of Incidence ABE and to the angle IBK its vertical angle EBF is equal and therefore the angle ABE is equal to the angle EBF Again the angle ADE is equal to the angle of Reflection GDK that is to its vertical angle EDF and therefore the two angles ABD and ADB of the triangle ABD are one by one equal to the two angles FBD and FDB of the triangle FBD Wherfore also the third angle BAD is equal to the third angle BFD which was to be proved Corollary 1. If the straight line AF be drawn it will be perpendicular to the straight line EK For both the angles at E will be equal by reason of the equality of the two angles ABE and FBE and of the two sides AB and FB Corollary 2. If upon any point between B and D there fall a straight line as AC whose reflected line is CH this also produced beyond C will fall upon F which is evident by the demonstration above 3 If from two points taken without a Circle two straight parallel lines drawn not oppositely but from the same parts fall upon the Circumference the lines reflected from them if produced they meet within the Circle will make an angle double to that which is made by two straight lines drawn from the Center to the points of Incidence Let the two straight parallels AB and DC in the 3d figure fall upon the Circumference BC at the points B and C and let the Center of the Circle be E and let AB reflected be BF and DC reflected be CG and let the lines FB and GC produced meet within the Circle in H and let EB and EC be connected I say the angle FHG is double to the angle BEC For seeing AB and DC are parallels and EB cuts AB in B the same EB produced will cut DC somewhere let it cut it in D let DC be produced howsoever to I and let the intersection of DC BF be at K. The angle therefore ICH being external to the triangle CKH will be equal to the two opposite angles CKH and CHK Again ICE being external to the triangle CDE is equal to the two angles at D and E. Wherefore the angle ICH being double to the angle ICE is equal to the angles at D and E twice taken and therefore the two angles CKH and CHK are equal to the two angles at D and E twice taken But the angle CKH is equal to the angles D and ABD that is D twice taken for AB and DC being parallels the altern angles D and ABD are equal Wherefore CHK that is the angle FHG is also equal to the angle at E twice taken which was to be proved Corollary If from two points taken within a circle two straight parallels fall upon the circumference the lines reflected from them shall meet in an angle double to that which is made by two straight lines drawn from the center to the points of Incidence For the parallels LB and IC falling upon the points B and C are reflected in the lines BH and CH and make the angle at H double to the angle at E as was but now demonstrated 4 If two straight lines drawn from the same point without a circle fall upon the circumference and the lines reflected from them being produced meet within the circle they will make an angle equal to twice that angle which is made by two straight lines drawn from the center to the points of Incidence together with the angle which the incident lines themselves make Let the two straight lines AB and AC in the 4th figure be drawn from the point A to the circumference of the circle whose center is D and let the lines reflected from them be BE and CG and being produced make within the circle the angle H also let the two straight lines DB and DC be drawn from the center D to the points of Incidence B and C. I say the angle H is equal to twice the angle at D together with the angle at A. For let AC be produced howsoever to I. Therefore the angle CH which is external to the triangle CKH will be equal to the two angles GKH and CHK Again the angle ICD which is external to the triangle CLD wil be equal to the two angles CLD and CDL But the angle ICH is double to the angle ICD and is therefore equal to the angles CLD and CDL twice taken Wherefore the angles CKH and CHK are equal to the angles CLD and CDL twice taken But the angle CLD being external to the triangle ALB is equal to the two angles LAB LBA consequently CLD twice taken is equal to LAB LAB twice taken Wherefore CKH CHK are equal to the angle CDL together with LAB and LBA twice taken Also the angle CKH is equal to the angle LAB once and ABK that is LBA twice taken Wherefore the Angle CHK is equal to the remaining angle CDL that is to the angle at D twice taken and the angle LAB that is the angle at A once taken which was to be proved Corollary If two straight converging lines as IC and MB fall upon the concave circumference of a circle their reflected lines as CH and BH will meet in the angle H equal to twice the angle D together with the angle at A made by the â—ncident lines produced Or if the Incident lines be HB and IC whose reflected lines CH and BM meet in the point N the angle CNB will be equal to twice the angle D together with the angle CKH made by the lines of Incidence For the angle CNB is equal to the angle H that is to twice the angle D together with the two angles A and NBH that is KBA But the angles KBA and A are equal to the angle CKH Wherefore the angle CNB is equal to twice the angle D together with the angle CKH made by the lines of Incidence IC and HB produced to K. 5 If two straight lines drawn from one point fall upon the concave circumference of a circle and the angle they make be lesse then twice the angle at the center the lines reflected from them and meeting within the circle will make an angle which being added to the angle of the incident lines will be equal to twice the angle at the center Let the two Lines AB and AC in the 5th figure drawn from the point A fall upon the concave circumference of the circle whose center is D let their reflected Lines BE and CE meet in the point E also let the angle A be less then twice the angle D. I say the angles A and E together taken are equal to twice the angle D. For let the straight Lines AB and EC cut the straight Lines DC and DB in the points G and H and the angle BHC will be equal to the two angles EBH and E also the same angle BHC will be equal to the two angles D and DCH
and in like manner the angle BGC will be equal to the two angles ACD A the same angle BGC will be also equal to the two angles DBG and D. Wherefore the four angles EBH E ACD and A are equal to the four angles D DCH DBG and D. If therefore equals be taken away on both sides namely on one side ACD and EBH and on the other side DCH and DBG for the angle EBH is equal to the angle DBG and the angle ACD equal to the angle DCH the remainders on both sides will be equal namely on one side the angles A and E and on the other the angle D twice taken Wherefore the angles A and E are equal to twice the angle D. Corollary If the angle A be greater then twice the angle D their reflected â—â—ines will diverge For by the Corollary of the third Proposition if the angle A be equal to twice the angle D the reflected Lines BE and CE will be parallel and if it be lesse they will concurre as has now been demonstrated and therefore if it be greater the reflected Lines BE and CE will diverge and consequently if they be produced the other way they will concurre and make an angle equal to the excesse of the angle A above twice the angle D as is evident by the fourth Article 6 If through any one point two unequal chords be drawn cutting one another either within the circle or if they be produced without it and the center of the circle be not placed between them and the Lines reflected from them concurre wheresoever there cannot through the point through which the former Lines were drawn be drawn another straight Line whose reflected Line shall passe through the point where the two former reflected Lines concurre Let any two unequal chords as BK and CH in the 6th Figure be drawn through the point A in the circle BC and let their reflected Lines BD and CE meet in F and let the center not be between AB and AC and from the point A let any other straight Line as AG be drawn to the circumference between B and C. I say GN which passes through the point F where the reflected Lines BD and CE meet will not be the reflected Line of AG. For let the arch BL be taken equal to the arch BG and the straight Line BM equal to the straight Line BA and LM being drawn let it be produced to the circuÌ„mference in O. Seeing therefore BA and BM are equal and the arch BL equal to the arch BG and the angle MBL equal to the angle ABG AG and ML will also be equal and producing GA to the circumference in I the whole lines LO and GI will in like manner be equal But LO is greater then GFN as shall presently be demonstrated and therefore also GI is greater then GN Wherefore the angles NGC and IGB are not equal Wherefore the Line GFN is not reflected from the Line of Incidence AG and consequently no other straight Line besides AB and AC which is drawn through the point A and faâ—ls upon the circumference BC can be reflected to the point F which was to be demonstrated It remains that I prove LO to be greater then GN which I shall do in this manner LO and GN cut one another in P and PL is greater then PG. Seeing now LP PG PN PO are proportionals therefore the two Extremes LP and PO together taken that is LO are greater then PG and PN together taken that is GN which remained to be proved 7 But if two equal chords be drawn through one point within a circle and the Lines reflected from them meet in another point then another straight Line may be drawn between them through the former point whose reflected Line shall pass through the later point Let the two equal chords BC and ED in the 7th figure cut one another in the point A within the circle BCD and let their reflected Lines CH and DI meet in the point F. Then dividing the arch CD equally in G let the two chords GK and GL be drawn through the points A and F. I say GL will be the Line reflected from the chord KG For the four chords BC CH ED and DI are by supposition all equal to one another and therefore the arch BCH is equal to the arch EDI as also the angle BCH to the angle EDI the angle AMC to its vertical angle FMD and the straight Line DM to the straight Line CM and in like manner the straight Line AC to the straight Line FD and the chords CG and GD being drawn will also be equal as also the angles FDG and ACG in the equal Segments GDI and GCB Wherefore the straight Lines FG and AG are equal and therefore the angle FGD is equal to the angle AGC that is the angle of Incidence equal to the angle of Reflection Wherefore the line GL is reflected from the incident Line KG which was to be proved Corollary By the very sight of the figure it is manifest that if G be not the middle point between C and D the reflected Line GL will not pass through the point F. 8 Two points in the circumference of a circle being given to draw two straight Lines to them so as that their reflected Lines may be parallel or contain any angle given In the circumference of the circle whose center is A in the 8th figure let the two points B and C be given and let it be required to draw to them from two points taken without the circle two incident Lines so that their reflected Lines may first be parallel Let AB and AC be drawn as also any incident Line DC with its reflected Line CF and let the angle ECD be made double to the angle A and let HB be drawn parallel to EC and produced till it meet with DC produced in I. Lastly producing AB indefinitely to K let GB be drawn so that the angle GBK may be equal to the angle HBK and then GB will be the reflected Line of the incident Line HB I say DC and HB are two incident Lines whose reflected Lines CF and BG are parallel For seeing the angle ECD is double to the angle BAC the angle HIC is also by reason of the parallels EC and HI double to the same BAC Therefore also FC and GB namely the lines reflected from the incident lines DC and HB are parallel Wherefore the first thing required is done Secondly let it be required to draw to the points B C two straight lines of Incidence so that the lines reflected from them may contain the given angle Z. To the angle ECD made at the point C let there be added on one side the angle DCL equal to half Z and on the other side the angle ECM equal to the angle DCL and let the straight Line BN be drawn parallel to the straight line CM and let the angle KBO be made equal to the
