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February 10, 2020, 23:36 |
Acceleration term in 6 DOF solver
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#1 |
Member
MEK
Join Date: Oct 2016
Posts: 39
Rep Power: 10 |
cg_y = DT_CG(dt)[1]; y-displacement
cg_x = DT_CG(dt)[0]; x-displacement vel_y = DT_VEL_CG(dt)[1]; y-velocity vel_x = DT_VEL_CG(dt)[0]; x-velocity Fy = -ky*cg_y - vel_y*cy; x force term Fx = -kx*cg_x - vel_x*cx; y force term prop[SDOF_LOAD_F_Y] = Fy; prop[SDOF_LOAD_F_X] = Fx; i would like to ask about acceleration term as there are displacement and velocity but if i also want to incooperate acceleration term what do i have to do |
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February 11, 2020, 04:25 |
Displacement, velocity, and acceleration
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#2 |
Senior Member
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Do note that DT_CG is not displacement rather the position vector of the center of gravity. Its variation across space would define displacement.
DT_VEL_CG is the velocity vector. As far as acceleration is concerned, a separate macro is not required since this is a discrete solver. You have to ensure that the velocity is changing over time; that is the acceleration. If you wish to apply an instantaneous acceleration, then that is defined in terms of the force and not acceleration as such. So, you can add another force.
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February 11, 2020, 06:05 |
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#3 |
Member
MEK
Join Date: Oct 2016
Posts: 39
Rep Power: 10 |
thank you so much for clarifying these things but still i am unable to understand how i would be able to add (MX doubble dot) if i dont know its value here solver compute it for me based on the movement of object.
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February 11, 2020, 06:21 |
No calculation of acceleration
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#4 |
Senior Member
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Solver does not compute . It computes only x and .
May I know the objective why you want to calculation the acceleration? As suggested earlier, if you wish to apply acceleration that implies applying a force.
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Regards, Vinerm PM to be used if and only if you do not want something to be shared publicly. PM is considered to be of the least priority. |
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February 11, 2020, 08:12 |
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#5 |
Member
MEK
Join Date: Oct 2016
Posts: 39
Rep Power: 10 |
Well my intentions to compute acceleration is due to coupled equations that I want to solve as attached in the image of first post maybe if u look into the equations in the image things will be clearer.
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February 11, 2020, 08:38 |
Macros
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#6 |
Senior Member
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As far as I understand, you appear to be mixing two things. For an SDOF solver, Fluent does not require you to solve extra equation, however, you can provide extra forces. Fluent (most likely) does not include the last terms in the equations you have, so, you can use the following macros to calculate extra forces and then add those to the forces that you already might have. But be cautious with what is being done because Fluent does not expect extra velocities or displacements to be provided by the user in SDOF solver; only the forces (both linear and angular).
DT_ACC_CG for linear acceleration DT_ALPHA_CG for angular acceleration
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