The correction for the immersion of the sides of the bell is of greater magnitude, and has an important practical significance. Let H be the total height in inches of the side of the gasholder, _h_ the height in inches of the top of the sides of the gasholder above the water-level, and _w_ = the weight of the sides of the gasholder in lb.; then, for any position of the bell, the proportion of the total height of the sides immersed (H - _h_)/H, and the buoyancy is (H - _h_)/H x _w_/S + pi/4_d^2_, in which S = the specific gravity of the material of which the bell is made. Assuming the material to be mild steel or wrought iron, having a specific gravity of 7.78, the buoyancy is (4_w_(H - _h_)) / (7.78Hpi_d^2_) lb. per square inch (_d_ being inches and _w_ lb.), which is equivalent to (4_w_(H - _h_)) / (0.03604 x 7.78Hpi_d^2_) = (4.54_w_(H - _h_)) / (H_d^2_) inches of water. Hence the complete formula for acetylene gasholders is: _p_ = 35.333W / _d^2_ - 4.54_w_(H - _h_) / H_d^2_ It follows that _p_ varies with the position of the bell, that is to say, with the extent to which it is filled with gas. It will be well to consider how great this variation is in the case of a typical acetylene holder, as, if the variation should be considerable, provision must be made, by the employment of a governor on the outlet main or otherwise, to prevent its effects being felt at the burners. Now, according to the rules of the "Acetylen-Verein" (_cf._ Chapter IV.), the bells of holders above 53 cubic feet in capacity should have sides 1.5 mm. thick, and crowns 0.5 mm. thicker. Hence for a holder from 150 to 160 cubic feet capacity, supposing it to be 4 feet in diameter and about 12 feet high, the weight of the sides (say of steel No. 16 S.W.G. = 2.66 lb. per square foot) will be not less than 12 x 4pi x 2.66 = 401 lb. The weight of the crown (say of steel No. 14 S.W.G. = 3.33 lb. per square foot) will be not less than about 12.7 x 3.33 = about 42 lb. Hence the total weight of holder = 401 + 42 = 443 lb. Then if the holder is full, _h_ is very nearly equal to H, and _p_ = (35.333 x 443) / 48^2 = 6.79 inches. If the holder stands only 1 foot above the water-level, then _p_ = 6.79 - (4.54 x 401 (144 - 12)) / (144 x 48^2) = 6.79 - 0.72 = 6.07 inches. The same result can be arrived at without the direct use of the second member of the formula: