Some of the ramparts of Coehorn, and all of those which Carnot proposed for his fortresses, are formed of earth unsupported by revet menta ; and even the opposite side of the ditch, instead of being faced with a steep wall, is by the latter engineer cut with a gentle slope from level of the natural ground to the bottom of the ditch. But the fortifications of Coehorn are provided with wet ditches, which prevent the besiegers from getting to the foot of the rampart by surprise and in those of (pant a high detached wall covered by a countergur;rd of earth puts it out of the power of the enemy, while that wall stands, to get across the ditch. Without such obstacles the unrevetod rampart would afford great facilities to the enemy in an effort to carry the fortress by aesault. Its exterior slope must form at moat an angle of 45* with the horizon, that the earth may support itself, and con sequently it may be easily ascended ; and any palisades or other impedimenta which the defenders might place on it would soon be displaced or destroyed by the batteries of the enemy. Besides these evils, the exterior slope, from its brehdth, occupies a great portion of ground it consequently obliges the engineer to contract the apace enclosed within the works, and thus to sacrifico in some measure the convenience both of the inhabitants and the garrison.
In order to investigate the conditions of stability in revetment walls, let a a o be a vertical section through the mass of earth retained by the wall ; a c being the slope which earth is supposed to assume when unsupported, and let SENN be a section of the wall, r e being the level of the bottom of the ditch, and al m being the bottom of the foundation. Imagine G to be the centre of gravity of the section E C 13 ; draw c t. parallel to n c and c 1 parallel to the horizon : then, by the resolution of forces, e and G a will have to one another the same proportion that the weight of the unsupported prism of earth (of any thickness) beans to its horizontal pressure. Let w be that weight ; then K U K 0 "`,- will express that pressure, and .w e will be the momentum 1 L or power by which the earth tends to overturn the wall about al.
Imagine the vertical lines q to be drawn ; then the form and dimensions of the part a u e of the wall are known; and let it be required to find the breadth el N of the rectangular part a N, so that the resistance of the whole shall be equal to the momentum of the supported earth. Suppose the centre of gravity of s as Q to be found, and let it be vertically over a. The centre of gravity of the rectangular part is vertically over b, the middle of Q N ; and let Q b be represented by r. Then if g be the specific gravity of the wall, we have by mechanics, A SI Q.un.e+se.u6.2 for the resistance of the wall; quently equating this expression with the above momentum of the earth, the value of .e, and therefore of Q ST, can be found. But great uncertainty exists respecting the position of the line of rupture & c, from our ignorance of the allowance to be made for the effect of friction on the tendency of the earth to slide downwards. Experiments
have led to the opinion that this effect is equal to half the pressure of the earth perpendicularly upon the inclined plane which it would assume if unsupported ; and that value is frequently adopted. Coulomb showed that the angle which the line of rupture makes with the vertical is half the angle which the line of natural slope makes with the vertical.
In order to find the magnitude which the triangle E D c should have when the supported earth exerts the greatest pressure against the wall, the following process may be used ; the earth above a D being at present, for simplicity, supposed to be removed. Imagine c to be the centre of gravity of that triangle, and the vertical line c a to be drawn; then it may represent the weight of the unsupported earth, and let it be resolved into the pressures represented by c r and t a, the former per pendicular to the slope, and the latter coincident with it. Imagine u s to be drawn to represent the reaction of the wall s Y e, and let it. be resolved into the forces represented by s a and it 11, perpendicular to and coincident with the slope, respectively. Then, r 11 representing the force with which the prism of earth would tend, if without friction, to elide down D C, a a represents the reaction by which the wall resists that force ; while o 1 and s a represent the pressure and reaction per pendicular to D C. Consequently, the friction being supposed to be equal to half the pressure, we have (at +a it) for the effect of friction; and in the case of equilibrium, IH=RII+i(GI-1-311).
Let E c =A, E D = r, a s=r, and let g be the specific gravity of the hg: earth ; then expresses the weight of the prism whose section is E D c, and whose thickness is unity, and which was represented by G and the triangles a I /I, II s a being similar to E C D, we get by propor tions I= R =11, G I = and s R= P . These values 2cD C 2cD C D being substituted in the above equation, the value of the pressure rt s or is will be found to be g . h . Now this quantity is to be 2 2 zs-A a maximum ; therefore making its differential relatively to z equal to zero, the value of z will be found to be 618/1; whence p=.1903g If this equation be differentiated relatively to h, the result will express the horizontal pressure against an elementary portion of the will at a variable height (which represent by A) above c : therefore multiplying by b and integrating, we get '1272 g for the whole force exerted by the earth to overturn about is a wall whose height se is represented by h, when that force is a maximum. When there is a parapet above D, its weight, expressed by the product of the area of the section multiplied by g, must be added to the above expression for the weight of the prism E D C in the preceding investigation, in order to obtain the value of the expression which is represented by a it.