| The Outward. | The Inward. |
| 1.3} | 4.1} |
| 7.2} | 7.0} The sum 19. |
| 9.0} The sum 24. | 5.2} |
| 5.3} | 2.1} |
| 0.0} | |
[Near five pages are occupied in describing the use of this screw, and the form and manner of making it; then follows:]
The true inclination of the screw being found, together with the certain quantity of water which every helix does contain; it is further considerable, that the water by this instrument does ascend naturally of itself, without any violence or labor; and that the heaviness of it does lie chiefly upon the centers or axis of the cylinder, both its sides being of equal weight (said Ubaldus); so that, it should seem, though we suppose each revolution to have an equal quantity of water, yet the screw will remain with any part upwards, according as it shall be set, without turning itself either way; and, therefore, the least strength being added to either of its sides should make it descend, according to that common maxim of Archimedes—any addition will make that which equiponderates with another to tend downwards.
But now, because the weight of this instrument and the water in it does lean wholly upon the axis, hence is it (said Ubaldus) that the grating and rubbing of these axes against the sockets wherein they are placed, will cause some ineptitude and resistency to that rotation of the cylinder; which would otherwise ensue upon the addition of the least weight to any one side; but (said the same author) any power that is greater than this resistency which does arise from the axis, will serve for the turning of it round.
These things considered together, it will hence appear how a perpetual motion may seem easily contrivable. For, if there were but such a water-wheel made on this instrument, upon which the stream that is carried up may fall in its descent, it would turn the screw round, and by that means convey as much water up as is required to move it; so that the motion must needs be continual, since the same weight which in its fall does turn the wheel is, by the turning of the wheel, carried up again.
Or, if the water, falling upon one wheel, would not be forcible enough for this effect, why then there might be two or three, or more, according as the length and elevation of the instrument will admit; by which means the weight of it may be so multiplied in the fall that it shall be equivalent to twice or thrice that quantity of water which ascends; as may be more plainly discerned by the following diagram:
Where the figure L M, at the bottom, does represent a wooden cylinder with helical cavities cut in it, which at A B is supposed to be covered over with tin plates, and three water-wheels upon it, H I K; the lower cistern, which contains the water, being C D. Now, this cylinder being turned round, all the water which from the cistern ascends through it, will fall into the vessel at E, and from that vessel being conveyed upon the water-wheel H, shall consequently give a circular motion to the whole screw. Or, if this alone should be too weak for the turning of it, then the same water which falls from the wheel H, being received into the other vessel F, may from thence again descend on the wheel I, by which means the force of it will be doubled. And if this be yet unsufficient, then may the water which falls on the second wheel I, be received into the other vessel G, and from thence again descend on the third wheel at K; and so for as many other wheels as the instrument is capable of. So that, besides the greater distance of these three streams from the center or axis by which they are made so much heavier, and besides that the fall of this outward water is forcible and violent, whereas the ascent of that within is natural—besides all this, there is thrice as much water to turn the screw as is carried up by it.
But, on the other side, if all the water falling upon one wheel would be able to turn it round, then half of it would serve with two wheels, and the rest may be so disposed of in the fall as to serve unto some other useful delightful ends.
When I first thought of this invention, I could scarce forbear, with Archimedes, to cry out εὕρηκα, εὕρηκα {heurêka, heurêka}; it seeming so infallible a way for the effecting of a perpetual motion that nothing could be so much as probably objected against it; but, upon trial and experience, I find it altogether insufficient for any such purpose, and that for these two reasons:
1. The water that ascends will not make any considerable stream in the fall.
2. This stream, though multiplied, will not be of force enough to turn about the screw.
1. The water ascends gently, and by intermissions; but it falls continually, and with force; each of the three vessels being supposed full at the first, that so the weight of the water in them might add the greater strength and swiftness to the streams that descend from them. Now, this swiftness of motion will cause so great a difference betwixt them that one of these little streams may spend more water in the fall than a stream six times bigger in the ascent, though we should suppose both of them to be continuate; how much more, then, when as the ascending water is vented by fits and intermissions, every circumvolution voiding so much as is contained in one helix; and, in this particular, one that is not versed in these kind of experiments may be easily deceived.
