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Horologii Oscillatorii Christian Huygens' Horologium Oscillatorium ; Part Two B. 28 Translated and annotated by Ian Bruce. HOROLOGII OSCILLATORII Part TWO (continued). Concerning the descent of bodies under gravity and the motion of these on a cycloid. PROPOSITION XV. For a given point on a cycloid, to draw a line through that point tangent to the cycloid. Let ABC be the cycloid, and B some given point on the curve, through which it is required to draw the tangent. Around the axis AD of the cycloid is drawn the generating circle AED, and BE is drawn parallel to the base cycloid, which crosses the said circle at E, and AE is joined, and to this finally HBN is constructed parallel through B. I say that this line is a tangent to the circle at B. A Indeed some point on the cycloid different from B is considered, and in the first place some higher point such as H is taken, and through H is drawn a line parallel to the base of the cycloid, which crosses the cycloid in L, the circle AED in K, and the line AE in M. Therefore, since KL is equal to the arc KA, and since the line KM is less than the arc KE, the line ML is less than the arc AE, that is to say, to the line EB, or MH; hence it is apparent that the point H lies outside the cycloid. In the next situation a point N is taken on the line HN below B, and through N is sent forth, as before, a line parallel to the base, which crosses the cycloid in Q, with the circle AED in O, and with the line AE produced in P. Therefore since OQ is equal to the arc OA; moreover OP is greater than the arc OE; and PQ is less than the arc EA, that is, the line EB, or PN. Thus it is again apparent that the point N lies outside the cycloid. Therefore when some point besides B is taken on the line HBN, it shall be outside the cycloid, and it is agreed that the line NG through the that point B is a tangent; QED. I have doubted whether I should relinquish this demonstration here in place of that found by the most distinguished Wren and published in Wallis's book de Cycloide, since my demonstration is not greatly different from that. Moreover, it is possible to absolve the proposition to be agreed upon from a general construction, which can be applied not only to the cycloid but for other curves arising from the rotation of some figure ; provided Christian Huygens' Horologium Oscillatorium ; Part Two B. 29 Translated and annotated by Ian Bruce. the figure is concave in the same part [as the generating circle], and arises from these that are called geometrical. [p. 40] [Huygens now proceeds to show that the line CA is normal to the tangent at A, and the neighbouring points N and B on the cycloid lie outside this circle.] If indeed the curve NAB, arisen from the rotation of a figure along the straight edge LD; truly by describing the point N taken on the circumference of the figure OL. And it is required that the tangent to the curve is drawn at the point A. The line CA is drawn from the point C, where the figure is a tangent to the line at the point describing the line at A : since the point of contact can always be found, if indeed the problem can be reduced to that in order that two lines parallel to each other can be drawn, of which the one may cross through a point describing the revolution of the given figure and the other is a tangent to the figure, and which stand apart from each other as much as the given point A from the line LD : I say that CA crosses the curve at right angles, or the circumference MAF described with centre C and with radius CA, touches the curve at the point A, thus the perpendicular to AC drawn through A is a tangent to the curve at that point. Indeed CB is drawn first to the point of the curve B, which is more distant than the point A from the level line LD, and it is understood that the rotating figure is in the position BED, with the describing point at B, with the contact to the ruler line at D, than with the point of the curve at C, when the point describing the curve was at A, and C is now raised to the point E ; and EC and EB can be joined, and KH is a tangent to the figure at E, crossing the ruler at H. Therefore, since the line CD is equal to the arc of the curve ED; truly the sum of EH and HD is greater than the length of the arc ED ; from which EH will be greater than CH. Hence the angle ECH is greater than CEH, and hence the angle ECL is smaller than the angle CEK. And on adding the angle KEB (equal to LCA) to KEC, there is the angle CEB: and on taking away from ECL the angle LCB, there is the angle ECB. Therefore the angle CEB is always greater than the angle ECB. And thus in the triangle CEB, the side CB is greater than EB, but EB appears to be equal to CA, since it arises from the same figure moved. [p. 41]. Therefore CB is also greater than CA, or CF. From which it is understood that the point B lies outside the circumference MAF. Again let a point N on the curve be taken between the line LD and the point A. Since the describing point is at N, the position of the [rolling] figure can be put at VL, and the Christian Huygens' Horologium Oscillatorium ; Part Two B. 30 Translated and annotated by Ian Bruce. point of contact L, truly the point which before was a tangent at C, shall now be raised to V : and CN, NV, VC, and VL are joined. Therefore VN is equal to CA [chords of equal length]; indeed it is CA itself translated to VN. Since now the line LC is equal to the arc of the curve LV, and hence is greater than the straight line LV, in the triangle CLV, the angle LVC is greater than LCBV. Whereby by adding the angle LVN to LVC, the whole angle NVC shall certainly be made greater than LCV, and hence generally greater than the angle NCV, which is a part of LCV. Therefore in the triangle CVN, the side CN is greater than the side VN, which is equal to CA, and hence CN is greater than CA also, or CM. Thus it is apparent that the point N falls outside the circle MAF, which hence is the tangent at the point A. Q.e.d. Moreover, it is the same construction and demonstration too, for a curve arising by describing a point, which is taken either inside or outside the curve of the rolling figure. Except that, in this latter case, a certain part of the curve falls below the ruler, thus there is some difference arising from the demonstration Indeed let A be a point, through which the tangent is drawn, in a given part of the curve NAB, which has dropped below the ruler CL, obviously described by a point N taken outside the rolling figure, but having a particular position in the same plane of this figure. [p.42]. Therefore by finding a point C, where the rotating figure is tangential to the ruler CD when the describing point is at A, the straight line CA is drawn. I say that this line crosses the NAB at right angles, or the circumference of the circle described with radius CA, and with centre C, is a tangent to the curve NAB at the point A. Moreover, it will be shown the touch the circle externally, since in the part of the curve above the ruler CD it will touch from within. Indeed with the same position and description for everything from before, the angle ECH is again shown to be greater than the angle CEH; and with the angle HCB added to the angle ECH, the angle is ECB; and from the angle CEH is taken HEB, which is equal Christian Huygens' Horologium Oscillatorium ; Part Two B. 31 Translated and annotated by Ian Bruce. to DCA, and the angle shall be CEB. Therefore the angle ECB is generally greater than CEB. Hence in triangle ECB, the side EB is greater than CB. But EB is equal to CA, or CF. And hence CF is greater than CB: and thus the point F on the circumference is beyond the curve NAB and more distant from the centre. Likewise it will be shown again that the angle LVC is greater than the angle LCV. Whereby CVP, which with LVC is equal to two right angles, is less than VCD. And by adding to VCD the angle DCN, the angle is equal to VCN; and by taking the angle PVN from the angle CVP, the angle is CVN. Therefore the angle VCN is generally greater than CVN. Thus in the triangle CVN, the side VN is greater than CN. Moreover, VN is itself equal to CA or CM. And hence CM is greater than CN, and thus the point on the circumference M lies more distant from the centre C, beyond the curve NAB.
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