Examples
4.2 Examples 3–6: SETTCOR function for tool environments
Tool Compensation (W1)
4-14 Function Manual, 08/2005 Edition, 6FC5397-0BP10-0BA0
Example 9:
Let machine data:
MD20360 $MC_TOOL_PARAMETER_DEF_MASK (definition of tool parameters)
contain the value 2 (tool length must be evaluated with factor 0.5 in the diameter axis), let X
be diameter axis:
N10 def real _LEN[11]
N20 def real _CORVAL[3]
N30 $TC_DP1[1,1]= 500
N40 $TC_DP2[1,1]= 2
N50 $TC_DP3[1,1]= 3.
N60 $TC_DP4[1,1]= 4.
N70 $TC_DP5[1,1]= 5.
N80 _CORVAL[0] = 1.
N90 _CORVAL[1] = 1.
N100 _CORVAL[2] = 1.
N110 t1 d1 g18 g0 x0 y0 z0 ; ==> MCS position X1.5 Y5 Z4
N120 r1 = settcor(_CORVAL, "g", 1, 1)
N130 t1 d1 x0 y0 z0 ; ==> MCS position X2.5 Y6 Z5
N140 r3 = $TC_DP3[1,1] ; = 5. = (3.000 + 2. * 1.000)
N150 r4 = $TC_DP4[1,1] ; = 5. = (4.000 + 1.000)
N160 r5 = $TC_DP5[1.1] ; = 6. = (5.000 + 1.000)
N170 m30
The compensation of the tool length is to be 1 mm in each axis (N80 to N100).
1 mm is thus added to the original length in lengths L2 and L3.
Twice the compensation value (2 mm) is added to the original tool length in L1, in order to
change the total length by 1 mm as required. If the positions approached in blocks N110 and
N130 are compared, it can be seen that each axis position has changed by 1 mm.
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