in a straight line perpendicular to its Superficies in that point in which it is pressed Let ABCD in the first figure be a hard Body and let another Body falling upon it in the straight line EA with any inclination or without inclination press it in the point A. I say the Body so pressing not penetrating it will give to the part A an endeavour to yeild or recede in a straight Line perpendicular to the line AD. For let AB be perpendicular to AD and let BA be produced to F. If therefore AF be coincident with AE it is of it self manifest that the motion in EA will make A to endeavour in the line AB Let now EA be oblique to AD and from the point E let the straight line EC be drawn cutting AD at right angles in D and let the rectangles ABCD and ADEF be completed I have shewn in the 8th Article of the 16th Chapter that the Body will be carried from E to A by the concourse of two Uniform Motions the one in EF and its parallels the other in ED and its parallels But the motion in EF and its parallels whereof DA is one contributes nothing to the Body in A to make it endeavour or press towards B and therefore the whole endeavour which the Body hath in the inclined line EA to pass or press the Straight line AD it hath it all from the perpendicular motion or endeavour in FA. Wherefore the Body E after it is in A will have onely that perpendicular endeavour which proceeds from the motion in FA that is in AB which was to be proved 7 If a hard Body falling upon or pressing another Body penetrate the same its endeavour after its first penetration will be neither in the inclined line produced nor in the perpendicular but sometimes betwixt both sometimes without them Let EAG in the same â— figure be the inclined line produced and First let the passage through the Medium in which EA is be easier then the passage through the Medium in which AG is As soon therefore as the Body is within the Medium in which is AG it will finde greater resistance to its motion in DA and its parallels then it did whilest it was above AD and therefore below AD it will proceed with slower motion in the parallels of DA then above it Wherefore the motion which is compounded of the two motions in EF and ED will be slower below AD then above it and therefore also the Body will not proceed from A in EA produced but below it Seeing therefore the endeavour in AB is generated by the endeavour in FA if to the endeavour in FA there be added the endeavour in DA which is not all taken away by the immersion of the point A into the lower Medium the Body will not proceed from A in the perpendicular AB but beyond it namely in some straight line between AB and AG as in the line AH Secondly let the passage through the Medium EA be less easie then that through AG. The motion therefore which is made by the concourse of the motions in EF and FB is slower above AD then below it and consequently the endeavour will not proceed from A in EA produced but beyond it as in AI. Wherefore If a hard Body falling which was to be proved This Divergency of the Straight line AH from the straight line AG is that which the Writers of Opticks commonly call Refraction which when the passage is eaâ—ier in the first then in the second Medium is made by diverging from the line of Inclination towards the perpendicular and contrarily when the passage is not so easie in the first Medium by departing farther from the perpendicular 8 By the 6th Theoreme it is manifest that the force of the Movent may be so placed as that the Body moved by it may proceed in a way almost directly contrary to that of the Movent as we see in the motion of Ships For let AB in the 2d figure represent a Ship whose length from the prow to the poop is AB and let the winde lie upon it in the straight parallel lines CB DE and FG and let DE and FG be cut in E and G by a straight Line drawn from B perpendicular to AB also let BE and EG be equal and the angle ABC any angle how small soever Then between BC and BA let the straight line BI be drawn and let the Sail be conceived to be spred in the same line BI and the winde to fall upon it in the points L M and B from which points perpendicular to BI let BK MQ and LP be drawn Lastly let EN and GO be drawn perpendicular to BG and cutting BK in H and K and let HN and KO be made equal to one another and severally equal to BA I say the Ship BA by the winde falling upon it in CB DE FG and other lines parallel to them will be carried forwards almost opposite to the winde that is to say in a way almost contrary to the way of the Movent For the Winde that blowes in the Line CB will as hath been shewn in the 6th Article give to the point B an endeavour to proceed in a straight line perpendicular to the straight line BI that is in the straight line BK and to the points M and L an endeavour to proceed in the straight lines MQ and LP which are parallel to BK Let now the measure of the time be BG which is divided in the middle in E let the point B be carried to H in the time BE. In the same time therefore by the wind blowing in DM FL and as many other lines as may be drawn parallel to them the whole Ship will be applyed to the straight line HN. Also at the end of the second time EG it will be applyed to the straight line KO Wherefore the Ship will always go forwards and the angle it makes with the winde will be equal to the angle ABC how small soever that angle be and the way it makes will in every time be equal to the straight line EH I say thus it would be if the Ship might be moved with as great celerity sidewayes from BA towards KO as it may be moved forwards in the line BA But this is impossible by reason of the resistance made by the great quantity of water which presseth the side much exceeding the resistance made by the much smaller quantity which presseth the prow of the Ship so that the way the Ship makes sidewayes is scarce sensible and therefore the point B will proceed almost in the very line BA making with the winde the angle ABC how acute soever that is to say it will proceed almost in the straight line BC that is in a way almost contrary to the way of the Movent which was to be demonstrated But the Sayl in BI must
the same Quality as Every Man is a Living Creature Some Man is a Living Creature or No Man is Wise Some Man is not Wise. Of these iâ— the Universal be true the Particular will be true also Contrary are Universal Propositions of different Quality as Every Man is happy No Man is happy And of these if one be true the other is false also they may both be false as in the example given Subcontrary are Particular Propositions of different Quality as Some Man is learned Some Man is not learned which cannot be both false but they may be both true Contradictory are those that differ both in Quantity and Quality as Every Man is a Living Creature Some Man is not a Living Creature which can neither be both true nor both false 18 A Proposition is said to follow from two other Propositions when these being granted to be true it cannot be denyed but the other is true also For example let these two Propositions Every Man is a Living Creature and Every Living Creature is a Body be supposed true that is that Body is the Name of Every Living Creature and Living Creature the Name of Every Man Seeing therefore if these be understood to be true it cannot be understood that Body is not the name of Every man that is that Every Man is a Body is false this Proposition will be said to follow from those two or to be necessarily inferred from them 19 That a true Proposition may follow from false Propositions may happen sometimes but false from true never For if these Every Man is a Stone and Every Stone is a Living Creature which are both false be granted to be true it is granted also that Living Creature is the name of Every Stone and Stone of Every Man that is that Living Creature is the Name of Every Man that is to say this Proposition Every Man is a Living Creature is true as it is indeed true Wherefore a true Proposition may sometimes follow from false But if any two Propositions be true a false one can never follow from them For if true follow from false for this reason onely that the false are granted to be true then truth from two truths granted will follow in the same manner 20 Now seeing none but a true Proposition will follow from true and that the understanding of two Propositions to be true is the cause of understanding that also to be true which is deduced from them the two Antecedent Propositions are commonly called the Causes of the inferred Proposition or Conclusion And from hence it is that Logicians say the Premisses are Causes of the Conclusion which may passe though it be not properly spoken for though Understanding be the cause of Understanding yet Speech is not the cause of Speech But when they say the Cause of the Properties of any thing is the Thing it self they speake absurdly Eor example if a Figure be propounded which is Triangular Seeing every Triangle has all its angles together equal to two right angles from whence it follows that all the angles of that Figure are equal to two right angles they say for this reason that that Figure is the Cause of that Equality But seeing the Figure does not it self make its angles and therefore cannot be said to be the Efficient-Cause they call it the Formall-Cause whereas in deed it is no Cause at all nor does the Property of any Figure follow the Figure but has its Being at the same time with it only the Knowledge of the Figure goes before the Knowledge of the Properties and one Knowledge is truly the Cause of another Knowledge namely the Efficient-Cause And thus much concerning Proposition which in the Progress of Philosophy is the first Step like the moving towards of one Foot By the due addition of another Step I shall proceed to Syllogisme and make a compleat Pace Of which in the next Chapter CHAP. IV. Of Syllogisme 1 The Definition of Syllogisme 2 In a Syllogisme there are but three Termes 3 Major Minor and Middle Term also Major and Minor Proposition what they are 4 The Middle Terme in every Syllogisme ought to be determined in both the Propositions to one and the same thing 5 From two Particular Propositions nothing can be concluded 6 A Syllogisme is the Collection of two Propositions into one Summe 7 The Figure of a Syllogisme what it is 8 What is in the mind answering to a Syllogisme 9 The first Indirect Figure how it is made 10 The second Indirect Figure how made 11 How the third Indirect Figure is made 12 There are many Moods in every Figure but most of them Uselesse in Philosophy 13 An Hypotheticall Syllogisme when equipollent to a Categoricall 1. A Speech consisting of three Propositions from two of which the third followes is called a SYLLOGISME and that which followes is called the Conclusion the other two Premisses For example this Speech Every man is a Living Creature Every Living Creature is a Body therefore Every Man is a Body is a Syllogisme because the third Proposition follows from the two first that is if those be granted to be true this must also be granted to be true 2 From two Propositions which have not one Terme common no Conclusion can follow and therefore no Syllogisme can be made of them For let any two Premisses A man is a Living Creature A Tree is a Plant be both of them true yet because it cannot be collected from them that Plant is the Name of a Man or Man the Name of a Plant it is not necessary that this Conclusion A Man is a Plant should be true Corollary Therefore in the Premisses of a Syllogisme there can be but three Termes Besides there can be no Terme in the Conclusion which was not in the Premisses For let any two Premisses be A Man is a Living Creature A Living Creature is a Body yet if any other Terme be put in the Conclusion as Man is two footed though it be true it cannot follow from the Premisses because from them it cannot be collected that the Name Two footed belongs to a Man and therefore againe In every Syllogisme there can be but three Termes 3 Of these Termes that which is the Predicate in the Conclusion is commonly called the Major that which is the Subject in the Conclusion the Minor and the other is the Middle Term as in this Syllogisme A Man is a Living Creature A Living Creature is a Body therefore A Man is a Body Body is the Major Man the Minor and Living Creature the Middle Term. Also of the Premisses that in which the Major Terme is found is called the Major Proposition and that which has the Minor Term the Minor Proposition 4 If the Middle Terme be not in both the Premisses determined to one and the same singular thing no Conclusion will follow nor Syllogisme be made For let the Minor Terme be Man the Middle Terme Living Creature and the Major Term
