" It'll just take one more pebble."
"Whatever are you doing with those buckets?"
The speakers were Hugh and Lambert. Place, the beach of Little Mendip. Time I:30 p.m. Hugh was floating a bucket in another a size larger, and trying how many pebbles it would carry without sinking. Lambert was lying on his back, doing nothing.
For the next minute or two Hugh was silent, evidently deep in thought. Suddenly he started. "I say, look here, Lambert!" he cried.
"If it's alive, and slimy, and with legs, I don't care to," said Lambert.
"Didn't Balbus say this morning that, if a body is immersed in liquid it displaces as much liquid as is equal to its own bulk?" said Hugh.
"He said things of that sort," Lambert vaguely replied.
"Well, just look here a minute. Here's the little bucket almost quite immersed: so the water displaced ought to be just about the same bulk. And now just look at itl" He took out the little bucket as he spoke, and handed the big one to Lambert. "Why, there's hardly a teacupful! Do you mean to say that water is the same bulk as the little bucket?"
"Course it is," said Lambert.
"Well, look here again!" cried Hugh, triumphantly, as he poured the water from the big bucket into the little one. "Why, it doesn't half fill it!"
"That's its business," said Lambert. "If Balbus says it's the same bulk, why, it is the same bulk, you know."
"Well, I don't believe it," said Hugh.
"You needn't," said Lambert. "Besides, it's dinner-time. Come along." They found Balbus waiting dinner for them, and to him Hugh at once propounded his difficulty.
"Let's get you helped first," said Balbus, briskly cutting away at the joint. "You know the old proverb, 'Mutton first, mechanics afterwards' ?"
The boys did not know the proverb, but they accepted it in perfect good faith, as they did every piece of information, however startling, that came from so infallible an authority as their tutor. They ate on steadily in silence, and, when dinner was over, Hugh set out the usual array of pens, ink, and paper, while Balbus repeated to them the problem he had prepared for their afternoon's task.
"A friend of mine has a flower-garden--a very pretty one, though no great size "
"How big is it?" said Hugh.
"That's what you have to find out!' Balbus gaily replied. "All I tell you is that it is oblong in shape--just half a yard longer than its width--and that a gravel-walk, one yard wide, begins at one corner and runs all round it."
"Joining into itself?" said Hugh.
"Not joining into itself, young man. Just before doing that, it turns a corner, and runs round the garden again, alongside of the first portion, and then inside that again, winding in and in, and each lap touching the last one, till it has used up the whole of the area."
"Like a serpent with corners?" said Lambert.
"Exactly so. And if you walk the whole length of it, to the last inch, keeping in the centre of the path, it's exactly two miles and half a furlong. Now, while you wind out the length and breadth of the garden, I'll see if I can think out that sea-water puzzle."
"You said it was a flower-garden?" Hugh inquired, as Balbus was leaving the room.
"I did," said Balbus.
"Where do the flowers grow?" said Hugh. But Balbus thought it best not to hear the question. He left the boys to their problem, and, in the silence of his own room, set himself to unravel Hugh's mechanical paradox.
"To fix our thoughts," he murmured to himself, as, with hands deep-buried in his pockets, he paced up and down the room, "we will take a cylindrical glass jar, with a scale of inches marked up the side, and fill it with water up to the 10-inch mark: and we will assume that every inch depth of jar contains a pint of water. We will take a solid cylinder, such that every inch of it is equal in bulk to half a pint of water, and plunge 4 inches of it into the water, so that the end of the cylinder comes down to the 6-inch mark. Well, that displaces 2 pints of water. What becomes of them? Why, if there were no more cylinder, they would lie comfortably on the top, and fill the jar up to the 12-inch mark. But unfortunately there is more cylinder, occupying half the space between the 10-inch and the 12-inch marks, so that only one pint of water can be accommodated there. What becomes of the other pint? Why, if there were no more cylinder, it would lie on the top, and fill the jar up to the 13-inch mark. But unfortunately--Shade of Newton!" he exclaimed, in sudden accents of terror. "When does the water stop rising?"
A bright idea struck him. "I'll write a little essay on it," he said.
