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| author | Klaus Thoden <kthoden@mpiwg-berlin.mpg.de> |
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| date | Thu, 02 May 2013 11:08:12 +0200 |
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<?xml version="1.0"?> <archimedes xmlns:xlink="http://www.w3.org/1999/xlink" > <info> <author>Pappus Alexandrinus</author> <title>Mathematical Collection, Book 8</title> <date>1970</date> <place>Cambridge</place> <translator>D. Jackson</translator> <lang>en</lang> <cvs_file>pappu_coll8_095_en_1970.xml</cvs_file> <cvs_version></cvs_version> <locator>095.xml</locator> </info> <text> <front></front> <body> <chap n="0"><p><s id="id.0.0.00.01">Translation </s><s id="id.0.0.00.02">Pappus's introduction to the science of Mechanics in which he mentions the Science of the Centre of Gravity, how great Weight is raised witha small Amount of Power, how two Lines are produced in continued Proportion beween two Lines, how a Circle is drawn which is equal to the Circle of a Cylinder of unknown Base and other Geometrical Matters besides, which are needed by Carpenters. He (also) mentions in it the five Powers by which Weight is lifted. God is all-knowing and all-wise. </s></p></chap> <chap n="1"><p type="head"><s id="id.1.0.00.01">Section 1 </s></p> <p><s id="id.1.0.01.01">The science of Mechanics, Hermodorus my son, in so far as it is useful in many great affairs of men, has become a worthy subject for philosophers' interest and a worthy object of aspiration for all those concerning themselves with the mathematical sciences. </s><s id="id.1.0.01.02">This is because this science is almost the first to approach knowledge of the nature of the matter of what is in the world. </s><s id="id.1.0.01.03">For, since the stability and decay of bodies together with their movements, fall into the category of universals, this science establishes with reasons such of them as are in their natural state, and compels some of them to move from the natural places peculiar to them, imparting to them movements contrary to their own by means of the devices which lie within its scope, and which are reached through ideas derived from matter itself. </s></p> <p><s id="id.1.0.02.01">Some of Heron's followers who have studied the science of mechanics have stated that this is in part theoretical and in part practical the former being derived from geometry, arithmetic, astronomy, and the natural sciences, while the latter is brought to perfection through the technique of blacksmiths and coppersmiths, builders, carpenters and sculptors, and their skill and subtlety in their work. </s></p> <p><s id="id.1.0.03.01">A man who had studied the sciences which we have just listed since his youth and had become expert in knowledge of the crafts which we have mentioned, and having in addition an excellent mobility in his nature, would, according to these people, excel in finding out mechanical contrivances, and he it was whom they used to call "chief-artificer". </s><s id="id.1.0.03.02">But those who were unable to combine these many sciences and crafts which we have mentioned were advised, if they wanted to do some work on mechanics, to learn the craft needed in that operation and appropriate to it. </s></p></chap> <chap n="2"><p type="head"><s id="id.2.0.00.01">Section 2 </s></p> <p><s id="id.2.0.01.01">They say that the crafts needed more than any others in human affairs, and which are related to the field of mechanics in which the so-called "chief-artificer" is the chief and foremost exponent, are: (firstly) what is called in Greek the craft of "manjana" this being what the ancients also used to call the craft of "mechanics". </s><s id="id.2.0.01.02">For the masters of this craft raise great weights aloft by means of their devices, contrary to the weight's natural motion, with very little power. </s><s id="id.2.0.01.03">Then there is the craft which makes the tools which are needed as a matter of particular urgency in war and its practitioners are also called "mechanicians". </s><s id="id.2.0.01.04">For among the instruments they manufacture are those known in Greek as "Qatabaltiqu", which can be used to shoot arrows, stones, iron and similar things for a considerable distance. Together with these two crafts is that which the Greeks particularised by the name of "mechanics", in so far as its practitioners raise water from considerable depths with very little effort by water-wheels which they construct. </s></p> <p><s id="id.2.0.02.01">The ancients also apply the name "mechanician" to those people who construct remarkable things, in some of which recourse is had to wind, after the pattern of Heron's so-called "pneumatic" machines, and in some of which sinews and ropes are used in such a way that their movements are made to resemble those of living animals, after the method of Heron in his account of his "Automata" and "Aequilibria", and Archimedes in his account of his "Floating Bodies". </s><s id="id.2.0.02.02">Another example of these are the water-clocks, such as Heron mentions in his account of water-machines. </s></p> <p><s id="id.2.0.03.01">It is clear that this sort of science has something in common with the science of sundials. </s><s id="id.2.0.03.02">Also included in mechanics are those spheres that are constructed to resemble the heavens, which are made to move in a regular and circular motion as would be their motion in water. </s></p></chap> <chap n="3"><p type="head"><s id="id.3.0.00.01">Section 3 </s></p> <p><s id="id.3.0.01.01">According to some, the man who knew the causes of all these things and the rules governing them was Archimedes the Syracusan. </s><s id="id.3.0.01.02">For he alone of those in this world of ours had so diverse a nature as to apply himself to all things, and he was prolific in ideas as was mentioned by Geminus the mathematician in the book which he wrote "On the arrangement of mathematics." </s><s id="id.3.0.01.03">Carpos of Antioch has claimed somewhere in his writings that Archimedes the Syracusan only wrote one book on mechanics this being his work on the construction of spheres, and he did not think that he need write anything on any branch other than this. </s><s id="id.3.0.01.04">However the ancients honoured him with respect to the science of mechanics, and they held his nature to be so remarkable that he continues to have great fame and praise among all people. </s><s id="id.3.0.01.05">As for points dominated by geometrical and arithmetical ideas, he took great pains in writing down even the meanest of these. </s><s id="id.3.0.01.06">For it appears to us as though he had such an affection for these sciences which we have mentioned, that he was not able to mix anything else with them. </s></p> <p><s id="id.3.0.02.01">This Carpos and others besides him have put geometry to suitable use in some crafts. </s><s id="id.3.0.02.02">For geometry can usefully be employed in connection with bodies in many crafts because of what they have in common with it since, it being like a parent to all arts, it can be used with advantage in the construction of tools and also in the craft of the chief-artificer. </s><s id="id.3.0.02.03">Since then there is common ground between it and the science that deals with sundials, with machines and with pictures of houses, its use there is advantageous. </s></p> <p><s id="id.3.0.03.01">It is then clear that it precedes all of them and is adorned by them. </s></p></chap> <chap n="4"><p type="head"><s id="id.4.0.00.01">Section 4 </s></p> <p><s id="id.4.0.01.01">Now since the science and craft of mechanics are in this situation and are divided into these parts, I consider it is fitting that I should write of those points where there is need for geometrical reasoning in the movement of heavy objects, which are mentioned by the ancients in their books together with those points which we ourselves have discovered easily, and that we should deal with this more concisely, clearly and better than those who have given an account of this before us. </s><s id="id.4.0.01.02">For example, there is when we have a given weight moved by a given power on a horizontal plane and another plane inclined to the first at a given angle, and we wish to find how much force will move the given weight on the inclined plane. </s><s id="id.4.0.01.03">This is something needed by the mechanicians called manjanariu, for if they add to the power that they find (which is the given power) another power provided by men, they can be confident of raising that weight. </s><s id="id.4.0.01.04">Another case is when there are two unequal given straight lines between which we wish to find two lines in continued proportion. </s><s id="id.4.0.01.05">By this we are enabled to add to and to take from any given shooting instrument and any shape something which is in a given ratio to it. </s></p> <p><s id="id.4.0.02.01">When something in the form of a wheel is given with a given number of sheets of teeth and we wish to place in conjunction with it a wheel with a given number of teeth, we can find by this means the diameter of the wheel placed in conjunction, and that is useful in many engineering problems in consequence of the conjunction of the "sheets" of wheels. </s><s id="id.4.0.02.02">The use of that will be explained in its own place together with other useful matters which are part of the science of the chief artificer and engineer, after we have first dealt with what concerns the knowledge of the centre of gravity. </s></p></chap> <chap n="5"><p type="head"><s id="id.5.0.00.01">Section 5 </s></p> <p><s id="id.5.0.01.01">As for what should be known for this subject about "the heavy" and "the light" together with the reason for the upward or downward movement of bodies and the meaning of "up" and "down" as well as their defining limits, this is something there is no need to mention now, in that Ptolemy mentioned these matters in his mathematical books. </s><s id="id.5.0.01.02">However, the definition of the centre of gravity of every single body is the principle and basis of the science of gravitational centres from which is derived the remaining sections of the craft of Mechanics, and so it is essential that we say what it is and give the meaning of this term, for in our view the other parts of this science will thus be clear. </s></p> <p><s id="id.5.0.02.01">We say that the centre of gravity of any body is a point within it, such that if one imagines the body to be suspended from it, it will remain still when displaced, and preserve that posture which it had at first without altering its position in any direction while it is being moved. </s><s id="id.5.0.02.02">Such a point is not only found in bodies of regular form but also in those of irregular form, as is made clear in the following way. </s></p></chap> <chap n="6"><p type="head"><s id="id.6.0.00.01">Section 6 (Figure 1) </s></p> <p><s id="id.6.0.01.01">I. Let ABGD be a vertical plane in the direction of the centre of the world towards which we see that everything which has weight inclines and let the straight line AB be parallel to the horizontal plane. </s><s id="id.6.0.01.02">If then some body which has weight is placed on line AB so as to be cut in every way by the plane when produced, it will have at some time a fixed position and it will not shift nor alter its position; and if, when this is so, we imagine the plane ABGD produced it will divide the body placed upon it into two equiponderant parts like things suspended in equilibrium round the plane. </s><s id="id.6.0.01.03">Furthermore should the weight be moved so that it touches the line AB at a different point, and moves away from the position it was in first, and is released, it will remain fixed and not shift. </s><s id="id.6.0.01.04">If then the plane ABGD is produced in a straight line, it will divide the weight into two equiponderant parts and meet the first plane which itself does the same thing. </s><s id="id.6.0.01.05">If it does not meet this plane there must be two weights which are in themselves both in equilibrium and not in equilibrium. </s></p></chap> <chap n="7"><p type="head"><s id="id.7.0.00.01">Section 7 (Fig. 2) </s></p> <p><s id="id.7.0.01.01">II. Having said this let the straight line AB be at right angles to the plane of the horizon and therefore clearly in the direction of the centre of the earth. </s><s id="id.7.0.01.02">Here too let the weight be positioned at point A, similar to the positioning mentioned before. </s><s id="id.7.0.01.03">When this has been done, the weight will stop at some time at point A and will keep its position, if it is able to remain fixed on the line which passes through this point. </s><s id="id.7.0.01.04">Then if it is fixed in this position and the line AB is produced, a part of this line will be in the figure posited. </s><s id="id.7.0.01.05">Let us imagine that this part is fixed in the body and that another part of the weight is placed on the line in such a position as to be at rest on it. </s><s id="id.7.0.01.06">I say now that when the line AB is produced it will meet the first line [? within the body?]