Thence the horizontal pressure against P _p_, and = (P _p_ . P R . R D .) / (P G), that is to say P _p_ . P R = R _r_ . P G, the given horizontal pressure is found to be = R _r_ . R D = (_b_ - _x_) _d_ _x_, and which, multiplied by R D, giving _b_ - _x_, becomes the momentum of the pressure relatively to M N = (_b_ - _x_)² _d_ _x_, and the sum of the momenta of pressure exercised upon the indefinite arc, O P = _f_ (_b_ - _x_)² _d_ _x_ = -(1/3)(_b_ - _x_)³ + the side. And since acting together such momenta equal _x_, there comes the side = (1/3)_b_³; and as the already-given sum of the momenta = (1/3)(_b_³ - (_b_ - _x_)³) = _b² x_ - _b x_² + (1/3)_x_³. Whence, taking _x_ = 2_a_, the sum of all the momenta of the horizontal pressure exercised on the whole semi-circumference O L F of the wheel, will be = 2_b_²_a_ - 4_b_ _a_² + (8/3)_a_³, and dividing that sum by the whole horizontal pressure, that is to say by _f_(_b_ - _x_)_d_ _x_ = (1/2)(_b_² - (_b_ - _x_)²) = _b_ _x_ - (1/2)_x_² = 2_b_ _a_ - 2_a_², gives _x_ = 2_a_, we have the formula (2_b_² - 4_b_ _a_ + (8/3)_a_³) / (2_b_ _a_ - 2_a_²) = (_b_² - 2_b_ _a_ + (4/3)_a_²) / (_b_ - _a_) = ((_b_ - _a_)² + (1/3)_a_²) / (_b_ - _a_) = _b_ - _a_ + ((2/3)_a_²) / (_b_ - _a_), which represents the distance of the level M N from the result of all the horizontal pressure against the circumference, which distance exceeds D C, and consequently the direction of the result passes from below the centre C of the wheel to a distance from the said centre, which is = ((1/3)_a_²)/(_b_ - _a_). If this distance be multiplied by the result of all the horizontal pressure, that is, by 2_a_.(_b_ - _a_); there is obtained (2/3)_a_³ for the momentum of the force which tends to make the wheel revolve from L towards O. This being established, it is known that the force which causes the half of the wheel F L G to revolve vertically to the top (calling _g_ the specific gravity of the wheel) is = (1 - _g_) F C O L, and which force passes through the center of gravity of F L O. And consequently the gravity of any circular segment divided by the half of the radius, is distant from the centre of the circle by a quantity equal to the twelfth of the cube of the chord divided by the segment; and therefore the centre of gravity of the semicircle F C O L, will be distant from the centre C by the quantity (1/12)8_a_³/(E C O L) = (2/3)_a_³/(E C O L). Consequently the momentum of this force tending to make the wheel revolve from O towards L will be = (2/3_a_³)/(E C O L) . (1 - _g_)(E C O L) = 2/3(1 - _g_)_a_³.