When machining a spherical surface on a CNC lathe, the shape error influencing factors and elimination methods are as follows. The error caused by the deviation of the tool tip from the spindle axis and its elimination method (taking the car ball as an example). As shown in Fig. 1, Y is the distance that the turning tool deviates from the X axis, Dt is the theoretical diameter of the AA section, D is the required ball diameter, and R=D/2 is the circular interpolation radius of the CNC lathe tool. In the AA profile, the tool circular interpolation curve shows an ellipse with a major axis of D and a minor axis of D1. The error is d=D-D1=D-2[(D/2)2-âˆ†Y2]1â„2 (1 In actual production, the product drawing generally proposes the diameter accuracy of the ball being machined. If the precision of the processing ball bearings is generally within Â± 0.005mm, that is d = 0.01mm. To ensure this accuracy, you must control âˆ†Y. From the formula (1) to know âˆ†Y=Â±1â„2(2Dd-d2)1â„2 (2)
Fig.1 Schematic diagram of the deviation of the tool tip from the X axis
If D=80mm, then when calculating the ball bearing, the calculated |âˆ†Y|â‰¤0.63mm. The counter knife method is shown in Figure 2. The value on the dial indicator is the âˆ†Y value. Tool circular interpolation error and its elimination method (in the car sphere as an example) As shown in Figure 3, âˆ†X is the tool circular interpolation center deviation from the Y axis distance, D is the required ball diameter, D1 is Actual machining diameter on XOY plane: D/2 is the circular interpolation radius of the tool. It can be seen that in the XOY plane, the error d=D1-D=2âˆ†X: in the XOZ plane, an elliptical sphere having a major axis diameter D1 and a minor axis diameter D, the error d=D1-D=2âˆ†X Using point-by-point comparison method to eliminate the influence of tool center circular interpolation error.
Figure 2 Schematic diagram
Fig. 3 Effect of tool circular interpolation center circle (inner spherical surface)
(a) 2âˆ†X>0, positive compensation âˆ†X
Set D as the required inner ball diameter for roughing, and 1 to 1.5mm semi-finishing for the roughing, ie A1=D-(1 to 1.5). Comparing the actual size of the rough inside diameter with the circular interpolation diameter A1 in the program, the error of the circular interpolation center of the tool from the center of the spindle is 2âˆ†X. If 2âˆ†X>0, the positive X-axis compensation âˆ†X, if 2âˆ†X<0, the negative X-axis compensation âˆ†X (Figure 4). Semi-rough car left 0.5mm fine car spare, that is, A2 = D-0.5, and then measure, compare, tool compensation method as above, until the car out of the required spherical surface. The outer spherical surface of the vehicle is the same as the spherical surface of the vehicle, and the compensation direction is the same. The difference is that the tool is installed in the opposite direction. Since the pulse equivalent of the stepper motor can reach 0.01, 0.005, and 0.001mm, the accuracy of the circular interpolation curve is corresponding to Â±0.01, Â±0.005, Â±0.001mm. According to the demand of the parts being processed, the corresponding CNC lathe can be selected to satisfy the requirements. The actual needs of production.
(b) âˆ†X<0, negative compensation âˆ†X
Figure 4 Tool Compensation