. </s><s id="id.7.0.01.07">And, if it does not meet it, then it has to be possible for there to be two planes on these two lines which do not meet within the body, each one of them being a plane which passes through line AB and divides the weight into two parts, in themselves both in equilibrium and not in equilibrium, which is impossible. </s><s id="id.7.0.01.08">Thus the lines which we mentioned must meet within the body. </s><s id="id.7.0.01.09">Likewise if other parts of the weight are placed at point A so that it remains fixed and AB is then produced, it will meet the lines taken first. </s><s id="id.7.0.01.10">It is clear from this that these lines which we have posited will cut one another at one single point and this point is called the centre of gravity. </s><s id="id.7.0.01.11">We can see then that if we imagine the weight to be suspended at its centre, it will not change or alter its position at all but will remain fixed in its original state, keeping to its former positioning every time it is moved. </s><s id="id.7.0.01.12">For all the planes which cut it will divide the weight into parts equal in inclination to one another; nor will there be any reason for change since the parts of it which border the point are fixed, however positioned, in equal inclination to one another. </s></p></chap> <chap n="8"><p type="head"><s id="id.8.0.00.01">Section 8 </s></p> <p><s id="id.8.0.01.01">The sum of what concerns the knowledge of the centre of gravity is then for the most part as we have given it. </s><s id="id.8.0.01.02">You can learn the basic principles through which this science is acquired if you look at Archimedes' book "On Aequilibria", and at Heron's work "On Mechanics", while here we shall set out in order those points connected with this that most people find unclear, among which is the following. </s></p></chap> <chap n="9"><p type="head"><s id="id.9.0.00.01">Section 9 (Fig.3) </s></p> <p><s id="id.9.0.01.01">III. We make ABG a triangle and let its sides be divided in equal ratios at points H T K, such that the ratio of BT to TG is as the ratio of AH to HB and as that of GK to KA; let us join lines HT, TK, KH. </s><s id="id.9.0.01.02">Now I maintain that the centre of gravity of the two triangles ABG HTK is one and the same. </s><s id="id.9.0.01.03">For we bisect each of the two lines BG GA at the two points D H and join lines AD BH; then point Z is the centre of gravity of ABG. </s><s id="id.9.0.01.04">For if this triangle stands on a plane perpendicular to the horizontal, on the line AD in it, it will not incline to either of the two sides because triangle ABD is equivalent to triangle AGD. </s><s id="id.9.0.01.05">Further, if this triangle stands on the line BH in it, on the perpendicular plane which we have mentioned, it will not incline in either direction, because triangles ABH BGH are equivalent. </s><s id="id.9.0.01.06">If the triangle is in equilibrium on each of the lines AD BH the point common to these two lines is the centre of gravity, this being point Z, which must be imagined as being placed in the middle of triangle ABG as we said before. </s><s id="id.9.0.01.07">Clearly it is necessary that the parts of the triangle be equal in thickness and weight; and it is clear that AZ is twice ZD and that BZ is twice ZH because the ratio of AG to GH is as that of AB to DH, which is as the ratio of BZ to ZH and as that of AZ to ZD, since the two triangles DZH ABZ are equiangular, as also are the two triangles GDH ABG. </s></p> <p><s id="id.9.0.02.01">When we join line DH it cuts line TK at point L. </s></p> <p><s id="id.9.0.03.01">Since the ratio of BT to TG (if we make DT a mean between them) is composed of the ratio of BT to TD and DT to TG; also componendo the ratio of BG to GT is as that of GA to AK; furthermore the ratios of the halves of the antecedents to the consequents are equal, that is, the ratio of DG to GT is as that of HA to AK, and convertendo, the ratio of GD to DT is as that of AH to HK; and GD is equal to BD and AH is equal to GH; thus the ratio of BD to DT is as that of GH to HK; componendo, the ratio of BT to TD is as that of GK to KH; </s></p> <p><s id="id.9.0.04.01">Then the ratio of AH to HB is composed of the ratio of GK to KH and the ratio of DT to TG; also, as will be shown, the ratio of DL to LH is composed of these two ratios. </s><s id="id.9.0.04.02">TL is equal to LK. </s></p> <p><s id="id.9.0.05.01">Now the ratio of AH to HB is as that of DL to LH and lines AB DH are parallel; lines AD BH are joined and intersect at point Z. </s><s id="id.9.0.05.02">The line which passes through points H Z L is a straight line as we shall shortly show. </s></p> <p><s id="id.9.0.06.01">Since the ratio of BZ to ZH is as that of HZ to ZL and BZ is twice ZH, then HZ is twice ZL. </s><s id="id.9.0.06.02">Now line HL divides triangle HTK into two halves, and HZ is twice ZL. </s><s id="id.9.0.06.03">Point Z is therefore the centre of gravity of triangle HTK and it is also the centre of gravity of triangle ABG. </s></p></chap> <chap n="10"><p type="head"><s id="id.10.0.00.01">Section 10 (Fig. 4) </s></p> <p><s id="id.10.0.01.01">IV. Let us now clarify the points whose account we have postponed. </s><s id="id.10.0.01.02">Let the ratio of GH to HK be as that of GD to DT and let the two lines DH TK be joined and let one of them cut the other at point L. </s></p> <p><s id="id.10.0.02.01">I say that TL is equal to LK and that the ratio of DL to LH is composed of the ratio of DT to TG and that of GK to KH. </s></p> <p><s id="id.10.0.03.01">From the point G let a line be drawn parallel to line TK this being GZ, and let it meet line DH when produced at point Z. </s><s id="id.10.0.03.02">As DL LH are two lines, with LZ being another line, then the ratio of DL to LH is composed of the ratio of DL to LZ and that of ZL to LH; further the ratio of DL to LZ is as that of DT to TG because the line ZG is parallel to the line KT. </s><s id="id.10.0.03.03">Also the ratio of ZL to LH is as that of GK to KH because triangles GHZ HKL are equiangular. </s><s id="id.10.0.03.04">Thus the ratio of DL to LH is composed of the ratio of DT to TG and that of GK to KH; in the same way it will also be clear that the ratio of KL to LT is composed of the ratio of KH to HG and that of GD to DT, for we can draw from point G a line GM parallel to the line DH; let it meet the line KT when produced at point M. </s><s id="id.10.0.03.05">As KL LT are two lines with LM being another line, the ratio of KL to LT is composed of the ratio of {...}L to LM and that of LM to LT; but the ratio of KL to LM is as that of {...}H to HG because HD is parallel to GM, and the ratio of LM to LT is as that of GD to DT because triangles DTL GTM are equiangular. </s><s id="id.10.0.03.06">Therefore the ratio of KL to LT is composed of the ratio of KH to HG (which is as the ratio of DT to GD) and the ratio of GD to DT. </s></p> <p><s id="id.10.0.04.01">But the ratio of DT to DG when combined with DG to DT produced the ratio of equality. Thus the ratio of KL to LT is also that of equality. </s><s id="id.10.0.04.02">Thus KL is equal to LT. </s></p></chap> <chap n="11"><p type="head"><s id="id.11.0.00.01">Section 11 (Fig. 5) </s></p> <p><s id="id.11.0.01.01">V. The remaining point of our postponed discussion is as follows: let there be line AB parallel to line GD. </s><s id="id.11.0.01.02">Let the ratio of GT to TD be as that of AZ to ZB; we join the two lines AG BD; let them intersect at point H. </s></p> <p><s id="id.11.0.02.01">I say that the line which passes through the points Z H T is a straight line. </s><s id="id.11.0.02.02">If this is not the case let the straight line be ZHH. </s><s id="id.11.0.02.03">Now since the ratio of ZH to HH is as that of ZA to GH and the ratio of ZB to HD is as that of ZH to HH, the ratio of ZB to HD is as that of AZ to GH. </s><s id="id.11.0.02.04">Then alternando the ratio of AZ to ZB (which is as the ratio of GT to TD) is as that of GH to HD which is impossible. </s></p> <p><s id="id.11.0.03.01">So the line which passes through the points ZHT is straight. </s></p></chap> <chap n="12"><p type="head"><s id="id.12.0.00.01">Section 12 (Fig. 6) </s></p> <p><s id="id.12.0.01.01">VI. In a given rectangular parallelogram AG we wish to draw a line DG such that when the trapezium ABGD is suspended from point D the two lines AD BG are horizontal. </s><s id="id.12.0.01.02">We then produce the straight line coming from point D and passing through the centre of gravity of the trapezium ABGD; it is clear that this line is perpendicular to the horizon, and to the line BG. </s><s id="id.12.0.01.03">Let this then be the line DL; let it be bisected at point H and from point H let us draw a line HZ parallel to BG and join line GH. </s><s id="id.12.0.01.04">Also let us divide it in two at point T so that GT is, twice TH; we bisect HZ at point H, and join the points HT by the line TKH; now H is the centre of gravity of the parallelogram BD. Point T is the centre of gravity of the triangle GDL; then the centre of gravity of the whole trapezium is on the line TH and is also on DL; therefore point K is then the centre of gravity of the trapezium ABGD: the centre of gravity of the parallelogram BD is H and the centre of gravity of the triangle GDL is T. </s><s id="id.12.0.01.05">Therefore the ratio of the parallelogram BD is to the triangle GDL as TK is to KH. </s><s id="id.12.0.01.06">Further, if we imagine the weight of the parallelogram BD to be concentrated at the point H while the whole weight of the triangle GDL is concentrated at T, TH becomes like a balance with the weights which we have mentioned at its two ends. </s><s id="id.12.0.01.07">If HT is divided at K so that the weight at point H is equal to that at T, by which I mean that the weight of the parallelogram BD is equal to that of the triangle GDL, then the ratio of the line TK to line KH is one which corresponds to the weights at the ends of the balance and the point at which the weight is in equilibrium is K; when the trapezium ABGD is suspended at point K it will be in equilibrium </s></p> <p><s id="id.12.0.02.01">Let then perpendiculars be dropped from the points H, T to the line BG, these being HM TN. </s><s id="id.12.0.02.02">Now the ratio of the parallelogram BD to the triangle GDL is the same as that of TK to KH, and the ratio of that parallelogram which we mentioned to the triangle is the same as the ratio of BL to half of LG; and the ratio of TK to KH is the same as that of NL to LM because the parallel lines HM, HL, TN cut the two lines HKT, MLN; the ratio of BL to half LG is as the ratio of NL to LM; therefore the area contained by BL and LM is equal to that contained by half of GL and LN. </s><s id="id.12.0.02.03">The line LM is half of line BL; and the square of LB is equal to GL multiplied by LN; thus the ratio of GL to LB is as that of BL to LN, and the ratio of GL to LN is equal to that of the square of GL to that of the square of BL. </s><s id="id.12.0.02.04">But GL is three times LN (because GH is also three times HT, for GT is twice TH); thus the square of GL is three times the square of BL and B G are given; so point L is given; because of this point D is given. </s><s id="id.12.0.02.05">Thus when we have divided BG at point L in such a way that the square of GL is three times the square of LB, point D is that from which the trapezium is suspended. </s><s id="id.12.0.02.06">The line BG is to be divided in the way I shall describe. </s></p></chap> <chap n="13"><p type="head"><s id="id.13.0.00.01">Section 13 (Fig. 7) </s></p> <p><s id="id.13.0.01.01">VII. We wish to divide a given straight line into two parts so that the greater is three times the smaller in power. </s><s id="id.13.0.01.02">Let the given straight line be AD; we divide it at point G into two parts such that AG is three times GD; let there be described on AD a semi-circle ABD. </s><s id="id.13.0.01.03">From point G we draw GB as a perpendicular to AD; we make the ratio of AH to HD be as the ratio of AG to GB. </s></p> <p><s id="id.13.0.02.01">I say that AH is three times DH in power. </s><s id="id.13.0.02.02">Now because BG is the mean proportional of the two lines AG GD, the ratio of AG to GD is as that of the square of AG to the square of BG, and this is as the ratio of the square of AH to the square of DH. </s><s id="id.13.0.02.03">Thus AH is three times DH in power. </s><s id="id.13.0.02.04">We can similarly also divide the line AD and any given line into two parts so that one part is in a given ratio to the other part in power. </s></p></chap> <chap n="14"><p type="head"><s id="id.14.0.00.01">Section 14 (Fig. 8) </s></p> <p><s id="id.14.0.01.01">VIII. When the position of the straight lines AB AG is given, point B is given, the line GD is drawn, and AB is cut at D, the proportion of AG to DB being known, the centre of gravity of the triangle AGD is on a straight line of given position. </s><s id="id.14.0.01.02">Let AG be bisected at point H; let us join the line DH and cut it at point Z so that HZ is a third of DH. </s><s id="id.14.0.01.03">Point