Henry Gray (1825–1861). Anatomy of the Human Body. 1918.  
 Δ A D V = work diagram, in reality the hypothenose is not straight but has a concave curve. The Δ has the same area as the rectangle A M M’ V. 
A M = the average tension. 
Work = A M x A V kilogrammeters if the size of the ordinate as expressed in kilograms and the abscissa in meters. 

 FIG. 366– No caption. (See enlarged image)

Although the muscle works with a changing tension, yet the accomplishment is the same as if it were contracting with the tension of the midposition. 
In reality the amount of work is somewhat greater since even in extreme contraction the muscle still retains a certain amount of tension so that the maximum amount of work is more nearly like A D X. We know that a muscle may have an extreme actual shortening of about 80 per cent. of its length when the tendon of insertion is cut. 
The trapezoid A D S V represents more nearly the amount of work, but since there are only approximate values and A D S V is not much larger than A M M’ V, we may use the latter. 
Only the tension and amount of shortening are needed to determine the amount of work of the muscle. Neither the lever arm nor the fiber angle in pinnate muscles need be considered. 
The diagram Fig. 367 shows that the lever arm is of no importance for determining the amount of work the muscle performs. 
J B and J B^{1} = two bones jointed at J. C D and E F = the direction of the pull of two muscles of equal crosssection, each having a muscle tension of 1000 gms. 
The centers of the attachments are such that perpendiculars J c and J e to C D and E F are equal to 40 and 23 mm. respectively, J c = 40 mm. and J e = 23 mm. The static moments are equal to 1000 x 40 and 1000 x 23, therefore the first muscle can hold a much larger load (L) on the bone J B^{1} at H^{1} (100 mm. from J) than the second muscle whose load can be designated as L^{1}. 
Equilibrium exists for the first muscle if
L x 100 = 1000 x 40 or L = 1000 x 40/100 = 400 gms. 
For the second muscle L^{1} x 100 = 1000 x 23.
L^{1} = 1000 x 23/100 = 230 gms. 
If we suppose J B to be fixed and J B^{1} to move in the plane of the paper about J and the muscle C D to shorten 5 mm. C d = C D − 5 mm. and with the tension of 1000 gms., J B^{1} will take the position J B^{2} and the load (L) will be lifted from H^{1} to H^{2}. 

