The trajectory of the breakaway point and the fallback point and the maximum escape angle and the minimum spherical radius

1. The trajectory of the point of departure and the point of return
The ball load in the mill consists of several layers of balls, each of which has a break point A i and a fall point B i . The coordinates of the A i points of each layer of the ball are different, but since they are all out of the point, they all have the same geometric condition. Similarly, the coordinates of each fall point B i are different, but they all conform to the same geometric condition. Finding the two geometric conditions, we find the connection between the two turning points, that is, the trajectory of the break point and the fall point.
The following work methods are known:
Here, when n is the given timing, a is a constant.
The formula (1) is a circular curve equation in which the center O of the mill is the pole and the coordinate axis OY is the polar axis, and the radius of the circle is . Since each layer of balls has a separation angle a i and a spherical layer radius R i and conforms to the above relationship, each of the A i is on a circle having a center of O 1 and a radius. This circle is the trajectory of each point A i , as shown in Figure 1. The distance from the falling point B i to the center of the mill is R i , and the polar angle θ between the pole axis OY and the polar axis OY is known as
Point B i is also on the circular motion trajectory, so there is still a polar coordinate equation
then , R = 0, the curve represented by this equation (ie, Basheer spiral) will pass through the center of the mill (ie pole) O. The curve represented by the formula (2) is used to traject the trajectory of the point B i .

Figure 1   Trajectory of the point of departure ( A i ) and the point of return ( B i ) [next]
2. Maximum escape angle and minimum spherical radius
It is obvious from the above figure that the closer to the ball layer in the center of the mill, the closer the trajectory of the detachment point and the trajectory of the falling point are, and the point at the center O of the mill is at a point. From the point of view of the phenomenon, the closer to the center of the mill, the circular motion and the parabolic motion interfere with each other, so that the two are almost inseparable. Therefore, the radius R 2 of the innermost spherical layer must have a limit value, which is smaller than it, and the spherical layer has no obvious circular motion and parabolic motion. This limit is called the minimum sphere radius ( Rmin ) and can be derived as follows.
then , X B has a limit value. then,

Solve this equation and get the two a values ​​as and respectively. From the relationship between the angle β and the angle a, when a = 26 ° 44 ' , β = -9 ° 48 ' , that is, the fall point is above the OX axis and to the right of OY. When a = 73 ° 44 ' , β = 131 °12 ' , the fall point is below the OX axis and to the left of the OY axis. From


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These two formulas indicate that the angle of departure of the outermost sphere is only related to the rate of rotation, and that the angle of departure of the innermost sphere is related to the rate of rotation and the rate of loading (marked with K).
According to the above situation, in order to ensure that the innermost ball can also be in a parabolic state (that is, all the ball layers are thrown away), the ball loading rate and the rotation rate must have a certain relationship. Moreover, this relationship must have a critical point. After this critical point, the rotation speed of the mill is not enough to cause the inner ball of the halo to be thrown off, and the steel ball is in a state of diarrhea. Here, the curve drawn by the calculation result is shown as the following figure. The figure shows the relationship between the ball loading rate, the rotation rate and the radius of the sphere, and also shows the limits of the sag and the drop determined by this relationship.


Figure 2 The relationship between K , ф and φ and the boundary of the parabolic condition
Figure 3 Areas and ball load-cut area in the mill [next]
4. Each area in the mill and the ball load-cut area
(1) The area where the steel balls are distributed in the mill and the grinding effect of each area
After analyzing the motion law of the steel ball in the mill in detail, the geometry of the steel ball distribution can be drawn more accurately, as shown in Fig. 3. It is apparent from the figure that the inside of the mill is divided into four different areas.
1 steel ball as a circular motion zone - the part of the solid line in the picture, the steel balls are all circular motion, the ore is clamped between the steel balls to be ground and stripped.
2 The steel ball is used as a parabolic drop zone—the part of the figure that depicts the virtual shadow line, indicating that the steel has fallen into the area. In the process of the falling of the steel ball, the ore is not worn until the falling foot reaches the foot indicated by the trajectory BB 2 of the falling point, and the steel ball has an impact on the ore.
3 kidney-shaped area - close to the center of the mill, the circular motion and parabolic motion of the steel ball have been difficult to distinguish clearly. In the area where the shadow is shaped like a kidney, the steel ball only vibrates and the grinding effect is weak. When there are more balls and the rotation speed is not enough to make them move actively, the kidney-shaped area is larger and the honing effect is also poor.
4 blank area - the crescent-shaped part outside the parabolic zone is where the steel ball is not, and of course there is no grinding effect. When the speed is insufficient, the steel ball is not far away, and the blank area is larger. The speed is too high, the steel ball is thrown far, and the blank area is small,
However, the steel ball directly lining the board will cause severe wear and wear, and the grinding effect is also poor.
The partitions in the mill are not only so obvious, but they can be quantitatively calculated. The ball load-cut area described below can be explained.
(2) When the cutting area of ​​the ball load rotates, there is a space of the ball, a part of the ball is distributed as a circular motion, and the other part is distributed as a ball which is parabolically dropped, and is taken from a section perpendicular to the long axis of the mill. The area occupied by all the moving balls is Ω, and the area occupied by the ball in the circular motion part is Ω 1 , and the area occupied by the ball as the parabolic motion is Ω 2 , then
Ω=Ω 1 +Ω 2 ( 7 )
Loading rate under dynamic
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Take a layer of ball, its radius of the sphere is R c , the angle of departure is a C , the angle of fall is β C , the angle of the center of the sphere is ф C , as can be seen from Figure 3
Integrate the above formula in the range of R 2 and R 1 to obtain
After Ω and Ω 1 are obtained, Ω 2 can be calculated. For example, a grinder has an inner radius of R, a rotational speed of 76%, a suitable ball loading rate of 40%, and a ball load-cut area Ω, Ω 1 , Ω 2 . According to Equation 2-26, Ω = 0.4πR 2 . If you take the ball in the middle to calculate, . The innermost sphere can be used as a limit condition for parabolic falling. It can be known from equation (10).



Obviously, it is known that the ball is too small, the Ω is very small, and there are not many grinding parts in the mill. Although the ball is suitable, the rotation speed is too low, there is almost no Ω 2 , and it becomes a diarrhea state. The grinding and peeling action is superior to the impact. Only when the ball loading rate and the rotation rate are suitable, it is guaranteed that the throwing state occurs, and there is a large Ω 2 , so that the impact effect is sufficient.

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