Z will then be the centre of gravity of the triangle AGD as has been explained above. </s><s id="id.14.0.01.04">Now the line ZH is drawn parallel to the line AH, and let a third of AB be the line AT. </s><s id="id.14.0.01.05">Now the line AH is also a third of AD because HZ is a third of HD; the remaining line HT is a third of BD, and the ratio of BD to AG is given; now the ratio of AG to ZH is given because AG is three times ZH, because DH is one and a half times DZ, its ratio to it being the same as that of AH to ZH; and AG is twice AH. And so the ratio of HT to HZ is given; as the angle ZHT is given because the angle at point A is given; then the angle HTZ is given. </s><s id="id.14.0.01.06">The point T is given and so the position of the line TZ is given, and the centre Z is on this line. </s></p></chap> <chap n="15"><p type="head"><s id="id.15.0.00.01">Section 15 </s></p> <p><s id="id.15.0.01.01">This and similar matters are what we wished to mention. </s></p> <p><s id="id.15.0.02.01">As for the things we need to use in Mechanics, these are as follows: </s></p> <p><s id="id.15.0.03.01">We wish to draw an inclined plane with its inclination towards a given point in a given non-inclined plane with parallel sides (that is to say a horizontal one), and such that its inclination is equal to a given angle. </s><s id="id.15.0.03.02">Now first let the given plane, ABGD, have its sides equal and parallel, and the given angle at which we wish to incline the plane be HZH. </s><s id="id.15.0.03.03">We erect at the points A B D of the positioned plane lines at right angles to it AT BK DL. </s><s id="id.15.0.03.04">Now let point G be the point at which we wish the plane to verge; we join AG and let line ZH be equal to it; we draw HH on ZH at right angles to it and make AT equal to HH. If we suppose that the line TG is joined, then the angle TGA is the angle of inclination of the plane we desire. </s><s id="id.15.0.03.05">From B we drop a perpendicular BM on AG and make ZN equal to GM. </s><s id="id.15.0.03.06">NS is placed on ZN at right angles; BK and DL are both made equal to NS; we join lines TK TL and produce them so that they meet lines AD AB produced, at points F and Q. </s><s id="id.15.0.03.07">That they meet is clear, since they go out from less than two right angles. </s><s id="id.15.0.03.08">Now lines TK GL are in the same plane the inclination of which is the angle TGA (which is equal to angle HZH); for if we suppose the line MO parallel to line AT and we join line OK, then MO is equal to NS because triangles ZNS MOG are equiangular; line KO is equal to line MB and is parallel to it, and plane KBMO has parallel sides and stands at right angles to the positioned plane. </s><s id="id.15.0.03.09">Because points F G Q lie in two planes - ABGD and KTLG - simultaneously, points F G Q are on one straight line FGQ which is the common section of the two planes mentioned. </s><s id="id.15.0.03.10">Thus also it is clear that points KOL lie on the common section of the plane KTLG and of the plane which passes through points K O L parallel to the plane ABGD. </s><s id="id.15.0.03.11">Then the line which passes through points K O L is parallel to line FGQ. </s><s id="id.15.0.03.12">Now because the ratio of AF to FD is as that of TA to LD and the ratio of AQ to QB is as that of AT to BK and DL is equal to BK then AF is equal to AQ and angle AFQ is equal to angle AQF; also angle FAG is equal to angle QAG. </s><s id="id.15.0.03.13">Therefore the remaining angle AGF is equal to angle AGQ; they are then both right angles, and line FQ is bisected by line AG at right angles and it stands on it at right angles. </s><s id="id.15.0.03.14">Line MO also stands at right angles to the plane ABGD; then line OG is at right angles to QF as has been made clear in the book on spheres. </s><s id="id.15.0.03.15">Then both OGQ and OGF are right angle and plane KTLG inclines from plane ABGD at the given angle HZH. </s></p></chap> <chap n="16"><p type="head"><s id="id.16.0.00.01">Section 16 (Fig. 9) </s></p> <p><s id="id.16.0.01.01">Further let AB be greater than AD; let the rest be as above. </s></p> <p><s id="id.16.0.02.01">I say that AGF is an acute angle. </s></p> <p><s id="id.16.0.03.01">Because the ratio of AF to FD is as that of AT to DL and the ratio of AQ to QB is as that of TA to BK and DL is equal to BK, the ratio of AF to FD is as that of AQ to QB and, separando, the ratio of AD to DF is as that of AB to BQ and, alternando, the ratio of AD to AB is as that of DF to BQ. </s><s id="id.16.0.03.02">Now line AD is smaller than line AB and so line FD is also smaller than line BQ. </s><s id="id.16.0.03.03">The whole of AF is then smaller than AQ; therefore angle AQF is smaller than angle AFQ and angle AFQ is greater than angle AQF. </s><s id="id.16.0.03.04">Now angle GAF is also greater than angle GAQ; and so the remaining angle AGF of triangle AGF is smaller than the remaining angle AGQ of triangle AGQ; thus the angle AGF is acute. </s></p> <p><s id="id.16.0.04.01">The planes mentioned verge at a point between G F determined by a perpendicular drawn from point A to the line GF falling on it. </s><s id="id.16.0.04.02">As this is known it is then possible for us to incline one plane to another at a given angle; it is also possible for us to tell the angle at which the inclining plane inclines, I mean that we can tell the plane's angle of inclination from the horizontal. </s></p></chap> <chap n="17"><p type="head"><s id="id.17.0.00.01">Section 17 (Fig. 10) </s></p> <p><s id="id.17.0.01.01">If a given weight is moved by a given force in the horizontal plane, and if there is a plane with a given angle of inclination to the positioned plane, how do we find what force is necessary to move the weight in the inclined plane? </s></p> <p><s id="id.17.0.02.01">Let the positioned plane be the one which passes through the line MN, and the plane which passes through the line MK be at a given angle of inclination KMN; let the given weight be A and let a force G move it in the positioned plane. </s><s id="id.17.0.02.02">Let there also be a sphere with centre H equal in weight to the body A; let it be placed on the plane which passes through the line MK touching it at point L, as has been made clear in the third theorem of the Book of Spheres. </s><s id="id.17.0.02.03">If the line HL is joined, it is perpendicular to the plane, and this has been made clear in the fourth theorem of the Book of Spheres. </s><s id="id.17.0.02.04">Thus, the line H{?} is perpendicular to KM; the plane KMHL is produced. </s><s id="id.17.0.02.05">Let the resultant section in the sphere be the circle LHS; let us draw from centre a line HH parallel to MN; from point L let us draw a perpendicular to it, this being LZ. </s><s id="id.17.0.02.06">Because the angle HTL is given, since it is equal to the given acute angle KMN and the angle at Z is given, angle HLZ which is equal to angle HTL is given, because the angles of triangle HTL are equal to those of triangle HLZ; thus the triangle HLZ is of given shape; then the ratio of HZ to ZH is given. </s><s id="id.17.0.02.07">Now we let the ratio of HZ to ZH be as that of weight A to weight B and as the ratio of force G to force D. </s><s id="id.17.0.02.08">The force which moves A is G and that which moves B in the same plane is D. </s><s id="id.17.0.02.09">Because the ratio of line HZ to line ZH is as that of weight A to weight B, if we place weights A B with centres of gravity H H they are in equilibrium when suspended from the point Z and LZ is perpendicular to the horizontal. </s><s id="id.17.0.02.10">If the centre of gravity of A is at point H or the sphere which has this centre is placed there instead, and weight B is placed on centre H, it is in equilibrium with the sphere, so that the sphere does not descend because of the plane's inclination but remains stationary on it and does not incline, staying in it's place as if it were placed on the positioned plane. </s><s id="id.17.0.02.11">Now when the sphere is placed on the plane force G moves it; so force G plus the force which moves weight B, namely D, moves it in the inclined plane; force D is given so the force which will move the sphere equal in weight to A in the given plane is given. </s></p></chap> <chap n="18"><p type="head"><s id="id.18.0.00.01">Section 18 (Fig. 11) </s></p> <p><s id="id.18.0.01.01">We have made this clear geometrically, but we have an example of this which affords us clarification of both operation and proof. </s><s id="id.18.0.01.02">We let weight A be two hundred talents and let force G (which is the amount supplied by forty men) move it in the horizontal plane. </s><s id="id.18.0.01.03">Now let angle KMN which is equal to angle HTL be two thirds of a right angle, and the remaining angle ZLT be one third of a right angle. </s><s id="id.18.0.01.04">Now angle HLT is a right angle so angle HLZ is two thirds of a right angle; by the measure in which four right angles make 360 angle HLZ is then 60, and by the measure in which 2 right angles are 360 then this angle is 120; now when the right-angled triangle HLZ is circumscribed by a circle, the arc on HZ is 120 by the measure in which a circle comprises 360; and HZ is 104 approximately by the measure in which HL, which is the diameter of the circle, is 120; that is clear from the tables of chords in the circle laid down by Ptolemy in the first section of his book on Mathematics. </s></p> <p><s id="id.18.0.02.01">Thus the ratio of HL (which is equal to HH) to HZ is as that of 120 to 104 and the remaining ratio of HZ to ZH is as that of 16 to 1{?} which is as the ratio of weight A to weight B and force G to force D. </s></p> <p><s id="id.18.0.03.01">Now weight A is two hundred talents and its moving force is the power of forty men; then weight B will be 1300 talents and force D will be that of 260 men; for the ratio of 16 to 104 is as that of 200 to 1300 and as that of 40 to 260. </s><s id="id.18.0.03.02">Weight A then (which is 200 talents), which is moved by the force of 40 men in the horizontal plane, is itself moved by all the men we mentioned (and they were 300) in a plane inclined to the horizontal at angle KMN which is two thirds of a right angle. </s></p></chap> <chap n="19"><p type="head"><s id="id.19.0.00.01">Section 19 (Fig. 12) </s></p> <p><s id="id.19.0.01.01">In this way we learn how to move a given weight with a given force. </s><s id="id.19.0.01.02">It is said that this section of Mechanics is one of Archimedes' discoveries and that when he discovered it he said, "Give me a place to stand that I may move the world for you!" </s><s id="id.19.0.01.03">Heron of Alexandria gave a most clear exposition of this operation in his book called "The Drawing of Weight" in which he makes use of a lemma proven in his books on Mechanics where he also mentions the five powers which are: isfin, lever, screw, compound pulley, and axle with wheels. </s><s id="id.19.0.01.04">These are the things by which, in general, a given weight is moved by a given power, I mean each of these powers. </s></p> <p><s id="id.19.0.02.01">In his book "Barulcus", however, he explains how a given weight is moved by a given power derived from positioning toothed wheels, when the ratio of the wheel's diameter to that of the axle is five to one, and when the weight to be moved is 1000 talents and the motive power 5 talents. </s></p></chap> <chap n="20"><p type="head"><s id="id.20.0.00.01">Section 20 </s></p> <p><s id="id.20.0.01.01">Let us now show this in the ratio of 2 to 1; so let the weight to be moved be 160 talents instead of 1000 and the motive force be 4 instead of the previous 5; I mean that the force supplied by the man who does the moving be a force whereby he can alone draw four talents without any mechanical device. </s></p> <p><s id="id.20.0.02.01">Now let the box which he calls "Glossocomus" be ABGD and let there be an axle HZ mounted at right angles in its two parallel long walls, which turns easily. </s><s id="id.20.0.02.02">Let there be connected to this axle a wheel HT with teeth like arrow-heads. </s><s id="id.20.0.02.03">Let its diameter be double the diameter of HZ which is an axle at right angles to the two walls. </s><s id="id.20.0.02.04">Let the inside of the box be square throughout its length being capable of housing the wheels. </s><s id="id.20.0.02.05">Let the wheels be adjusted and let such of it as is on the two sides of the wheel be round or curved. </s><s id="id.20.0.02.06">If ropes are attached to the weight to be drawn, are passed through a hole or wide aperture in side AB, are then wound round axle HZ on both sides of the wheel HT, and wheel HT is turned, it will turn the axle it is on. </s><s id="id.20.0.02.07">Let its ends be pivoted in iron boxes which move; these are placed in the walls AD BG which we have mentioned. </s><s id="id.20.0.02.08">The ropes tied in the wall w{...