Radius of Curvature. 83. The circle AmB, fig. 22, which coincides with a curve or curved surface through an indefinitely small space on each side of m the point of contact, is called the curve of equal curvature, or the osculating circle of the curve MN, and om is the radius of curvature. In a plane curve the radius of curvature r, is expressed by M A D fig. 22. √(d3x)2+(d2y)2+ (d2z)2 ds being the constant element of the curve. Let the angle com be represented by 0, then if Am be the indefinitely small but constant element of the curve MN, the triangles com and ADm are similar; hence mA : mD :: om : mc, or ds: In the same manner cos 0= dy ds ; also d. sin ◊ = ; but these evidently become dr ds But d. cos 0= de sin e, and do = d.cos 0 dx d dy ds d2x de = + , dy and de dzy dx Now if om the radius of curvature be represented by r, then moA being the indefinitely small increment de of the angle com, we have r: ds :: 1: de; for the sine of the infinitely small angle is to be considered as coinciding with the arc: hence de = and adding one to each side of the last equation, it becomes (day) ds = √ (d2x)2 + (d3y)2 dxds But it has been shown that r= ; hence in a plane curve the dzy radius of curvature is ds2 T= the projection of the radius of curvature on x o y, and it is evident that a similar expression will be found for the projection of the radius of curvature on each of the other co-ordinate planes. In fact } √(d2x)2 +(d2y) is the sagitta of curvature nE; for for the arc being indefinitely small, the tangent may be considered as coinciding with it. Thus the three projections of the sagitta of curvature of the surface, or curve of double curvature, are } √(a2x)2 + (dy)*; } √(d*x)2+(d°z)2; } √(ď2y)2+(d°z)3 ; hence the sum of their squares is } √(d2x)2 + (d2y)2 + (d2z)* ; and the radius of curvature of a surface, or curve of double curvature, is Pressure of a Particle moving on a curved Surface. 84. If the particle be moving on a curved surface, it exerts a pressure which the surface opposes with an equal and contrary pressure. Demonstration. For if F be the resulting force of the partial accelerating forces X, Y, Z, acting on the particle at m, it may be resolved into two forces, one in the direction of the tangent mT, and the other in the normal mN, fig. 12. The forces in the tangent have their full effect, and produce a change in the velocity of the particle, but those in the normal are destroyed by the resistance of the surface. If the particle were in equilibrio, the whole pressure would be that in the normal; but when the particle is in motion, the velocity in the tangent produces another pressure on the surface, in consequence of the continual effort the particle makes to fly off in the tangent. Hence when the particle is in motion, its whole pressure on the surface is the difference of these two pressures, which are both in the direction of the normal, but one tends to the interior of the surface and the other from it. The velocity in the tangent is variable in consequence of the accelerating forces X, Y, Z, and becomes uniform if we suppose them to cease. Centrifugal Force. 85. When the particle is not urged by accelerating forces, its motion is owing to a primitive impulse, and is therefore uniform. In this case X, Y, Z, are zero, the pressure then arising from the velocity only, tends to the exterior of the surface. And as v the velocity is constant, if ds be the element of the curve described in the time dt, then therefore ds is constant; and when this value of dt is substituted in equation (7), in consequence of the values of X', Y', Z', in the end for by article 81 the particle may be considered as free, whence The first member of this equation was shown to be the pressure of the particle on the surface, which thus appears to be equal to the square of the velocity, divided by the radius of curvature. 86. It is evident that when the particle moves on a surface of unequal curvature, the pressure must vary with the radius of cur vature. 87. When the surface is a sphere, the particle will describe that great circle which passes through the primitive direction of its motion. In this case the circle AmB is itself the path of the particle; and in every part of its motion, its pressure on the sphere is equal to the square of the velocity divided by the radius of the circle in which it moves; hence its pressure is constant. 88. Imagine the particle attached to the extremity of a thread assumed to be without mass, whereof the other extremity is fixed to the centre of the surface; it is clear that the pressure which the particle exerts against the circumference is equal to the tension of the thread, provided the particle be restrained in its motion by the thread alone. The effort made by the particle to stretch the thread, in order to get away from the centre, is the centrifugal force. Hence the centrifugal force of a particle revolving about a centre, is equal to the square of its velocity divided by the radius. 89. The plane of the osculating circle, or the plane that passes through two consecutive and indefinitely small sides of the curve described by the particle, is perpendicular to the surface on which the particle moves. And the curve described by the particle is the shortest line that can be drawn between any two points of the surface, consequently this singular law in the motion of a particle on a surface depends on the principle of least action. With regard to the Earth, this curve drawn from point to point on its surface is called a perpendicular to the meridian; such are the lines which have been measured both in France and England, in order to ascertain the true figure of the globe. 90. It appears that when there are no constant or accelerating forces, the pressure of a particle on any point of a curved surface is equal to the square of the velocity divided by the radius of curvature at that point. If to this the pressure due to the accelerating forces be added, the whole pressure of the particle on the surface will be obtained, when the velocity is variable. 91. If the particle moves on a surface, the pressure due to the centrifugal force will be equal to what it would exert against the M fig. 24. m curve it describes resolved in the direction of the normal to the surface in that point; that is, it will be equal to the square of the velocity divided by the radius of the osculating circle, and multiplied by the sine of the angle that the plane of that circle makes with the tangent plane to the surface. Let MN, fig. 24, be the path of a particle on the surface; mo the radius of the osculating circle at m, and mD a tangent to the surface at m; then om being radius, oD is the sine of the inclination of the plane of the osculating circle on the plane that is tangent to the surface at m, the centrifugal force is equal to v2 x oD om If to this, the part of the pressure which is due to the accelerat |