Henry Gray (1825–1861). Anatomy of the Human Body. 1918.
pages 369
| will find that muscle pull can be resolved into two components, a turning component and a friction or pressure component as shown in Fig. 369. |
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FIG. 369– No caption. (See enlarged image) |
| D F = the fixed bone from which the muscle takes its origin. |
| D K = the movable bone. |
| O I = a line from the middle of origin to the middle of insertion. |
| I M = size and direction of the muscle pull. |
| If the parallelogram is constructed with I t and M b ⊥ to D K, then I t = the turning component and I b = the component which acts against the joint. |
| The size of the two components depends upon the insertion angle φ. The smaller this angle the smaller the turning component, and the nearer this angle φ is to 90° the larger the turning component. |
| I t = I M x sin φ |
| I b = I M x cos φ |
| If φ = 90° cos φ = 0, sine φ = 1 hence I b = 0 and I t = I m |
| If φ = 0° cos φ = 1, sine φ = 0 hence I b = 1 and I t = 0 |
| With movements of the bone D K the angle of insertion is continually changing, and hence the two components are changing in value. |
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FIG. 370– No caption. (See enlarged image) |
| If, for example, the distance from origin 0 to the joint D is greater than from D to I, as in the Brachialis or Biceps muscles, the turning component increases until the insertion angle φ = 90°, which is the optimum angle for muscle action, while the pressure component gradually decreases. If the movement continues beyond |