W. J. (William James) Lewis.

A treatise on crystallography online

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Hence the second and third indices of the face are equal, for the
parameters are equal. But the axis OX will be met at a different
distance ; and on the same side of the origin as Y and 0Z y if the
face is less steeply inclined to the equatorial plane than the corre-
sponding face of the fundamental rhombohedron, i.e. if V m lies
nearer to than V. The axis of X will be met on the opposite
side of the origin, if V m is further away from than F. The face
is, therefore, (Jdl) where h and I have the same, or opposite signs

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according as the pole of the face is nearer to C, or more remote
from it, than r (100).

Let Fig. 340 represent part of a section of the rhombohedra mR
and R in the plane 2 containing OX and the polar edges
V m M, V t M. The coign A is on the line drawn parallel \
to OM through r,, where 0r t = V m -r 3. Also, V»k ■ \
is the polar face-diagonal and meets OX in L, OM \ \l,x
in B and V t M in L r Adopting V t M as parameter, y >P M
then 0L- Y t M+h. But, from the similar triangles :°^r
V,LJ m and 0LV m , we have r 'fc A

FA _ P,P m _ mc'+ c m + 1

OZ-OP™" mc " m * ; ' ^

From the similar triangles OLE and MBL ti we have Fig- 340.

since (Art. 31) 0£=BAf.

Adding (28) and (29), we have

'-^-^♦"^Jr <*»•

The face through V m \ of Fig. 340 meets the adjacent plane 2'
in the polar edge V m \ which, since OV m >OV,

must be prolonged to meet OY at K on the nega- ^

tive side of the origin. Fig. 341 represents part A

of a section of R and m/2 in 2'. But, since / :

YM t is parallel to OY, the triangles V m K0 and V/i

P w if P are similar : we therefore have l^r o\

y M yy™ _ mc-c _m-\ / •* \

OK~OV n ~ mc " m ( '' / .< \ I

In (31) the lengths are all regarded as posi- ^ \j

tive; but I is YM' + -0K. Hence, «

i^M'_JVM,J^-m Fl °- 341 -
OZ " OX " m '
The Millerian symbol (AZZ) is therefore given by the ratios:

* - * - JL /32V

2m+l "1-w 1-m W ''

and, if the Millerian symbol is known, the Naumanman symbol mR
is found from

-ra < 33 >-

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The same equations are found when 0V m is taken to be less
than OF.

36. Equations (32) and (33) can also be obtained from the
a.r. of the four poles {Cgrm}\
where C, Fig. 342, is the pole
(111), gWl), r(100) and 0?{2ll}.
For, in Fig. 338, OBV m is the
inclination to the pinakoid of the
face of mR through F m A, and is
therefore the angle Cg. Hence
tan Cg = tan OBV m =OV m + OB

Again, D = A Cr = t\OBV;
and UmD = t&nOBV=c + b.
Dividing the former tangent by
the latter, we have

t&nCg + UnD = m (34).

But, from the a.r. {Cgrm} of the four poles given above, we

sin Cg
sin Cr

sm mg
sin mr

tan Cg -r tan Cr =






' h + 2l


since mg = 90° - Cg, and mr = 90° - Cr.

The angles on the left sides of (34) and (35) being the same, the
terms on the right must be equal.


w =


h + 2l'

2m + 1 l-m'

These are the results given in (32) and (33).

The special forms given in Art. 34 are immediately deduced
from the above equations ; for, if m = 0, h = I, and the face is (III) or
(111). If m = oo, then h + 2l = 0; and the face is (2ll) or (211),
and the form is the hexagonal prism {2 II}.

1 A different type has here been used to represent the pole (2lT) to avoid
confusing it with the Naumannian index m.

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37. The Millerian symbol for the inverse rhombohedron — mR
is found from equations (32) by changing the sign of m: it can
also be obtained from the geometry of the figure as follows.

