Page:Newton's Principia (1846).djvu/551

 and moon's declinations to any given time; D and E the mean apparent diameters of the sun and moon: and, supposing F and G to be their apparent diameters to that given time, their forces to raise the tides under the equator will be, in the syzygies $$\scriptstyle \frac{VG^3}{2RE^3}L+\frac{TF^3}{2RD^3}S$$; in the quadratures, $$\scriptstyle \frac{VG^3}{2RE^3}L-\frac{TF^3}{2RD^3}S$$. And if the same ratio is likewise observed under the parallels, from observations accurately made in our northern climates we may determine the proportion of the forces L and S; and then by means of this rule predict the quantities of the tides to every syzygy and quadrature.

At the mouth of the river Avon, three miles below Bristol (p. 450 to 453), in spring and autumn, the whole ascent of the water in the conjunction or opposition of the luminaries (by the observation of Sturmy) is about 45 feet, but in the quadratures only 25. Because the apparent diameters of the luminaries are not here determined, let us assume them in their mean quantities, as well as the moon's declination in the equinoctial quadratures in its mean quantity, that is, 23½°; and the versed sine of double its complement will be 1682, supposing the radius to be 1000. But the declinations of the sun in the equinoxes and of the moon in the syzygies are of no quantity, and the versed sines of double the complements are each 2000. Whence those forces become L + S in the syzygies, and $1682/2000$ L - S in the quadratures; respectively proportional to the heights of the tides of 45 and 25 feet, or of 9 and 5 paces. And, therefore, multiplying the extremes and the means, we have 5L + 5S = $15138/2000$L - 9S, or L = $28000/5138$S = 5$5/11$S.

But farther; I remember to have been told that in summer the ascent of the sea in the syzygies is to the ascent thereof in the quadratures as about 5 to 4. In the solstices themselves it is probable that the proportion may be something less, as about 6 to 5; whence it would follow that L is 5$1/6$S [for then the proportion is $$\scriptstyle\frac{1682}{2000}L+\frac{1682}{2000}S:L-\frac{1682}{2000}S::6:5$$]. Till we can more certainly determine the proportion from observation, let us assume L = 5⅓S; and since the heights of the tides are as the forces which excite them, and the force of the sun is able to raise the tides to the height of nine inches, the moon's force will be sufficient to raise the same to the height of four feet. And if we allow that this height may be doubled, or perhaps tripled, by that force of reciprocation which we observe in the motion of the waters, and by which their motion once begun is kept