Page:A Treatise on the Steam Engine (1847).djvu/112

 $$\therefore c=\tfrac{asr+brs-a^2s}{br}=\tfrac{(ar+br-a^2)s}{br}$$

or, $$c=\tfrac{(a+b)r-a^2}{br} ds$$ ....(F)

It may be observed that, if c in formulas (D) and (E), is taken equal to nothing they become the same as formulas (B) and (C).

Having now investigated formulas for calculating the proportions of all the parallel motions we proposed to examine, we shall conclude by collecting these formulas together for the greater convenience of reference, and subjoin practical rules deduced from them for the use of those who prefer rules to algebraic formulas.

The sum of the lengths of the radius bar, and the part of the beam (C G) which works it = a = Horizontal distance of the centre of the radius bar (H) from the main centre, plus half the versed sine of the arc described by extremity of the beam (D).

Radius of side lever or engine beam, (D C) = b Length of radius bar (H F) = r Length of side rod (P D) = s Part of side rod above the centre of radius bar (P Q) = c Part of connecting link intercepted between the beam and the point E (G E) = p The remaining part of the connecting link (F E) = p’  Length of that part of the beam that works the radius bar (G C) = r’ 

$$r=\tfrac{r' \cdot p}{p'}$$

$$p'=\tfrac{r'(p+p')}{(r+r')}$$ ... Figs. 1 and 2 ... (A)