*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.

*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”.

Old King Cole. Vindex.

*Problem*. — An oblong garden, half 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, 60½.

*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, 60½.

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 60½”! 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 60.25 x 2 = 120½. His first assertion is only true of a square garden. His second is irrelevant, since 60.25 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 60½. 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+½ as its length, and makes her first “coil” the sum of x-½, x-½, x-1, x-1, i.e. 4x-3: but the fourth term should be x-1½, so that her first coil is ½ a yard too long. Her second coil is the sum of x-2½, x-2½, x-3, x-3: here the first term should be x-2 and the last x-3½: 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 5½. 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.”

Vindex.

Alumnus Etonae. Old King Cole.

Dinah Mite. Magpie. Janet. Taffy.

http://ebooks.adelaide.edu.au/c/carroll/lewis/tangled/answers9.html

Last updated Monday, December 22, 2014 at 10:48