# lattice hcp command

Dear all,

I would like to have a hcp lattice of titanium by the parameters of a=b=2.95 A , c= 4.686 A . But it is written in the user manual that this command should be for example like lattice hex 0.85 . But by this command, how can I make the two parameters of a hcp lattice structure clear for the program?
what does this 0.85 show?

regards,
Soroush

The 0.85 example is for LJ reduced units where
it represents a reduced density. If you want
real or metal units, then you put the lattice
constant. Note that hcp has only one degree
of freedom (the lattice constant), with the c/a
ratio fixed at sqrt(3). If you want some other
ratio, then you can use the lattice custom
option.

Steve

Dear steve,

I have a=2.95 and c=4.686 so the c/a ratio is 1.588. Now, is it correct and enough to write this command?: lattice hex 1.588 . this command defines the c/a ratio but not the exact size of a and c!
a part of the input script by this command is below:

units metal
boundary p p p
lattice hex 1.588
region box block 30 30 70 units box
create_box 1 box
create_atoms 1 region box
mass 1 47.867
#Foils EAM
pair_style eam/alloy
pair_coeff * * Ti.SET Ti

# initial velocities

velocity all create 300 5812775 units box
fix 1 all nvt temp 300 300 0.01 drag 0.5

# equilibrium

run 50000

Soroush

No - as I said the definition of hcp in LAMMPS is
a sqrt(3) ratio, and you specify the a.

If you want another c/a ratio then it is a custom
lattice and you can use the custom option.

Also you keep using "hex" in your emails,
but hex is a 2d lattice in LAMMPS. Hcp is
3d.

You don't need to guess about any of this.
The lattice doc page explains it in great
detail.

Steve