which make equal Angles on either side of the Perpendicular be produced to the Tangent they will be equal 12 There is in Euclide a Definition of Straight-lined Parallels but I do not find that Parallels in general are any where defined and therefore for an Universal definition of them I say that Any two lines whatsoever Straight or Crooked as also any two Superficies are PARALLEL when two equal straight lines wheresoever they fall upon them make always equal Angles with each of them From which Definition it follows First that any two straight lines not enclined opposite wayes falling upon two other straight lines which are Parallel and intercepting equal parts in both of them are themselves also equal and Parallel As if AB and CD in the third Figure enclined both the same way fall upon the Parallels AC and BD and AC and BD be equal AB and CD will also be equal and Parallel For the Perpendiculars BE and DF being drawn the Right Angles EBD and FDH will be equal Wherefore seeing EF and BD are parallel the angles EBA and FDC will be equal Now if DC be not equal to BA let any other straight line equal to BA be drawn from the point D which seeing it cannot fall upon the point C let it fall upon G. Whereore AG will be either greater or less then BD and therefore the angles EBA and FDG are not equal as was supposed Wherefore AB and CD are equal which is the first Again because they make equal Angles with the Perpendiculars BE and DF therefore the Angle CDH will be equal to the Angle ABD and by the Definition of Parallels AB and CD will be parallel which is the second That Plain which is included both wayes within parallel lines is called a PARALLELOGRAM 1 Corollary From this last it follows That the Angles ABD and CDH are equal that is that a straight line as BH falling upon two Parallels as AB and CD makes the internal Angle ABD equal to the external and opposite Angle CDH 2 Coroll And from hence again it follows that a straight line falling upon two Parallels makes the alternate Angles equal that is the Angle AGF in the fourth figure equal to the Angle GFD For seeing GFD is equal to the external opposite Angle EGB it will be also equal to its vertical Angle AGF which is alternate to GFD 3 Coroll That the internal Angles on the same side of the line FG are equal to two Right Angles For the Angles at F namely GFC and GFD are equal to two Right Angles But GFD is equal to its alternate Angle AGF Wherefore both the Angles GFC and AGF which are internal on the same side of the line FG are equal to two Right Angles 4 Coroll That the three Angles of a straight-lined plain Triangle are equal to two Right Angles and any side being produced the external Angle will be equal to the two opposite internal Angles For if there be drawn by the Vertex of the plain Triangle ABC figure 5. a Parallel to any of the sides as to AB the Angles A and B will be equal to their alternate Angles E F the Angle C is common But by the 10th Article the three Angles E C and F are equal to two Right Angles and therefore the three Angles of the Triangle are equal to the same which is the first Again the two Angles B and D are equal to two Right Angles by the 10th Article Wherefore taking away B there will remain the Angles A and C equal to the Angle D which is the second 5 Coroll If the Angles A and B be equal the sides AC and CB will also be equal because AB and EF are parallel And on the contrary if the sides AC and CB be equal the Angles A and B will also be equal For if they be not equal let the Angles B and G be equal Wherefore seeing GB and EF are parallels and the Angles G and B equal the sides GC and CB will also be equal and because CB and AC are equal by supposition CG and CA will also be equal which cannot be by the 11th Article 6 Coroll From hence it is manifest that if two Radii of a Circle be connected by a straight Line the Angles they make with that connecting Line will be equal to one another and if there be added that segment of the Circle which is subtended by the same line which connects the Radii then the Angles which those Radii make with the circumference wil also be equal to one another For a straight line which subtends any Arch makes equal Angles with the same because if the Arch and the Subtense be divided in the middle the two halves of the segment wil be congruous to one another by reason of the Uniformity both of the Circumference of the Circle and of the straight Line 13 Perimeters of Circles are to one another as their Semidiameters are For let there be any two Circles as in the first figure BCD the greater and EFG the lesser having their common Center at A and let their Semidiameters be AC and AE I say AC has the same proportion to AE which the Perimeter BCD has to the Perimeter EFG For the magnitude of the Semidiameters AC and AE is determined by the distances of the points C and E from the Center A and the same distances are acquired by the uniform motion of a point from A to C in such manner that in equal times the distances acquired be equal But the Perimeters BCD and EFG are also determined by the same distances of the points C and E from the Center A and therefore the Perimeters BCD and EFG as well as the Semidiameters AC and AE have their magnitudes determined by the same cause which cause makes in equal times equal spaces Wherefore by the 13 Chapter and 6th Article the Perimeters of Circles and their Semidiameters are Proportionals which was to be proved 14 If two straight Lines w ch cōstitute an Angle be cut by straight-lined Parallels the intercepted Parallels will be to one another as the parts w ch they cut off frō the Vertex Let the straight lines AB and AC in the 6 figure make an Angle at A be cut by the two straight-lined Parallels BC and DE so that the parts cut off from the Vertex in either of those Lines as in AB may be AB and AD. I say the Parallels BC and DE are to one another as the parts AB and AD. For let AB be divided into any number of equal parts as into AF FD DB and by the points F and D let FG and DE be drawn Parallel to the base BC and cut AC in G and E and again by the points G and E let other straight lines be drawn Parallel to AB and cut BC in H and I. If now the point A be understood to be moved uniformly over AB and in the
same time B be moved to C and all the points F D and B be moved uniformly and with equal Swiftness over FG DE and BC then shall B pass over BH equal to FG in the same time that A passes over AF and AF and FG will be to one another as their Velocities are and when A is in F D will be in K when A is in D D will be in E and in what manner the point A passes by the points F D and B in the same manner the point B will pass by the points H I and C the straight lines FG DK KE BH HI IC are equal by reason of their Parallelisme and therefore as the velocitie in AB is to the velocity 〈◊〉 BC so is AD to DE but as the velocity in AB is to the velocity in BC so is AB to BC that is to say all the Parallels will be severally to all the parts cut off from the Vertex as AF is to FG. Wherefore AF. GF AD. DE AB BC are Proportionals The Subtenses of equal Angles in different Circles as the straight lines BC and FE in the 1 figure are to one another as the Arches which they subtend For by the 8th Article the Arches of equal Angles are to one another as their Perimeters are and by the 13th Art the Perimeters as their Semidiameters But the the Subtenses BC and FE are parallel to one another by reason of the equality of the Angles which they make with the Semidiameters and therefore the same Subtenses by the last precedent Article will be proportional to the Semidiameters that is to the Perimeters that is to the Arches which they subtend 15 If in a Circle any number of equal Subtenses be placed immediatly after one another and straight lines be drawn from the extreme point of the first Subtense to the extreme points of all the rest The first Subtense being produced will make with the second Subtense an external Angle double to that which is made by the same first Subtense and a Tangent to the Circle touching it in the extreme point thereof and if a straight line which subtends two of those Arches be produced it will make an external Angle with the third Subtense triple to the Angle which is made by the Tangent with the first Subtense and so continually For with the Radius AB in the 7th figure let a circle be described in it let any number of equal Subtenses BC CD DE be placed also let BD BE be drawn by producing BC BD BE to any distance in G H and I let them make Angles with the Subtenses which succeed one another namely the external Angles GCD and HDE Lastly let the Tangent KB be drawn making with the first Subtense the Angle KBC I say the Angle GCD is double to the Angle KBC and the Angle HDE triple to the same Angle KBC For if AC be drawn cutting BD in M and from the point C there be drawn LC perpendicular to the same AC then CL and MD will be parallel by reason of the right Angles at C and M and therefore the alterne Angles LCD and BDC wil be equal as also the Angles BDC and CBD will be equal because of the equality of the straight lines BC and CD Wherefore the Angle GCD is double to either of the Aâ—gles CBD or CDB and therefore also the Angle GCD is double to the Angle LCD that is to the Angle KBC Again CD is parallel to BE by reason of the equality of the Angles CBE and DEB and of the straight lines CB and DE and therefore the Angles GCD and GBE are equal and consequently GBE as also DEB is double to the Angle KBC But the external Angle HDE is equal to the two internal DEB and DBE and therefore the Angle HDE is triple to the Angle KBC c. which was to be proved 1 Corollary From hence it is manifest that the Angles KBC and CBD as also that all the Angles that are comprehended by two straight lines meeting in the circumference of a Circle and insisting upon equal Arches are equal to one another 2 Coroll If the Tangent BK be moved in the Circumference with Uniform motion about the Center B it will in equal times cut off equal Arches and will pass over the whole Perimeter in the same time in which it self describes a semiperimeter about the Center B. 3 Coroll From hence also we may understand what it is that determines the bending or Curvation of a straight line into the circumference of a Circle namely that it is Fraction continually encreasing in the same manner as NuÌ„bers from One upwards encrease by the continual addition of Unity For the indefinite straight Line KB being broken in B according to any Angle as that of KBC again in C according to a double Angle and in D according to an Angle which is triple and in E according to an Angle which is quadruple â—o the first Angle and so continually there will be described a figure which will indeed be rectilineal if the broken parts be considered as haâ—ing magnitude but if they be understood to be the least tâ—aâ— can 〈◊〉 â—â—at is as so many Points then the figure described will â—ot be rectilineal but a Circle whose Circumference wâ—… broken line 4 〈…〉 been said in this present Article it may 〈…〉 Angle in the center is double to an Angle in the Circumference of the same Circle if the intercepted Arches be equal For seeing that straight Line by whose motion an Angle is determined passes over equal Arches in equal times as well from the Center as from the Circumference and while that which is from the Circumference is passing over half its own Perimeter it passes in the same time over the whole Perimeter of that which is from the Center the Arches w ch it cuts off in the Perimeter whose Center is A wil be double to those which it makes in its own Semiperimeter whose Center is B. But in equal Circles as Arches are to one another so also are Angles It may also be demonstrated that the external Angle made by a Subtense produced and the next equal Subtense is equal to an Angle from the Center insisting upon the same Arch As in the last Diagram the Angle GCD is equal to the Angle CAD For the external Angle GCD is double to the Angle CBD and the Angle CAD insisting upon the same Arch CD is also double to the same Angle CBD or KBC 16 An Angle of Contingence if it be compared with an Angle simply so called how little soever has such proportion to it as a Point has to a Line that is no proportion at all nor any quantity For first an Angle of coÌ„tingence is made by coÌ„tinual flexion so that in the generation of it there is no circular motion at all in which consists the nature of an Angle simply so called and therefore it cannot be