Let Fig. 343 be part of a section in the plane 2 of the
rhombohedra R and — mR) and let the polar face-
diagonal V m y meet OX in G and the lower polar edge
V t M of R in G r Then from the similar triangles
V m OG and V m VG,, we have

OG~OV m " mc ~ m K >'

Again, from the similar triangles OBG and MBG t>

we have


:. adding (36) and (37),

= 1


m — 1 , 2m
+ 1 =




Fig. 344 represents a section of the same rhombohedron in the
plane 2', where V ni M t is the polar edge of the face
through V m y of Fig. 343 and meets OY at F on
the negative side of the origin.

From the similar triangles V m FO, V m MV, we


VM f VV m= mc + c = m+l

OF 0V m mc m K h

Hence, the intercepts on the axes are OG, — OF,

-OF-, or


2m-:T 1+m ' 1+m "
If now the upper face parallel to that taken is
denoted by (hl f l), then the face through V m y is (hl} t )\ and the
intercepts it makes on the axes are in the ratios: —

h I




1 - 2m l+m'

m =



It is clear that these expressions differ from those for mR given

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in (32) and (33) only in the sign of m. Hence, expressions (32)
and (33) may be taken to apply to both cases, provided care is
taken to give the correct sign to m. Expression (33) serves also as
a ready test for determining whether a rhombohedron [hll] is a
direct or an inverse one. Thus, when the indices of {011} are
introduced in (33), we have for m the value — £; similarly, {Til}
gives m = — 2; and {311} gives to = 4.

It follows therefore that the simplest way of drawing any
rhombohedron, the Millerian symbol of which is given, is to find m
from equation (33). The apices V m and V m at distances mc from
the origin can then be marked off on the vertical axis; and the
drawing can be made in the manner described in Art. 31.

38. The symbols of the faces of the rhombohedron {100} are
given in table d of class II ; those of {hll}> which may be either a
direct or inverse rhombohedron, are :

hU Ihl llh III III Tlh (j).

These rhombohedra are geometrically common to classes II, III
and IV.

39. Since the median edges of rhombohedra and scalenohedra
(Art. 40) are inclined to the equatorial plane and cross it in zigzag
fashion, such forms can be quickly drawn in the following simple
manner. The eye is supposed to be situated in the equatorial
plane, and the triad axis to be in the paper : any horizontal plane
is then reduced to a straight line.

Two straight lines VOV, and FOF i9 Fig. 345, are first drawn at


Fio. 346.

right angles to one another. On FOF t any six equal lengths
OS=ST=TF=OS~ST = T t F t> are marked off. Lengths OK and OF,,

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equal to the distances of the rhombohedral apices from the origin
are determined on the vertical line, and the length intercepted
between the apices is then trisected in the points t and t r Through
the points of trisection lines are drawn parallel to FOF t to inter-
sect the verticals through T, F 9 &c, in the coigns /a, /x„, <fcc.
If OV=c, and Vfi /t , Vfi /y &c., are drawn as in the figure, we obtain
the fundamental rhombohedron jR={100}. To draw the rhombo-
hedron mR, we take 0V m = 0V m =mc> trisect V m V m in t and t,,
and join the apices V mi V m to the coigns A, X t/9 &c, in which the
horizontal lines through t, and t meet the verticals through, T, F, <fec,
the median coigns being taken in the same order as that adopted for
R. The inverse rhombohedron - R = {T22} is given by joining V
to the points of intersection of the horizontal line t^ tl with the
verticals through S, T &nd F /9 and V t to the median coigns on the
verticals through T t > S t , F: the figure is then easily completed.
The rhombohedron - mR is obtained in a similar manner.

All that is now needed to draw any scalenohedron mRn (Art.
40) is to take F n , V n on the triad axis at distances mnc from the
origin, and join them to the median coigns A of mR.

To find the unit of length OA" corresponding to the arbitrary
length OS, a circle is described with centre at S, and radius = SS t :
the point A where the circle cuts the vertical /x^,A is at unit
distance from ; and OA" = OA. The length V = c for any par-
ticular substance is therefore OA cos 30° tan D.

Since the horizontal plane is reduced to a line the method is not
adapted to represent crystals in which lines parallel to the equatorial
plane, but inclined to one another, have to be shown.

