<div dir="ltr"><div>If you give positions with respect to hexagonal axis, you need to set "rhombohedral=.false.". In this way you obtain 2 Ba, 2 Ti, 6 O, which is correct for the rhombohedral (also known as trigonal) cell.<br><br></div>Note that the celldm parameters (or A and C) should be those for the rhombohedral latttice, not for the hexagonal one.<br><br><br></div><div class="gmail_extra"><br><div class="gmail_quote">On Tue, Sep 19, 2017 at 8:16 AM, Amar Singh <span dir="ltr"><<a href="mailto:amarsingh122014@rediffmail.com" target="_blank">amarsingh122014@rediffmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><span class="">Thanks for the suggestion <i>Paolo Giannozzi. </i><br></span>Removing ibrav working fine partially. Space group R3C (no. 161) itself generated six equivalent positions for Ba and Ti, which is correct according to the known structure. Though, even generated O atoms are six in numbers, which should be 18. <span class=""><br>I also confirmed the positions of all atoms by another software (Diamond) and it is able to generate 18 O atoms along with six Ba & Ti. Would be thankful for any suggestion on that.<br></span>Following the section of script I am using.<br><br> &system<span class=""><br> celldm(1) = 10.578,<br> celldm(3) = 2.475,<br> nat = 3,<br> ntyp = 3,<br> ecutwfc = 40<br> space_group = 161<br><br>ATOMIC_POSITIONS crystal_sg<br>Ba 0.00000 0.00000 0.26640 1 1 1<br>Ti 0.00000 0.00000 0.01568 1 1 1<br>O 0.1338 0.3363 0.08333 1 1 1<br><br></span>thanks<span class="HOEnZb"><font color="#888888"><br>AS<br></font></span><br>______________________________<wbr>_________________<br>
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