AFLOW Prototype: AB3_mC32_8_4a_12a-001
This structure originally had the label AB3_mC32_8_4a_12a. Calls to that address will be redirected here.
If you are using this page, please cite:
D. Hicks, M.J. Mehl, M. Esters, C. Oses, O. Levy, G.L.W. Hart, C. Toher, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 3, Comp. Mat. Sci. 199, 110450 (2021). (doi=10.1016/j.commatsci.2021.110450)
Links to this page
https://aflow.org/p/QVN8
or
https://aflow.org/p/AB3_mC32_8_4a_12a-001
or
PDF Version
Prototype | TaTi$_{3}$ |
AFLOW prototype label | AB3_mC32_8_4a_12a-001 |
ICSD | none |
Pearson symbol | mC32 |
Space group number | 8 |
Space group symbol | $Cm$ |
AFLOW prototype command |
aflow --proto=AB3_mC32_8_4a_12a-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak \beta, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak z_{6}, \allowbreak x_{7}, \allowbreak z_{7}, \allowbreak x_{8}, \allowbreak z_{8}, \allowbreak x_{9}, \allowbreak z_{9}, \allowbreak x_{10}, \allowbreak z_{10}, \allowbreak x_{11}, \allowbreak z_{11}, \allowbreak x_{12}, \allowbreak z_{12}, \allowbreak x_{13}, \allowbreak z_{13}, \allowbreak x_{14}, \allowbreak z_{14}, \allowbreak x_{15}, \allowbreak z_{15}, \allowbreak x_{16}, \allowbreak z_{16}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}+x_{1} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ | = | $\left(a x_{1} + c z_{1} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{1} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ta I |
$\mathbf{B_{2}}$ | = | $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\left(a x_{2} + c z_{2} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{2} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ta II |
$\mathbf{B_{3}}$ | = | $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\left(a x_{3} + c z_{3} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{3} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ta III |
$\mathbf{B_{4}}$ | = | $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\left(a x_{4} + c z_{4} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{4} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ta IV |
$\mathbf{B_{5}}$ | = | $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\left(a x_{5} + c z_{5} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{5} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti I |
$\mathbf{B_{6}}$ | = | $x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\left(a x_{6} + c z_{6} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{6} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti II |
$\mathbf{B_{7}}$ | = | $x_{7} \, \mathbf{a}_{1}+x_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\left(a x_{7} + c z_{7} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{7} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti III |
$\mathbf{B_{8}}$ | = | $x_{8} \, \mathbf{a}_{1}+x_{8} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ | = | $\left(a x_{8} + c z_{8} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{8} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti IV |
$\mathbf{B_{9}}$ | = | $x_{9} \, \mathbf{a}_{1}+x_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $\left(a x_{9} + c z_{9} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{9} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti V |
$\mathbf{B_{10}}$ | = | $x_{10} \, \mathbf{a}_{1}+x_{10} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $\left(a x_{10} + c z_{10} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{10} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti VI |
$\mathbf{B_{11}}$ | = | $x_{11} \, \mathbf{a}_{1}+x_{11} \, \mathbf{a}_{2}+z_{11} \, \mathbf{a}_{3}$ | = | $\left(a x_{11} + c z_{11} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{11} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti VII |
$\mathbf{B_{12}}$ | = | $x_{12} \, \mathbf{a}_{1}+x_{12} \, \mathbf{a}_{2}+z_{12} \, \mathbf{a}_{3}$ | = | $\left(a x_{12} + c z_{12} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{12} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti VIII |
$\mathbf{B_{13}}$ | = | $x_{13} \, \mathbf{a}_{1}+x_{13} \, \mathbf{a}_{2}+z_{13} \, \mathbf{a}_{3}$ | = | $\left(a x_{13} + c z_{13} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{13} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti IX |
$\mathbf{B_{14}}$ | = | $x_{14} \, \mathbf{a}_{1}+x_{14} \, \mathbf{a}_{2}+z_{14} \, \mathbf{a}_{3}$ | = | $\left(a x_{14} + c z_{14} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{14} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti X |
$\mathbf{B_{15}}$ | = | $x_{15} \, \mathbf{a}_{1}+x_{15} \, \mathbf{a}_{2}+z_{15} \, \mathbf{a}_{3}$ | = | $\left(a x_{15} + c z_{15} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{15} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti XI |
$\mathbf{B_{16}}$ | = | $x_{16} \, \mathbf{a}_{1}+x_{16} \, \mathbf{a}_{2}+z_{16} \, \mathbf{a}_{3}$ | = | $\left(a x_{16} + c z_{16} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{16} \sin{\beta} \,\mathbf{\hat{z}}$ | (2a) | Ti XII |