**4. ****Determination of the Circumferential Stress Forces**

## 4.1 Development of equations relating to the circumferential Stress Forces

Based on Kepler’s laws of rotation (Appendix 6) and the arguments as set out in Section 5 let us assume as a working hypothesis, that the Earth can be modelled as a rotating body where its COM is offset from the principal axis of rotation. For the purposes of this paper, two approaches are considered to determine the principal forces associated with an unbalanced rotating body.

**Fig 17A**

**Fig 17****b**

**Fig 19**

## 4.2 **Model 1****:**** Rigid body dynamics**

**Model 1**

**:**

**Rigid body dynamics**

As discussed in Section 4, the Pacific plate has all the appearances of being in compression while the almost diametrically opposed African plate appears to be in tension. The simplest model is to consider the Earth as an eccentrically rotating solid body such as unbalanced flywheel. Although this model (Fig 17A) and enumerated in Appendix 4 accounts for the compressive and tensile stresses developed in the outer rim it does not describe the circumferential forces which are thought to be linked to the tectonic forces resulting in plate movement. This model which describes the circumferential stress forces also describes the situation that would occur if the lithosphere were treated as a thin shell sphere subjected to an internal pressure with a developed ‘vertical force P’. Fig C within Fig 19 is shown as an aid to understand the terms involved. In this case the area of maximum stress would be along the diameter of the shell at right angles to the force. The area resisting this developed force is described by the thickness of the thin shell multiplied by the mean diameter. The propagation of the crack that is now the Mid-Atlantic Ridge is in this area of maximum stress and would have occurred after separation had begun.

**Fig 18A & 18B**

Models used for the calculation of the differential circumferential stress forces required to move the crust to the mantie. The movement to the lighter side is independent of the hard rotation

**4.3 ****Model 2: Outer Rim able to slide relative to the main body**

**4.3**

**Model 2: Outer Rim able to slide relative to the main body**

In order to determine the forces postulated as being responsible for tectonic movement, the model used is one in which the thin crust can slide relative to the solid body at the crust/mantle interface. By way of illustration, Fig 18A shows that if an unbalanced disc with an outer annular ring containing fluid is rotated about its principal axis, the liquid will move to the ‘lighter’ side. This action would also give a plausible explanation to account for the sea level in Pacific Ocean being permanently higher^{37 }than that of the Atlantic and Indian Oceans. This situation is noted by the difference in tidal heights either side of Panama. The mean level of the tidal heights is also affected by weather patterns, salinity and possibly Coriolis forces. Fig. 18B shows an analogous situation with the sliding continental plates.

If we consider the crust as being able to move relative to the mantle, albeit it over a long geological time span, then force vector diagrams (Figs 17B & 19) can be constructed by making the following assumptions:

The crust is a thin shell that is able to slide relative to the mantle

The forces due to eccentricity are superimposed on the stress caused by the general rotation and gravity and

The stress, which is of interest for the purposes of tectonic movement, is the differential stress owing to this eccentricity.

By approaching the problem in terms of a thin shell moving relative to the mantle, it is possible to consider which increments of the tensile force are responsible for putting the Pacific Basin under compression (crumpled profile, Ring of Fire) and the African Plate under tension (Rift Valley). The calculations to derive the expression of the circumferential stress at the surface of the Earth are based on the consideration of the eccentrically induced loads on the thin crust as detailed in Appendix 3. The term ‘radius of eccentricity’ was introduced to denote the distance between the centre of mass and the major axis of rotation. From Appendix 3 the following relationship was derived:

Total circumferential force (F) acting on the crust = MRω^{2}Eπ/4 **Equation (2)**

If we take into equation 2, an element of crust to be 1000 metres thick with an average density of 2.8x10^{3 }kgm-^{3}, then for a 1000m column of cross-section 1m x 1 m, the mass per unit area of crust (m^{2}) is 1000 x 1 x1 x 2.8x10^{3} = 2.8x10^{6} kg.

If the radius of the Earth (r) = 6400 km (6.4x10^{6} m), the rotational velocity of the Earth at the equator (ω) = 7.27x10^{-5} radians sec^{-1} and the Radius of Eccentricity at the Core (E) = 1 km. we get:

F= 2.8 x10^{6} x 6.4x10^{6}x (7.27x10^{-5})^{2} x 10^{3 }x π /4 = 6.64x10^{7} N.

Since the magnitude of the circumferential stress is Force/Area, this becomes 6.64 x 10^{7} / 1 x10^{3} = 6.64 x 10^{-2} Nmm^{-2}.