But, secondly, though there were so great a disproportion, yet, notwithstanding, the force of these outward streams might well enough serve for the turning of the screw, if it were so that both its sides would equiponderate the water being in them (as Ubaldus had affirmed). But now, upon farther examination, we shall find this assertion of his to be utterly against both reason and experience. And herein does consist the chief mistake of this contrivance; for the ascending side of the screw is made, by the water contained in it, so much heavier than the descending side, that these outward streams, thus applied, will not be of force enough to make them equiponderate, much less to move the whole, as may be more easily discerned by this figure:
Where A B represents a screw covered over, C D E one helix or revolution of it, C D the ascending side, E D the descending side, the point D the middle; the horizontal line C F showing how much of the helix is filled with water, viz., of the ascending side, from C the beginning of the helix, to D the middle of it; and on the descending side, from D the middle, to the point G, where the horizontal does cut the helix. Now, it is evident that this latter part, D G, is nothing near so much, and consequently not so heavy as the other, D C; and thus is it in all the other revolutions, which, as they are either more or larger, so will the difficulty of this motion be increased. Whence it will appear that the outward streams which descend must be of so much force as to countervail all that weight whereby the ascending side in every one of these revolutions does exceed the other. And though this may be effected by making the water-wheels larger, yet then the motion will be so slow that the screw will not be able to supply the outward streams.
There is another contrivance to this purpose, mentioned by Kircher de Magnete, 1, 2, p. 4, depending upon the heat of the sun and the force of winds; but it is liable to such abundance of exceptions that it is scarce worth the mentioning, and does by no means deserve the confidence of any ingenious artist.
Thus have I briefly explained the probabilities and defects of those subtle contrivances whereby the making of a perpetual motion has been attempted. I would be loath to discourage the enquiry of any ingenious artificer by denying the possibility of effecting it with any of these mechanical helps; but yet (I conceive) if those principles which concern the slowness of the power in comparison to the greatness of the weight were rightly understood and thoroughly considered, they would make this experiment to seem, if not altogether impossible, yet much more difficult than otherwise, perhaps, it will appear. However, the inquiring after it cannot but deserve our endeavors, as being one of the most noble amongst all these mechanical subtilties. And, as it is in the fable of him who dug the vineyard for a hidden treasure, though he did not find the money, yet he thereby made the ground more fruitful, so, though we do not attain to the effecting of this particular, yet our searching after it may discover so many other excellent subtilties as shall abundantly recompense the labor of our inquiry.
And then, besides, it may be another encouragement to consider the pleasure of such speculations, which do ravish and sublime the thoughts with more clear angelical contentments. Archimedes was generally so taken up in the delight of these mathematical studies of this familiar siren (as Plutarch styles them) that he forgot both his meat and drink, and other necessities of nature; nay, that he neglected the saving of his life, when that rude soldier, in the pride and haste of victory, would not give him leisure to finish his demonstration. What a ravishment was that, when, having found out the way to measure Hiero's crown, he leaped out of the bath, and (as if he were suddenly possessed) ran naked up and down, crying εὕρηκα, εὕρηκα {Greek: heurêka, heurêka}! It is storied of Thales that, in his joy and gratitude for one of these mathematical inventions, he went presently to the Temple, and there offered up a solemn sacrifice; and Pythagoras, upon the like occasion, is related to have sacrificed a hundred oxen; the justice of Providence having so contrived it, that the pleasure which there is in the success of such inventions should be proportioned to the great difficulty and labor of their inquiry.
The Paradoxical Hydrostatic Balance
The following was contributed to an English scientific journal in 1831, the name of the author of the article is unknown to us, but here is what he wrote:
This hydrostatic balance, like the compound balance of Desaguliers, may be introduced to illustrate the impossibility of perpetual motion by a weight removed from the centre of a wheel.
Take the hollow-rimmed wheel A B; let it be air-tight and half filled with water. Let C be the axle; at B place a hollow ball loaded to near sinking. Such a wheel, however fine its axle may be, or however well lubricated, will not make a single revolution, though the weight B occupies that part at which every deluded perpetual-motionist is desirous it should be placed; concluding that, by such an arrangement, the production of another Orffyrean wheel must be inevitable.
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.