angle NBK which being done BO will be the Line of Reflection from the Line of Incidence NB. Lastly from the incident Line LC let the reflected Line CO be drawn cutting BO at O and making the angle COB I say the angle COB is equal to the angle Z. Let NB be produced till it meet with the straight line LC produced in P. Seeing therefore the angle LCM is by construction equal to twice the angle BAC together with the angle Z the angle NPL which is equal to LCM by reason of the parallels NP and MC will also be equal to twice the same angle BAC together with the angle Z. And seeing the two straight lines OC and OB fall from the point O upon the points C and B and their reflected lines LC and NB meet in the point P the angle NPL will be equal to twice the angle BAC together with the angle COP But I have already proved the angle NPL to be equal to twice the angle BAC together with the angle Z. Therefore the angle COP is equal to the angle Z Wherefore Two points in the circumference of a Circle being given I have drawn c. which was to be done But if it be required to draw the incident Lines from a point within the circle so that the Lines reflected from them may contain an angle equal to the angle Z the same method is to be used saving that in this case the angle Z is not to be added to twice the angle BAC but to be taken from it 9 If a straight line falling upon the circumference of a circle be produced till it reach the Semidiameter and that part of it which is intercepted between the circumference and the Semidiameter be equal to that part of the Semidiameter which is between the point of concourse and the center the reflected Line will be parallel to the Semidiameter Let any Line AB in the 9th figure be the Semidiameter of the circle whose center is A and upon the circumference BD let the straight Line CD fall and be produced till it cut AB in E so that ED and EA may be equal from the incident Line CD let the Line DF be reflected I say AB and DF will be parallel Let AG be drawn through the point D. Seeing therefore ED and EA are equal the angles EDA and EAD will also be equal But the angles FDG and EDA are equal for each of them is half the angle EDH or FDC Wherefore the angles FDG and EAD are equal and consequently DF and AB are parallel which was to be proved Corollahy If EA be greater then ED then DF and AB being produced will concurre but if EA be less then ED then BA and DH being produced will concurre 10 If from a point within a circle two straight Lines be drawn to the Circumference and their reflected Lines meet in the Circumference of the same circle the angle made by the Lines of Reflection will be a third part of the angle made by the Lines of Incidence From the point B in the 10th figure taken within the circle whose center is A let the two straight lines BC and BD be drawn to the circumference and let their reflected Lines CE and DE meet in the circumference of the same circle at the point E. I say the angle CED will be a third part of the angle CBD Let AC and AD be drawn Seeing therefore the angles CED and CBD together taken are equal to twice the angle CAD as has been demonstrated in the 5th article and the angle CAD twice taken is quadruple to the angle CED the angles CED and CBD together taken will also be equal to the angle CED four times taken and therefore if the angle CED be taken away on both sides there will remain the angle CBD on one side equal to the angle CED thrice taken on the other side which was to be demonstrated Coroll Therefore a point being given within a Circle there may be drawn two Lines from it to the Circumference so as their reflected Lines may meet in the Circumference For it is but trisecting the Angle CBD which how it may be done shall be shewn in the following Chapter CHAP. XX. Of the Dimension of a Circle and the Division of Angles or Arches 1 The Dimension of a Circle neer determined in Numbers by Archimedes and others 2 The first attempt for the finding out of the Dimension of a Circle by Lines 3 The second attempt for the finding out of the Dimension of a Circle from the consideration of the nature of Crookedness 4 The third attempt and some things propounded to be further searched into 5 The Equation of the Spiral of Archimedes with a straight Line 6 Of the Analysis of Geometricians by the Powers of Lines 1 IN the comparing of an Arch of a Circle with a Straight Line many and great Geometricians even from the most ancient times have exercised their wits and more had done the same if they had not seen their pains though undertaken for the common good if not brought to perfection vilified by those that envy the prayses of other men Amongst those Ancient Writers whose Works are come to our hands Archimedes was the first that brought the Length of the Perimeter of a Circle within the limits of Numbers very litle differing from the truth demonstrating the same to be less then three Diameters and a seventh part but greater then three Diameters and ten seventy one parts of the Diameter So that supposing the Radius to consist of 10000000 equal parts the Arch of a Quadrant will be between 15714285 and 15 04225 of the same parts In our times Ludovicus Van Cullen Willebrordus Snellius with joint endeavour have come yet neerer to the truth and pronounced from true Principles that the Arch of a Quadrant putting as before 10000000 for Radius differs not one whole Unity from the number 15707963 which if they had exhibited their arithmetical operations and no man had discovered any errour in that long work of theirs had been demonstrated by them This is the furthest progress that has been made by the way of Numbers and they that have proceeded thus far deserve the praise of Industry Nevertheless if we consider the benefit which is the scope at which all Speculation should aime the improvement they have made has been little or none For any ordinary man may much sooner more accurately find a Straight Line equal to the Perimeter of a Circle and consequently square the Circle by winding a small thred about a given Cylinder then any Geometrician shall do the same by dividing the Radius into 10000000 equal parts But though the length of the Circumference were exactly set out either by Numbers or mechanically or onely by chance yet this would contribute no help at all towards the Section of Angles unless happily these two Problemes To divide a given Angle according to any proportion assigned and To finde a
equal that no inequality can be discovered between them I will therefore leave this to be further searched into For though it be almost out of doubt that the Straight Line BP and the arch BMD are equal yet that may not be received without demonstration and means of Demonstration the Circular Line admitteth none that is not grounded upon the nature of Flexion or of Angles But by that way I have already exhibited a Straight Line equal to the Arch of a Quadrant in the First and Second aggression It remains that I prove DT to be equal to the Sine of 45 degrees In BA produced let AV he taken equal to the Sine of 45 degrees and drawing and producing VH it will cut the arch of the Quadrant CNA in the midst in N and the same arch again in O and the Straight line DC in T so that DT will be equal to the Sine of 45 degrees or to the straight line AV also the Straight line VH will be equal to the straight line HI or the Sine of 60 degrees For the square of AV is equal to two squares of the Semiradius and consequently the square of VH is equal to three Squares of the Semiradius But HI is a mean proportional between the Semiradius and three Semiradii and therefore the square of HI is equal to three Squares of the Semiradius Wherefore HI is eqval to HV But because AD is cut in the midst in H therefore VH and HT are equal and therefore also DT is equal to the Sine of 45 degrees In the Radius BA let BX be taken equal to the Sine of 45 degrees for so VX will be equal to the Radius and it will be as VA to AH the Semiradius so VX the Radius to XN the Sine of 45 degrees Wherefore VH produced passes through N. Lastly upon the center V with the Radius VA let the arch of a circle be drawn cutting VH in Y which being done VY will be equal to HO for HO is by construction equal to the Sine of 45 degrees and YH will be equal to OT therefore VT passes through O. All which was to be demonstrated I will here add certain Problemes of which if any Analyst can make the construction he will thereby be able to judge clearly of what I have now said concerning the dimension of a Circle Now these Problems are nothing else at least to sense but certain symptomes accompanying the construction of the first and third figure of this Chapter Describing therefore again the Square ABCD in the fifth figure and the three Quadrants ABD BCA and DAC let the Diagonals AC BD be drawn cutting the arches BHD CIA in the middle in H and I the straight lines EF and GL dividing the square ABCD into four equal squares and trisecting the arches BHD and CIA namely BHD in K and M and CIA in M and O. Then dividing the arch BK in the midst in P let QP the Sine of the arch BP be drawn and produced to R so that QR be double to QP and connecting KR let it be produced one way to BC in S and the other way to BA produced in T. Also let BV be made triple to BS and consequently by the second article of this Chapter equall to the arch BD. This construction is the same with that of the first figure which I thought fit to renew discharged of all lines but such as are necessary for my present purpose In the first place therefore if AV be drawn cutting the arch BHD in X and the side DC in Z I desire some Analyst would if he can give a reason Why the straight lines TE and TC should cut the arch BD the one in Y the other in X so as to make the arch BY equal to the arch YX or if they be not equal that he would determine their difference Secondly if in the side DA the straight line Da be taken equal to DZ and Va be drawn Why Va and VB should be equal or if they be not equal What is the difference Thirdly drawing Zb parallel and equal to the side CB cutting the arch BHD in c and drawing the straight line Ac and producing it to BV in d Why Ad should be equal and parallel to the straight line aV and consequently equal also to the arch BD. Fourthly drawing eK the Sine of the arch BK taking in eA produced ef equal to the Diagonal AC and connecting fC Why fC should pass through a which point being given the length of the arch BHD is also given and c and why fe and fc should be equal or if not why unequal Fifthly drawing fZ I desire he would shew Why it is equal to BV or to the arch BD or if they be not equal What is their difference Sixtly granting fZ to be equal to the arch BD I desire he would determine whether it fall all without the arch BCA or cut the same or touch it and in what point Seventhly the Semicircle BDg being completed Why gI being drawn and produced should pass through X by which point X the length of the arch BD is determined And the same gI being yet further produced to DC in h Why Ad which is equal to the arch BD should pass through that point h. Eighthly upon the Center of the square ABCD which let be k the arch of the quadrant EiL being drawn cutting eK produced in i Why the drawn straight line iX should be parallel to the side CD Ninthly in the sides BA and BC taking Bl and Bm severally equal to half BV or to the arch BH and drawing mn parallel and equal to the side BA cutting the arch BD in o Why the straight line wich connects Vl should pass through the point o Tenthly I would know of him Why the straight line which connects aH should be equal to Bm or if not how much it differs from it The Analyst that can solve these Problemes without knowing first the length of the arch BD or using any other known Method then that which proceeds by perpetual bisection of an angle or is drawn from the consideration of the nature of Flexion shall do more then ordinary Geometry is able to perform But if the Dimension of a Circle cannot be found by any other Method then I have either found it or it is not at all to be found From the known Length of the Arch of a Quadrant and from the proportional Division of the Arch and of the Tangent BC may be deduced the Section of an Angle into any given proportion as also the Squaring of the Circle the Squaring of a given Sector and many the like propositions which it is not necessary here to demonstrate I will therefore onely exhibit a Straight line equal to the Spiral of Archimedes and so dismiss this speculation 5 The length of the Perimeter of a Circle being found that Straight line is also found which