} are wound on the axle move the weight. </s><s id="id.20.0.02.09">To move wheel HT a force equivalent to more than 80 talents is needed because the diameter of the wheel is double that of the axle. </s><s id="id.20.0.02.10">This lemma has been proved by Heron in his books on Mechanics, together with many other matters which are useful in the affairs of the world. </s></p></chap> <chap n="21"><p type="head"><s id="id.21.0.00.01">Section 21 </s></p> <p><s id="id.21.0.01.01">But when the given power available to us is not one which can move 80 talents but 4 talents, we put another axle KL parallel to axle HZ. </s><s id="id.21.0.01.02">Let a toothed wheel MN be mounted on it. </s><s id="id.21.0.01.03">Let its teeth engage the teeth of wheel HT, which will obtain when the ratio of the number of teeth on HT to the number of teeth on MN is the same as the ratio of the diameter of wheel HT to the diameter of wheel MN. </s><s id="id.21.0.01.04">How this takes place will be shown later. </s></p> <p><s id="id.21.0.02.01">The wheel MN then is known. </s><s id="id.21.0.02.02">Let wheel SO also be mounted on axle KL having a diameter twice that of wheel MN. </s><s id="id.21.0.02.03">Thus the power which will move the weight by means of wheel SO needs to be that which moves forty talents as 80 talents is twice 40 talents. </s></p></chap> <chap n="22"><p type="head"><s id="id.22.0.00.01">Section 22 </s></p> <p><s id="id.22.0.01.01">Further we place beside wheel SO another wheel RSh with a diameter twice that of wheel FQ but in such a way that its teeth do not engage those of wheel MN. </s><s id="id.22.0.01.02">The power required to move the weight by means of wheel RSh must be one capable of moving 20 talents. </s><s id="id.22.0.01.03">But the given power is one which will move 4 talents. </s><s id="id.22.0.01.04">It is then necessary to place another toothed wheel of this kind such as wheel TTh teside wheel RSh. </s><s id="id.22.0.01.05">Let there then be mounted on the same axle as wheel TTh a toothed wheel KhDh; let the ratio of the diameter of KhDh to that of TTh be as the ratio of 2 to 1. </s><s id="id.22.0.01.06">The motive force required to move the weight by means of wheel KhDh is what can move 10 talents. </s><s id="id.22.0.01.07">We then also mount another toothed wheel SZ alongside wheel KhDh, and on this axle let there be a saw-toothed wheel LaLb, with the ratio of its diameter to that of wheel SZ being that of the 10 talents to the number of talents moved by the given force, which is 4. </s></p></chap> <chap n="23"><p type="head"><s id="id.23.0.00.01">Section 23 </s></p> <p><s id="id.23.0.01.01">This done, suppose box ABGD raised in the air and stationary; we attach the weight to axle HZ and the four talents to wheel LaLb; neither of the attachments will descend although the axles move easily and the wheels engage one another, but they will remain in equilibrium after the manner of a balance, I mean the four talents and the weight of 160 talents. </s><s id="id.23.0.01.02">If we increase one of them by a small weight the side which has been increased in weight will incline and descend. </s><s id="id.23.0.01.03">For example, if we add some weight to the force that consists of the 4 talents it will overcome the 160-talent weight and draw it. </s></p></chap> <chap n="24"><p type="head"><s id="id.24.0.00.01">Section 24 </s></p> <p><s id="id.24.0.01.01">Instead of this increase let there be alongside wheel LaLb a screw DGh the grooving of which coincides with the saw-teeth of wheel LaLb. </s><s id="id.24.0.01.02">How this is done is explained in Heron's book on Mechanics; we shall also explain it later. </s><s id="id.24.0.01.03">Now it is necessary that the screw should turn easily. </s><s id="id.24.0.01.04">Let the screw have 2 ends, one outside the box by the wall GD and the other inside. </s><s id="id.24.0.01.05">Let the outside end be square and have a handle WY. </s><s id="id.24.0.01.06">If we grasp the handle and turn the screw, wheel LaLb turns and the adjoining wheel SZ turns with it. </s><s id="id.24.0.01.07">Therefore the wheel lying next to it turns, this being wheel KhDh. </s><s id="id.24.0.01.08">Then the adjoining wheel TTh turns and with it wheel RSh which lies next to it. </s><s id="id.24.0.01.09">The wheel adjoining that FQ, also turns as does wheel HT lying next to it. </s><s id="id.24.0.01.10">With this wheel the axle HZ on which it is mounted also turns, and it is round axle HZ that the ropes tied to the weight are wound; thus we move the weight. </s><s id="id.24.0.01.11">It is clear that it will move as a further force has been added to the force we have, which is that derived from the handle which, when it is turned, describes a circle larger than the screw's circle of revolution. This has been explained in Archimedes' book "On Balances" and in the books of Heron and Philon on Mechanics, namely that large circles overcome small circles with the same centre of revolution. </s></p></chap> <chap n="25"><p type="head"><s id="id.25.0.00.01">Section 25 </s></p> <p><s id="id.25.0.01.01">We have mentioned most of such matters as embrace mechanical method; but there are many divisions and types in the science of instruments, for such of what is known in the science of mechanics, sundials and water machines as is known by reasoning, clarifies such instrumental constructions as are composed in this craft. </s><s id="id.25.0.01.02">Much of this is outside the craft of Mechanics. </s><s id="id.25.0.01.03">Matters are worked out in it which are difficult to work out using geometrical methods and they are sorted out instrumentally, and so working them is easier. </s></p> <p><s id="id.25.0.02.01">Now the problem known as the Delian problem which is three-dimensional cannot be worked out geometrically because it is impossible to draw a conic section in a plane; when however this is worked out instrumentally it can be done by hand more easily than in any other way. </s><s id="id.25.0.02.02">It is also possible in this way to do what we now intend, which is to find a cube which is double a cube; with this apparatus is found not only the double but also one which is in any ratio to it. </s></p></chap> <chap n="26"><p type="head"><s id="id.26.0.00.01">Section 26 (Fig. 13) </s></p> <p><s id="id.26.0.01.01">A semi-circle ABG is drawn and from centre D we erect a perpendicular DB on diameter LG; we then make a ruler pass through point A and, with fixed axis at point A, let it revolve round a centre so that the ruler rotates between points B G. </s><s id="id.26.0.01.02">Having done this we wish to find two cubes in a given ratio to one another. </s></p> <p><s id="id.26.0.02.01">The ratio of BD to DH is made equal to the given ratio. </s><s id="id.26.0.02.02">We join line GM and produce it to Z. </s><s id="id.26.0.02.03">The ruler must be rotated between G B until such as lies between lines ZH HB is equal to such as lies between line BH and arc BKG. </s><s id="id.26.0.02.04">When we move the ruler it is possible for us to do this. </s><s id="id.26.0.02.05">Let this have occurred when the ruler is on line AT and let HT be equal to TK. </s><s id="id.26.0.02.06">I say that the ratio of the cube on BD to the cube on DT is the ratio we required, which is that of BD to DH. </s><s id="id.26.0.02.07">We complete the circle and join the line KD producing it to L. </s><s id="id.26.0.02.08">Then we join LH being parallel to line BD because line KT is equal to line TH and line KD is equal to line DL. </s><s id="id.26.0.02.09">We join the two lines AL LG. </s><s id="id.26.0.02.10">Because angle HAL is a right angle, as it is in a semi-circle, and AM is a perpendicular, the ratio of the square of line LM to the square of line AM which is as the ratio of GM to MA is as that of the square of line AM to the square of line MH; we make the ratio of AM to MH an aggregate and the ratio compounded of GM to MA and the ratio of AM to MH (which is as the ratio of GM to MH) is as the ratio compounded from the ratio of the square of line AM to the square of line MH and the ratio of AM to MH but the ratio compounded from the ratio of the square of line AM to the square of line MH and the ratio of AM to MH is as the ratio of the cube of line AM to the cube of line MH; then the ratio of GM to MH is as that of the cube of line AM to the cube of line MH. </s><s id="id.26.0.02.11">But the ratio of GM to MH is as that of GD to DH, which is as that of BD to DH, and the ratio of AM to MH is as that of AD to DT, which is as that of BD to DT. </s><s id="id.26.0.02.12">Thus the ratio of BD to DH which is equal to the given ratio is as that of the cube of line BD to the cube of line DT, which is what we wished to show. </s></p> <p><s id="id.26.0.03.01">An instrumental problem on the cylinder. </s></p></chap> <chap n="27"><p type="head"><s id="id.27.0.00.01">Section 27 </s></p> <p><s id="id.27.0.01.01">XIII. The problems arising in that section of Mechanics which is the science of instruments present difficulties when geometry cannot be used to solve them. </s><s id="id.27.0.01.02">Examples of this are: such things as are described on one distance; the question put by chief-artificers on the cylinder with defective bases. </s><s id="id.27.0.01.03">They pose this question when they say: when a part of the surface of a right cylinder is given, and no part of the circumference of the bases is sound, how can the thickness of the cylinder be found, I mean the diameter of the circle on which the cylinder has been constructed? </s></p> <p><s id="id.27.0.02.01">It is found in this way. </s></p></chap> <chap n="28"><p type="head"><s id="id.28.0.00.01">Section 28 (Fig. 14) </s></p> <p><s id="id.28.0.01.01">Two points A B are marked on the given surface; these are made centres and at any one distance on the surface a mark G is made. </s><s id="id.28.0.01.02">Taking A B again as centres point D is marked off at a greater distance. </s><s id="id.28.0.01.03">Taking these two points as centres we mark off at another distance point H, at another distance point Z and at another distance point H. </s><s id="id.28.0.01.04">The points G D H Z H are in one plane because these points are the apexes of isosceles triangles and the lines drawn from them. </s><s id="id.28.0.01.05">The line joining points A B, which is the base of the triangles, is bisected by perpendiculars to line AB; these five perpendiculars are in one plane, and it is clear that points G D H Z H are also in one plane. </s><s id="id.28.0.01.06">We can place them in any plane we wish as follows: we make in the plane a triangle TKL of the three lines which join points G D H; in it we also make a triangle KLM of the three lines joining points D H Z, and from the three lines joining points H Z H a triangle MLN; then triangles TKL, KLM, LMN take the place of triangles GDH, DHZ, HZH. </s><s id="id.28.0.01.07">If we draw through points T K L M N an ellipse, then the smaller of the two axes is a diameter of the circle on which the cylinder was constructed. </s></p></chap> <chap n="29"><p type="head"><s id="id.29.0.00.01">Section 29 (Fig. 15) </s></p> <p><s id="id.29.0.01.01">XLV. We wish to construct an ellipse to pass through points T K L M N. </s><s id="id.29.0.01.02">Let there then be constructed the ellipse, and it is TKLMN. </s><s id="id.29.0.01.03">We join lines MK NT; let them first of all be parallel, and let each be bisected at points A B; we join line AB. </s><s id="id.29.0.01.04">We produce it to points H Z of the ellipse, and then line HZ is a diameter of the ellipse; this is explained in the 28th figure of Apollonius's book on Cones. </s><s id="id.29.0.01.05">This line is given in position, and both the points A B are given in position. </s><s id="id.29.0.01.06">We draw from point L a line LS parallel to diameter HZ, and we join lines SK LM; let them meet TN at points F H; then, what results from the intersection of LM with TN is given. </s><s id="id.29.0.01.07">Because the ratio of the plane contained by SD and DL to the plane contained by MD and DK is as the ratio of the plane contained by SG and GL to the plane contained by HG and GF (by similar triangles), and is as the ratio of the plane contained by SG and GL to the plane contained by NG and GT (which is explained in the figure from the section of Apollonius's book on Cones), then the plane contained by HG and GF is equal to the plane contained by NG and GT. </s><s id="id.29.0.01.08">The plane contained by NG and GT is given because lines NG GT are given, and so the plane contained by HG and GF is given; also point {?} is given and so point F is given. </s><s id="id.29.0.01.09">Point K is given, and so the position of line KFS is given. </s><s id="id.29.0.01.10">Similarly LGS and hence point S are given, and point S is on the boundary of the ellipse. </s><s id="id.29.0.01.11">We join lines NS LT; let them meet diameter HZ at points R Q. </s><s id="id.29.0.01.12">Then the ratio of the plane contained by NG and GT to the plane contained by SG and GL is as that of the plane contained by NA and AT to the plane contained by Q{?