Vietnam or Thailand ? Vote for the TOP Country of the Week !
Updated: May 27, 2025
After this, I drew a semi-circle, containing ten yards in a semi-diameter, and twenty yards in the whole, driving down two rows; of strong stakes, not 6 inches from each other.
The power that two wheels of different magnitudes have upon each other is in the same proportion as if the semi-diameter of the two wheels were joined together and made into that kind of lever I have described, suspended at the part where the semi-diameters join; for the two wheels, scientifically considered, are no other than the two circles generated by the motion of the compound lever.
Since the sums of the times along AK, KB, and along AC, CB are equal, if from the former sum one deducts the time along KB, and if from the other one deducts the time along RB, there will remain the time along AK as equal to the time along the two parts AC, CR. Consequently in the time that the light has come along AK it will also have come along AC and will in addition have made, in the medium from the centre C, a partial spherical wave, having a semi-diameter equal to CR. And this wave will necessarily touch the circumference KS at R, since CB cuts this circumference at right angles.
The common petticoat has not above four yards in the circumference; whereas this over our heads had more in the semi-diameter; so that, by allowing it twenty-four yards in the circumference, the five millions of woollen petticoats, which, according to Sir William Petty, supposing what ought to be supposed in a well-governed state, that all petticoats are made of that stuff, would amount to thirty millions of those of the ancient mode: a prodigious improvement of the woollen trade! and what could not fail to sink the power of France in a few years.
This wave is then represented by the circumference SNR, the centre of which is A, and its semi-diameter equal to two-thirds of CB. Then if one considers in order the other pieces H of the wave AC, it appears that in the same time that the piece C reaches B they will not only have arrived at the surface AB along the straight lines HK parallel to CB, but that, in addition, they will have generated in the diaphanous substance from the centres K, partial waves, represented here by circumferences the semi-diameters of which are equal to two-thirds of the lines KM, that is to say, to two-thirds of the prolongations of HK down to the straight line BG; for these semi-diameters would have been equal to entire lengths of KM if the two transparent substances had been of the same penetrability.
"Certainly. Then you subtract the Dip and the Refraction in Altitude, take the sun's semi-diameter from the Nautical Almanac, and add the Parallax. Do you follow me?" "Perfectly; it sounds the easiest thing. But I don't wish to hear the remarks of the Admiralty when they see the result."
But the minor diameter of this Ellipse, Gg, will bear to Qq the proportion which has been defined previously, Article 27, between CG and the major semi-diameter of the spheroid, CP, namely, that of 98,779 to 105,032.
Then also as 3 to 2 will the sum of the one set be to the sum of the other; that is to say, TF to AS, and DE to AK, and BE to SK or DV, supposing V to be the intersection of the curve EK and the ray FO. But, making FB perpendicular to DE, the ratio of 3 to 2 is also that of BE to the semi-diameter of the spherical wave which emanated from the point F while the light outside the transparent body traversed the space BE. Then it appears that this wave will intersect the ray FM at the same point V where it is intersected at right angles by the curve EK, and consequently that the wave will touch this curve.
Then as N is to CQ so is VC to CD, as appears by the method of finding the refraction which we have shown above, Article 31; but as VC is to CD, so is VB to DS. Then as N is to CQ, so is VB to DS. Let ML be perpendicular to CL. And because I suppose the eyes Rr to be distant about a foot or so from the crystal, and consequently the angle RSr very small, VB may be considered as equal to the semi-diameter CQ, and DP as equal to CL; then as N is to CQ so is CQ to DS. But N is valued at 156,962 parts, of which CM contains 100,000 and CQ 105,032.
About the centre C, with semi-diameter CG, let the circumference gRG be described, cutting the ray RC at R; and let RV be the perpendicular on CG. Then as the line N is to CG let CV be to CD, and let DI be drawn parallel to CM, cutting the Ellipse gMG at I; then joining CI, this will be the required refraction of the ray RC. Which is demonstrated thus.
Word Of The Day
Others Looking