EXPLANATIONS OF THE PLATES.
[Plate I.—Figure 1.]
Of the Twelve Houses.—The 1st, 4th, 7th, and 10th houses—angular.—These are of more durable signification than the others, denoting the wife or husband—a situation under Government, &c. &c.
The twelve houses have signification of all the various concerns of human life, and of nature at large.
For Example.—When the cusp of the first house is well aspected by, or has the presence of Jupiter or Venus, and these are not afflicted by the aspects of evil planets, they preserve life in infancy, and give health, and often an agreeable person.
But if their rays or presence (says Varley) should be thrown on the cusp of the second house, then the native will have success in concerns of property. The Sun in this house helps to disperse property; and if he be peregrine, that is, in the sign of a contrary nature to his own, where he has no dignities, and is without reception, then the native's property is dispersed in vainglorious expenses; but if the Sun be in Leo, his property in general will be ample enough to admit of instant acts of bounty and benevolence.
In a similar strain, Astronomers particularize the remaining eleven houses. It would be impossible, in any reasonable space, to describe further the operations of the planets in the several houses thus assigned to them.
As to when the Planets are most powerful.—Barrett says:—The planets are powerful when they are ruling in a house, or in exaltation, or triplicity, or term, or face, without combustion of what is direct in the figure of the heavens; but we must take care that they are not in the bounds or under the dominion of Saturn or Mars. The angles of the ascendant, and 10th and 7th are fortunate; as also the lord of the ascendant, and place of the Sun and Moon.
The Moon is powerful if she be in her house, in exaltation, in triplicity, in face, or in degree convenient for the desired work, &c. &c.
[Figure 2.]
Varley's Table of Signs, Houses, Exaltation, and Triplicity.
The falls of the Planets are opposite to their Exaltations, and their Detriments opposite to their Houses.
| Aries and Scorpio are the house of Mars | ♂ |
| Taurus and Libra are the house of Venus | ♀ |
| Gemini and Virgo are under the dominion of Mercury | ☿ |
| Cancer is the house of the Moon | ☽ |
| Leo is the house of the Sun | ☉ |
| Sagittarius and Pisces are the houses of Jupiter | ♃ |
| Capricorn is the house of Saturn | ♄ |
| And Aquarius is governed by the Herschel Planet | ♅ |
[Plate II.]
This table gives the usual symbols employed for indicating the several planets, and which are still retained in Astronomy for simplicity of expression, but which Astrologers venerate as possessing a cabalistic character.
Associated with these symbols are the names of certain principal angels, spirits, or demons, forming, however, but a small proportion of such airy nothings.
The Astrological Symbols were also employed by the Alchemists to indicate the seven metals then known.
[Plate III.]—Squaring the Circle.
Mr. James Smith, of Liverpool, the most laborious among recent workers in this field of enquiry, claiming to have propounded several simple and exact methods, offers the following as sufficiently demonstrative:—
I construct my diagrams in the following way:—I draw two straight lines at right angles, making O the right angle. From the point O, in the direction OA, I mark off four equal parts together equal to OA, and from O, in the direction of OB, I mark off three of such equal parts together, equal to OB, and join AB. It is obvious, or rather self-evident, that AOB is a right-angled triangle, of which the sides that contain the right angle are in the ratio of 4 to 3, by construction. With A as centre and AB as interval, I describe the circle X, produce AO and BO to meet and terminate in the circumference of the circle at the points G and C, and join AC, CG, and BG, producing the quadrilateral ACGB. I bisect AG at F, and with O as centre, and OF as interval, describe the circle Z. The line OF is the line that joins the middle points of the diagonals in the quadrilateral ACGB; and it follows that,
{AG2 + CB2 + 4(OF2)} = {AC2 + CG2 + BG2 + AB2.}
When AO = 4, we get the following equation:—
{52 + 62 + (4 × 1'52)} = {52 + sqrt(102) + sqrt(102) + 52,} or, {25 + 36 + 9}={25 + 10 + 10 + 25} = 70. From the points B and C, I draw straight lines at right angles to AB and AC, and therefore tangential to the circle X, to meet AG produced at D, and join BD and CD, producing the quadrilateral ACDB. I bisect AD at E, and with O as centre, and EO as interval describe the circle XY, and with E as centre, and EA or ED as interval describe the circle Y.
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, Gn2 + Gm2 = 3-1/8 (AB2); that is, {8'752 + 1'252} = 3-1/8 (52), 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.
