lattice angles, polygons, and cones, Habilitation thesis, Technische Universitaet Graz, 2009:

http://www.geometrie.tugraz.at/karpenkov/hab/habilitation.pdf

See Theorem 2.2, p. 5, with an illustration on p. 6.   For an overview of properties of lattice trigonometric functions, see Section 1.1, starting on page 164.    The proof of the first part of Theorem 2.2 is given, starting on page 191 and the proof of the second part of Theorem 2.2 is given, starting on page 194.    This hibilitation document also includes an article on irrational lattice angles (attached as an appendix).

For me, the most interesting part of this habilitation document is the appended article:

O. Karpenkov, Classification of lattice-regular lattice convex polytopes, arXiv 2006.

See, also, additional papers by O. Karpendov at

https://www.researchgate.net/profile/Oleg_Karpenkov

Many thanks, James. I am not sure wheter I can tranfer this into a "language" I am able to follow :-) but I will see how far I will come. Finally I can still contact O. Karpenkov.

Dear Gert;  ı concentrate on the reciprocal lattice representation,  where every triplet formed by the permutations  of  plus and  minus  hkl  numbers corresponds to a distinct  point  in  3D- space.  In the present case this is equal to =  6! / 3!  = 120 points.  The number of combinations of them into doublet is  equal to      (120-1)! =119!  if  we don't worry about the plus minus directorial character of these  angles  then we have; 

[6!/3! -1]!  /2 = 119!/2=  5.574586...  e196  /2   ?  EQD  

Dear Tarik, if you consider a general plane given by {hkl} the mximum number of permutation results in 48 possible vectors. From each vector you can now determine the angles, i.e. there are 48 angles (inclusive 0° for itself). Because of symmetry you don't have to do this for each vector so that these 48 angles are the maximum number. However, also because of symmetry the number of different angles is smaller if the lattice is high symmetric and the combination of h, k and l is somehow specific. So, for cubic symmetry and for {110} you will only find 0, 60, 90, 120, and 180. If one considers complementary angles as equivalent the number reduces to the obvious 0°, 60° and 90°, i.e. practically only 60° and 90° are listet in tables.

My question is, how many angles results for any kin of {hkl}? Using cubic symmetry (or an orthonormal lattice), for {hk0} we have usually 6 angles (after "reduction" as written above), but for {h 2h 0} as special case og {hk0} there are only 5 different angles. A similar effect appears for {hhl}. The number of different angles is 5 whereas for {hh 3h} there are only 4 different angles. My question is, is there a formular which enables to predict this number of different angles?    

It becomes perhaps more challenging for planes of different {hkl}1 and {hkl}2.

Dear Gert,   finally  after reading your last comments I have anticipated that you are interested in the number of distinct angles formed as a subset of the whole possible angles formed  out of the given set  of {hkl}  planes of form. I was only trying to  figure out that number for the cubic system first. 

We should not forget the plane of form is defined as a group of planes which are connected by the symmetry elements (operations) associated for a given lattice system that determined by the minimum symmetry elements.   That means hkl set also includes negative counter parts of   hkl for cubic systems,  hk only for the tetragonal sistem, etc. Also I know that reciprocal lattice of  FCC  is  BCC or visa versa.

Looking the reciprocal lattice, one can easily see that   (110) :101 =60, (110 ):(-110)=90;  (110): (-101)=120;  (110):(-1-10)=180  EQD  etc.

What does EQD mean? 

Dear Gert;   EQD:   This is a mathematical terminology   extensively used by physicists  as a short hand expression in the class rooms  of USA  Universities  that means  'Quoted Equation  Derived' . Regards

Funny. If I "google" it, it never appears as hit. But you mixed up the letters: QED and not EQD, right? I only know QED as latin terms: Quod erat demonstrandum.

Dear Gert;  You are right  QED  is the proper abbreviations.  I have a habit to change the things around little bit.  Don't forget  Latin 'erat'  means anything to be proven, and it doesn't imply mathematical equation or relationship what soever.

Google entry:  Erat translates from Latin to English as he is, she is, or it is.  

Coming back to my original question: Do you know any general equation how I can determine the number of different angles between symmetry equivalent directions (in reciprocal space)? 

Dear Gert,  I don't  know any one, who has  worked out with that problem.  But one may find relationships such as  by working out with  the cubic, tetragonal and orthorhombic systems using the fact that reciprocal basic vector set is an orthogonal three-pot vectors set.  Regards

Tetragonal and orthorhombic can be challenging as well because of the arbitrary c/a ratio, but already for cubic would be fine if it could be explainable. For all other the orthogonality doesn't make a big difference (as far as I understand the problem until now).

For cubic it is a simple questions. nevertheless the solution is not easy to me. The question is easy: h1h2+k1k2+l1l2=const. for two permutations of h2k2l2  where the numbers are the same but have to be exchanged and perhaps multiplied by -1. I only have no clever mathematic describtion for an equation.

The combination of N elements in groups of K is: C(N,K) = [N(N-1)(N-2)...(N-K+1)]/K!

