2 5 7 11 21 31 41 51 61
- --- -- ---- ---------- ---------- ---------- ---------- ----------
|X| | |FCON|XXXXXXXXXX| | | | |
- --- -- ---- ---------- ---------- ---------- ---------- ----------
| | | | | | |
| | | | | | |_(4) Constant friction moment (F10.0)
| | | | | |
| | | | | |_(3) Friction coefficent for axial force (F10.0)
| | | | |
| | | | |_(2) Friction coefficent for overturning moment (F10.0)
| | | |
| | | |_(1) Friction coefficient for transverse force (F10.0)
| | |
| | |_Compulsory Data Record Keyword(A4)
| |
| |_Optional User Identifier(A2)
|
|_Compulsory END on last data record in data category(A3)
(1), (2), (3), (4) Coulomb friction always uses a local axis system for calculation with the X-axis being the axis of the instantaneous relative rotational velocity of the two parts of the articulation. For a hinge this will always be the axis of the hinge; for a ball joint it will vary as the two structures move relative to one another. The frictional moment is given by
where:
ε = 0 if the relative rotational velocity is less than 0.001 rad/s, 1 otherwise.
k1 - k4 are coefficients. Note that these are not conventional dimensionless friction coefficents, as used in the equation F = μR. These coefficients are factors to be applied to the appropriate forces to give frictional moments, and they must include effects of the bearing diameter etc.
k1, k2 and k3 must not be negative. k1 and k3 have dimensions of length, and the maximum value
allowed is
where 
is the acceleration due to gravity expressed in the current
unit system and 
is the acceleration due to gravity expressed in the [kg,
meter, second] unit system. k2 is non-dimensional and has a maximum value of
0.025.
This moment is transformed into articulation, structure or global axes as appropriate before output.