Perpetual Motion

by Percy Verance; Henry Dircks

Discussion by P. Gregorio Fontana

P. Gregorio Fontana was professor of higher mathematics at the Royal University of Pavia, in the Province of Lombardy, Italy. In 1786 he published what he designated "Examination of a New Argument in Favor of Perpetual Motion." In part he says:

1. A vertical wheel (Fig. 2) divided in two halves by a vertical plane which passes through its diameter F O, has the half F P O immersed in water under the level M N, and the other half wholly out of the water, being cut off in F O by a peculiar mechanism from all communication with the reservoir, the exterior half of the wheel being F Q O; this turns freely round on an axle passing through the centre C. Now the wheel being specifically lighter than the water, the immersed part F P O comes with a continual rotation to the top with a force equal to the excess of the weight of a volume of water corresponding to the immersed portion, over the weight of the immersed portion; which rotation passing through the centre of gravity of the exterior part, and consequently out of the centre C, obliges the wheel to turn around C.

Such being the case, the question to be asked is whether the wheel has itself a perpetual motion, as may be judged at first sight.

2. To reply adequately, it is at first necessary to know what effect is produced on the wheel by the horizontal pressure which the water exercises on the semi-circumference F L O.

Having taken for this purpose, a part P p, and having drawn to the diameter the ordinate P. R, p r, and marked the radius P C, and from it P G perpendicular to the radius C L, which determines the quadrant O L, the distance of the lowest point O from the level of the water will be = b, the semi-diameter of the wheel = a, C R = x, and the specific gravity of the water = 1; the perpendicular pressure against the part P p = P p . R D, which resolved in two, one horizontal P R, the other vertical P G, gives the proportion

PG : PR :: Pp . RD : (Pp . PR . RD) / (PG).

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)b³; and as the already-given sum of the momenta = (1/3)(b³ - (b - x)³) = b² x - b x² + (1/3)x³. Whence, taking x = 2a, the sum of all the momenta of the horizontal pressure exercised on the whole semi-circumference O L F of the wheel, will be = 2b²a - 4b a² + (8/3)a³, and dividing that sum by the whole horizontal pressure, that is to say by f(b - x)d x = (1/2)(b² - (b - x)²) = b x - (1/2)x² = 2b a - 2a², gives x = 2a, we have the formula

(2b² - 4b a + (8/3)a³) / (2b a - 2a²) = (b² - 2b a + (4/3)a²) / (b - a) = ((b - a)² + (1/3)a²) / (b - a) = b - a + ((2/3)a²) / (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)a²)/(b - a).

If this distance be multiplied by the result of all the horizontal pressure, that is, by 2a.(b - a); there is obtained (2/3)a³ 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)8a³/(E C O L) = (2/3)a³/(E C O L). Consequently the momentum of this force tending to make the wheel revolve from O towards L will be = (2/3a³)/(E C O L) . (1 - g)(E C O L) = 2/3(1 - g)a³.

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 (2/1a³)/(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 2/3g a³. Thus the whole momentum to make the wheel turn from O to L, will be 2/3(1 - g)a³, + 2/3g a³ = 2/3a³, 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)a³ 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 2/3g a³, 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 = 2a(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 = 2b a - 4b a + 8/3a³.

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.

Subtracting this sum from the preceding, there remains the sum of the momenta acting on the zone of the half-wheel from the exterior to the interior = 2b² a - 4b a² + 8/3a³ - 2(b - λ)² (a - λ) + 4(b - λ) (a - λ)² - 8/3(a - λ)³ - 2b² λ - 4b a λ + 4a² λ - 2a λ² + 2/3λ³ = 2λ (b(b - a) - b a + 2a² - aλ + 1/3λ²) = 2λ ((b - a)(b - a) + a² - aλ + 1/3λ²) Then dividing this sum of the momenta by the sum of the pressure there will be 2λ(((b - a)(b - a) + a² - aλ + 1/3λ²)/(2λ(b - a))) = b + p (a(a² - aλ + 1/3λ²)/(b - a)) the distance of the center of the pressure from the level of the fluid, that is, to the distance of the result of all the pressure from that level. From this it is evident that the center of pressure falls under the center of the wheel, C, to the distance (a² - aλ + 1/3λ²)/(b - a) .

Whence multiplying this distance by the result of the pressure, or by 2λ(b - a), we obtain 2λ(a² - aλ + 1/3λ²) to express the momentum of the horizontal pressure of the water, directed to make the wheel turn from L to O.

Now the momentum with which the vertical impulse of the fluid tends to make the semicircle F C O L turn from O to L (supposing the wheel not with a simple zone, but with a circular plane) is = 2/3a³. Likewise the momentum of the impulse of the fluid to cause the internal semicircle V C I G from O to L is - 2/3(a - λ)³. Then taking this second momentum from the first, the momentum of the zone from the fluid V G I O L F to give the wheel an impulse from O to L will be = 2/3(a³ - (a - λ)³) = 2λ(a² - aλ + 1/3λ²) which is precisely the momentum with which the horizontal pressure of the fluid to impress on the wheel an impulse in the opposite direction, that is to say from L to O. Consequently from the pressure of the fluid the wheel cannot have any motion around its center.

The weight of the wheel itself, by which the half-zone immersed in the water tends to make the wheel turn from L to O, and the half which is out of the water, to make it turn in the reverse direction, such a weight, I say, cannot induce any motion of rotation, and both halves remain in equilibrium around the center C.