We proceed to indicate the positions of the planes of symmetry in such
a projection as Fig. 345, and to prove that 0A is the unit length on V.

Suppose the rhombohedron represented in Fig. 345 to be turned
through 90° about the line F0F n and to be projected on the paper: it
will then be given by Fig. 346. During the rotation the coigns remain
in the vertical planes through the points S, T, F, &c. ; and the parallel
vertical plane through the eye and centre may be represented by EE f .
The dyad axes (of which Ob is alone shown) now lie in the paper, and
bisect pairs of opposite median edges.

Let EOn=6; then ^^=60°-^, and Fii„0=n„0E,=60 o +e.

From the right-angled triangle SOfi, we have
08=0fi' sin B;
from the triangle T0& 0T= Op sin (60° - B) ;
and from the triangle F0p tt , 0F= 0f*„ sin (60° + B).

L. c. 25

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But Op=Ofi'=Oti„;

and ST= OT- OS= Ofi {sin (60° - 6) - sin 3} ;

TF= OF- 0T= Op {sin (60° + 3) - sin (60° - 3)}.

Also TF=ST=0S.

.-. sin(60 o + ^)-sin(60°-^)=sin(60°-^)-sin^=sin^...(42).

The first and last terms of (42) are equal for any value of 3, for
sin(6O°+0)-sin(6O o -0)=2cos6O o sin0=sin0;
since 2 cos 60°= 1.

The value of 3, corresponding to the orientation of the rhombohedron
in the two figures, has therefore to be found by combining the first and
second terms, or the second and third. Taking the first pair of sides in
(42), we have

sin(60 o +^)+sin^=2sin(60 o -^)=2cos(30°+^);
since 60° - 6 = 90° - (30° + 3).

sin(0O o +^)+sin^=sin(3O o +^+3O o )+sin(3O o +^-3O°)

=2 sin (3O°+0) cos 30°.
.-. 2sin(3O o +0)co83O o =2cos(3O°+0);

and cot (3O°+0)= cos 30° (43).

Therefore, by computation,

30° + 0=49° 6-4'; and 0=19° 64'.

Hence, in Fig. 345, the plane 2,
containing V,p and the parallel axis OX,
is inclined to the paper at an angle of
49° 6*4' to the right front ; the planes 2,
and 2„ containing OF and OZ respec-
tively, are inclined, to the left front at
angles of 70° 536' and 10° 6Z& respec-

We have now to determine the unit
length OA" on the vertical axis VV t of
Fig. 345 which corresponds to the arbi-
trary length OS taken on the horizontal

line. In Fig. 346 Ob is a dyad axis, and its length is that denoted
by a. But

.-. 7 T / d=7 T / 0cot7 T / d0=!7 T / 0cot(30 o +^)=(from(43)) 7 T / 0cos30°=V37 T / 0-r2.

Also, S,T = T,0+2;

.-. S t V~T t V+T t S*=T t 0*(Z+\)±l=T t 0*.
.-. S,B=T,0=S t S=S t F t .

Hence d lies on a circle described with centre at S, and radius =» SS, « 20S.
To obtain 0A" in Fig. 345, a circle is described with centre at S t and
radius =20S: the point A, in which the circle cuts the vertical through T n
is at distance a from 0. OA" is then cut off equal to 0A ; and V is found
by multiplying this length by c, or cos 30° tan D.

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The acalenohedron, {hkl\.

40. By the law of rational indices, faces belonging to a crystal
can be drawn through an edge of any of its forms to meet any non-
parallel zone-axis at distances which are commensurable multiples
of the intercept made on this axis by one of the faces. The form
through the edge of which the new faces have to be drawn will be
called the avanlicury form. Thus a pair of possible faces of a
crystal of this class can be drawn through a median edge of any
rhombohedron mR to meet the triad axis at points V nt V n 9 where
V n = V n = nO V m . But by the character of the symmetry, similar
pairs of faces must be drawn through each of the remaining median
edges. Since the points V n and V n on the triad axis are equally
distant from the centre of the rhombohedron and opposite median
edges are parallel, it follows that the form, Fig. 347, has twelve
faces arranged in pairs which are parallel. Further, each face is
a scalene triangle having its polar edges in two of the planes of
symmetry 2, 2,, 2„; and the
angles over the three edges are
unequal. The form is therefore
called the scalenohedron of the
rhombohedral system; and as
its faces will be shown to meet
the Millerian axes of reference
at unequal distances, its symbol
will be {hkl\. The symbol mRn
was used by Naumann to denote
the form. In this symbol the
number m preceding R indicates
the auxiliary rhombohedron, the
median coigns A, A,, <fec, of which
have to be joined to V* and V n .
The number n following R in-
dicates the multiple of 0V m re-
quired to give 0V n , which is
therefore mnc.