} and AR (by similar triangles), and as the ratio of the plane contained by NA and AT to the plane contained by HA and AZ, which is explained in the figure from the section of Apollonius's book on Cones. </s><s id="id.29.0.01.13">Thus the plane contained by QA and AR will be equal to the plane contained by HA and AZ. </s><s id="id.29.0.01.14">The plane contained by QA and AR is given because both the lines QA AR are given; then the plane contained by HA and AZ is given. </s><s id="id.29.0.01.15">Similarly it is shown that the plane contained by HB and BZ is given also. </s><s id="id.29.0.01.16">AB is given; then line HZ is therefore given as we shall explain later; the diameter HZ is then of given size. </s><s id="id.29.0.01.17">It is evident that the conjugate diameter is given because the ratio of the transverse HZ to the latus rectum is given, being the same as the ratio of the plane contained by HA and AZ to the square of AN. </s></p></chap> <chap n="30"><p type="head"><s id="id.30.0.00.01">Section 30 (Fig. 16) </s></p> <p><s id="id.30.0.01.01">XV. Let us now explain the point whose discussion I postponed. </s></p> <p><s id="id.30.0.02.01">Let the planes contained by AG and GB and AD and DB be given; and let GD be given. </s></p> <p><s id="id.30.0.03.01">I say that AB is also given. </s></p> <p><s id="id.30.0.04.01">Let the plane contained by DG and GH be equal to the plane contained by AG and GB and the plane contained by GD and DZ equal to the plane contained by AD and DB; then the ratio of GH to HA is as that of AZ to ZD, because both of these ratios is as the ratio of GB to BD. </s><s id="id.30.0.04.02">The plane contained by GH and ZD is then equal to the plane contained by HA and AZ; and the plane contained by HA and AZ is given. </s><s id="id.30.0.04.03">Thus point A is given, and similarly point B. </s></p></chap> <chap n="31"><p type="head"><s id="id.31.0.00.01">Section 31 (Fig. 17) </s></p> <p><s id="id.31.0.01.01">XVI. Further, let us not make lines NT MK joining the given points parallel, and we join lines NK MT: let them intersect at point Sh. </s><s id="id.31.0.01.02">We produce from point L a line LTTh parallel to line MT. </s><s id="id.31.0.01.03">Let the ratio of the plane contained by TTh and TL to the plane contained by KT and NT be as the ratio of the plane contained by NSh and ShK to the plane contained by MSh and ShT. </s><s id="id.31.0.01.04">Now the plane contained by NT and TK is given, and so the plane contained by LT and TTh is also given, and line LT is given; thus line TTh is given. </s><s id="id.31.0.01.05">This leads back to the preceding theorem; I mean that we need to draw an ellipse on the five points N M L Th T; let us then draw that ellipse NMThLT with lines MT LTh being parallel. </s></p></chap> <chap n="32"><p type="head"><s id="id.32.0.00.01">Section 32 (Fig. 18) </s></p> <p><s id="id.32.0.01.01">XVII. It is easy for us having found any two conjugate diameters of the ellipse to find the axes of the ellipse instrumentally. </s><s id="id.32.0.01.02">We do this as follows: </s></p> <p><s id="id.32.0.02.01">We let the conjugate diameters of the ellipse which has been found be AB GD and let GD be greater than AB; let them bisect, one another at point H. </s><s id="id.32.0.02.02">Through point A we draw line ZAH parallel to line GD. </s><s id="id.32.0.02.03">The plane contained by HA and AT is made equal to the square of DH, and HT is bisected at point K; this then lies between points A T because DH is greater than HA. </s><s id="id.32.0.02.04">We draw from point K a line KL perpendicular to HT, cutting ZH at L. </s><s id="id.32.0.02.05">If we take point L as centre and describe with distance H a circle which cuts line ZH, let it cut it at points Z and H; we join lines HH HZ and produce them. </s><s id="id.32.0.02.06">To these lines we erect two perpendiculars AM AN; each of the squares of HO and HF is made equal to the plane contained by HH and HM, and each of the squares of HQ and HR equal to the plane contained by ZH and HN. </s><s id="id.32.0.02.07">Thus the axes of the ellipse are found, being OF QR; this is explained in Figure 37 of the first section of Apollonius's book on Cones. </s><s id="id.32.0.02.08">The smaller of them is equal to the thickness of the cylinder as we stated previously. </s></p></chap> <chap n="33"><p type="head"><s id="id.33.0.00.01">Section 33 </s></p> <p><s id="id.33.0.01.01">If a sphere is suspended and its position is given we wish to know on what point of a positioned plane it will fall if lowered perpendicularly, and the shortest perpendicular which can be drawn between a point on the surface of the sphere and a point in the positioned plane. </s><s id="id.33.0.01.02">There is a preliminary matter to be dealt with first, which is: if we have a suspended circle of given position which is not at right angles to the plane, we wish to find the common section of the two planes in the positioned plane and the inclination of one plane to the other. </s></p></chap> <chap n="34"><p type="head"><s id="id.34.0.00.01">Section 34 (Fig. 19) </s></p> <p><s id="id.34.0.01.01">We let the suspended circle be ABG. </s><s id="id.34.0.01.02">Three points A B G are marked on it, from which perpendiculars to the positioned plane are produced as I describe: we draw a line GD from point G to the positioned plane. </s><s id="id.34.0.01.03">Let this line if it veers from its position fall on two other points H Z of the plane. </s><s id="id.34.0.01.04">The centre of the circle which passes through points D H Z is made point K; the perpendicular dropped from point G to the positioned plane falls on point K, and GK is given. </s><s id="id.34.0.01.05">Let us then drop from points A B, two perpendiculars to the positioned plane BT AL; we join the two lines KL TL and produce them. </s><s id="id.34.0.01.06">We let the ratio of KM to ML be as that of GK to AL, and the ratio of TO to OL be as that of BT to AL. </s><s id="id.34.0.01.07">Then points M O are given. </s><s id="id.34.0.01.08">It is possible for us to arrange it so that, among the perpendiculars we have taken, one, say AL, should be the smallest. </s><s id="id.34.0.01.09">Lines MAG BAO are then straight and are in the plane of circle ABG. </s><s id="id.34.0.01.10">Thus the section common to this plane and to the positioned plane is given and is MO. </s><s id="id.34.0.01.11">We then drop from point A a perpendicular AN to MO. </s><s id="id.34.0.01.12">From point N we also erect a perpendicular NS on MO. </s><s id="id.34.0.01.13">Angle ANS is then given and is the angle at which the planes are inclined to one another. </s></p></chap> <chap n="35"><p type="head"><s id="id.35.0.00.01">Section 35 (Fig. 20) </s></p> <p><s id="id.35.0.01.01">XIX. Having explained this preliminary matter, let there be a sphere in suspended position; we wish to find on which point in the positioned plane it will fall when lowered perpendicularly; also to find the shortest line between the surface of the sphere and the position plane, which is perpendicular to that plane. </s><s id="id.35.0.01.02">We let the suspended sphere be one which has as its centre point H on which is drawn ABG, one of the great circles lying in it. </s><s id="id.35.0.01.03">The plane in which this is, is either perpendicular to the positioned plane or not and we know whether it is perpendicular or not in that any three points are marked on the circumference of the circle, and from them perpendiculars are dropped to the given plane as we mentioned in the preceding figure. </s><s id="id.35.0.01.04">If one straight line passes through the points on which the perpendiculars fall, then the planes cut one another at right angles. </s></p></chap> <chap n="36"><p type="head"><s id="id.36.0.00.01">Section 36 (Fig. 20) </s></p> <p><s id="id.36.0.01.01">Firstly let the plane in which the circle lies be perpendicular to the positioned plane; we draw from points A G perpendiculars AD GH to the positioned plane; the perpendiculars AD GH may or may not be equal; firstly let them be equal; line DH, which joins the points, where the perpendiculars fall, is bisected at point Z. </s><s id="id.36.0.01.02">Z is then the point on the plane which we require. </s><s id="id.36.0.01.03">If we bisect arc ABG at point B, point B is the point which we required on the surface of the sphere, I mean the one lying over point Z on the plane. </s><s id="id.36.0.01.04">Line BZ is the shortest perpendicular which can be drawn between the surface of the sphere and the positioned plane. </s></p></chap> <chap n="37"><p type="head"><s id="id.37.0.00.01">Section 37 (Fig. 21) </s></p> <p><s id="id.37.0.01.01">Further, we do not posit equal perpendiculars but make the smaller one AD; we make the ratio of HT to TD as that of GH to AD. </s><s id="id.37.0.01.02">When we produce HD, the line joining points G A meets it in point T in the positioned plane, line AT is given and angle ATD is also given. </s><s id="id.37.0.01.03">When that is so, we make a circle with diameter KL equal to the great circle ABG; LM is made equal to AT and angle KMN is made equal to angle ATD; we drop perpendiculars to MN from points K L, LO KN; from the centre of the circle a perpendicular RQF is dropped and arc AB is made equal to arc LQ; then line DZ is equal to line OF and line DH is bisected at point Z. </s><s id="id.37.0.01.04">Point Z is the point on which the sphere falls when lowered, and point B is that point on the surface of the sphere which falls on the plane; the shortest perpendicular which can be drawn between the sphere and the positioned surface is on perpendicular BZ which is equal to FQ which is given. </s></p></chap> <chap n="38"><p type="head"><s id="id.38.0.00.01">Section 38 (Fig. 22) </s></p> <p><s id="id.38.0.01.01">Further we do not posit circle ABG to be in a plane which lies at right angles to the positioned plane, and we make the common section of the two planes DT. </s><s id="id.38.0.01.02">Two points A G on the ends of the diameter of circle ABG are marked such that the line AG connecting them, when produced, meets the common section DT it is possible for us to do this because line DT is in the plane of circle ABG. </s><s id="id.38.0.01.03">Let it meet it at point T; then line AT is given and the angle at point T is also given. </s></p></chap> <chap n="39"><p type="head"><s id="id.39.0.00.01">Section 39 </s></p> <p><s id="id.39.0.01.01">From centre H we produce a perpendicular to DT, HBD; this is done as follows: a circle KhZS is made equal to the great circle ABG; let its diameter be ZH; HK is made equal to AT and angle HKM equal to angle ATD; from centre O we draw a perpendicular to KM, OLM; arc AB is also made equal to arc HL, and line KM will then be equal to TD; line DB will be equal to ML and BD will be perpendicular to DT; when produced it will pass through centre H because of the similarity of the figures. </s><s id="id.39.0.01.02">From point D and in the positioned plane we produce a perpendicular to DT, DY; line TD is then at right angles to the plane in which are lines HD DY, and circle ABG also stands at right angles to plane HDY. </s><s id="id.39.0.01.03">When then plane HDY is produced, from its section there is obtained a great circle at right angles to great circle ABG, passing through its poles and points B O. </s><s id="id.39.0.01.04">Thus if we take a pole F of circle ABG and draw a circle which passes through points F B O, this circle is one of the great circles which lie in the sphere and is in the plane in which are H D Y. </s><s id="id.39.0.01.05">Let B F O be on this circle. </s><s id="id.39.0.01.06">Further, we draw a circle through QRSh with diameter QSh equal to BO; QTh is made equal to BD and angle QThS equal to angle BDY; we produce from centre L of circle QRSh a perpendicular to ThS LNS; we also make arc BT of circle BFO equal to arc QN, and DY equal to ThS. </s><s id="id.39.0.01.07">If we join line YT it will be equal to NS, and if we produce it it will pass through centre H, and will be perpendicular to the positioned plane because it is perpendicular to line YD; point Y is given and is the point on which the sphere falls; point T is the place on the sphere which falls on the plane, and it is given. </s><s id="id.39.0.01.08">The shortest perpendicular which can be drawn between the sphere and the positioned plane is line TY and it is given. </s></p></chap> <chap n="40"><p type="head"><s id="id.40.0.00.01">Section 40 </s></p> <p><s id="id.40.0.01.01">If there is a sphere of given position with a positioned point outside it, how do we find that point on the surface of the sphere through which the line passes which connects the given point with the centre the sphere? </s><s id="id.40.0.01.02">This is clear, for, if the line drawn from the given point to the surface of the sphere is rotated, it describes a circle on the surface. </s><s id="id.40.0.01.03">The pole of this circle is the required point. </s></p></chap> <chap n="41"><p type="head"><s id="id.41.0.00.01">Section 41 (Fig. 23) </s></p> <p><s id="id.41.0.01.01">Further, we position a sphere and let there be outside it two given points; we wish to find the two points at which the line joining the two given points cuts the surface of the sphere. </s></p> <p><s id="id.41.0.02.01">Let there be a sphere with