A Quantitative Computational Method for Landslip Orientation of a Body Isolated from Bed Rock in Mountain by Using Spatial Analytic Geometry

The paper is going to give a quantitative computational method for “Landslip Orientation of a body isolated from bed rock in Mountain” only with spatial analytic geometry. The paper gives computational formulae in proper order for only landslip plane, just two landslip planes and just three landslip planes, and gives numerical examples. And the paper gives a general computational model for landslip orientation of m landslip planes. The author puts forward “Un-allowed Straight Angle Law”, “Critical Straight Angle Law” and “Allowed Straight Angle Law”. Finally, the author gives a project expression of a landslip plane on unit sphere.


Introduction
In paper [1], Professor Shi Gen Hua found that the landslip of an isolated rock body is only able to produce along finite planes.If one parallel moves the finite planes passed through origin O, then he will find on the unite sphere centered O they form a polygon constructed by their up sides.Let S denote the lowest point (or points) of the polygon, if the perpendicular height of point S is lower than the equatorial plane of the sphere, then OS denotes the landslip orientation.In this paper, in order to convenient to engineers, the author provides mainly a quantitative computational method to solve OS.Therefore we suppose that 1 2 , , ,    and n  are n possible landslip planes, where n is a natural number.Their equations are as follows1 : For convenient and united, we let x axis point to east, y axis to north, and z axis perpendicularly to up orientation.
Professor Shi originally used stereographic projection 2to transfer the planes onto the equatorial plane, and he applied topologic method to prove the existence of the orientation of landslip.Furthermore, Shi prevented landslip by piling to withstand landslip along the steepest decent orientation.

To Analyze the Landslip Orientation in the Case of Only One Landslip Plane
The basic supposition and deduction of compute formulas for the Case of Only One Landslip Plane are as follows.
Here suppose there is only one landslip plane , whose equation is: z ax by   , where     Let s denote the steepest ascent orientation, then it holds  .For the signs of   , a b , we have the following eight cases: , , 0, , , , ,0 , , , 0, , ,              (5) , in the fourth quadrant.Let angle  denote the steepest ascent orientation.
The steepest decent angle is 180 denotes the angle interval on horizontal plane, in which does not allow slip, called un-allowed interval.However the open interval denotes the angle interval on horizontal plane, in which allows slip, called allowed interval.We explain basic properties for the case of Only One Landslip Plane.
Because the normal of the decent slip plane is , and the steepest ascent orientation is   . The allowed interval is , 2 2 The intersectional line L of them is their boundary line.
Because the line L is horizontal, the weight is not able to move the isolated rock.However, if there is wind force, water force or earth force to move it, possibly it is able to produce horizontal movement.So the line L is critical line.
We have known that the normal of the slip plane  is , whose horizontal project is ab.The horizontal project orientation of intersectional line L is ba, which is the critical orientation between  and T  .The horizontal project of  is resolved to be However, the slip orientation 0 s after normalization of horizontal project had not been mentioned in Shi's paper [1].
The first case of plane division is as follows.
. And the angle inter- is un-allowed to slip.

Now 45
  , the slip orientation points from northeast to south-west.In the practice, at the risk area, the support orientation should point from south-west to north-east according to the horizontal angle , and to the angle of gradient  by piling to withstand landslip.

To Analyze the Other Seven Cases for the Plane Division in the Case of Only One Landslip Plane
The second case of plane division is as follows.When  is ascent gradient,  is ascent angle.The un-allowed The allowed angle interval is The third case of plane division is as follows.When , where  is ascent support gradient,  is ascent angle.The un-allowed in- The fifth case of plane division is as follows.When , where  is ascent support gradient,  is ascent angle.The un-al- The seventh case of plane division is as follows.When ,  denotes the ascent gradient of supporting,  is the angle of ascent supporting.The un-allowed interval is

The Case of Just Having Two Slip Planes
Now we suppose that 1  and 2  are two possible slip planes.If 1  and 2  are parallel each other, and because the two planes have a common point: , they must be coincident, and turn to be the case of only one slip plane, as stated in the former section.Therefore we only discuss the case of that the two slip planes 1  and 2  have an intersectional line.Suppose the equations of the two slip planes are as follows; Suppose , and the normal of where Let 1,2,0 l denote the slip orientation under normalizetion horizontal project, then it holds the following formula Or in detail, when 1,2 0 h  , it holds while the slip horizontal angle is 1,2 1,2