The above method of derivation of the general form of the class
is similar to that by which the tetrakis-hexahedron {hkO} of the
cubic system was, in Chap, xiv., Art. 18, derived from the cube.
It affords a ready means of drawing the form when the Naumannian
indices m and n have been determined.


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From the method of construction it follows that the median
coigns of the scalenohedron lie in the horizontal planes trisecting
at t and r t the length V m V m intercepted between the apices of the
auxiliary rhombohedron mR. Since the polar edges of all rhombo-
hedra join the apices to the points M in the equatorial plane, then,
(Art. 33), t,A = t,A" = <fcc. = 20Jf -=- 3 = 4acos 30°-r 3, a constant
distance independent of the values of m and n. Hence the median
coigns of all scalenohedra lie in the vertical edges of the prism {OTl},
when its faces are drawn through the points A, A\ A" y <fcc.; OA
being the unit of length a on the dyad axis in Fig. 332.

41. The limits of the series of scalenohedra having the same
median edges as the auxiliary rhombohedron mR are given by
making: (i) n = l, when the scalenohedron becomes identical with
mR itself 3 (ii) n = oo , when all the faces become parallel to the
triad axis. The form then becomes the hexagonal prism, the faces
of which truncate the median edges of the rhombohedron, and are
therefore perpendicular each to one of the dyad axes : this prism
was shown in Art. 27 to be {OTl}.

42. We proceed to find the intercepts made by the faces on
the axes of . reference ; and hence to determine the connection
between the Millerian symbol {hM\ and the Naumannian symbol
mRn. Incidentally, expressions will be obtained which will enable
us to transform the above symbols to those involving four axes of
reference. Rhombohedral crystals are
still frequently referred to a set of four
axes ; but such axes are better suited to
represent the forms of hexagonal crystals,
and their discussion will be postponed to
the next chapter.

Let V*M t be the face (hlk). In
Fig. 348 two faces and the axes OX and
07 are alone shown. The polar edges
F*A, V n \ lie in the plane 2 and meet
the axis OX in the points S and £', and
the horizontal line OM in the points H
and H'. The similar polar edges V n X /9
V n \ lie in 2,, and meet 7 in the points
T and T ti and the horizontal line 0M t in

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K and K r The median edge A\ meets the dyad axis at A tn where
OA = a. and A X = A X .

By a semi-revolution about 0A lt the two faces change places,
the edges V n \ t and V % \ changing places, also V n \ and V n \.
Hence, OT t on OY t is equal to 0£' measured on OX> and like-
wise OT - OS. Furthermore, since the same rotation interchanges
OM and OM lf we have 0K= OH, and 0K t = OH'.

Hence, to find the line HA tl K n in which the face V n Xk t meets
the equatorial plane, it suffices to determine the points H and H'
in which the edges F*X, V n \ meet OM. The points S and S',
and the intercepts OS and 02 7 , are then easily found.

Let Fig. 349 represent a section of the scalenohedron and
of the rhombohedron mR by the plane 2, which contains the polar
edges V n Xj F m X, and OX parallel to the edge V ,M of the fundamen
tal rhombohedron. Now 0r t = V m h- 3 ■= mc -f- 3, and t, k = 20 M -r- 3.

From the similar triangles OHV* and
t\V*) we have

mnc + mc + 3 Sn + l /AAX



OM 3w+l
Off" 2n



From the similar triangles 0H'V n ,

r\V n> we have

t,A _ t, V n __ mnc — mc-r3_3n— 1 , . „.