centre B and let the two given points outside it be A G. </s><s id="id.41.0.02.02">Let the points at which the lines joining A G to point B cut the surface of the sphere be points D H. </s><s id="id.41.0.02.03">A great circle DHZH is drawn passing through these two points, this being one of the circles that can be described on the sphere; then lines AD GH are given for the reasons we explained just previously, and, because half the sphere's diameter is given, both ADB and BHG are given. </s><s id="id.41.0.02.04">Line AG joining the two given points is given. </s><s id="id.41.0.02.05">We establish then a triangle TKL from the three lines AB AG GB. With centre T we draw a circle RMNO equal to circle DHZH; if this circle cuts line KL it is plain that the line joining points A G also cuts the sphere, and if the circle does not cut line KL then line AG does not cut the sphere; so let the circle cut line KL at points M N. </s><s id="id.41.0.02.06">Arc DH is made equal to arc RM and arc HZ equal to arc ON; it is then clear that points H Z are the points at which line AG cuts the surface of the sphere, and they are given. </s></p></chap> <chap n="42"><p type="head"><s id="id.42.0.00.01">Section 42 (Fig. 24) </s></p> <p><s id="id.42.0.01.01">There are also matters that are of use in affairs, which are particularised as instrumental, especially when the so-called analytic method reduces them to something simple and it is possible in them to avoid following a single analogical approach. </s><s id="id.42.0.01.02">An example of this is when we wish to construct hexagons in a given circle, one of which embraces the centre while the remaining six are constructed on the sides of the hexagon in the middle, the sides opposite those on which they are constructed being chords which fit in the circle. </s></p> <p><s id="id.42.0.02.01">Let the given circle be a circle with its centre the point H and let this centre have around it a hexagon with itself in the middle; and let its side be TK; let it be that when a hexagon is constructed on side TK of this hexagon, its side MN which is opposite side TK is a chord of the circle; we join line HK. </s><s id="id.42.0.02.02">HK is then connected with line KL, which is a side of the hexagon, in a straight line because angle HKT is two thirds of a right angle and angle TKL is a right angle and a third. </s><s id="id.42.0.02.03">We join line HN. </s><s id="id.42.0.02.04">Because lines HK KL are equal, line HL is then twice line LN and the angle at L is given, because it is a right angle and a third. </s><s id="id.42.0.02.05">Thus the shape of triangle HLN is given and so the ratio of HN to NL is given; line HN is given and so line NL is given, this being the side of the hexagon. </s></p></chap> <chap n="43"><p type="head"><s id="id.43.0.00.01">Section 43 </s></p> <p><s id="id.43.0.01.01">The instrumental construction is as follows: </s></p> <p><s id="id.43.0.02.01">We make one sixth of the diameter of a circle, AG, and on it is constructed a segment of a circle which receives two thirds of a right angle, and it is arc ABG; we make GH four of that measure by which is five. </s><s id="id.43.0.02.02">We draw BH, and make it a tangent. </s></p> <p><s id="id.43.0.03.01">I say that AB when joined is equal to TK which is the side of the hexagon. </s><s id="id.43.0.03.02">We draw BG and produce it; we cut off from it BD equal to line AB so triangle ABD is equilateral. </s><s id="id.43.0.03.03">Line AZ is made equal to half the diameter of the circle. </s><s id="id.43.0.03.04">Because the ratio of AH to HG is as the of 9 to 4, the ratio of the square of AB to the square of BG will also be equal to this ratio. </s><s id="id.43.0.03.05">Line AB is then equal to one and a half times BG, and is equal to BD. </s><s id="id.43.0.03.06">Thus line BG is twice GD. </s><s id="id.43.0.03.07">Let ZG also be twice GA; then line BZ when joined is equal to twice AD which is equal to AB. </s><s id="id.43.0.03.08">Now HL is twice LN and they contain equal angles; triangle ABZ is then similar to triangle HNL. </s><s id="id.43.0.03.09">Line AZ is equal to line NH; and so line AB is equal to line NL which is equal to line TK. </s><s id="id.43.0.03.10">This can be stated in another way which is easier than this. </s></p></chap> <chap n="44"><p type="head"><s id="id.44.0.00.01">Section 44 (Fig. 25) </s></p> <p><s id="id.44.0.01.01">Line AZ is made equal to half the diameter of the given circle. </s><s id="id.44.0.01.02">We take a third of this, AG, on which is constructed a segment of a circle ABG which receives two thirds of a right angle. </s><s id="id.44.0.01.03">Let GH be four of that measure by which AG is five. </s><s id="id.44.0.01.04">We draw from point H a line HB tangential to the segment of the circle. </s><s id="id.44.0.01.05">We join lines AB ZB and also line BG. </s><s id="id.44.0.01.06">We produce it to D, and make BD equal to AB; we join line AD. </s><s id="id.44.0.01.07">Because then lines HGA HB have been drawn to the circle, one cutting it and the other being tangential to it, the plane contained by AH and HG is equal to the square of HB. </s><s id="id.44.0.01.08">The ratio of BH to GH is then as that of AH to HB and so the angles of triangle GBH are equal to the angles of triangle ABH. </s><s id="id.44.0.01.09">The ratio of AH to AB is then as the ratio of HB to BG and the ratio of the square of line AH to the square of line HB is as that of the square of line BG. </s><s id="id.44.0.01.10">But the ratio of the square of line AH to the square of line HB is as that of AH to HG, which is as the ratio of 9 to 4. </s><s id="id.44.0.01.11">The ratio of AH to HG is as that of the square of line BD to the square of line BG, because AB is equal to BD and so the ratio of the square of line BD to the square of line BG is as the ratio of 9 to 4. </s><s id="id.44.0.01.12">Line BD is then equal to one and a half times line BG and line BG is twice line GD. </s><s id="id.44.0.01.13">Line GZ is furthermore twice line GA and so the ratio of ZG to GA is the same as that of BG to GD. </s><s id="id.44.0.01.14">The angles at point G are equal; thus the angle at D is equal to angle ZBG, and the angle at Z is equal to angle GAD; thus the ratio of line ZB to line BG is as that of line AD to line DG. </s><s id="id.44.0.01.15">Alternando the ratio of ZB to AD is as that of BG to GD. </s><s id="id.44.0.01.16">Line BG is twice line GD and so line ZB is twice line AD, which is equal to AB. </s><s id="id.44.0.01.17">The angle at point D is two thirds of a right angle and so angle ZBG is two thirds of a right angle. </s><s id="id.44.0.01.18">The whole angle ABZ is thus a right angle and a third. </s><s id="id.44.0.01.19">If then we construct a circle on centre H with half its diameter equal to AZ, draw from its centre line HL making line HL equal to line ZB, and construct on it angle HLN equal to angle ZBA, joining line HN, the angles of triangle HLN will be equal to the angles of triangle AZB. </s><s id="id.44.0.01.20">Line AZ is equal to line HN and so line NL is equal to AB. </s><s id="id.44.0.01.21">It is clear that seven hexagons are constructed in the circle from lines equal to line AB as we desired. </s></p> <p><s id="id.44.1.00.01">Section A (Figs. 26, 27) </s></p> <p><s id="id.44.1.01.01">XXV. We wish to show how to draw a perpendicular from a given point on a given straight line, if we have a given distance which must not be exceeded when we draw circles. </s></p> <p><s id="id.44.1.02.01">Now let the given straight line be AB and let G be the given point on it. </s><s id="id.44.1.02.02">When we wish to draw a perpendicular on AB from this point, we take any two equal lines GD GH on either side of point G. </s></p> <p><s id="id.44.1.03.01">Let both of them be shorter than the given line. </s></p> <p><s id="id.44.1.04.01">We make points D and H centres and at the distance of a line greater than line GH and equal to the given distance, we describe two arcs intersecting at Z. </s><s id="id.44.1.04.02">We join line ZG and it is then perpendicular to AB. </s></p> <p><s id="id.44.1.05.01">The proof of that is a clear one. </s></p> <p><s id="id.44.1.06.01">This can be done in another way. </s></p> <p><s id="id.44.1.07.01">We make the point from which we wish to erect the perpendicular, the end A of the line. </s><s id="id.44.1.07.02">We make point A a centre. </s><s id="id.44.1.07.03">Another point D is marked on the line; we make points A D centres and describe on them at distances equal to AD, arcs intersecting at point H. </s><s id="id.44.1.07.04">We join line HD and produce it to Z. </s><s id="id.44.1.07.05">We make HZ equal to HD and join line ZA. </s><s id="id.44.1.07.06">This will be a perpendicular on AB. </s></p> <p><s id="id.44.1.08.01">The proof of that is clear in that we join line AH. </s></p> <p><s id="id.44.2.00.01">Section B (Figs. 28, 29) </s></p> <p><s id="id.44.2.01.01">We wish to bisect line AB; when again the given distance which must not be exceeded is one distance. </s></p> <p><s id="id.44.2.02.01">We separate off from line AB at both its ends two lines AG BD equal to the given distance. </s><s id="id.44.2.02.02">If line GD is less than the distance or equal to it we make points G D centres and draw at that distance arcs intersecting on both sides of line AB at H T. </s><s id="id.44.2.02.03">We place a ruler on points H T marking a point at the place where it cuts line AB. </s><s id="id.44.2.02.04">Line AB is bisected at that point. </s></p> <p><s id="id.44.2.03.01">If line GD is greater than the given distance, then we separate off from it two lines DZ GH equal to the given distance. </s><s id="id.44.2.03.02">If line HZ is smaller than the given distance or equal to it, we make points Z H centres and draw two arcs at the given distance intersecting at points H T. </s><s id="id.44.2.03.03">We continue in this way until line A B is bisected for us. </s><s id="id.44.2.03.04">Let it then be bisected at point K. </s><s id="id.44.2.03.05">We can bisect line AB in another way. </s><s id="id.44.2.03.06">We separate off from AB two lines AZ BH equal to the given distance; we make A Z centres and describe two arcs at distance AZ intersecting at G. </s><s id="id.44.2.03.07">We also make points B H centres and at distance BH describe two arcs intersecting at point D. </s><s id="id.44.2.03.08">Let the line joining points G D cut line AB at point H. </s><s id="id.44.2.03.09">It is then clear that line AB is bisected at point H because the line joining points G A is equal to the line joining points B D and parallel to it. </s></p> <p><s id="id.44.3.00.01">Section C (Fig. 30) </s></p> <p><s id="id.44.3.01.01">Now if we wish to divide AB into three equal parts, we take two points G D as we did previously, and we join the two lines AG BD and produce them. </s><s id="id.44.3.01.02">We separate off from them, produced, two lines GK DL equal to the given distance. </s><s id="id.44.3.01.03">We join GL DK, and we shall then have divided AB into three, equal parts at points Z H. </s></p> <p><s id="id.44.3.02.01">If we wish to divide line AB into four or more equal parts we separate off from each of the two lines AG BD when produced, lines equal to the given distance, and numbering one less than the number of parts into which we wish to divide line AB; just as when we wished to divide it into three we separated off from them two lines which were AG GK and BD DL. </s><s id="id.44.3.02.02">When lines join the corresponding points line AB is divided into equal parts of the required number. </s><s id="id.44.3.02.03">The proof of that is evident because the lines joined are parallel. </s></p> <p><s id="id.44.4.00.01">Section D (Figs. 31, 32) </s></p> <p><s id="id.44.4.01.01">We wish to add to line AB a line equal to it in the same straight line with it, and the distance that we use is one given distance. </s></p> <p><s id="id.44.4.02.01">We separate off two lines AD BH equal to the given distance; we make points A D centres and we construct at the given distance two arcs intersecting at point Z. </s><s id="id.44.4.02.02">Similarly we make points B H centres and describe at that given distance two arcs intersecting at point T. We join line AZ and produce it in a straight line. </s><s id="id.44.4.02.03">We then make ZH equal to the given distance and join HT which we produce so that it meet AB produced at point K. </s><s id="id.44.4.02.04">Then BK will be equal to AB because AK is twice KB. </s><s id="id.44.4.02.05">For AH is twice BT and when line BT is joined it is parallel to line AH. </s></p> <p><s id="id.44.4.03.01">We also wish to draw a line equal to AB in the same straight line with its end at point G. </s><s id="id.44.4.03.02">Now BG is bisected at point D and DH is made equal to AD and connected to it in a straight line as we explained in the figure before this. </s><s id="id.44.4.03.03">Then AB will be equal to GH. </s><s id="id.44.4.03.04">Because then AD is equal to DH and GD is equal to DB, AG remains equal to BH; and when we make GH common, the whole of AB is equal to the whole of GH. </s></p> <p><s