In the case of an axial moment, the square root of the resulting mean square is called the radius of gyration of the system about the axis in question.
Which we shall meet with presently as the ellipsoid of gyration at G.
Since I~=Ii., I~=o, we deduce 100=3/4Ma2, ~ =4MaZ; hence the value of the squared radius of gyration isfora diameter 3/4ai, and for the axis of symmetry 3/4af.
From its axis (0), if the radius of gyration about a longitudinal axis through G, aiid 0 the inclin - ation of OG to the vertical, FIG.
If M be the total mass, k the radius of gyration (~ ii) about the axis, we have sin 0, (3)
The radius of gyration of the section is 2a 2.
After a certain discount for friction and the recoil of the gun, the net work realized by the powder-gas as the shot advances AM is represented by the area Acpm, and this is equated to the kinetic energy e of the shot, in foot-tons, (I) e d2 I + p, a in which the factor 4(k 2 /d 2)tan 2 S represents the fraction due to the rotation of the shot, of diameter d and axial radius of gyration k, and S represents the angle of the rifling; this factor may be ignored in the subsequent calculations as small, less than I %.
The square of the radius of gyration with respect to a diameter is ia2.
The formula (16) expresses that the squared radius of gyration about any axis (Ox) exceeds the squared radius of gyration about a parallel axis through G by the square of the distance between the two axes.
The squares of the radii of gyration about the principal axes at P may be denoted by k,i+k32, k,f + ki2, k12 + k,2 hence by (32) and (35), they are rfOi, r2Oi, r20s, respectively.
This is called the ellipsoid of gyration at 0; it was introduced into the theory by J.
It possesses thi property that the radius of gyration about any diameter is half thi distance between the two tangents which are parallel to that diameter, In the case of a uniform triangular plate it may be shown that thi momental ellipse at G is concentric, similar and similarly situatec to the ellipse which touches the sides of the triangle at their middle points.
If k be the radius of gyration about p we find k2 =2Xarea AHEDCBAXONap, where a$ is the line in the force-diagram which represents the sum of the masses, and ON is the distance of the pole 0 from this line.
If K be the radius of gyration about a parallel axis through G, we have kf=K2+h2 by If (16), and therefore i=h+K1/h, whence GO.GP=K2.
Where K is the radius of gyration about the axis of symmetry, a is the constant distance of G from the plane, and R, F are the normal and tangential components of the reaction of the plane, as shown in fig.
Let W be the weight of a flywheel, R its radius of gyration, ai its maximum, aj its minimum, and A=~1/8(a1+ai) its mean angular velocity.
R is called the radius of gyration of the body with regard to an axi:
The velocity of a liquid particle is thus (a 2 - b 2)/(a 2 +b 2) of what it would be if the liquid was frozen and rotating bodily with the ellipse; and so the effective angular inertia of the liquid is (a 2 -b 2) 2 /(a 2 +b 2) 2 of the solid; and the effective radius of gyration, solid and liquid, is given by k 2 = 4 (a 2 2), and 4 (a 2 For the liquid in the interspace between a and n, m ch 2(0-a) sin 2E 4) 1 4Rc 2 sh 2n sin 2E (a2_ b2)I(a2+ b2) = I/th 2 (na)th 2n; (8) and the effective k 2 of the liquid is reduced to 4c 2 /th 2 (n-a)sh 2n, (9) which becomes 4c 2 /sh 2n = s (a 2 - b 2)/ab, when a =00, and the liquid surrounds the ellipse n to infinity.
The moment of inertia of the body about the axis, denoted by But if is the moment of inertia of the body about a mean diameter, and w the angular velocity about it generated by an impluse couple M, and M' is the couple required to set the surrounding medium in motion, supposed of effective radius of gyration k', If the shot is spinning about its axis with angular velocity p, and is precessing steadily at a rate about a line parallel to the resultant momentum F at an angle 0, the velocity of the vector of angular momentum, as in the case of a top, is C i pµ sin 0- C2µ 2 sin 0 cos 0; (4) and equating this to the impressed couple (multiplied by g), that is, to gN = (c 1 -c 2)c2u 2 tan 0, (5) and dividing out sin 0, which equated to zero would imply perfect centring, we obtain C21 2 cos 0- (c 2 -c 1)c2u 2 sec 0 =o.