Balbus's Essay
"When a solid is immersed in a liquid, it is well known that it displaces a portion of the liquid equal to itself in bulk, and that the level of the liquid rises just so much as it would rise if a quantity of liquid had been added to it, equal in bulk to the solid. Lardner says precisely the same process occurs when a solid is partially immersed: the quantity of liquid displaced, in this case, equalling the portion of the solid which is immersed, and the rise of the level being in proportion."Suppose a solid held above the surface of a liquid and partially immersed: a portion of the liquid is displaced, and the level of the liquid rises. But, by this rise of level, a little bit more of the solid is of course immersed, and so there is a new displacement of " second portion of the liquid, and a consequent rise of level. Again, this second rise of level causes a yet further immersion, and by consequence another displacement of liquid and another rise. It is self-evident that this process must continue till the entire solid is immersed, and that the liquid will then begin to immerse whatever holds the solid, which, being connected with it, must for the time be considered a part of it. If you hold a stick, six feet long, with its ends in a tumbler of water, and wait long enough, you must eventually be immersed. The question as to the source from which the water is supplied--which belongs to a high branch of mathematics, and is therefore beyond our present scope--does not apply to the sea. Let us therefore take the familiar instance of a man standing at the edge of the sea, at ebb-tide, with a solid in his hand, which he partially immerses: he remains steadfast and unmoved, and we all know that he must be drowned. The multitudes who daily perish in this manner to attest a philosophical truth, and whose bodies the unreasoning wave casts sullenly upon our thankless shores, have a truer claim to be called the martyrs of science than a Galileo or a Kepler. To use Kossuth's eloquent phrase, they are the unnamed demigods of the nineteenth century." (**Note by the writer--For the above essay I am indebted to a dear friend, now deceased.)
"There's a fallacy somewhere," he murmured drowsily, as he stretched his long legs upon the sofa. "I must think it over again." He closed his eyes, in order to concentrate his attention more perfectly, and for the next hour or so his slow and regular breathing bore witness to the careful deliberation with which he was investigating this new and perplexing view of the subject.
§ 1. The Buckets
Problem.--Lardner states that a solid, immersed in a fluid, displaces an amount equal to itself in bulk. How can this be true of a small bucket floating in a larger one?
Solution.--Lardner means, by "displaces", "occupies a space which might be filled with water without any change in the surroundings." If the portion of the floating bucket, which is above the water, could be annihilated, and the rest of it transformed into water, the surrounding water would not change its position: which agrees with Lardner's statement.
Five answers have been received, none of which explains the difficulty arising from the well-known fact that a floating body is the same weight as the displaced fluid. Hecla says that "Only that portion of the smaller bucket which descends below the original level of the water can be properly said to be immersed, and only an equal bulk of water is displaced." Hence, according to Hecla, a solid whose weight was equal to that of an equal bulk of water, would not float till the whole of it was below "the original level" of the water: but, as a matter of fact, it would float as soon as it was all under water. Magpie says the fallacy is "the assumption that one body can displace another from a place where it isn't", and that Lardner's assertion is incorrect, except when the containing vessel "was originally full to the brim". But the question of floating depends on the present state of things, not on past history. Old King Cole takes the same view as Hecla. Tympanum and Vindex assume that "displaced" means "raised above its original level", and merely explain how it comes to pass that the water, so raised, is less in bulk than the immersed portion of bucket, and thus land themselves --or rather set themselves floating--in the same boat as Hecla.
I regret that there is no Class List to publish for this Problem.
§ 2. Balbus's Essay
Problem.--Balbus states that if a certain solid be immersed in a certain vessel of water, the water will rise through a series of distances, two inches, one inch, half an inch, etc., which series has no end. He concludes that the water will rise without limit. Is this true?
Solution.--No. This series can never reach 4 inches, since, however many terms we take, we are always short of 4 inches by an amount equal to the last term taken.
Three answers have been received--but only two seem to me worthy of honours.