id="id.44.5.00.01">Section E (Figs. 33, 34, 35) </s></p> <p><s id="id.44.5.01.01">XXXI. Further, we wish to draw a line equal to line AB, not in a straight line with it, but in a straight line with AG, and with point A as its end. </s><s id="id.44.5.01.02">We make each of AT AK equal to the given distance and we produce two lines AB AG as far as points Z H; we separate off from them lines equal to the given distance, namely AD DZ AH HH. </s><s id="id.44.5.01.03">We join lines HZ DH. </s><s id="id.44.5.01.04">Let them intersect at point M. </s><s id="id.44.5.01.05">We join line AM,, and also line TB. </s><s id="id.44.5.01.06">Let it cut AM produced at point N. </s><s id="id.44.5.01.07">Let the line which joins K N and is produced in a straight line cut the line TG at point G. </s><s id="id.44.5.01.08">Then AG is equal to AB. </s><s id="id.44.5.01.09">The proof of that is evident. </s></p> <p><s id="id.44.5.02.01">XXXIII. Let the given line now be AB. </s><s id="id.44.5.02.02">We wish to draw a line GD equal to it and in a straight line. </s><s id="id.44.5.02.03">DZ is made equal to DB. </s><s id="id.44.5.02.04">We have explained previously how this is done. </s><s id="id.44.5.02.05">Because AD is equal to DH and DB is equal to DZ, the whole of AB is equal to the whole of HZ. </s><s id="id.44.5.02.06">Let DG be equal to HZ; then line DG is equal to line AB. </s></p> <p><s id="id.44.5.03.01">XXXIV. Now let there be two lines AB GD in whatever position and let the shorter of them be AB. </s><s id="id.44.5.03.02">We wish to separate off from GD a line the end of which will be point G and which will be equal to AB. </s><s id="id.44.5.03.03">So we join AG and produce it to H; let AZ be equal to AB. </s><s id="id.44.5.03.04">We can construct this by means of the construction we mentioned before this one. </s><s id="id.44.5.03.05">We make GH equal to line HZ; this is constructed by means of the previous construction. </s><s id="id.44.5.03.06">Then line AB is equal to line GH. </s></p> <p><s id="id.44.6.00.01">Section F (Fig. 36) </s></p> <p><s id="id.44.6.01.01">XXXV. By means of such as we have already mentioned, namely lines drawn using only one distance, we say that it is possible for us to use that same length and construct with it a triangle whose sides are equal to three given lines of which two are greater than the third. </s><s id="id.44.6.01.02">Now that is not easy to do using the construction mentioned by Euclid in his Elements, since we have only one small pair of compasses with which to work. </s><s id="id.44.6.01.03">So let the given lines be AB G D. </s><s id="id.44.6.01.04">We wish to construct a triangle the sides of which will be equal to these lines. </s><s id="id.44.6.01.05">First let lines G D be equal and AB be bisected at H. </s><s id="id.44.6.01.06">We produce from H a perpendicular HZ on AB. </s><s id="id.44.6.01.07">Let HZ be equal to one of the lines G D. </s><s id="id.44.6.01.08">We join line AZ and make HH equal to the given distance. </s><s id="id.44.6.01.09">We draw from H a perpendicular HT on HZ and we draw from it (at point T) a line TK at right angles; TK is then equal to the given distance. </s><s id="id.44.6.01.10">Because G D are greater than AB each of the lines G D which is equal to HZ is greater than AH, and the ratio of ZH to HA is as that of TK to KA; and line TK is greater than line KA. </s><s id="id.44.6.01.11">If then we make point A a centre and draw an arc at the given distance which is equal to HH and TK cutting TK when produced, let it cut it at L. </s><s id="id.44.6.01.12">We join line AL and produce it so that it joins line HZ when produced at point M, and we join line BM. </s><s id="id.44.6.01.13">Then the sides of the triangle AMB are equal to lines AB G D because the ratio of MA to AL is as that of ZA to AT, in that line TL is parallel to base MZ. </s><s id="id.44.6.01.14">The ratio of ZA to AT is as that of ZH to HH and so the ratio of MA to AL is as the ratio of ZH to HH. </s><s id="id.44.6.01.15">Line AL is equal to line HH and so line AM is equal to line HZ which is equal to G. </s><s id="id.44.6.01.16">In the same way MB is equal to D because lines AH HM are equal to lines HM HB and the angles at point H are right angles. </s><s id="id.44.6.01.17">We have then constructed a triangle ABM the sides of which are equal to lines AB G D. </s></p> <p><s id="id.44.7.00.01">Section G (Fig. 37) </s></p> <p><s id="id.44.7.01.01">XXXVI. Further we make lines AB G D of different lengths. </s><s id="id.44.7.01.02">Let the largest of them be AB and the smallest G; we draw a line of whatever position, MH. </s><s id="id.44.7.01.03">We make HQ equal to D and we make QR QM equal to G. </s><s id="id.44.7.01.04">Then we draw HZ at right angles to line HM; let it be equal to line HR. </s><s id="id.44.7.01.05">We draw MN at right angles to HM; let it be equal to AB. </s><s id="id.44.7.01.06">We join line HN. </s><s id="id.44.7.01.07">Let it meet ZH which is perpendicular to HZ at point H. </s><s id="id.44.7.01.08">We make BO equal to ZH and bisect AO at S. </s><s id="id.44.7.01.09">We make ST perpendicular to AB and we make line AT equal to line G which is the smallest of the three lines. </s><s id="id.44.7.01.10">We join line TB. </s><s id="id.44.7.01.11">Then triangle ATB has sides equal to lines AB G D which are the given lines. </s></p> <p><s id="id.44.7.02.01">The proof of that is as follows: </s></p> <p><s id="id.44.7.03.01">We complete the drawing of the planes with parallel sides; and because MQ is equal to QR, and we add RH to both of them, the plane contained by MH and HR together with the square of QR, which is equal to the square of MQ, is equal to the square of line HQ. </s><s id="id.44.7.03.02">Now the difference between the squares of lines HQ QM is equal to plane MZ with parallel sides because RH is equal to HZ; but plane MN is equal to the completed figure HTKL. </s><s id="id.44.7.03.03">Thus the difference between the squares of lines HQ QM is equal to the plane contained by HT and KL. </s><s id="id.44.7.03.04">The square of line MN is greater than the square of line HQ because [MN] is equal to AB which is the greatest of the three lines. </s><s id="id.44.7.03.05">Thus the square of line MN, which is the square of line TK of plane HTKL which has parallel sides [is greater than the plane HTKL which has parallel sides] and so line TK is greater than line TH; AB is then greater than HT which is equal to ZH. </s><s id="id.44.7.03.06">BO is equal to ZH and AO is bisected at S. </s><s id="id.44.7.03.07">From S is drawn a line ST at right angles. </s><s id="id.44.7.03.08">Because MH is equal to two of the sides of the triangle with the given sides, MH is greater than MN. </s><s id="id.44.7.03.09">ZH is greater than HR because triangles MNH HZH have equal sides, for angle MNH is equal to external angle ZHH. </s><s id="id.44.7.03.10">Line HT which is equal to line HZ is equal to line BO. </s><s id="id.44.7.03.11">Line MR is equal to line HR. </s><s id="id.44.7.03.12">Thus line BO is greater than line HR. </s><s id="id.44.7.03.13">Line AB is smaller than line MH because MN is smaller than MH and the remainder of line AO is smaller than line MR and line AS is smaller than line QR which is equal to G. </s><s id="id.44.7.03.14">Thus it is possible to draw a line from point A to ST which will be equal to G; we draw it as we said and it is AT; we join TB. </s><s id="id.44.7.03.15">It is then equal to D because the difference between the squares of lines HQ QM is the plane contained by MH and HR. </s><s id="id.44.7.03.16">The difference between the squares of lines BS SA is the plane contained by AB and BO; and the plane contained by MH and HR is equal to the plane contained by AB and BO and this is equal to plane HZ which in turn is equal to plane HL. </s><s id="id.44.7.03.17">The plane contained by AB and BO is equal to plane HL, and the difference between the squares of HQ QM is equal to that between the squares of BS SA. </s><s id="id.44.7.03.18">The difference between the squares of BS SA is equal to that between the squares of TB TA. </s><s id="id.44.7.03.19">Now the square of line QM is equal to the square of line TA because AT is equal to QR; and so the square of line HQ is equal to the square of line TB, and line HQ is equal to line TB; and it was equal to D. </s><s id="id.44.7.03.20">So line TB equals line D and the sides of triangle ATB are equal to the given lines AB G D. </s><s id="id.44.7.03.21">We only employed one length in the whole of this. </s></p></chap> <chap n="45"><p type="head"><s id="id.45.0.00.01">Section 45 (Fig. 38) </s></p> <p><s id="id.45.0.01.01">Let us now describe how the wheels which we mounted together are constructed. </s><s id="id.45.0.01.02">We make two wheels A B, turned on a lathe, which are mounted alongside one another. </s><s id="id.45.0.01.03">Now let the ratio of the diameter of wheel A to the diameter of wheel B be as the ratio of the number of teeth on wheel A to the number of teeth on wheel B, for the two wheels which are placed next to one another are like this because the ratio of circumference to circumference is the same as that of diameter to diameter. </s><s id="id.45.0.01.04">We shall explain this later. </s><s id="id.45.0.01.05">The teeth of wheel A total 60 and those of wheel B 40. </s></p> <p><s id="id.45.0.02.01">I say that the ratio of the speed of movement of wheel A to the speed of movement of wheel B is as the ratio of the number of teeth on wheel B to the number of teeth on wheel A. </s></p> <p><s id="id.45.0.03.01">Because then wheels A B are mounted alongside one another the number of B's teeth moved is the same as the number of A's teeth moved. </s><s id="id.45.0.03.02">When then wheel B makes one complete revolution, wheel A is moved the amount of 40 teeth. </s><s id="id.45.0.03.03">When however wheel B is turned through 60 revolutions, I mean the same number of revolutions as A has teeth, then wheel B moves 2.400 teeth which is equal to the total of the number of A's teeth multiplied by the number of B's teeth. </s><s id="id.45.0.03.04">We shall likewise also make clear that when wheel A is turned 40 times, I mean as the number of teeth on B, wheel A moves 2.400 teeth, that is the total of the teeth of wheel A multiplied by the teeth of wheel B. </s><s id="id.45.0.03.05">Now when wheel A turns 40 times, which is as the number of teeth on wheel B, wheel B turns 60 times which is as the number of teeth on wheel A. </s><s id="id.45.0.03.06">Thus the ratio of the speed of movement of wheel A to that of wheel B is equal to the ratio of the number of teeth on wheel B to the number of teeth on wheel A. </s></p></chap> <chap n="46"><p type="head"><s id="id.46.0.00.01">Section 46 (Fig. 39) </s></p> <p><s id="id.46.0.01.01">We shall now show that the ratios of the circumferences of circles to one another are the same as the ratios of their diameters to one another. </s></p> <p><s id="id.46.0.02.01">We make two circles be AB GD; let their diameters be AB GD. </s><s id="id.46.0.02.02">I say that the ratio of the circumference of circle AB to that of GD is as the ratio of diameter AB to diameter GD. </s><s id="id.46.0.02.03">The proof of this is as follows: </s></p> <p><s id="id.46.0.03.01">The ratio of circle AB to circle GD is as the ratio of the square of line AB to the square of line GD. </s><s id="id.46.0.03.02">But circle AB is four times the rectangular plane contained by diameter AB and the circumference circle AB. </s><s id="id.46.0.03.03">Circle GD is four times the rectangular plane contained by the diameter of circle GD and its circumference, for the total of half the diameter of the circle multiplied by its circumference is twice the area of the circle, as Archimedes pointed out, and as we explained in the commentary on the first section of the Almagest. </s><s id="id.46.0.03.04">We also explained it in a separate theorem. </s><s id="id.46.0.03.05">The ratio then of the rectangular plane contained by the diameter of circle [AB and its circumference to the rectangular plane contained by the diameter of circle] GD and its circumference is the same as the ratio of the square on line AB to the square on line GD. </s><s id="id.46.0.03.06">Alternando, the ratio of the plane formed by the multiplication of the diameter of circle AB by its circumference to the square of line AB is as the ratio of the plane formed by the multiplication of the diameter of circle GD by its circumference to the square on diameter GD. </s><s id="id.46.0.03.07">The ratio then of the circumference of circle AB to diameter AB is the same as the ratio of the circumference of circle GD to diameter GD. </s><s id="id.46.0.03.08">Alternando the ratio of the circumference of circle AB to the circumference of circle GD is as the ratio of diameter AB to diameter GD. </s></p></chap> <chap n="47"><p type="head"><s id="id.47.0.00.01">Section 47 (Fig. 40) </s></p> <p><s id="id.47.0.01.01">When a wheel is given with a given number of teeth, and we mount alongside it another wheel with a given number of teeth, how is the diameter of the