 | Mansfield Merriman - Geodesy - 1903 - 274 pages
...residual errors is to be made a minimum in order to find the most probable values of an observed quantity. The radius of gyration with respect to an axis through the center of gravity bears also an analogy to the probable error. Thus the Method of Least Squares may be justified... | |
 | Mansfield Merriman - Geodesy - 1899 - 274 pages
...residual errors is to be made a minimum in order to find the most probable values of an observed quantity. The radius of gyration with respect to an axis through the center of gravity bears also an analogy to the probable error. Thus the Method of Least Squares may be justified,... | |
 | William Brandon Gordon - Mechanics - 1914 - 296 pages
...weight of the pendulum, d\/dP= -Mgn smy/(Mtf+Mh*), (409) k being the radius of gyration of the pendulum with respect to an axis through the center of mass and parallel to the axis of suspension. Making the same approximation as for the simple pendulum and integrating twice, we have from Eq. (409),... | |
 | Halsey Dunwoody - Graphic statics - 1917 - 390 pages
...a horizontal axis, under the action of gravity only, is (k2 •=7r\r » • •.*.... \<jj./ &A A: being the radius of gyration with respect to an axis...parallel to the axis of suspension. This value of ^ is closely approximate when the arc of oscillation is small. Square the above equation for T, multiply... | |
 | James Ellsworth Boyd - Mechanics, Applied - 1921 - 442 pages
...the center of gravity and then to transfer from the center of gravity to the second axis. If fc0 is the radius of gyration with respect to an axis through the center of gravity, /o = mkl. If k is the radius of gyration with respect to the axis OO', fc2 = - = mkl + md*... | |
 | James Ellsworth Boyd - Strength of materials - 1924 - 456 pages
...diameter. Solve also for an axis 20 inches and for an axis 40 inches from the center. 6. If fco is the radius of gyration with respect to an axis through the center of gravity and k is the radius of gyration with respect to a parallel axis at a distance d from the center... | |
 | James Ellsworth Boyd - Strength of materials - 1924 - 456 pages
...a diameter. Solve also for an axis 20 inches and for an axis 40 inches from the center. 5. If k0 is the radius of gyration with respect to an axis through the center of gravity and k is the radius of gyration with respect to a parallel axis at a distance d from the center... | |
 | John Presti, George Wetzel, James Colaizzi - Study Aids - 1999 - 504 pages
...where / is the moment of inertia of a rigid body about an axis, Icm is the moment of inertia about an axis through the center of mass and parallel to the axis of 7, h is distance between these two parallel axes, and M is the mass of the body. (a) A uniform sphere... | |
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... the radius of gyration with respect to an axis through the center of mass and parallel to the axis of suspension. 