But moreover a certain momentum will be derived from the other half F Q O of the wheel, which being out of the water, tends by its own weight downwards with a force = _g_ . (E C O Q) = _g_ . (E C O L), which multiplied by the distance (2/1_a_³)/(E C O L) of the centre of gravity of the semicircle F Q O from the centre of the wheel gives as a momentum of force tending to turn the wheel from O to L the quantity 2/3_g_ _a_³. Thus the whole momentum to make the wheel turn from O to L, will be 2/3(1 - _g_)_a_³, + 2/3_g_ _a_³ = 2/3_a_³, that is to say the same that is found to turn the wheel in the opposite direction, viz., from L to O, and thence the wheel remains perfectly motionless. 3. Cor. I. If the wheel were specifically heavier than the water, one would not be able to conceive in that case any motion from L to O, as seemed probable in the former supposition. Since, then, the momentum of the force, which turns vertically downwards the portion of the wheel F C O L, and tends to make it revolve from L to O is = 2/3(_g_ - 1)_a_³ to which momentum should be added a certain portion of the horizontal pressure, that is to say 2/3, and thus is obtained the whole momentum 2/3_g_ _a_³, tending to cause the wheel to turn from L to O; and to which momentum precisely, is equal such of the weight of the half F C O Q as tends to give to the wheel a contrary revolution, that is, from O to L. 3. Cor. II. If the wheel in place of being a circular plane were a zone bounded by two concentric peripheries (Fig. 3), then from the sum of the horizontal pressure of the water against the exterior periphery should be taken the sum of the opposite horizontal pressure against the other interior semi-periphery of the zone. So calling _a_ the greater radius of the zone, and λ its breadth, the sum of the first horizontal pressure is = 2_a_(_b_ - _a_) and the sum of the second = 2(_a_ - λ)(_b_ - λ) - (_a_ - λ) = 2(_a_ - λ)(_b_ - _a_). Then subtract the latter from the former and there remains 2(_b_ - _a_)λ for the sum of the whole pressure, which acts upon the zone (_sic_) of the half of the wheel immersed in the fluid in a direction tending from the outside to the interior of the wheel. Moreover the sum of the momenta of all the horizontal pressure on the exterior circumference relatively to the level M N is = 2_b_ _a_ - 4_b_ _a_ + 8/3_a_³. And similarly the sum of the momenta of the horizontal pressure opposite, on the interior semi-circumference, relatively to the given level is = 2(_b_ - λ)² - (_a_ - λ) - 4(_b_ - λ) × (_a_ - λ)² + 8/3(_a_ - λ)³.