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Old   May 8, 2020, 10:10
Default UDF - Question
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Hi i am trying to understand the define grid motion udf given in the manual.

Can someone explain what is 0.230 "pow (NODE_X(v)/0.230, 0.5)" doing here ? . Why do we want to divide the node x by that number ?

node motion based on simple beam deflection equation
compiled UDF
************************************************** ********/
#include "udf.h"

Thread *tf = DT_THREAD(dt);
face_t f;
Node *v;
real NV_VEC(omega), NV_VEC(axis), NV_VEC(dx);
real NV_VEC(origin), NV_VEC(rvec);
real sign;
int n;

/* set deforming flag on adjacent cell zone */

sign = -5.0 * sin (26.178 * time);

Message ("time = %f, omega = %f\n", time, sign);

NV_S(omega, =, 0.0);
NV_D(axis, =, 0.0, 1.0, 0.0);
NV_D(origin, =, 0.0, 0.0, 0.152);

v = F_NODE(f,tf,n);

/* update node if x position is greater than 0.02
and that the current node has not been previously
visited when looping through previous faces */
if (NODE_X(v) > 0.020 && NODE_POS_NEED_UPDATE (v))
/* indicate that node position has been update
so that it's not updated more than once */

omega[1] = sign * pow (NODE_X(v)/0.230, 0.5);
NV_VV(rvec, =, NODE_COORD(v), -, origin);
NV_CROSS(dx, omega, rvec);
NV_S(dx, *=, dtime);
NV_V(NODE_COORD(v), +=, dx);

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Old   May 8, 2020, 12:21
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That's just an equation for a cantilever motion. You would anyway use your own equation.

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Old   May 8, 2020, 15:37
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Thanks Vinerm.
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