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Henry Gray (1825–1861). Anatomy of the Human Body. 1918.

pages 367

If the second muscle likewise shortens 5 mm. then E f = E F — 5 mm., and with the tension of 1000 gms. the bone J B1 will take the position J B3 and the weight or load (L1) will be lifted from H1 to H3. The question now is to prove that the work done is the same in both cases, namely, 5 x 1000 grammillimeters. If so, 400 x H1 H2 = 230 x H1 H3 = 5000 grammillimeters.

Since the two radii C d and C d’ are very long as compared with the arc d d’ we may consider this short arc as a line ?? to C D at d’, likewise the arc f f’ may be considered as a straight line ?? to E F. In the same manner we can consider the short arcs F f, D d, H1 H2 and H1 H3 ?? to the line J B1. The sides D d’ and F f’ of the Δ D d d’ and F f f’ are each 5 mm.

The lever arm D J = 60 mm. and J F = 30 mm.

FIG. 367– No caption. (See enlarged image)

The Δ D d d’ is similar to the Δ D c J

hence  D d : 5 :: 60 : 40  D d = 300/40 also     H1 H2 : D d :: 100 : 60 H1 H2 : 300/40 :: 100 : 60   H1 H2 = 300/24

hence F f : 5 :: 30 : 23    F f = 150/23 also H1 H3 : F f :: 100 : 30 H1 H3 :150/23::100:30    H1 H3 = 1500/69 &#133 400 x 300/24 = 230 x 1500/69 = 5000

Thus we see that the work of the two muscles depends on the size of the contraction and on the tension and not on the lever arm in very small contractions or in