polygon No. 2, that there will only remain within it the area of one circular ring, as per cal. No. 25, in excess of the circular area of the hexagon. Now we shall find, by cal. No. 13, that this area is exactly represented by the 6 arcs of the inner circle as forming of the 6 triangles which they cover, and on separating the area of these 6 arcs from that of the 6 plain magnitudes cut off the squares, we will thus leave attached to the area of the 18 triangles above them 12 small triangular forms, value as per cal. No. 24, and exactly equivalent to that cut off the polygon No. 2, as per Fig. 8 and per cal. No. 29; by which means you will plainly see that the exact area of the 18 triangles is perfectly preserved while being thus absolutely converted into a circular. ring the height of a triangle-and containing an area, as per cal. No. 31, equal to 18 or 15 of the entire circle.

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And with a view to exhibit a still more perfect and complete illustration and proof of my second hypothesis, I purpose exhibiting the aforesaid 18 triangles, not only increased to the area of 18 circular rings, but also changed into the perfect form of 18 separate and distinct arched triangles-see Fig. 9.

First, in order to form them into 18 circular rings, it will be necessary to add 18, or 33 triangles, to their collective area. Now, in order to do this, we will commence with the area of the 12 outside arcs, cut off the value of the circle by that of the first polygon, as per cal. No. 36, and next include the area cut off by that of the second polygon, as per cals. Nos. 20 or 37, and next the area of the angles cut off the latter polygon, as per cals. Nos. 29 or 38, which, when taken together, as appears per cals. Nos. 36, 37, and 38, will exactly make up the area of the 3 triangles, as per cal. No. 39. And it will be noted that each and all the items which make up this area lie outside of the 15 circular rings, and embrace all in that direction which the circular boundary encloses. Now, by cal. No. 39 we find that this amount is equivalent to 3 circular rings, or of the circular area, so that it may be added to the 1 to make ; or it may also be formed into 18 separate arcs; so that, taking the whole series as they now stand to each other, from the first cal. No. 39=33 triangles, cal. No. 31-18 triangles, cal. No. 32=1} triangles, as the value of the arcs of the 6 triangles represented by cal. No. 33, all together making, per cal. No. 34, 28 triangles, or 24 circular rings, as the entire contents of the circle of 8 inches diameter.

15

24

18 24

3

Now, in conclusion, I would simply call attention to the fact

of their of the

that in forming the 18 outside arcs, it will be observed that, as only 12 of the 18 triangles are projected outward, the 6 turned inwards will require (owing to their curved form) one-half of the 48 small arcs of which the 12 large outside a cs are composed, the area of each being equivalent to 4 of the small ones of the inner circle. I have shown how they may be applied on the face of the figure. But allowing the areas of those 12 arcs to remain in their position outside of the circle, we will thus have before us a perfect circular ring, solely composed of 18 triangles and their 18 arcs; the latter representing exactly area in addition, thus together making 18 circular rings, or entire circular area. In proof of my third hypothesis, I think that it will be only necessary to refer to cal. No. 7, which represents the area of the 6 outside arcs as, or of the hexagon, as per cal. No. 6. And, again, the same amount will appear represented by cal. No. 9 with half its value, equal to of the hexagon, added as per cal. No. 13, together making, as per cal. No. 11, 16, or circular rings, containing the area of 7 triangles, and thus making of the entire circle. And to show cal. No. 12, which represents

4

6

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that it is so, I would refer you to the area of the hexagon, as per cal. No. 6, to which is added of that amount, as per cal. No. 13, making together, as per cal. No. 14, in like manner, 7 triangles, or the area of the entire circle, exactly equal to cal. No. 11. And these facts I presume will be regarded as a complete proof of my hypothesis No. 3.

The special object of my previous investigations being solely confined to the ascertainment of the true area of a circle of 8 inches diameter, by a process of circular ring measurement, without any reference to its boundary value, this being a natural consequent; say the area and diameter being given, to find the circumference, divide the area by one-fourth the diameter-the result will be the circumference; or the circumference and the diameter being given, to find the area, multiply the circumference by one-fourth the diameter-the result will give the area. And this matter, taken in connection with the various facts already described, I trust will be regarded not only as affording the most satisfactory proofs of my 1st, 2nd, and 3rd hypotheses, as found by the circular ring mode of enquiry, but also, at same time, serving to establish on the clearest evidence the true area and boundary of a circle of 8 inches diameter.

RESEARCH BY ANGULAR MEASUREMENTS.

I NOW purpose bringing forward some additional evidence in confirmation of that already given in relation to the values of areas and boundaries, the present proofs having been obtained by a distinctly different mode of enquiry from that of the former one, namely, from the angular measurement of the circular area; and the results so found will appear to be still further confirmed, or I should rather say completely established, by what I regard as the most important and interesting portion of all my investigations, namely, the discovery of the true circular boundary, simply obtained as the square root of a square of three figures.