Example for Just Two Slip Planes
  , the ascent angle 1 actg 2 0.955317 We have supposed that the intersectional line of 1  and 2   And the decent gradient Refer to Equation (15), because 1,2 0 h  , it holds: while the steepest slip horizontal angle is . And the un-allowed interval is   3.92699075 3.46334,5.49779 Very fortunately, here we have 1,2 1 5 4 In dealing with the landslip of an isolated rock in a tunnel, we should note not only the caving in along the slip orientation 0 s of the intersectional line To sum up, the damage orientations are three: The slip orientation along the intersectional line The steepest horizontal normalized slip orientation along  1 : The steepest horizontal normalized slip orientation along  2 ;   Fortunately, it holds here: 1,2,0 1,0   l s .If having not done through repeated computations, only by intuition, it is difficult to see the result.In the paper [1] of Professor Shi Gen Hua, he claimed that the slip orientation 1,2,0 l of the intersectional line of the most damage one, the author also had the view.However the slip orientations 1,0 s and 2,0 s had not be mentioned, which should be noted also.
According to the data of the problem, the most damage orientations are three: 1) The slip orientation 1,2,0 l along the intersectional line 1,2 L ; 2) The steepest slip orientation s .That needs a detail analysis for three orientations, please refer to the following minimal analysis.
Because the section appears fortunate case: 1,2,0 1,0   l s , we must note special cases which appear occasionally like this kind.

Checking computations: Substituting
 and 2  .The following is to find Allowed Slip Pyramid.Refer to Figure 1 which is the first minimal angle.However The ascent angle is actg 2 0.955317

Example for Just Three Slip Planes
Suppose the first landslip plane 1  is: and the allowed interval is Suppose the second landslip plane The un-allowed interval is   and the allowed interval is   10 10 10 , 2.03444,5.17604 2 2 It follows that the orientation 1,10 l of intersectional line 1,10 L is denoted by On one hand, we must note the first paragraph in Section 6, the steepest decent slip orientation of 1  is   The allowed interval is The allowed interval is   10 10 10 , 2.03444,5.17604 2 2 Because of the intersection of Suppose the third landslip plane 2  is: The un-allowed interval is and the allowed interval is It follows that the orientation 2,10 l of intersectional line 2,10 L is denoted by The allowed interval is   The allowed interval is   10 10 10 , 2.03444,5.17604 2 2 The intersection of We have supposed that the intersectional line of 1  and 2 And the decent slip gradient is Refer to Equation (15), because of 1,2 0 h  , it holds The steepest decent slip horizontal angle is Fortunately, here the decent slip orientation of 1,2 , which is coincide with 0 s in section 2, above Equation (8).Note that here 1  is namely the plane  in the example in section 2, when 1 a b   .The steepest decent angle is The allowed interval is The un-allowed interval is   1,2 1 2 3.92699075 3.46334,5.49779 For three planes 1  , 2  and 10  , the allowed interval is   1.095159 0.34851 4 Along counter clockwise, from 10  to 1  to 2  , now 2,10  is the steepest decent gradient.How to analyze?At first we list the main data for planes as follows    The angle of ray  T is denoted by .Refer to the Figure 2, the fifth boundary surface of the pyramid is defined by the horizontal plane which is from OQ to OP according to clockwise.
To sum up, the allowed decent slip pyramid degenerates to be a four edge pyramid, whose boundary surface begin with the edge OP, through 2  to the edge under 1 2T, and through 1  to the edge 1,10 L under 1 10T, and through 10  arrived OQ, and according to clockwise, through horizontal plane to the edge OP.
Through minimal computation, the final conclusion is as follows: in the four edge pyramid of the section,  We should firstly reinforce along the orientation, i.e. from south-west to north-east, with ascent angle actg 2 0.955317 54 45'      , to reinforce.Minimal value, i.e. the steepest decent slip orientation is along the edge 1,2 l which is under the ray 1 2T

Discussions for the Case of m Slip Planes
According to Section 1, let 1 2 , , ,    and m  be m possible slip planes, where m is a natural number.Their equations are expressed by .For convenient and unification, the x axis points to east, the y axis points to north, and the z axis points to up.
Suppose 1 i j m    Among them, we choose an element i  , whose equation is . And the intersectional line i L of the contour plane z c Let i s denote the steepest ascent orientation, it follows Let i s denote the horizontal project of the steepest ascent orientation i s , which is also the horizontal project of normal i n , i.e. it holds   This is a deputation of plane i  . Let , then ,0 i s is able to be expressed by