weakned by little and little But this cannot be done but by the long continuance of action or by actions often repeated and therefore Custome begets that Facicility which is commonly and rightly called Habit and it may be defined thus HABIT is Motion made more easie and ready by Custome that is to say by perpetual endeavour or by iterated endevours in a way differing from that in which the Motion proceeded from the beginning and opposing such endeavours as resist And to make this more perspicuous by example We may observe that when one that has no skill in Musique first puts his hand to an Instrument he cannot after the first stroke carry to his hand to the place where he would make the second stroke without taking it back by a new endeavour and as it were beginning again pass from the first to the second Nor will he be able to go on to the third place without another new endeavour but he will be forced to draw back his hand again and so successively by renewing his endeavour at every stroke till at the last by doing this often and by compounding many interrupted motions or endeavours into one equal endeavour he be able to make his hand go readily on from stroke to stroke in that order and way which was at the first designed Nor are Habits to be observed in living creatures only but also in Bodies inanimate For we find that when the Lath of a Crossbow is strongly bent and would if the impediment were removed return again with great force if it remain a long time bent it will get such a Habit that when it is loosed and left to its own freedome it will not onely not restore it self but will require as much force for the bringing of it back to its first posture as it did for the bending of it at the first CHAP. XXIII Of the Center of Equiponderation of Bodies pressing doâ—â—ards in straight Parallel Lines 1 Definitions and Suppositions 2 Two Plains of Equiponderation are nâ—â— parallel 3 The Center of Equiponderation is in every Plain of Equiponderation 4 The Moments of equal Ponderants are to one another as their distances from the center of the Scale 5,6 The Moments of unequal Ponderants have their proportion to one another compounded of the proportions of their Waights and distances from the center of the Scale reciprocally taken 7. If two Ponderants have their Moments and Distances from the Center of the Scale in reciprocal proportion they are equally poised and contrarily 8 If the parts of any Ponderant press the Beam of the Scale every where equally all the parts cut out off reckoned from the Center of the Scale will have their Moments in the same proportion with that of the parts of a Triangle cut off from the Vertex by straight Lines parallel to the base 9 The Diameter of Equiponderation of Figures which are deficient according to commensurable proportions of their altitudes and bases divides the Axis so that the part taken next the vertex is to the other part as the complete figure to the deficient figure 10 The diameter of Equiponderation of the Complement of the half of any of the said deficient figures divides that line which is drawn through the vertex parallel to the base so that the part next the vertex is to the other part as the complete figure to the Complement 11 The Center of Equiponderation of the half of any of the desicient figures in the first row of the Table of the 3d. Article of the 17th Chapter may be found out by the numbers of the second row 12 The center of Equiponderation of the half of any of the figures in the second row of the same Table may be found out by the numbers of the fourth row 13 The Center of Equiponderation of the half of any of the figures in the same Table being known the Center of the Excess of the same figure above a Triangle of the same altitude and base is also known 14 The Center of Equiponderation of a solid Sector is in the Axis so divided that the part next the Vertex be to the whole Axis want half the Axis of the portion of the Sphere as 3 to 4. 1 Definitions 1 A Scale is a straight line whose middle point is immoveable all the rest of its points being at liberty and that part of the Scale which reaches from the center to either of the waights is called the Beam 2 Equiponderation is when the endeavour of one Body which presses one of the Beams resists the endeavour of another Body pressing the other Beam so that neither of them is moved and the Bodies when neither of them is moved are said to be Equally poised 3 Waight is the aggregate of all the Endeavours by which all the points of that Body which presses the Beam tend downwards in lines parallel to one another and the Body which presses is called the Ponderant 4 Moment is the Power which the Ponderant has to move the Beam by reason of a determined situation 5 The plain of Equiponderation is that by which the Ponderant is so divided that the Moments on both sides remain equal 6 The Diameter of Equiponderation is the common Section of the two Plains of Equiponderation and is in the straight line by which the waight is hanged 7 The Center of Equiponderation is the common point of the two Diameters of Equiponderation Suppositions 1 When two Bodies are equally pois'd if waight be added to one of them and not to the other their Equiponderation ceases 2 When two Ponderants of equal magnitude and of the same Species or matter press the Beam on both sides at equal distances from the center of the Scale their Moments are equal Also when two Bodies endeavour at equal distances from the center of the Scale if they be of equal magnitude and of the same Species their Moments are equal 2 No two Plains of Equiponderation are parallel Let A B C D in the first figure be any Ponderant whatsoever and in it let E F be a Plain of Equiponderation parallel to which let any other Plain be drawn as G H. I say G H is not a Plain of Equiponderation For seeing the parts A E F D and E B C F of the Ponderant A B C D are equally pois'd and the weight E G H F is added to the part A E F D and nothing is added to the part E B C F but the weight E G H F is taken from it therefore by the first Supposition the parts A G H D and G B C H will not be equally pois'd and consequently G H is not a Plain of Equiponderation Wherefore No two Plains of Equiponderation c. Which was to be proved 3 The Center of Equiponderation is in every Plain of Equiponderation For if another Plain of Equiponderation be taken it will not by the last Article be parallel to the former Plain and therefore both those Plains will
together taken so that the part next the Vertex be triple to the other part or to the whole straight line as 3 to 4. For let A B C in the 9th fig. be the Sector of a Sphere whose Vertex is the ceâ—ter of the Sphere A whose Axis is A D and the circle upon B C is the common base of the portion of the Sphere and of the Cone whose Vertex is A the Axis of which portion is E D and the halfe thereof F D and the Axis of the Cone A E. Lastly let A G be ¾ of the straight line A F. I say G is the center of Equiponderation of the Sector A B C. Let the straight line F H be drawne of any length making right angles with A F at F and drawing the straight line A H let the triangle A F H be made Then upon the same center A let any arch I K be drawne cutting A D in L and its chord cutting A D in M and dividing M L equally in N let N O be drawne parallel to the straight line F H and meeting with the straight line A H in O. Seeing now B D C is the Spherical Superficies of the portion cut off with a plain passing through B C and cutting the Axis at right angles and seeing F H divides E D the Axis of the portion into two equal parts in F the center of Equiponderation of the Superficies B D C will be in F by the 8th article and for the same reason the center of Equiponderation of the Superficies I L K K being in the straight line A C will be in N. And in like manner if there were drawne between the center of the Sphere A and the outermost Spherical Superficies of the Sector arches infinite in number the centers of Equiponderation of the Sphericall Superficies in which those arches are would be found to be in that part of the Axis which is intercepted between the Superficies it selfe and a plaine passing along by the chord of the arch and cutting the Axis in the middle at right angles Let it now be supposed that the moment of the outermost sphericall Superficies B D C is F H. Seeing therefore the Superficies B D C is to the Superficies I L K in proportion duplicate to that of the arch B D C to the arch I L K that is of B E to I M that is of F H to N O let it be as F H to N O so N O to another N P and again as N O to N P so N P to another N Q and let this be done in all the straight lines parallel to the base F H that can possibly be drawn between the base and the vertex of the triangle A F H. If then through all the points Q there be drawn the crooked line A Q H the figure A F H Q A will be the complement of the first three-siâ—ed figure of two Meanes and the same will also be the moment of all the Sphericall Superficies of which the Solid Sector A B C D is compounded and by consequent the moment of the Sector it selfe Let now F H be understood to be the semidiameter of the base of a right Cone whose side is A H and Axis A F. Wherfore seeing the bases of the Cones which passe through F and N and the rest of the points of the Axis are in proportion duplicate to that of the straight lines F H and N O c. the moment of all the bases together that is of the whole Cone will be the figure it self A F H Q A and therefore the center of Equiponderation of the Cone A F H is the same with that of the solid Sector Wherefore seeing A G is ¾ of the Axis A F the center of Equiponderation of the Cone A F H is in G and therefore the center of the solid Sector is in G also and divides the part A F of the Axis so that A G is triple to G F that is A G is to A F as 3 to 4 which was to be demonstrated Note that when the Sector is a Hemisphere the Axis of the Cone vanisheth into that point which is the center of the Sphere and therefore it addeth nothing to half the Axis of the portion Wherefore if in the Axis of the Hemisphere there be taken from the center ¾ of halfe the Axis that is 3 â— of the Semidiameter of the Sphere there will be the center of Equiponderation of the Hemisphere CHAP. XXIV Of Refraction and Reflection 1 Definitions 2 In perpendicular Motion there is no Refraction 3 Things thrown out of a thinner into a thicker Medium are so refracted that the Angle Refracted is greater then the Angle of Inclination 4 Endeavour which from one point tendeth every way will be so Refracted at that the sine of the Angle Refracted will be to the sine of the Angle of Inclination as the Density of the first Medium is to the Density of the second Medium reciprocally taken 5 The sine of the Refracted Angle in one Inclination is to the sine of the Refracted Angle in another Inclination as the sine of the Angle of that Inclination is to the sine of the Angle of this Inclination 6 If two lines of Incidence having equal Inclination be the one in a thinner the other in a thicker Medium the sine of the angle of Inclination will be a Mean proportional between the two sines of the Refracted angles 7 If the angle of Inclination be semirect and the line of Inclination be in the thicker Medium and the proportion of their Densities be the same with that of the Diagonal to the side of a Square and the separating Superficies be plain the Refracted line will be in the separating Superficies 8 If a Body be carried in a straight line upon another Body and do not penetrate the same but be reflected from it the angle of Reflexion will be equal to the Angle of Incidence 9 The same happens in the generation of Motion in the line of Incidence 1 Definitions 1 REFRACTION is the breaking of that straight Line in which a Body is moved or its Action would proceed in one and the same Medium into two straight lines by reason of the different natures of the two Mediums 2 The former of these is called the Line of Incidence the later the Refracted Line 3 The Point of Refraction is the common point of the Line of Incidence and of the Refracted Line 4 The Refracting Superficies which also is the Separating Superficies of the two Mediums is that in which is the point of Refraction 5 The Angle Refracted is that which the Refracted Line makes in the point of Refraction with that Line which from the same point is drawn perpendicular to the separating Superficies in a different Medium 6 The Angle of Refraction is that which the Refracted line makes with the Line of Incidence produced 7 The Angle of Inclination is
