 | Sir Richard Phillips - Physics - 1832 - 286 pages
...Observation 1. The centre of gravity of a circle, sphere, or any regular body, is its centre. The centre of gravity of a parallelogram, is at the intersection of its diagonals. 2. To find the centre of gravity of a triangle, draw lines from any two angles to bisect their opposite... | |
 | Building - 1905 - 874 pages
...distance of the center of gravity of a triangle from the base is equal to one-third of the altitude. The center of gravity of a parallelogram is at the intersection of its two diagonals; consequently, it is midway between its sides. The center of gravity of an irregular... | |
 | Correspondence schools and courses - 1905 - 1026 pages
...distance of the center of gravity of a triangle from the base is equal to one-third of .the altitude. The center of gravity of a parallelogram is at the intersection of its two diagonals; consequently, it is midway between its sides. The center of gravity of an irregular... | |
 | Agriculture - 1905 - 854 pages
...distance of the center of gravity of a triangle from the base is equal to one-third of the altitude. The center of gravity of a parallelogram is at the intersection of its two diagonals; consequently, it is midway between its sides. The center of gravity of an irregular... | |
 | James Joseph Sylvester - Mathematics - 1908 - 751 pages
...to that of qualiform figures of the same kind. Thus, to take a most simple example, since the centre of gravity of a parallelogram is at the intersection of its diagonals, it must be and is true that the centre of gravity of a quadrilateral whose density at any point varies... | |
 | 1910 - 528 pages
...two-thirds of its length from the vertex. Thus, the center of gravity O in the figure is at a distance AO from A, equal to two-thirds of the length of the line...gravity of a parallelogram is at the intersection of ita diagonals, as AB and CD in Fig. 7. A parallelogram is a figure having four sides, the opposite... | |
 | Building - 1910 - 636 pages
...distance of the center of gravity of a triangle from the base is equal to one-third of the altitude. The center of gravity of a parallelogram is at the intersection of its two diagonals; consequently, it is midway between its sides. F10. 33 The center of gravity of an irregular... | |
 | International Correspondence Schools - Civil engineering - 1913 - 456 pages
...3 FIG. 4 axis, the intersection of the two axes will be the center of gravity of the section; thus, the center of gravity of a parallelogram is at the intersection of the diagonals and that of a circle or an ellipse is at the geometrical center of the figure. The center... | |
 | John Anthony Miller, Scott Barrett Lilly - Mechanics, Analytic - 1915 - 328 pages
...Therefore, the center of gravity is at the intersection of the medians. (c) The student may prove that the center of gravity of a parallelogram is at the intersection of the diagonals of the parallelogram. (d) The center of gravity of a regular hexagon is at its center.... | |
 | Thomas J. Foster - Coal mines and mining - 1916 - 1230 pages
...any other axis, the intersection of the two axes will be "the center of gravity of the section; thus, the center of gravity of a parallelogram is at the intersection of the diagonals and that of a circle or an ellipse is at the geome- . _ trical center of the figure.... | |
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The perpendicular distance of the center of gravity of a triangle from the base is equal to one-third of the altitude. The center of gravity of a parallelogram is at the intersection of its two diagonals; consequently, it is midway between its sides. 