0H'~0V n ~~ ^ " ~3^~ W;

. Oif 3n-l
" OH'' 2n '"


Fig. 349.

The trace of the face in the equatorial plane may be taken to
coincide with the line HKL t of Fig. 332, p. 372. The distance
0L t on OM" is then given by the fact that the triangle K t 0L t is
made up of K t 0H and H0L /9 of which the angles at are 120°
and 60°. Hence,

OK . OL sin 120° = 0K 4 . OH sin 60° + OH. OL sin 60° ;
. OM OM OM 3w+l 3n-l 1

: ( 48 );

0Z " OH OK,



since OK = Otf'.

43. To find A, £, Z in terms of m and n.

The Millerian indices A, Z, £, are given by h = V t M •*- 0$,

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1= V e M' + 0T, k= VM"+0U\ if, as before, VM is the para-
meter, and OS, 0T ( and OU are the intercepts on the axes, due
attention being paid to the directions in which they are measured.
By a semi-revolution about OA /t the polar
edge V n \T, Fig. 348, is brought to the
position V n XH f which meets OX in S\
The intercept 0T t , measured in the nega-
tive direction along OY t is equal to OS'
measured on OX. Hence

l = -V,M + 0S'.

Since OX is parallel to V t M, we can,
by pairs of similar triangles, find V^M-r- OS
and V t M± OS', and therefore the indices
h and L Let, in Fig. 349, / be the point
of -intersection of V t M and V n \, and f
that of V M and V n k. From the similar
triangles OHS, MHf, we have

Fio. 349.

fM HM OM-OH />( Ku3n+l f n + 1 ,, QX

'__- _ __._ = (from (45))-^— -1= ^ ...(49).

0S~ OH



And from the similar triangles OtfF*, V t fV n , we have

r / _ F, V* _ mnc + c _ wn + 1

Off "OF* mnc " mn


Adding (49) and (50), we find

, _ V t M _n+\ mn + 1 3mw + m + 2
0$ 2ro mn 2ran


Again, from the similar triangles H'f'M, H'OS', we have
fM H'M OM-OH' /17u 3n-l n-1

and from the similar triangles S'O V n , f V t V n ,

VJ' __V f V % _ mnc-c _ mn-1
OS' ~~ 0V n mnc mn

\ adding (52) and (53), we have
. F M _ n - 1 mn -

1 3mn - m — 2






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Again, if, in Fig. 350, V % L t is the trace of (hlk) in the
plane 2„, and if it meets OZ and
V t M" produced in U and G, then
from the similar triangles UOL /y

M"G M"L _ OM"
OL ~ OL.


+ 1


= (from (48)) - + 1


And from the similar triangles
V n 0U, V % V t G, we have

V t G F V* _ mn+\
OU " OV» ~ mn

Fig. 360.


Subtracting (56) from (55), we have

M"V t \ mn+l _2(m-l)

OU n mn 2mn


n mn

The above equation gives the numerical value of the ratio
V t M u + OU. Since, however, OU fa measured in the negative
direction, the index

k = -M"r,+ OU=2(l-m) + 2mn (58).

Hence from (51), (54) and (58),

* = L = *__ ... . .(59)

m + 2 + Smn m+2-3mn 2(l-m) * '*

44. Equations (59) suffice for finding both m and n when h 9
k and I are given. For, if we add or subtract the numerators, and
also the denominators, of any of the ratios, we obtain a ratio equal
to any one of them. Thus, taking the first and second terms, we
have by addition

A 4- 7 1c

— trr = rt jz v = (doubling the terms of the latter and sub-

2(n» + 2) 2(1 -m) x °

tracting from the former) — « = (by addition)

m —

h + l-2k

h+k+l "

Again, subtracting the numerator and denominator of the second

member of (59) from those of the first, we have

h-l k , , . .h + k + l

= (from above)

§mn 2(l-w»)"




h + l-2k'


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Knowing then the indices h> k, I of one face of the scalenohedron,
we can find m and n ; and can then draw the form by the method
given in Art. 40.