Consequently if N is the number of symmetry-equivalent (hkl) planes the number of angles between pairs of planes (K =2) is C(N,2) = [N(N-1)]/2

I think, however, that you mean the number of DIFFERENT angles. I have not an answer in this case, but one must take into account that each angle is repeated M times, if M is the multiplicity of the symmetry operations.

You are right with M, but this I already excluded when I only take one (hkl) as reference direction and calculate the angles what finally tells you that for m-3m 48 possible solutions can appear. However, since I don't distingiush between phi and 180°-phi, the maximum number reduces to 24, actually 23 since 0° (or 180°) are trivial solutions. The present result is a bit surprizing: usually a general {hkl} like {239} generates 16 angles, {238} and {235} generate 15, {236} and {234} give 14 different angles. It might be that some of the angles are very similar so that my current graphic solution is not really exact (although I am actually interested in an practical application so that only the practically differentiatable angles matter).       

Only as example: let's use {236}. Then (236) has the same angle to (623), (362), (3-26) and (-326) since in all cases 2*h+3*k+6*l=36.

Dear Gert;  I obtained a formula for the multiplicity factor cubic xtals assuming that  hkl are distinct numbers. The formulas for other five special cases  of  hkl indices will be reported due time.

3!  for  p(hkl) +  3!  for  p(h- k- l-)  + 3! for p(h- kl) + 3! for  p(hk- l) + 3! for p(hk l-)

+ 3!for  p(h-k- l)  +3! for p(hk-l-)  + 3! for p(h- k l-)  =  3! x 8 = 48   where   p  is the permutation operator.   Best regards

Many thanks, Tarik. I tried it yesterday as well but did not come to a final answer. i hope your equation will help me (or solve already the entire problem).  

Dear Gert,  This is the results I obtained. Now I can work with angular connections between the planes for a given form using symmetry operations which define the form in the reciprocal space in my mind.  For the high Miller indices even though there is no problem in calculation but the selection of unique angles is cumbersome!!!

Zhhl =3!/2! for p(hhl) + 3!/2! for p(h- h- l-) + 3! /2! for p(h- hl) + 3! /2!p(hh -l) + 3!/2! p(hh l-)+ 3! /2!p(h-h- l) +3!/2! p(hh-l-) + 3!/2! p(h- h l-) = 3!/2! x 8 = 24 where p is the permutation operator.

Z0kl= 3! for p(0kl) + 3! for p(0 k- l) + 3! for p(0 k l-) + 3! p(0k- l-)  = 3! x 4 = 24

Z0kk= 3!/2! for p(0kk) + 3!/2! for p(0 k- k-) + 3!/2 for p(0 k- k) + 3! p(0k k-) = 3! /2!x 4 = 12

Zhhh =3!/3! for p(hhh) + 3!/3! for p(h- h- h-) + 3! /3! for p(h- hh) + 3! /3!p(hh -h) + 3!/3! p(hh h-)+ 3! /3!p(h-h- h) +3!/3! p(hh-h-) + 3!/3! p(h- h h-) = 8

Z00l=3!/2! for p(00l) + 3!/2! for p(0 0l-) = 3!/2!x2=6    QED

Best Regards,  and Merry Christmas to ALL

Dear GERT; Take Z [001)} there are 3 four fold axis. [100], [010], [001]. Each zone axis has 4 {100} planes but only the six out of twelve planes are distinct as one expects. That number appears as multiplicity as reported previously. Here one has following distinct angles for each zone axis. {90,180,270,360} the angle between two planes belonging two different zones is either equal to zero or ninety degrees. This is due to the fact every zone shares its two members with another zone.

Z {011} = There 3 four fold axis [100], [010], [001]. Each zone axis has 4 planes {1+-1+-0}, {1+-01+-}, {01+-1+-} all these planes are distinct and multiplicity is 12. The angles between planes in same zone is 90o and distinct angles are {90,180,270,360}. The angles between planes belonging different zone are

{60, 120, 240,300} where cos (theta) = +-1/2.

Z{111} Here again 3 four fold axis [100],[010],[001, each axis has 4 planes {1+-1+-1+-}, {1+-1+-1+-},{1+-1+-1+-}. Out of 12 planes there are only 8 distinct planes as given previously. The angle between two planes in a given zone is arcos (1/3) =70.5288 regardless the zone axis but the angle between two planes belonging two different zone axis either is 70.5288 or arcos (-1/3) =109.4712.

Actually systematic fashion one can calculate not only the angular connections between distinct planes but also the set of distinct angles.

One can also make the following rather trivial statement; the angles theta between successive planes belongs to the same zone axis are all equal and that doesn't change for different zone axis as long as they belong to same symmetry operation group. But for the angular difference between planes belonging different zone takes either theta or the complimentary part (180 - theta). Regards

NOTE:  All this shows that in cubic systems up to now only the   3- four fold symmetry elements  play  major  role in dictating the angular difference between planes  belonging to different zone axis as long as they belong to same symmetry group.