Part 10

Now, to square the circle, or, in other words, to get exactly equal in superficial area to the circle X, I will show how to find it. From the point G draw a straight line--say G m--perpendicular to ED, making G m equal GD. Produce GA to a point n, making G n equal to 2AG - GD, and join n m. The square on n m will be the required square. (I have indicated this square by dotted lines.) For example:--If AO = 4, then AG = 5, and GD = 1'25; therefore {2 AG - GD} = {10 - 1'25} = 8'75 = Gn: and Gm = 1'25; therefore, Gn^2 + Gm^2 = 3-1/8 (AB^2); that is, {8'75^2 + 1'25^2} = 3-1/8 (5^2), or, {76'5625 + 1'5625} = {3'125 × 25}; and this equation=Area of the Circle X; and area of the square on n m :: and it follows, that the area of every circle, is equal to the area of a square on the hypotenuse of a right-angled triangle, of which the sides that contain the right angle are in the ratio of 7 to 1, and the sum of these two sides equal to the diameter of the circle. In many ways I have proved this fact, by practical or constructive geometry.

PLATE IV.

Duplication of the Cube.--In his "Young Geometrician; or, Practical Geometry without Compasses," 1865, Mr. Oliver Byrne's 40th Problem is as follows:--

Let AB be the side of a given cube BD. It is required to find AC, the side of another cube CE, so that the solid contents of the cube CE are double the solid contents of the cube BD.

Ancient and modern mathematicians (says Mr. Byrne) have in vain attempted to solve this problem geometrically, that is, by the ruler and compasses only.

Let AB = BG = GR = RQ = QP = QO = OR = VZ. The length of the shortest side of the lesser set square; a line of any other given length may be applied. Draw OP and VR parallel to it; then apply the set squares in close contact, the edge OV of OVT passing through the point O, while the points of V and Z of ZSV fall exactly on the lines RV, RZ. Then draw the line ZBC, cutting FA produced in C; then the cube on AC is double the cube on AB.

PLATE V.

Trisection of an Angle.--In his work entitled Young Geometrician, 1865, Mr. Oliver Byrne gives as the 39th Problem: To divide a given angle BAC into three equal angles:--

The line A m is made = p q, the least side of the lesser triangular ruler; by (II) p m is drawn parallel, and m n perpendicular to AB. Then both rulers are kept in motion, and at the same time in close contact, as represented in the figure, until p falls on the line p m, and n on the line m n; r nA passing through the angular point A.

Then the angle DAB is one-third of the angle CAB. Mr. Byrne asserts that this problem is not capable of solution by the straight line and circle. Mathematicians have in vain attempted to solve it geometrically, that is, by the ruler and compasses only.

PLATE VI.--FIGURE 1.

Perpetuum Mobile. Desaguliers demonstrated the absurdity of attempting to raise weights enclosed in a cellular wheel, simply by providing for their approach in succession nearer to the centre on the ascending side, while they should be projected further from the centre on the descending side. He remarks:--

Those who think the velocity of the weight is the line it describes, expect that that weight shall be overpoised, which describes the shortest line, and therefore contrive machines to cause the ascending weight to describe a shorter line than the descending weight.

For example, in the circle A B D a, the weights A and B being supposed equal, it is imagined that, if by any contrivance whatever, whilst the weight A describes the arc A a, the weight B is carried in any arc, as B b, so as to come nearer the centre in its rising, than if it went up the arc B D; the said weight shall be overpoised, and consequently, by a number of such weights, a perpetual motion produced.

Now the velocity of any weight is not the line which it describes in general, but the height that it rises up to, or falls from, with respect to its distance from the centre of the earth. So that when the weight describes the arc A a, its velocity is the line A C, which shows the perpendicular descent, and likewise the line B C denotes the velocity of the weight B, or the height that it rises to, when it ascends in any of the arcs B b, instead of the arc B D: so that, in this case, whether the weight B, in its ascent be brought nearer the centre or not, it loses no velocity, which it ought to do, in order to be raised up by the weight A.

Indeed, if the weight at B, could by any means spring as it were, or be lifted up to x, and move in the arc x b, the end would be answered, because then the velocity would be diminished, and become xC.

FIGURE 2.

In "The Life, Times, and Scientific Labours of the Marquis of Worcester," 1865, page 454, will be found a full account of the present diagram, which is intended to illustrate as far as possible, an approach to the probable construction of the wheel by the Marquis in the 56th article of his memorable "Century of Inventions."

If any likely-looking method, could, more than another, render hopelessness more hopeless, surely this mechanical demonstration must prove most efficient for that purpose. For here, we actually produce a wheel agreeing to the terms with which Desaguliers closes his demonstration, when he suggests the only likely method to effect the end proposed, namely, perpetual motion. We find the fallen weight is absolutely "lifted up" as he desires, and "moves in the arc" he describes, and yet although he declares that then "the end would be answered"--it absolutely is not answered in this instance.

It is not requisite to calculate throughout the effect of the Marquis's entire number of 40 weights; four will suffice, taking the vertical and horizontal spokes a a a a, showing two rings a and b; one, b, 12 inches within the other, so that the wheel being, as the Marquis says, 14 feet diameter, the inner ring will be 12 feet diameter. Now let each weight D be attached in the centre of a cord or chain a´, D, b´, 2 feet long, and then secure one end, as a´, so the extreme end of each spoke a´, and the other end of the cord, as b´, to place on one lesser ring, as at b, or 12 inches from each spoke.

We shall then find by admeasurement that the upper weight on the vertical spoke is 7 feet from the centre, and the lower weight 6 feet; while at the same time there appears to be a preponderance due to the superior length of the horizontal arm A´; but against this latter we have the rising weight b´D, 1 foot from the centre, which, added to the 6 feet on the horizontal spoke, neutralizes the hoped-for effect, and the wheel remains in statu quo.

THE END.

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LONDON. E. & F. N. SPON. 48, CHARING CROSS.]

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LONDON. E. & F. N. SPON. 48, CHARING CROSS.]

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LONDON. E. & F. N. SPON. 48, CHARING CROSS.]