Hi
Just a general warning about using Excel with trigonometric functions:
=sin(0) gives 0
=sin(90) gives 0.893997
=sin(180) gives -0.80115
In other words, I haven't gotten sine and cosine to work reliably with
Excel. It might be a good idea to check the calculations using some
other software, maybe just a good old hand calculator.
That was my five cents. :)
/Arto
On 2014-02-11 10:13, "Łukasz Kruszewski" wrote:
Ok - sorry, changed degrees into radians. But there is still a difference
for sulphur in aluminite.
Best regards,
Luke K.
OK, here it goes:
Uiso = 1/3 * [U22 + 1/sin^2(beta)*(U11 + U33 + 2U13cos(beta)] =
1/3 * [0.01562 + (1/sin^2(110.18))*(0.01937 + 0.01277 + 2*
(-0.00027)*cos(110.18)] =
1/3 * [0.01562 + (1/(0.9386)^2) * (0.03214 + (-0.00054)*(-0.345)] =
1/3 * [0.01562 + (1/0.88097) * 0.0323263] =
1/3 * [0.01562 + 1.135 * 0.0323263] =
1/3 * (0.01562 + 0.03669) = 1/3 * 0.05231 = 0.01726
The given Uiso for sulphur is, meanwhile, 0.01621.
I have also been trying to calculate Uiso for melanophlogite, which is
cubic, co the formula goes:
Uiso = 1/3* (U11 + U22 + U33).
Using this formula, I have obtained the value of 0.0339, which is exactly
the same as the one given in the corresponding CIF file.
However, calculating the Uiso using the GENERAL formula gives the value of
0.0287. Both values (0.0339 and 0.0287) were calculated using an EXCEL
spread sheet. So how is such a difference possible?
Best regards!
Luke Kruszewski
Dear Łukasz,
There must be something wrong with your calculation. A quick examination
shows that S must have a higher displacement parameter than Al and that
Ueq must be approximately 0.016.
I hope this helps.
Bob Gould
On 10/02/2014 21:50, "Łukasz Kruszewski" wrote:
Dear Rietveld friends,
I'm having some problems with calculating Uiso from the anisotropic
parameters. Here is an example - I copy first two sites of aluminite
(monoclinic):
_atom_site_aniso_label
_atom_site_aniso_U_11
_atom_site_aniso_U_22
_atom_site_aniso_U_33
_atom_site_aniso_U_12
_atom_site_aniso_U_13
_atom_site_aniso_U_23
S 0.01937 0.01562 0.01277 -0.00248 -0.00027 0.00954
Al1 0.01117 0.01169 0.01295 0.00006 -0.00012 0.00789
Using the formula for Uiso for the monoclinic system, which is:
1/3 * [U22 + 1/sin^2(beta)*(U11 + U33 + 2U13cos(beta)]
for the "S" site we obtain the value of 0,0054
for "Al": 0.01225
Meanwhile, the listed Uiso are:
_atom_site_label
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
_atom_site_U_iso_or_equiv
S 0.70076 0.37376 0.93018 0.01621
Al1 0.65581 0.45332 0.47633 0.01229
While the author-given value for Al is close to the one calculated
myself,
the one for S is much different. By the way, 0.01621 * 1/3 = 0.0054.
I could find it a simple mistake, I've found such a problem in case of
many other structures. And here's my kind question: am I missing
something?
Best regards!
--
Robert Gould
Tel.UK: +44 (0)131 667 7230 or +44 (0)796 040 3872
Canada: +1 519 387 8223
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