and the Properties of Straight Parallels 13 The Circumferences of Circles are to one another as their Diameters are 14 In Triangles Straight Lines parallel to the Bases are to one another as the parts of the Sides which they cut off from the Vertex 15 By what Fraction of a Straight Line the Circumference of a Circle is made 16 That an Angle of Contingence is Quantity but of a Different kinde from that of an Angle simply so called and that it can neither add nor take away any thing from the same 17 That the Inclination of Plains is Angle simply so called 18 A Solid Angle what it is 19 What is the Nature of Asymptotes 20 Situation by what it is determined 21 What is like Situation What is Figure and what are like Figures 1 BEtween two points given the shortest Line is that whose extreme points cannot be drawn further asunder withour altering the quantity that is without altering the proportion of that line to any other line given For the Magnitude of a Line is computed by the greatest distance which may be between its extreme points So that any one Line whether it be extended or bowed has alwayes one and the same Length because it can have but one greatest distance between its extreme points And seeing the action by which a Straight Line is made Crooked or contrarily a Crooked Line is made Straight is nothing but the bringing of its extreme points neerer to one another or the setting of them further asunder a CROOKED Line may rightly be defined to be That whose extreme points may be understood to be drawn further asunder and a STRAIGHT Line to be That whose extreme points cannot be drawn further asunder and comparatively A more Crooked to be That line whose extreme points are neerer to one another then those of the other supposing both the Lines to be of equal Length Now howsoever a Line be bowed it makes alwayes a Sinus or Cavity sometimes on one side sometimes on another So that the same Crooked Line may either have its whole Cavity on one Side onely or it may have it part on one side and part on other sides Which being well understood it will be easie to understand the following Comparisons of Straight and Crooked Lines First If a Straight a Crooked Line have their Extreme points common the Crooked Line is longer then the Straight Line For if the extreme points of the Crooked Line be drawn out to their greatest distance it wil be made a straight line of which that which was a Straight Line from the beginning will be but a part and therefore the Straight Line was shorter then the Crooked Line which had the same extreme points And for the same reason if two Crooked Lines have their extreme points common and both of them have all their cavity on one and the same side the outermost of the two will be the longest Line Secondly A Straight Line and a perpetually Crooked Line cānot be coincident no not in the least part For if they should then not onely some Straight Line would have its extreme points common with some Crooked Line but also they would by reason of their coincidence be equal to one another which as I have newly shewn cannot be Thirdly Between two points given there can be understood but one straight Line because there cannot be more then one least Interval or Length between the same points For if there may be two they will either be coincident and so both of them will be one Straight Line or if they be not coincident then the application of one to the other by extension will make the extended Line have its extreme points at greater distance then the other and consequently it was Crooked from the beginning Fourthly From this last it follows that two Straight Lines cannot include a Superficies For if they have both their extreme points common they are coincident and if they have but one or neither of them common then at one or both ends the extreme points will be disjoyned and include no Superficies but leave all open and undetermined Fifthly Every part of a Straight Line is a Straight Line For seeing every part of a Straight Line is the least that can be drawn between its own extreme points if all the parts should not constitute a Straight Line they would all together be longer then the whole Line 2 APLAIN or a Plain Superficies is that which is described by a Straight Line so moved that all the several points thereof describe several Straight Lines A straight line therefore is necessarily all of it in the same Plain which it describes Also the Straight Lines which are made by the points that describe a Plain are all of them in the Same Plain Moreover if any Line whatsoever be moved in a Plain the Lines which are described by it are all of them in the same Plain All other Superficies which are not Plain are Crooked that is are either Concave or Convex And the same Comparisons which were made of Straight and Crooked Lines may also be made of Plain and Crooked Superficies For First If a Plain and a Crooked Superficies be terminated with the same Lines the Crooked Superficies is greater then the Plain Superficies For if the Lines of which the Crooked Superficies consists be extended they will be found to be longer then those of which the Plain Superficies consists which cannot be extended because they are Straight Secondly Two Superficies wherof the one is Plain and the other continually Crooked cannot be coincident no not in the least part For if they were coincident they would be equal nay the same Superficies would be both Plain and Crooked which is impossible Thirdly Within the same terminating Lines there can be no more then one Plain Superficies because there can be but one least Superficies within the same Fourthly No number of Plain Superficies can include a Solid unless more then two of them end in a Common Vertex For if two Plains have both the same terminating Lines they are coincident that is they are but one Superficies and if their terminating Lines be not the same they leave one or more sides open Fifthly Every part of a Plain Superficies is a Plain Superficies For seeing the whole Plain Superficies is the least of all those that have the same terminating Lines and also every part of the same Superficies is the least of all those that are terminated with the same Lines if every part should not constitute a Plain Superficies all the parts put together would not be equal to the whole 3 Of Straightness whether it be in Lines or in Superficies there is but one kinde but of Crookedness there are many kindes for of Crooked Magnitudes some are Congruous that is are coincident when they are applyed to one another others are Incongruous Again some are 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 or Uniform that is have their parts howsoever taken congruous to one
same Chapter I have shewn that Whatsoever is at Rest will alwayes be at Rest unless there be some other Body besides it which by getting into its place suffers it no longer to remain at Rest. And that Whatsoever is Moved will alwayes be Moved unless there be some other Body besides it which hinders its Motion Tenthly In the 9 Chapter and 7 Article I have demonstrated that When any Body is moved which was formerly at Rest the immediate efficient cause of that Motion is in some other Moved and Contiguou◠Body Eleventhly I have shewn in the same place that Whatsoever is Moved will always be Moved in the same way and with the same Swiftness if it be not hindered by some other Moved and Contiguou◠Body 2 To which Principles I shall here add these that follow First I define ENDEAVOUR to be Motion made in less Space and Time then can be given that is less then can be determined or assigned by Exposition or Number that is Motion made through the length of a Point and in an Instant or Point of Time For the explayning of which Definition it must be remembred that by a Point is not to be understood that which has no quantity or which cannot by any means be divided for there is no such thing in Nature but that whose quantity is not at all considered that is whereof neither quantity nor any part is computed in demonstration so that a Point is not to be taken for an Indivisible but for an Undivided thing as also an Instant is to be taken for an Undivided and not for an Indivisible Time In like manner Endeavour is to be conceived as Motion but so as that neither the quantity of the Time in which nor of the Line in which it is made may in demonstration be at all brought into comparison with the quantity of that Time or of that Line of which it is a part And yet as a Point may be compared with a Point so one Endeavour may be compared with another Endeavour and one may be found to be greater or lesse then another For if the Vertical points of two Angles be compared they will be equal or unequal in the same proportion which the Angles themselves have to one another Or if a straight Line cut many Circumferences of Concentrick Circles the inequality of the points of intersection will be in the same proportion which the Perimeters have to one another And in the same manner if two Motions begin and end both together their Endeavours will be Equal or Unequal according to the proportion of their Velocities as we see a bullet of Lead descend with greater Endeavour then a ball of Wooll Secondly I define IMPETUS or Quickness of Motion to be the Swiftness or Velocity of the Body moved but considered in the several points of that time in which it is moved In which sense Impetus is nothing else but the quantity or velocity of Endeavour But considered with the whole time it is the whole velocity of the Body moved taken together throughout all the time and equal to the Product of a Line representing the time multiplyed into a Line representing the arithmetically mean Impetus or Quickness Which Arithmetical Mean what it is is defined in the 29th Article of the 13th Chapter And because in equal times the wayes that are passed are as the Velocities and the Impetus is the Velocity they go withal reckoned in all the several points of the times it followeth that during any time whatsoever howsoever the Impetus be encreased or decreased the length of the way passed over shall be encreased or decreased in the same proportion and the same Line shall represent both the way of the Body moved and the several Impetus or degrees of Swiftness wherewith the way is passed over And if the Body moved be not a point but a straight line moved so as that every point thereof make a several straight line the Plain described by its motion whether Uniform Accelerated or Retarded shall be greater or less the time being the same in the same proportion with that of the Impetus reckoned in one motion to the Impetus reckoned in the other For the reason is the same in Parallelograms and their Sides For the same cause also if the Body moved be a Plain the Solid described shall be still greater or less in the proportions of the several Impetus or Quicknesses reckoned through one Line to the several Impetus reckoned through another This understood let ABCD in the first figure of the 17th Chapter be a Parallelogram in which suppose the side AB to be moved parallelly to the opposite side CD decreasing al the way till it vanish in the point C and so describing the figure ABEFC the point B as AB decreaseth will therefore describe the Line BEFC and suppose the time of this motion designed by the line CD and in the same time CD suppose the side AC to be moved parallelly and uniformly to BD. From the point O taken at adventure in the Line CD draw OR parallel to BD cutting the Line BEFC in E and the side AB in R. And again from the point Q taken also at adventure in the Line CD draw QS parallel to BD cutting the Line BEFC in F and the side AB in S and draw EG and FH parallel to CD cutting AC in G and H. Lastly suppose the same construction done in all the points possible of the Line BEFC I sa◠that as the proportions of the Swiftnesses wherewith QF OE DB and all the rest supopsed to be drawn parallel to DB and terminated in the Line BEFC are to the proportions of their several Times designed by the several parallels HF GE AB and all the rest supposed to be drawn parallel to the Line of time CD and terminated in the Line BEFC the aggregate to the aggregate so is the Area or Plain DBEFC to the Area or Plain ACFEB For as AB decreasing continually by the line BEFC vanisheth in the time CD into the point C so in the same time the line DC continually decreasing vanisheth by the same line CFEB into the point B and the point D describeth in that decreasing motion the line DB equall to the line AC described by the point A in the decreasing motion of A B their swiftnesses are therefore equal Again because in the time GE the point O describeth the line OE and in the same time the point R describeth the line RE the line OE shall be to the line RE as the swiftness wherewith OE is described to the swiftness wherwith RE is described In like māner because in the same time HF the point Q describeth the Line QF and the point S the Line SF it shall be as the swiftness by which QF is described to the swiftness by which SF is described so the Line it self QF to the Line it self SF and so in all the Lines that can possibly be