 
,0 cos ,sin , , cos , sin Let ,0 i s denote the horizontal normalized project of the steepest ascent orientation, then it follows   ,0 cos ,sin The intersectional line of two planes i  and , , , where denote the decent slip orientation of horizontal project normalization, then it holds   , , , Or in detail, when , 0 , while the decent slip horizontal angle is , , , ,0 , , cos ,sin , 0 The steepest decent slip horizontal angle is 180 . The horizontal intersectional line i L is their boundary line.

"Un-Allowed Straight Angle Law" about m Planes
Now we draw the figure of the un-allowed straight angle law about m planes as follows.
In the Figure 3, we draw the m steepest ascent angle 1 , , , , , , There is an angle between two adjacent rays.Suppose all the angles are less than , i.e. less than a straight angle, then the un-allowed intervals 1 , , , , , ,  cover the whole circumference angle.Therefore, there is not any horizontal angle which is allowed decent slip.This may be simply called un-allowed straight angle law.

"Critical Straight Angle Law" about m Planes
Now we draw the figure of the critical straight angle law about m planes as follows.
In the Figure 4, we draw the m steepest ascent angles 1 , , , , , , This is called critical straight angle law.If there is wind force, water force or earth force to move it, possibly it is able to produce horizontal movement.

"Allowed Straight Angle Law" about m Planes
Now we draw the figure of the allowed straight angle law about m planes as follows.
In the Figure 5, we draw the m steepest ascent angle 1 , , , , , , There is an angle between two adjacent rays.No harm of generality, suppose along counter clockwise the angle mO is more than , i.e. more than a straight angle.The angle mOP is equal to a right angle, then the angle POQ which is more than zero, is the intersectional part of all allowed angles 1 , , , , , , compare the z values along these orientations on unit sphere.How to analytically compare?Please refer to Section 6: the example for just have three decent slip planes, in which we should give up some planes and orientations.After sifting we need only analytically compare the values on the left orientations, in which the orientation of the minimal z value is just the steepest decent slip angle formed by m planes 1 , , ,

Project Expression of Landslip Plane on Unit Sphere
If thinking of the theoretical beauty-ness, one should naturally use the unit spherical project , the steepest ascent orientation is

Inspiration from the Computation along One Orientation
, and the steepest decent The equation of unit sphere is 2

An Example of Practical Computation on the Project of Unit Sphere
We consider on plane z  0 the following ellipse whose long radius is equal to 1, and short radius is equal to cos .The orientation of long radius is , pointed to The orientation of short radius is , pointed to .
The equation of the ellipse in the coordinates  is as follows We plan to solve the equation of the ellipse in the coordinates xy.In order to doing this, we should find the corresponding formula of the coordinate transformation.Therefore we should firstly draw the hint figure of coordinate transformation.
The hint Figure 6 for coordinate transformation is as follows.
In the figure  denotes the angle from  axis to x axis, whose measure is equal to the angle from 2 The deduction of the formula for coordinate transformation: let It follows the formula Finally, we obtain The above elliptic Equation (93) changes into For the decent slip plane , the allowed interval is   , 2.67795,5.81954 2 2 . Now we divide T  into equidistance 10 n components.In practice the n should be , , or .In the section being only a hint, so we take 1 n  .Note

    
In the plane 0 z  for the ray with orientation , its , and its equation is as follows .Therefore, the computation is reasonable.

General Description for the Project Computation on Unit Sphere
The equation of decent slip plane  is zaxby, whose normal orientation is: . From ab, one ob-    x y z is namely the intersectional point of the plane  and the unit sphere along the orientation angle i  .
For the set   i  , after finding the corresponding three dimensional points, one obtains a downward semi circular plane.The solid angle between it and the horizontal plane is namely the domain of allowed decent orientation.
For the discussion of m planes, refer to Section 7, here we don't repeat it.