Tympanum says that the statement about the stick "is merely a blind, to which the old answer may well be applied, solvitur ambulando, or rather mergendo". I trust Tympanum will not test this in his own person, by taking the place of the man in Balbus's Essay! He would infallibly be drowned.
Old King Cole rightly points out that the series, 2, 1, etc., is a decreasing geometrical progression: while Vindex rightly identifies the fallacy as that of "Achilles and the Tortoise".
Class List. I.
Old King Cole. Vindex.
§ 3. The Garden
Problem.--An oblong garden, hall a yard longer than wide , consists entirely of a gravel walk, spirally arranged, a yard wide and 3630 yards long. Find the dimensions of the garden.
Answer.--60, 60 1/2.
Solution.--The number of yards and fractions of a yard traversed in walking along a straight piece of walk, is evidently the same as the number of square yards and fractions of a square yard contained in that piece of walk: and the distance trsversed in passing through a square yard at a corner, is evidently a yard. Hence the area of the garden is 3630 square yards: i.e. if x be the width, x(x+1/2)=3630. Solving this quadratic, we find x=60. Hence the dimentions are 60, 60 1/2.
Twelve answers have been received--seven right and five wrong.
C. G. L., Nabob, Old Crow, and Tympanum assume that the number of yards in the length of the path is equal to the number of square yards in the garden. This is true, but should have been proved. But each is guilty of darker deeds. C. G. L.'s "working" consists of dividing 3630 by 60. Whence came this divisor, O Segiel? Divination? Or was it a dream? I fear this solution is worth noting. Old Crow's is shorter, and so (if possible) worth rather less. He says the answer "is at once seen to be 60 x 60 1/2"! Nabob's calculation is short, but "as rich as a Nabob" in error. He says that the square root of 3630, multiplied by 2, equals the length plus the breadth. That is 60&#middot;25 x 2 = 120 1/2. His first assertion is only true of a square garden. His second is irrelevant, since 60&#middot;25 is not the square root of 3630! Nay, Bob, this will not do! Tympanum says that, by extracting the square root of 3630, we get 60 yards with a remainder of 30/60, or half a yard, which we add so as to make the oblong 60 x 60 1/2. This is very terrible: but worse remains behind. Tympanum proceeds thus: "But why should there be the half-yard at all? Because without it there would be no space at all for flowers. By means of it, we find reserved in the very centre a small plot of ground, two yards long by half a yard wide, the only space not occupied by walk." But Balbus expressly said that the walk "used up the whole of the area", O Tympanum! My tympa is exhausted: my brain is num! I can say no more,
Hecla indulges, again and again, in that most fatal of all habits in computation--the making two mistakes which cancel each other. She takes x as the width of the garden, in yards, and x+1/2 as its length, and makes her first "coil" the sum of x-1/2, x-1/2, x-1, x-1, i.e. 4x-3: but the fourth term should be x-1 1/2, so that her first coil is 1/2 a yard too long. Her second coil is the sum of x-2 1/2, x-2 1/2, x-3, x-3: these two mistakes cancel and this coil is therefore right. And the same thing is true of every other coil but the last, which needs an extra half-yard to reach the end of the path: and this exactly balances the mistake in the first coil. Thus the sum-total of the coils comes right though the working is all wrong.
Of the seven who are right, Dinah Mite, Janet, Magpie, and Taffy make the same assumption as C. G. L. and Co. They then solve by a quadratic. Magpie also tries it by arithmetical progression, but fails to notice that the first and last "coils" have special values.
Alumnus Etonae attempts to prove what C. G. L. assumes by a particular instance, taking a garden 6 by 5 1/2. He ought to have proved it generally: what is true of one number is not always true of others. Old King Cole solves it by an arithmetical progression. It is right, but too lengthy to be worth as much as a quadratic.
Vindex proves it very neatly, by pointing out that a yard of walk measured along the middle represents a square yard of garden, "whether we consider the straight stretches of walk or the square yards at the angles, in which the middle line goes half a yard in one direction and then turns a right angle and goes half a yard in another direction."
Class List. I.
Vindex.
II.
Alumnus Etonae. Old King Cole. III.
Dinah Mite. Magpie. Janet. Taffy.