every where triplicate of the proportions of AB to GE and of AB to HF c. the proportions of HF to AB and of GE to AB c. by the 16 Article of the 13 Chap. are triplicate of the proportions of QF to DB and of OE to DB c. and therefore the deficient figure ABEFC which is the aggregate of all the lines HF GE AB c. is triple to the complement BEFCD made of all the lines QF OE DB c. which was to be proved It follows from hence That the same complement BEFCD is ¼ of the whole Parallelogram And by the same method may be calculated in all other Deficient Figures generated as above declared the proportion of the Parallelogram to either of its parts as that when the parallels encrease fom a point in the same proportion the Parallelogram will be divided into two equal Triangles when one encrease is double to the other it will be divided into a Semiparabola and its Complement or into 2 and 1. The same construction standing the same conclusion may otherwise be demonstrated thus Let the straight line CB be drawn cutting GK in L through L let MN be drawn parallel to the straight line AC wherefore the Parallelograms GM and LD will be equal Then let LK be divided into three equal parts so that it may be to one of those parts in the same proportion which the proportion of AC to GC or of GK to GL hath to the proportion of GK to GE. Therefore LK will be to one of those three parts as the Arithmetical proportion between GK and GL is to the Arithmetical proportion between GK and the same GK want the third part of LK and KE will be somwhat greater then a third of LK Seeing now the altitude AG or ML is by reason of the continual decrease to be supposed less then any quantity that can be given LK which is intercepted between the Diagonal BC and the side BD will be also less then any quantity that can be given and consequently if G be put so neer to A in g as that the difference between Cg and CA be less then any quantity that can be assigned the difference also between Cl removing L to l and CB will be less then any quantity that can be assigned and the line gl being drawn produced to the line BD in k cutting the crooked line in e the proportion of Gk to Gl will still be triplicate to the proportion of Gk to Ge and the difference between k and e the third part of kl will be less then any quantity that can be given and therefore the Parallelogram eD will differ from a third part of the Parallelogram Ae by a less difference then any quantity that can be assigned Again let HI be drawn parallel and equal to ge cutting CB in P the crooked line in F and BD in I and the proportion of Cg to CH will be triplicate to the proportion of HF to HP and IF will be greater then the third part of PI. But again setting H in h so neer to g as that the difference between Ch and Cg may be but as a point the point P will also in p be so neer to l as that the difference between Cp and Cl will be but as a point and drawing hp till it meet with gk in i cutting the crooked line in f and having drawn eo parallel to BD cutting DC in o the Parallelogram fo will differ less from the third part of the Parallelogram gf then by any quantity that can be given And so it will be in all other Spaces generated in the same manner Wherefore the differences of the Arithmetical and Geometrical Means which are but as so many points B e f c. seeing the whole Figure is made up of so many indivisible Spaces will constitute a certain line such as is the line BEFC which will divide the complete Figure AD into two parts whereof one namely ABEFC which I call a Deficient Figure is triple to the other namely BDCEF which I call the Complement thereof And whereas the proportion of the altitudes to one another is in this case everywhere triplicate to that of the decreasing quantities to one another in the same manner if the proportion of the altitudes had been every where quadruplicate to that of the decreasing quantities it might have been demonstrated that the Deficient Figure had been quadruple to its Complement and so in any other proportion Wherefore a Deficient Figure which is made c. Which was to be demonstrated The same rule â—oldeth also in the diminution of the Bases of Cylinders as is demonstrated Chap. 15. Art 2. â— By this Proposition the magnitudes of all Deficient Figures when the proportions by which their bases decrease continually are proportionall to those by which their altitudes decrease may be compared with the magnitudes of their Complements and consequently with the magnitudes of their Complete Figures And they will be found to be as I have set them down in the following Tables in which I compare a Parallelogram with three-sided Figures and first with a straight lined triangle made by the base of the Parallelogram continually decreasing in such manner that the altitudes be alwayes in proportion to one another as the bases are and so the triangle will be equal to its Complement or the proportions of the altitudes and bases wil be as 1 to 1 and then the triangle will be half the Parallelogram Secondly with that three-sided Figure which is made by the continual decreasing of the bases in subduplicate proportion to that of the altitudes and so the Deficient Figure will be double to its Complement and to the Parallelogram as 2 to 3. Then with that where the proportion of the altitudes is triplicate to that of the bases and then the Deficient Figure will be triple to its Complement and to the Parallelogram as 3 to 4. Also the proportion of the altitudes to that of the bases may be as 3 to 2 and then the Deficient Figure will be to its Complement as 3 to 2 to the Parallelogram as 3 to 5 and so forwards according as more mean proportionals are taken or as the proportions are more multiplyed as may be seen in the following Table For example if the bases decrease so that the proportion of the altitudes to that of the bases be alwayes as 5 to 2 and it be demanded what proportion the Figure made has to the Parallelogram which is supposed to be Unity then seeing that where the proportion is taken five times there must be four Means look in the Table amongst the three-sided figures of four Means and seeing the proportion was as 5 to 2 look in the uppermost row for the number 2 and descending in the 2d Columne till you meet with that three-sided Figure you will finde 5 7 which shews that the Deficient Figure is to the Parallelogram as 5 â—
A F and A f be drawn A F therefore will be the sine of the angle of Inclination of the straight line A B and A f the sine of the angle of Inclination of the straight line A h which two Inclinations are by construction made equal I say as the density of the Medium in which are B C and b c is to the density of the Medium in which are B D and b d so is the sine of the angle Refracted to the sine of the angle of Inclination Let the straight line F G be drawn parallel to the straight line A B meeting with the straight line b B produced in G. Seeing therefore A F and B G are also parallels they will be equal and consequently the endeavour in A F is propagated in the same time in which the endeavour in B G would be propagated if the Medium were of the same density But because B G is in a thicker Medium that is in a Medium which resists the endeavour more then the Medium in which A F is the endeavour will be propagated less in B G then in A F according to the proportion which the density of the Medium in which A F is hath to the density of the Medium in which B G is Let therefore the density of the Medium in which B G is be to the density of the Medium in which A F is as B G is to B H and let the measure of the time be the Radius of the Circle Let H I be drawn parallel to B D meeting with the circumference in I and from the point I let I K be drawn perpendicular to B D which being done B H and I K will be equal and I K will be to A F as the density of the Medium in which is A F is to the density of the Medium in which is I K. Seeing therefore in the time A B which is the Radius of the Circle the endeavour is propagated in A F in the thinner Medium it will be propagated in the same time that is in the time B I in the thicker Medium from K to I. Therefore B I is the Refracted line of the line of Incidence A B and I K is the sine of the angle Refracted and A F the sine of the angle of Inclination Wherefore seeing I K is to A F as the density of the Medium in which is A F to the density of the Medium in which is I K it will be as the density of the Medium in which is A F or B C to the density of the Medium in which is I K or B D so the sine of the angle Refracted to the sine of the angle of Inclination And by the same reason it may be shewn that as the density of the thinner Medium is to the density of the thicker Medium so will K I the sine of the angle Refracted be to A F the sine of the Angle of Inclination Secondly let the Body which endeavoureth every way be in the thicker Medium at I. If therefore both the Mediums were of the same density the endeavour of the Body in I B would tend directly to L and the sine of the angle of Inclination L M would be equal to I K or B H. But because the density of the Medium in which is IK to the density of the Medium in which is L M is as BH to B G that is to A F the endeavour will be propagated further in the Medium in which L M is then in the Medium in which I K is in the proportion of density to density that is of M L to A F. Wherefore B A being drawn the angle Refracted will be C B A and its sine A F. But L M is the sine of the angle of Inclination and therefore again as the density of one Medium is to the density of the different Medium so reciprocally is the sine of the angle Refracted to the sine of the angle of Inclination which was to be demonstrated In this Demonstration I have made the separating Superficies B b plain by construction But though it were concave or convex the Theoreme would nevertheless be true For the Refraction being made in the point B of the plain separating Superficies if a crooked line as P Q be drawn touching the separating line in the point B neither the Refracted line B I nor the perpendicular B D will be altered and the Refracted angle K B I as also its sine K I will be still the same they were 5 The sine of the angle Refracted in one Inclination is to the sine of the angle Refracted in another Inclination as the sine of the angle of that Inclination to the sine of the angle of this Inclination For seeing the sine of the Refracted angle is to the sine of the angle of Inclination whatsoever that Inclination be as the density of one Medium to the density of the other Medium the proportion of the sine of the Refracted angle to the sine of the angle of Inclination will be compounded of the proportions of density to density and of the sine of the angle of one Inclination to the sine of the angle of the other Inclination But the proportions of the densities in the same Homogeneous Body are supposed to be the same Wherefore Refracted angles in different Inclinations are as the sines of the angles of those Inclinations which was to be demonstrated 6 If two lines of Incidence having equal inclination be the one in a thinner the other in a thicker Medium the sine of the angle of their Inclination will be a mean proportional between the two sines of their angles Refracted For let the straight line AB in the same 3d figure have its Inclination in the thinner Medium and be refracted in the thicker Medium in B I and let E B have as much Inclination in the thicker Medium and be refracted in the thinner Medium in B S and let R S the sine of the angle Refracted be drawn I say the straight lines R S A F and I K are in continual proportion For it is as the density of the thicker Medium to the density of the thinner Medium so R S to A F. But it is also as the density of the same thicker Medium to that of the same thinner Medium so AF to IK Wherefore R S. A F A F. I K are propoortionals that is R S A F and I K are in continual proportion and A F is the Mean proportional which was to be proved 7 If the angle of Inclination be semirect and the line of Inclination be in the thicker Medium and the proportion of the Densities be as that of a Diagonal to the side of its Square and the separating Superficies be plain the Refracted line will be in that separating Superficies For in the Circle A C in the 4th figure let the angle