Hence the circumferential tensile stress is = 6.64 x 10^{-2} Nmm^{-2}, 0.644 Bar or c. 9.8 lbs.in^{-2 }

In order to determine the mean forces to a 95% confidence level, uncertainty calculations (Howarth) have been applied to the consideration of:

(a) The density of the continental crust varying between 2.3x10^{3 }kgm^{-3} and 2.9x10^{3 }kgm^{-3}

(b) The density of the oceanic crust varying between 2.83x10^{3} kgm^{-3} and 2.89x10^{3} kgm^{-3}

(c) The Earth’s radius varying between 6.3567x10^{3} km and 6.3781x10^{3} km and

(d) The equatorial rotational velocity varying between 7.27x10^{-5} rad. sec^{-1} and 7.292x10^{–5} rad. sec^{-1}.

By assuming these values can be taken as endpoints of several uniform distributions, then by generating a random sample of 999 values (see Appendix 5) the following values are obtained

Mean Circumferential Stress (Continental Crust) = Fc = 72.97 (68.14, 76.74) x10^{6}N

Mean Circumferential Stress (Oceanic Crust) = Fo = 75.87 (74.97, 76.69) x10^{6}N

Thus, the differential circumferential forces created by placing the centre of mass of the Earth 1.0 km off-centre are large and cannot be ignored. The calculated circumferential forces if applied to the cross-sectional area of the South American plate are more than enough to push it over the Nazca plate.

Whereas Figs 18 & 19 shows the logic train used to develop Equation 2, Fig 21 displays its application. The graphical relationship between F and E is shown in Appendix 3. In a limiting case, if the ‘radius of eccentricity’ is zero, the rotating body will be balanced, and the differential circumferential forces (DCF) will be zero.

**Fig 20**

## 4.5 Relating the circumferential stress force to an everyday situation

**Fig 21: Actual Incline is 1:200**

In order to better understand the magnitude of the calculated circumferential stress in the continental crust, it is helpful to relate the model to more familiar applications. This is shown pictorially in Fig 22.

The stress value of 7.29x10^{-2} Nmm^{-2} if applied to a 1 tonne braked motor vehicle with a rear surface area of 1000 mm x 1300 mm=1.3x10^{6} mm^{2} will yield a push force of 94,770 N.

In Imperial units this equates to a push of 21,305 lbf (pounds force) or 9.5 tonf (tons force).

Rounded up and put more simply, this equates to the vehicle being pushed by 118 people each of whom weighs 180 pounds (81.8 kg) (see Fig 22). If the altitude of the Andes is taken as 5 km and the distance between the Peru-Chile trench and the Cordillera–Real is taken as c.1000 km, the incline is approx. 1:200. Therefore, the vehicle can be considered to be on a level surface for scaling purposes. Normally a 3 tonne hoist will easily pull the vehicle up a 1:3 incline onto a pick-up truck It is also worth noting that an upward acting net force of 2.37 x 10^{-2} N/mm^{2 }(3.5 psig) on a 60 metre wing span of an aircraft is sufficient to keep a large 350 tonne aircraft flying. A puff of wind with dynamic pressure as low as 0.135 x 10^{-2} N/mm^{2 }(0.2 psig) acting on the large surface area of a ship’s sail will cause a boat to move across water.

**Fig 28**

## 4.4 Coefficient of Friction

The estimation of the magnitude of the circumferential forces acting on an element of crust (see section 6.2 and Appendix 3) allows for the calculation of the coefficient of friction (μ) at the crust / mantle interface. The force diagram required for this calculation is depicted in Fig 28. The determination of the coefficient of friction will thus yield an indication of the material and topography at the crust/mantle interface.

Let F be the ‘moving force’ on the 1m x 1m x 1000m crustal column and W= the ‘vertical downward force’ exerted by the weight of that column.

As the ratio of the height of the Andes to the distance between the western edge of the South American shelf to the Cordillera Real is c.1:200 the inclined plane presented by the Nazca plate can be considered as being essentially flat.

If we take the radius of eccentricity to be 1 Km, then from eq. (2) as above we have

F= 6.64 x 10^{7} N and

W=2.8x10^{6}x 9.807 Kg

N= 2.745 x10^{7} N.

Hence μ= F/W= 6.64x10^{7} / 2.745 x10^{7} = 2.419.

Alternatively, if the radius of eccentricity is taken to be 0.5km then μ =1.35

The graphical relationship between μ (coefficient of friction) and the calculated force (F) and the corresponding radius of eccentricity is shown in Appendix 3.

The research programme by Morrow and Lockner^{32} on the cored rock samples from the Hayward Fault region in Northern California showed the coefficients of friction of these samples (sandstone, basalt, fine and coarse-grained gabbros, and keratophyre). These were calculated from the fracture angles which occurred under the applied axial loading and were shown to range between 0.5 and 0.9. At low applied loads of 32 and 64 MPa (to simulate depths of 2 and 4 km) the μ values for the coarse gabbro, basalt and keratophyre varied between 1.0 and 1.5. The calculated values of the coefficient of friction are not unrealistic and compare favourably with laboratory simulations.