It is more convenient, however, to use degrees for the angle *A, *inches for the lengths, pounds for the forces and grains for the masses. It is also desirable to have the constant C of such a value that *f=CA *where *f* is expressed in pounds and *A* is expressed in degrees. In order to use the units as just explained and still obtain the velocities and accelerations in feet per second and feet per second per second, respectively, the above formulas become:

v=[492*B*_{1}C(*A*'^{2}-*A*^{2})/(MR^{2}+m)]^{1/2}, *V=Rv, *

**a**=450,800*CA*/*(MR ^{2} + m*)

*-*7,868

*MRC*(

*A*'

^{2}-

*A*

^{2})Z/(

*MR*

^{2}+

*m*)

^{2},

**A**=450,800*CAR*/(MR^{2}+m)

+ 7,868*mC*(*A*'^{2}-*A*^{2})Z/(*MR*^{2} + *m*)^{2},

**f**=2.22*m***a** × 10^{-6}, **F**=4.44M**A** × 10^{-6},

**T**=**F**/2 sin *E*=*S***F**/2**P**=2.22*SM***A** × 10^{-6}/*P*

= (*f*-**f**)/sin *(A+E)*,

R=V/*v*=cos *A*+ (*Y/P*) sin *A*.

In order to show how these formulas may be used, they are now applied to a six-foot bow which has an eight-inch rigid middle section (*L*=4 inches), and which has a six-inch bracing height *(Ho= *6* *inches), and a draw *D*=27.5 inches.

From the equation *Ho*=(3*B*_{1}/4) sin *Ao*, *Ao*=14°-29'.

From the equation *S* = (3*B*_{1}/4) cos *Ao* + *B*_{1}/4 + *L, *S= 35.2373 inches.

The active bending portion of each limb, *B*_{1}=32 inches.

Let each limb have a constant thickness, *t=*0.6* *inch. Let the width at the dip equal *W*=1.5 inches. Let the width at any other point equal *w=Wx/B*_{1}*, *where *x *is the distance from the nock. This type bow is discussed in the January, 1932 issue of *Ye Sylvan Archer. *All sections of such a bow are equally stressed and the limbs bend in true arcs of circles.