nor can they go directly backwards against the force of the Movent it remayns therefore that they diffuse themselves upon the Superficies of that Body as towards O and P Which was to be proved 9 Compounded Circular Motion in which all the parts of the moved Body do at once describe Circumferences some greater others less according to the proportion of their several distances from the common Center carries about with it such Bodies as being not fluid adhere to the Body so moved and such as do not adhere it casteth forwards in a Straight Line which is a Tangent to the point from which they are cast off For let there be a Circle whose Radius is AB in the fourth figure and let a Body be placed in the Circumference in B which if it be fixed there will necessarily be carried about with it as is manifest of it self But whilest the motion proceeds let us suppose that Body to be unfixed in B. I say the Body wil cōtinue its motion in the Tangent BC. For let both the Radius AB and the Sphere B be conceived to consist of hard matter and let us suppose the Radius AB to be stricken in the point B by some other Body which falls upon it in the Tangent DB. Now therefore there will be a motion made by the concourse of two things the one Endeavour towards C in the Straight Line DB produced in which the Body B would proceed if it were not retained by the Radius AB the other the Retention it self But the Retention alone causeth no endeavour towards the Center and therefore the Retention being taken away which is done by the unfixing of B there will remain but one Endeavour in B namely that in the Tangent BC. Wherefore the Motion of the Body B unfixed will proceed in the Tangent BC which was to be proved By this demonstration it is manifest that Circular Motion about an unmoved Axis shakes off and puts further from the Center of its motion such things as touch but do not stick fast to its Superficies and the more by how much the distance is greater from the Poles of the Circular Motion and so much the more also by how much the things that are shaken off are less driven towards the Center by the fluid ambient for other Causes 10 If in a fluid Medium a Spherical Body be moved with simple Circular Motion and in the same Medium there float another Sphere whose matter is not fluid this Sphere also shall be moved with simple Circular Motion Let BCD in the 5th figure be a Circle whose Center is A and in whose Circumference there is a Sphere so moved that it describes with Simple Motion the Perimeter BCD Let also EFG be another Sphere of Consistent matter whose Semidiameter is EH and Center H and with the Radius AH let the Circle HI be described I say the Sphere EFG will by the Motion of the Body in BCD be moved in the Circumference HI with Simple Motion For seeing the Motion in BCD by the 4th Article of this Chapter makes all the points of the fluid Medium describe in the same time Circular Lines equal to one another the points E H and G of the Straight Line EHG will in the same time describe with equal Radii equal Circles Let EB be drawn equal and parallel to the Straight Line AH and let AB be connected which will therefore be equal and parallel to EH and therefore also if upon the Center B and Radius BE the arch EK be drawn equal to the arch HI and the straight Lines AI BK and IK be drawn BK and AI will be equal and they will also be parallel because the two arches EK and HI that is the two angles KBE and IAH are equal and consequently the Straight Lines AB and KI which connect them will also be equal and parallel Wherefore KI and EH are parallel Seeing therefore E and H are carried in the same time to K and I the whole Straight Line IK will be parallel to EH from whence it departed And therefore seeing the Sphere EFG is supposed to be of consistent matter so as all its points keep alwayes the same situation it is necessary that every other Straight Line taken in the same Sphere be carried alwayes parallel to the places in which it formerly was Wherefore the Sphere EFG is moved with simple Circular Motion which was to be demonstrated 11 If in a fluid Medium whose parts are stirred by a Body moved with Simple Motion there float annother Body which hath its Superficies either wholly hard or wholly fluid the parts of this Body shall approach the Center equally on all sides that is to say the motion of the Body shall be Circular and Concentrique with the motion of the Movent But if it have one side hard and the other side fluid then both those Motions shall not have the same center nor shall the floating Body be moved in the Circumference of a perfect Circle Let a Body be moved in the Circumference of the Circle KL MN in the 2d figure whose center is A. And let there be another Body at I whose Superficies is either all hard or all fluid Also let the Medium in which both the Bodies are placed be fluid I say the Body at I will be moved in the Circle IB about the Center A. For this has been demonstrated in the last Article Wherefore let the Superficies of the Body at I be fluid on one side and hard on the other And first let the fluid side be towards the Center Seeing therefore the Motion of the Medium is such as that its parts do continually change their places as hath been shewn in the 5th Article if this change of place be considered in those parts of the Medium which are contiguous to the fluid Superficies it must needs be that the small parts of that Superficies enter into the places of the small parts of the Medium which are contiguous to them And the like change of place will be made with the next contiguous parts towards A. And if the fluid parts of the Body at I have any degree at all of tenacity for there are degrees of tenacity as in the Aire and Water the whole fluid side will be lifted up a little but so much the less as its parts have less tenacity whereas the hard part of the Superficies which is contiguous to the fluid part has no cause at all of elevation that is to say no endeavour towards A. Secondly let the hard Superficies of the Body at I be towards A. By reason therefore of the said change of place of the parts which are contiguous to it the hard Superficies must of necessity seeing by Supposition there is no empty Space either come neerer to A or else its smallest parts must supply the contiguous places of the Medium which otherwise would be empty But this cannot be by reason of
the supposed hardness and therefore the other must needs be namely that the Body come neerer to A. Wherefore the Body at I has greater endeavour towards the center A when its hard side is next it then when it is averted from it But the Body in I while it is moving in the circumference of the Circle IB has sometimes one side sometimes another turned towards the center and therefore it is sometimes neerer sometimes further off from the center A. Wherefore the Body at I is not carried in the circumference of a perfect Circle which was to be demonstrated CHAP. XXII Of other Variety of Motion 1 Endeavour and Pressure how they differ 2 Two kinds of Mediums in which Bodies are moved 3 Propagation of Motion what it is 4 What motion Bodies have when they press one another 5 Fluid Bodies when they are pressed together penetrate one another 6 When one Body presseth another and doth not penetrate it the action of the pressing Body is perpendicular to the Superficies of the Body pressed 7 When a hard Body pressing another Body penetrates the same it doth not penetrate it perpendicularly unless it fall perpendicularly upon it 8 Motion sometimes opposite to that of the Movent 9 In a full Medium Motion is propagated to any distance 10 Dilatation and Contraction what they are 11 Dilatation and Contraction suppose Mutation of the smallest parts in respect of their situation 12 All Traction is Pulsion 13 Such things as being pressed or bent restore themselves have motion in their internal parts 14 Though that which carrieth another be stopped the Body carried will proceed 15 16 The effects of Percussion not to be compared with those of Waight 17 18 Motion cannot begin first in the internal parts of a Body 19 Action and Reaction proceed in the same Line 20 Habit what it is 1 I Have already in the 15th Chap. at the 2d Article defined Endeavour to be Motion through some Length though not considered as Length but as a Point Whether therefore there be resistance or no resistance the Endeavour will still be the same For simply to Endeavour is to Go. But when two Bodies having opposite Endeavours press one another then the Endeavour of either of them is that which we call Pressure and is mutual when their pressures are opposite 2. Bodies moved and also the Mediums in which they are moved are of two kinds For either they have their parts coherent in such manner as no part of the Moved Body will easily yeild to the Mouent except the whole Body yeild also and such are the things we call Hard Or else their parts while the whole remains unmoved will easily yeild to the Movent and these we call Fluid or Soft Bodies For the words Fluid Soft Tough and Hard in the same manner as Great and Little are used onely comparatively and are not different kinds but different degrees of Quality 3 To Do and to Suffer is to Move and to be moved and nothing is moved but by that which toucheth it and is also moved as has been formerly shewn And how great sover the distance be we say the first Movent moveth the last moved Body but mediately namely so as that the first moveth the second the second the third and so on till the last of all be touched When therefore one Body having opposite Endeavour to another Body moveth the same and that moveth a third and so on I call that action Propagation of Motion 4 When two fluid Bodies which are in a free and open Space press one another their parts will endeavour or be moved towards the sides not onely those parts which are there where the mutual contact is but all the other parts For in the first contact the parts which are pressed by both the endeavouring Bodies have no place either forwards or backwards in which they can be moved and therefore they are pressed out towards the sides And this expressure when the forces are equal is in a line perpendicular to the Bodies pressing But whensoever the formost parts of both the Bodies are pressed the hindermost also must be pressed at the same time for the motion of the hindermost parts cannot in an instant be stopped by the resistance of the formost parts but proceeds for some time and therefore seeing they must have some place in which they may be moved and that there is no place at all for them forwards it is necessary that they be moved into the places which are towards the sides every way And this effect followes of necessity not onely in Fluid but in Consistent and Hard Bodies though it be not alwayes manifest to sense For though from the compression of two stones we cannot with our eyes discerne any swelling outwards towards the sides as we perceive in two Bodies of wax yet we know well enough by reason that some tumor must needs be there though it be but little 5 But when the Space is enclosed and both the Bodies be fluid they will if they be pressed together penetrate one anoteer though differently according to their different endeavours For suppose a hollow Cylinder of hard matter well stopped at both ends but filled first below with some heavy fluid Body as Quicksilver and above with Water or Aire If now the bottome of the Cylinder be turned upwards the heaviest fluid Body which is now at the top having the greatest endeavour downwards and being by the hard sides of the vessel hindered from extending it selfe sidewayes must of necessity either be received by the lighter Body that it may sink through it or else it must open a passage through it selfe by which the lighter Body may ascend For of the two Bodies that whose parts are most easily separated will the first be divided which being done it is not necessary that the parts of the other suffer any separation at all And therefore when two Liquours which are enclosed in the same vessel change their places there is no need that their smallest parts should be mingled with one another for a way being opened through one of them the parts of the other need not be separated Now if a fluid Body which is not enclosed press a hard Body its endeavour will indeed be towards the internal parts of that hard Body but being excluded by the resistance of it the parts of the fluid Body will be moved every way according to the Superficies of the hard Body and that equally if the pressure be perpendicular for when all the parts of the Cause are equal the Effects will be equal also But if the pressure be not perpendicular then the angles of Incidence being unequal the expansion also will be unequal namely greater on that side where the angle is greater because that motion is most direct which proceeds by the directest Line 6 If a Body pressing another Body do not penetrate it it will nevertheless give to the part it presseth an endeavour to yeild and recede