A method for adjusting the curved roller table of the pinch roller shaft adjustment mechanism before shearing
A technology for adjusting the mechanism and the roller table, which is applied in the field of flying shears, can solve the problems affecting the working efficiency of the equipment, the long stroke of the roller table slider, and the large friction and loss, so as to reduce the probability of wear and jamming, speed and The effect of small stroke and extended service life
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Embodiment 1
[0025] In the embodiment of the present invention: a method for adjusting the curved roller table of the roller shaft adjustment mechanism of the pinch roller before shearing, including the left limit position of the roller shaft slider sliding in the roller table and the right limit position of the roller shaft slider sliding in the roller table Location;
[0026] see figure 1 , when the roller slider slides in the roller table and reaches the left limit position: in order to ensure the minimum sliding speed and the shortest length of the roller slider in the roller table, define the relative speed of the slider v r perpendicular to the drag velocity v e , the relative velocity v r always parallel to the rod L 1 , get a=180-(w 1 t+θ 1 +w 2 t+θ 2 ), θ 1 and θ 2 、w 1 and w 2 Rod L 1 and pole L 2 The initial angle and rotational angular velocity, get: v r =L 2 w 2 sin(w 1 t+θ 1 +w 2 t+θ 2 ), let the roller table profile curve be y r , then we can get y by in...
Embodiment 2
[0036] Taking an actual roller table adjustment mechanism as an example to elaborate the adjustment method of the above roller table curve: the schematic diagram of the double-scale flying shear adjustment mechanism of a high-speed steel feeding system is as follows Figure 3-4 as shown, image 3 , 4 Indicates that the adjustment mechanism is in two limit positions;
[0037] H=360,h 1 =25,L 1 =220, L 3 =203,L 2 =175.22,L 4 =275,L 6 =200.83, h 2 =20
[0038] From the calculation of the above known parameters, it can be seen that the rod L 1 and L 2 The angle turned from the start position to the end position is:
[0039] θ 12 -θ 11 =1.6695rad-1.4473rad=0.2222rad;
[0040] θ22 -θ 21 =1.6051rad-1.3805rad=0.2246rad;
[0041] Assuming that the whole angle adjustment time is 2s, the rod L 1 and L 2 The angular velocity is w 1 =0.1111rad / s,w 2 =0.1123rad / s, the initial (t=0) angles are:
[0042] θ 1 = θ 11 = 1.4473rad; θ 2 = θ 21 = 1.3805rad;
[0043] Substi...
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