AFLOW Prototype: AB3_hP24_149_acgi_3l-001
This structure originally had the label AB3_hP24_149_acgi_3l. Calls to that address will be redirected here.
If you are using this page, please cite:
D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comp. Mat. Sci. 161, S1-S1011 (2019). (doi=10.1016/j.commatsci.2018.10.043)
Links to this page
https://aflow.org/p/9DMX
or
https://aflow.org/p/AB3_hP24_149_acgi_3l-001
or
PDF Version
Prototype | OTi$_{3}$ |
AFLOW prototype label | AB3_hP24_149_acgi_3l-001 |
ICSD | 36055 |
Pearson symbol | hP24 |
Space group number | 149 |
Space group symbol | $P312$ |
AFLOW prototype command |
aflow --proto=AB3_hP24_149_acgi_3l-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}, \allowbreak x_{7}, \allowbreak y_{7}, \allowbreak z_{7}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | O I |
$\mathbf{B_{2}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}$ | (1c) | O II |
$\mathbf{B_{3}}$ | = | $z_{3} \, \mathbf{a}_{3}$ | = | $c z_{3} \,\mathbf{\hat{z}}$ | (2g) | O III |
$\mathbf{B_{4}}$ | = | $- z_{3} \, \mathbf{a}_{3}$ | = | $- c z_{3} \,\mathbf{\hat{z}}$ | (2g) | O III |
$\mathbf{B_{5}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (2i) | O IV |
$\mathbf{B_{6}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (2i) | O IV |
$\mathbf{B_{7}}$ | = | $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6l) | Ti I |
$\mathbf{B_{8}}$ | = | $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6l) | Ti I |
$\mathbf{B_{9}}$ | = | $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6l) | Ti I |
$\mathbf{B_{10}}$ | = | $- y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6l) | Ti I |
$\mathbf{B_{11}}$ | = | $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6l) | Ti I |
$\mathbf{B_{12}}$ | = | $x_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6l) | Ti I |
$\mathbf{B_{13}}$ | = | $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6l) | Ti II |
$\mathbf{B_{14}}$ | = | $- y_{6} \, \mathbf{a}_{1}+\left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{6} - 2 y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6l) | Ti II |
$\mathbf{B_{15}}$ | = | $- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6l) | Ti II |
$\mathbf{B_{16}}$ | = | $- y_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6l) | Ti II |
$\mathbf{B_{17}}$ | = | $- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{6} + 2 y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6l) | Ti II |
$\mathbf{B_{18}}$ | = | $x_{6} \, \mathbf{a}_{1}+\left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6l) | Ti II |
$\mathbf{B_{19}}$ | = | $x_{7} \, \mathbf{a}_{1}+y_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{7} + y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{7} - y_{7}\right) \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (6l) | Ti III |
$\mathbf{B_{20}}$ | = | $- y_{7} \, \mathbf{a}_{1}+\left(x_{7} - y_{7}\right) \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{7} - 2 y_{7}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{7} \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (6l) | Ti III |
$\mathbf{B_{21}}$ | = | $- \left(x_{7} - y_{7}\right) \, \mathbf{a}_{1}- x_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{7} - y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{7} \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (6l) | Ti III |
$\mathbf{B_{22}}$ | = | $- y_{7} \, \mathbf{a}_{1}- x_{7} \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{7} + y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{7} - y_{7}\right) \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ | (6l) | Ti III |
$\mathbf{B_{23}}$ | = | $- \left(x_{7} - y_{7}\right) \, \mathbf{a}_{1}+y_{7} \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{7} + 2 y_{7}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{7} \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ | (6l) | Ti III |
$\mathbf{B_{24}}$ | = | $x_{7} \, \mathbf{a}_{1}+\left(x_{7} - y_{7}\right) \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{7} - y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{7} \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ | (6l) | Ti III |