45. To find the symbols of all the faces of {hkl}.

The pair of faces passing through AA of Fig. 351 change places
after a semi-revolution about Oh„. This axis, being perpendicular
to OZj interchanges positive with equal negative lengths on it ; and
positive lengths on OX with equal negative ones on 0Y t and vice
versa. Hence, the face F*AA being (hlk), the face V n W, is (Ihl).
This is also obvious from the discussion in the two last articles
and Fig. 348.

The above two faces are associated with two parallel faces
drawn through the median edge parallel to AX,. The symbols of
the latter faces are, therefore, (hll) and (Ihk). The four faces are
necessarily tautozonal with the four rhombohedral faces meeting in
the same median edges, and with the two prism-faces (110) and
(110) truncating these edges.

Again, by rotations of 120° about the triad axis the above four
faces are brought into the positions of two other sets of four similar
faces. We have already seen that the triads of interchangeable
faces have their symbols in the
same cyclical order; hence the
faces of the scalenohedron {hkl},
Fig. 351, taken in order from
P*XX, and P*XX , have the fol-
lowing symbols:
hlk hkl khl Ihk Ikh klh\
Ihk III klh III hkl khl)


The pairs of faces in the columns
are interchangeable by semi-revo-
lutions about the dyad axes 8„,
8, and 8, respectively.

It is easy to see that the
above faces are symmetrically
placed with respect to the planes
of symmetry. Thus the pairs of
faces meeting in the edges P*X,
V n \ meet OX in S and £' respectively, where OS = V 4 M+h>

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0S,= V 4 M-r-l, for I is a negative number (Art. 43). But the
plane 2 bisects the angle between OY and 0Z> and the intercepts
on these axes are reciprocal reflexions. The face P*AA being (Mk),
the symbol of V n \\„ isjhkl); and the face V n te, being (Ihk) the
symbol of V n W M is (Ikh). Similar proofs can be applied to the
pairs which meet in polar edges V % \ , <fcc. \ and also to pairs of faces
which do not meet in edges, such as V n \\" (Jdh) and V n \'\, (khl).

The relations of the poles P of the scalenohedron {hkl} = mRn
g those of the inscribed

Fio. 352.


and of

auxiliary rhombohedron mR, are
shown in Fig. 352. The two
faces of each form which meet
in the edge XX, must be in a zone
with a^(llO), the prism-face per-
pendicular to the dyad axis Oh„' }
and this face would, if developed,
truncate the edge. It is clear,
however, that the two faces of
the scalenohedron make a smaller
angle with (110) than the two
faces of the rhombohedron g. Hence the pole P' lies between a H
and g. If P' is (KUc) and g in [a„P'] is (h f ll) ; then, by Weiss's

hk + l 4 h-l t (h + l) = 0.

This equation is satisfied by making h, = h — k + l, and l t = k. The
inscribed rhombohedron mR has therefore the symbols {h-k + l y
k, k}. But, in Art. 36, it was shown that, if mR is identical with
{hi I}, then m = (A — I) + (A + 21). Introducing into this expression
the values of h t and l t just found, we have

h-k + l- k _ h-2k + l _0-3k
h-k + l+2k ~ h + k + l "" ~0 " '
the same result as is given in (60).

The number n can now be found from equations (59).

47. Although, in the method of derivation employed in
Art. 40 and in all the succeeding Articles, the number m has been
supposed to be positive, the process is perfectly general ; and all the
relations hold true if m is negative and the inscribed rhombohedron
is the inverse form, —mR. All that is necessary is to make m
negative in equations (59) and in all equations into which m enters.

m =

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Such a scalenohedron may, when it is desired to indicate its precise
position with respect to the axes
of reference, be called an inverse
scalenohedron, — mRn.

It is clear therefore that,
for the same numerical values
of m and n, we have two tauto-
morphous scalenohedra, the posi-
tions of which in space are deter-
mined by the sign of m; and the
poles of which are shown in
Fig. 353.

Online LibraryW. J. (William James) LewisA treatise on crystallography → online text (page 34 of 53)