Complements for Vertical Plane
In view of theory, it is possible for the existing of a decent slip plane In the coordinates z, the equation of unit sphere is as follows This is an ellipse whose long radius is equal to 1 and on z axis, and whose short radius is equal to cos and on x axis.This is a formula expression in coordinate xz of  which is the project of the great circle of unit sphere on the plane 0   .We solve where is an ellipse with its long radius equal to 1 and on intersection line L of  and the contour plane z  constant C satisfies ax by c   .The normal of z c  is where  denotes the horizontal angle in the xy plane during the steepest ascent,  the ascent gradient.From ab, one may compute actg b a projects of n and s are equal to each other, i.e. equal to   , a b .The orientation of intersectional line of the contour plane z  C and  is project orientation is b a, which is perpendicular to horizontal project of s .From ab one obtains actg b a .We have supposed that  is an angle in xy horizontal plane for the steepest ascent case.If the signs of ab are , then actg b a   , in the first quadrant; if the signs of ab are , actg b a     , in the third quadrant.If the signs of ab are , then actg b a     , in the second quadrant.If the signs of ab are , from north to south.The support orientation should point from south to north.If  denotes the angle of gradient, then tg b the slip orientation is from north-west to south-east.The support orientation is from south-east to north-west.If  denotes the angle of gradient, is the gradient of ascent support,  is ascent angle.The fourth case of plane divisionis as follows.When 0 a  and 0 b  ,    , at x axis; 0 T   .The slip orientation is from west to east.The support orientation is from east to west.If  denotes the angle of gradient, the slip orientation is from south-west to north-east.The support orientation should be from north-east to south-west.If  denotes the angle of gradiis ascent support gradient,  is ascent angle.The sixth case of plane division is as follows.When 0 slip orientation is from south to north.The support orientation should be from north to south.If  denotes the angle of gradient, then tg b

1 ,0 s along 1  2  . Because the existence of 1 
; and 3) The steepest slip orientation 2,0 s along forms a bound for the orientation 2,0 s ; but the existence of 2  does not form a bound for the orientation 1,0 which falls in the fourth quadrant and beyond the scope of PO -1 2T, we do not take it.Therefore, in the section, the steepest slip horizontal angle is namely the other hand, through computation we get:, the steepest decent slip orientation along 10  is

2 .
77) Now we draw the orientation figure of horizontal project as follows We give explanations for the Figure Let the coordinates Oxy denote the horizontal project plane The straight angle under the line 2T-O-P is denoted by straight angle on the left of the line 10T-O-P is denoted by   10 2.03444,5.17604T   .The straight angle on the left of the line 1T-O-R is denoted by

Figure 2 .
Figure 2. The orientation figure of horizontal project.
decent slip orientation.


. Suppose they permute according to counter clockwise.

Figure 3 .Figure 4 .
Figure 3.The figure of un-allowed straight angle law.


. Suppose they permute according to counter clockwise.

Figure 5 .
Figure 5.The figure of allowed straight angle law.


. This is the most important orientation which need reinforce.
x axis.In the problem, refer to the first and second paragraphs of article 8.1, because of tg b a

tains actg b a .
Suppose  denotes the steepest ascent angle in the horizontal plane xy, then by using ab one

Let
0 z  , we consider the following ellipse, whose long radius is equal to 1, and short radius equal to cos.The orientation of long radius is , whose orientation denotes to the angle 2 T    .The orientation of short radius is , whose orientation points to .In the coordinates , the equation of the ellipse is as folfind the equation of the ellipse in the coordinates xy.Refer to the Figure 6, in which  denotes the angle from  axis to x axis.

0From
which is perpendicular to the horizontal plane.Suppose the equation of 0 turns into: 0 ax by   , i.e. ab it follows actg b a .Suppose  is the steepest ascent angle in horizontal plane xy, then from the signs of ab we are able to determine Let  denote the angle from  axis to x axis, whose value is equal to the angle from 2 into the coordinates xyz, the equations of  is as follows is in the fourth quadrant, and the slip orientation is from south-east to north-west.The support orientation should be from north-west to south-east.If  denotes the angle of gradi-   , at x axis, and T    .The slip orientation is from east to west.The support orientation should be from west to east.If  denotes the angle of gradient, then tg a , the slip horizontal angle is POR.The allowed slip three edge pyramid formed by two planes 1  and 2  is as follows: The first edge is OP.The first boundary surface is 2  .The second edge is 1,2 l which is under the ray 1 2T.The second boundary surface is 1 .The third edge is OR.The third boundary surface is the horizontal plane from OR to OP along clockwise.The following is Minimal Analysis.
falls in, or what axis the solution falls on, then we solve the value of ix .Then we compute: tg