I will now describe the mode by which I obtained both the results stated. First, to obtain the true area of the circle by angular measurement, to the height of the large triangle of the hexagon, as per cal. No. × 1, I added one-fifth of said height, as per cal. No. × 2, thus making together as per cal. No. x 3, and forming a perpendicular line extending from the centre to a short distance above the circular boundary—see Fig. 10. Now, when the height of this line is multiplied by 2 inches, or half the side of the large triangle (same as multiplying by 4 and taking half the area so found), it will thus give the true area of one sixth part of the polygon, as per cal. No. ×4; and this amount being further multiplied by 6, will give the true area of the entire circle, as per cal. No. x 5.

From this amount I will now deduct the area of the first polygon, as per cal. No. × 6, when there will then appear in the angular form, as occupying the space between the boundaries of the two polygons, exactly the area of the 12 arcs, as per cal. No. x7, and corresponding exactly to the extent of 80 figures with that already found by the circular ring measurement, as shown by cal. No. 36.

Yet I must here observe that the linear measurement of the angular boundary above stated, although found exactly capable of enclosing the true area of the circle, proves, nevertheless, to be considerably greater in extent than the true circular boundary, for, when measured as a boundary, it gives a greater area than it actually encloses.

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This I have proved, by taking the square of 2, or half the base of the outer angle, 4, and to this square I have added the square of its height, say 48, or 2 of 12, that being the square of the height of the hexagon triangle; so from the square of 4:48 I have extracted the square root (which gives the hypotenuse of the right angle, and which also forms one side of this outer angle), and extended this root to the extent of 80 decimals. Now, 12 times the extent of this square root will give the total value of the angular boundary that encloses the true area of the circle, as given in cal. No. x 13.

Now, I think it is clearly manifest that no angular boundary of any less dimensions could possibly include the same area, as any reduction of the angular boundary must necessarily reduce the area that it encloses, as the angles, evidently, must cut off the circle exactly the very same area that each includes in the apex of its angle above the circle, so that a boundary line less than that of the angular one, which would also be capable of enclosing the same area, must necessarily be the true circumference of the circle, provided that it bears the test-viz., that half such boundary multiplied by half the diameter, or the whole boundary multiplied by one-fourth the diameter, will give the true area as found by the angular

measurement.

Now, in order to find the boundary line required, I would here call your attention to the fact, that the angular measurement found in conformity with the 47 prop. of Euclid, B. 1, must necessarily include in the hypotenuse of each side of the six angles of which the whole boundary of the polygon is formed, say six squares of the height of the angle formed by and connecting its two sides, an excess the value of two squares, as the six angles collectively are in reality only equal to four right angles, and, consequently, the areas said angles enclose-see Euclid, prop. 32, B. 1 and cor. 7; so that, notwithstanding that the angular enclosure thus formed embraces the true area of the circle, yet its linear measurement, as a boundary, must therefore be greater than that forming the true circumference of the circle, by the value of these two squares. In proof of this fact, I have deducted one-third from the square of the height of each of the six triangles, thus reducing the linear measurement of each hypotenuse forming each of their respective sides by that amount, and thereby embracing within the area of the six large angles collectively the value of four squares only, instead of six of their height.

Now, to bring out this calculation, I added to the base of the square of the right angle, say 4, the square of its height, say '48-16, or onethird less, leaving 32 as its reduced square; so from their sum, 4·32, I extracted the square root to the extent of 80 figures, which gives exactly one-twelfth the linear boundary of the circle, as shown by cal. No. x8. Now, by multiplying this amount by 6, and regarding it as linear measure only, we find that the product represents exactly one-half of the true boundary of the circle, as per cal. No. x9.

Now, to prove this fact, we will next multiply this half boundary by half the diameter of the circle, say 4 inches, when we shall find its true area thus produced, as per cal. No. × 10, and constituting the very same amount as that given by the angular measurement for the true area of the circle, as per cal. No. x5, so that we must therefore reasonably conclude that the square root of 432, as per cal. No. ×8, multiplied by 12, as per cal. No. ×14, represents the true linear boundary of a circle of 8 inches diameter, as no other form of boundary of this linear measurement, save the circular, could possibly enclose this area.

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Now, before concluding this part of my investigation, I would desire to mention, that in the course of my previous investigation of this problem by the circular ring mode of enquiry, I had the pleasure of describing a very simple means of converting a series of 18 triangles into a perfect circular ring of 1 of the true area of the circle, without effecting any change whatever in their collective area, or even in the depth of their circular ring. I would now, in like manner, wish to describe a very simple mode of converting the true angular form and measurement into the true circular form and measurement, both in boundary and area, so that I have no doubt that its very simplicity will be interesting. On looking back to the place where the mode of calculating the respective areas of the angular and circular boundaries is given, one important fact may have been observed as common to both-viz., that the height of the perpendicular line, formed by adding one-fifth to the height of the triangle, when multiplied by 2, gives one-sixth the true area of the circle; while, on the other hand, two-twelfths, or one-sixth of the circular boundary, when multiplied by 2, gives exactly the same amount. Now this fact can only be accounted for on one condition, namely, that the line represented by cal. No. x3 must therefore be exactly equivalent to one-sixth of the true boundary of the circle, and on being compared with