wheel mounted alongside found? </s></p> <p><s id="id.47.0.02.01">We make A the given wheel, and let B be the number of its teeth; let this be 60. </s><s id="id.47.0.02.02">Let wheel G adjoin wheel A, and let the number of its teeth, which is 40, be D. </s><s id="id.47.0.02.03">We wish to find [the diameter of] wheel G. </s><s id="id.47.0.02.04">Because number B is the number of teeth in wheel A, and number D is the number of teeth in wheel G, and the teeth of wheel A as its circumference and the teeth of wheel G as its circumference, the ratio of number B to number D is as the ratio of the circumference of wheel A to the circumference of wheel G. </s><s id="id.47.0.02.05">The ratio of the circumference to the circumference is as that of the diameter to the diameter and the ratio of number B to number D is given, because it is the same as that of 60 to 40. </s><s id="id.47.0.02.06">The diameter of wheel A is given, and so the diameter of wheel G is given, for we let the ratio of the diameter of circle A to another line be as the ratio of the number 60 to the number 40; the circle described on that diameter will be equal to the wheel made. </s></p></chap> <chap n="48"><p type="head"><s id="id.48.0.00.01">Section 48 (Fig. 41) </s></p> <p><s id="id.48.0.01.01">We carry out its construction with instruments as follows: we draw any straight line HZ and we divide it into the same number of parts as the number of teeth on wheel A, which is 60. </s><s id="id.48.0.01.02">From point Z a perpendicular ZH is drawn to HZ. </s><s id="id.48.0.01.03">ZH is made equal to the diameter of wheel A; we join HH. </s><s id="id.48.0.01.04">Let HT be 40 by that measure by which HZ is 60, the former being the number of teeth on wheel G. </s><s id="id.48.0.01.05">Line TK is drawn from point T parallel to line ZH. </s><s id="id.48.0.01.06">Line TK will then be equal to the diameter of wheel G. </s><s id="id.48.0.01.07">The proof of this is clear. </s></p></chap> <chap n="49"><p type="head"><s id="id.49.0.00.01">Section 49 (Fig. 42) </s></p> <p><s id="id.49.0.01.01">How a screw of which the grooves are adjusted to the inclining teeth of a given wheel is constructed is explained as follows: </s></p> <p><s id="id.49.0.02.01">We suppose a cylinder of even thickness ADZH turned on a lathe; let its side be AH, and let a distance of revolution of the screw be AH. </s><s id="id.49.0.02.02">Let there be a sheet of brass, the part K [TH of which is a right-angled triangle with the angle at T the right angle, the remainder being the rectangle TKLM; and let] TH be equal to AB and TK equal to the circumference of the cylinder ADZH. </s><s id="id.49.0.02.03">We envelop the cylinder in the attached sheet, and fold it on it so that the plane TKLM with parallel sides will surround the cylinder, passing through points D H. </s><s id="id.49.0.02.04">Let point T be placed on point A, and point H on point B. </s><s id="id.49.0.02.05">This done we draw beside the folded side HK one turn of the screw, as line BD. </s><s id="id.49.0.02.06">Again moving the sheet so that point T is on point B and point H on point G, we draw along HK another turn of the screw so that we have drawn in all two turns. </s><s id="id.49.0.02.07">While we take point A to the position of point B, if uniform movement obtains, point B moves on the surface of the cylinder and returns to the region in which it was; the point which we said turns into line AH describes in its passage one turn of the screw. </s><s id="id.49.0.02.08">This has been explained by Apollonius of Perga. </s><s id="id.49.0.02.09">If we bisect AB, BG and so on until we come to the bisection adjacent to H, and draw the turns of the screw on the sheet one by one passing through those points; and if we take between those points what we wish that is suitable for cutting the groove of the screw, and then file from without and within in the groove of the screw, we shall obtain shapes which resemble the shape of the lentil in it. </s></p></chap> <chap n="50"><p type="head"><s id="id.50.0.00.01">Section 50 (Fig. 43) </s></p> <p><s id="id.50.0.01.01">Again, we suppose that at the edge of one of the two surfaces of the given wheel there is a circumference QTK of a circle with centre S. </s><s id="id.50.0.01.02">Let the distances between points Q T Sh be equal to one another, and equal to distance AB of the preceding diagram. </s><s id="id.50.0.01.03">Let it, for instance, be a 24th part of the circle, and from points Q T Sh we draw lines QO TO ShO facing centre S so as to meet circle MNFTh which is drawn on centre S. </s><s id="id.50.0.01.04">From the mid-points of the arcs which are between [points O O O, and they are] the points lettered M N F Th, we draw lines MQ NQ NT FT FSh ShTh to points Q T Sh; we draw from point Q of OQ on the curved surface of the wheel, a line QR opposite to it so that it reaches the other surface of the wheel. </s><s id="id.50.0.01.05">A circle KhD is drawn on the surface at the edge of the wheel just as we drew on the first surface. </s><s id="id.50.0.01.06">At point R we take an arc RKh, equal to half arc QT, also an arc KhD equal to arc QT. </s><s id="id.50.0.01.07">We also make arc DKh equal to arc TSh; we do likewise with the remainder of the circle, and join lines QKh TD ShH; we shall then have constructed the inclination of the teeth. </s><s id="id.50.0.01.08">Because circle QT equal to KhD we describe on the other surface of the wheel a circle equal to circle MN with its centre opposite point S. </s><s id="id.50.0.01.09">Towards its circumference lines are drawn from points Kh D facing its centre. </s><s id="id.50.0.01.10">We do as we did on the circumference of circle QTSh and the other side of the wheel will be drawn for us. </s><s id="id.50.0.01.11">The shapes which are between the lines are cut, as for instance the shape embraced by NQT and FDH and the portion opposite NQT on the other side; thus we shall have slanting teeth on the wheel. </s><s id="id.50.0.01.12">Each one of them enters the cut which is in the screw, because the distances which are between them, which are equal to QT, are equal to distance AB which we used in cutting the screw. </s><s id="id.50.0.01.13">It is clear that whenever the screw makes one turn one tooth of the wheel passes. </s><s id="id.50.0.01.14">Heron explained this at the beginning of his Mechanics. </s><s id="id.50.0.01.15">We ourselves have only included the construction of this that nothing should be lacking from this book. </s></p></chap> <chap n="51"><p type="head"><s id="id.51.0.00.01">Section 51 (Fig. 44) </s></p> <p><s id="id.51.0.01.01">We imagine a screw AB and the winding cut which is in the screw is AGDHZB, and we imagine that each one of them is a single turn. </s><s id="id.51.0.01.02">Let a toothed wheel HGHT be mounted with it, and let the bases of its teeth be HG GH HT; let them be formed to the grooves of the screw; let GH be in contact with the screw meshing with it thoroughly. </s><s id="id.51.0.01.03">The teeth do not mesh with the rest of the cut. </s><s id="id.51.0.01.04">If we rotate the screw so that point H is in position G, point G will move to where point H was, whenever the screw turns through one revolution. </s><s id="id.51.0.01.05">Tooth GH will move to position GH and tooth HT to position GH. </s><s id="id.51.0.01.06">Also when tooth HT moves to position GH and the screw revolves once, all the teeth turn. </s><s id="id.51.0.01.07">The situation regarding the rest of the teeth should be understood in the same way; the number of times the screw moves when the wheel revolves once, is according to the number of teeth on the wheel. </s></p></chap> <chap n="52"><p type="head"><s id="id.52.0.00.01">Section 52 </s></p> <p><s id="id.52.0.01.01">That is something which has been explained in the book On Drawing Weight. </s><s id="id.52.0.01.02">As regards the five powers which we mentioned and which we said had been mentioned by Heron, we shall treat of them briefly as an, aide-memoire for lovers of knowledge. </s><s id="id.52.0.01.03">In addition we wish to mention devices which are called "having one part", "having two parts", "having three parts" and "having four parts"; these are matters the need for which is essential. </s><s id="id.52.0.01.04">Perhaps those who study this science may not find the books in which these matters are mentioned for we have come across books on this which were defective and many of them have been extremely corrupt, with their beginnings and ends missing. </s></p> <p><s id="id.52.0.02.01">Since there are five powers by means of which a given weight is moved by a given force, we reckon it a matter of urgent need that their forms, names, and uses be explained. </s><s id="id.52.0.02.02">Heron and Philon have both produced information as to why these forces are related to one nature despite the difference in their forms. </s></p> <p><s id="id.52.0.03.01">The names of these powers then are: </s><s id="id.52.0.03.02">the axle with a wheel turning on it; the lever; the compound pulley; the wedge; that which is called the endless screw. </s></p></chap> <chap n="53"><p type="head"><s id="id.53.0.00.01">Section 53 </s></p> <p><s id="id.53.0.01.01">The axle with a wheel turning on it is made as follows: </s></p> <p><s id="id.53.0.02.01">it is necessary to take a strong square piece of wood, to plane its ends and smoothe them and they become rounded; two rings of brass should be placed on these ends, being fixed to the axle, so that, when they have been mounted in round holes in uprights like poles they turn and move easily. </s><s id="id.53.0.02.02">Let plates of brass be in the holes, placed round the rings which are on the ends of the axle. </s><s id="id.53.0.02.03">The piece of wood which we mentioned is called an "axle". </s><s id="id.53.0.02.04">Let a wheel be placed round the middle of the axle with a square hole in which the axle is fitted, so that axle and wheel move together. </s><s id="id.53.0.02.05">This wheel is called "baritrukhiun", which means "the surrounding wheel". </s><s id="id.53.0.02.06">Let there be two cuts like two notches in the axle on either side of the wheel which receive the ropes which are tied to the axle. </s><s id="id.53.0.02.07">In the edge of the surrounding wheel let there be as many holes as are considered needed, and these holes are needed so that spokes may be put in them which can be pulled and by means of which both wheel and axle are moved. </s><s id="id.53.0.02.08">We have now explained how this tool is constructed. </s><s id="id.53.0.02.09">Its use is as follows: when we wish to move a large weight with little power, we take the ropes tied to the weight and wind them round the two cuts both sides of the wheel. </s><s id="id.53.0.02.10">We then put the spokes into the holes which are in the surrounding wheel; we turn this wheel by means of these, by pulling the spoke downwards. </s><s id="id.53.0.02.11">The weight then moves easily with little power while the ropes are wound round the axle or round something else so that the whole of the rope should not lie round the axle. </s><s id="id.53.0.02.12">It is necessary that the size of this apparatus should be in accordance with the weight to be moved. </s><s id="id.53.0.02.13">The calculation then lies in accordance with the ratio of the weight to be moved to the power which moves it, as will be shown later. </s></p></chap> <chap n="54"><p type="head"><s id="id.54.0.00.01">Section 54 </s></p> <p><s id="id.54.0.01.01">The second power, which is produced by the lever, is worthy of consideration as a means by which great weight can be moved. </s><s id="id.54.0.01.02">When some people wished to move a great weight, needing in this primarily to raise it from the ground, and there was no place on the weight where they could obtain a purchase because all the parts of the base of the weight were on the ground, they dug a little beneath it, and introduced under it the end of a long piece of wood putting the end in the ground; they then depressed the other end after having placed under the piece of wood near the weight the stone which is called "that which is placed beneath the lever". </s><s id="id.54.0.01.03">When this method of moving something occurred to them as I have described, they considered moving great weight with this kind of operation. </s><s id="id.54.0.01.04">They called this piece of wood the lever be it square or round. </s><s id="id.54.0.01.05">The nearer the stone, under the lever, is to where the weight lies, the easier moving it becomes, as will be explained later. </s></p></chap> <chap n="55"><p type="head"><s id="id.55.0.00.01">Section 55 </s></p> <p><s id="id.55.0.01.01">The third power is produced by the apparatus known as the compound pulley. We use it when we wish to draw some weight; we attach ropes to the weight and draw it with a force equivalent to the weight to be carried. </s><s id="id.55.0.01.02">Then we untie the rope from the object to be carried, tie one of the ends to a firm fixed place, and pass the other end through a pulley tied to the weight; we draw this end. </s><s id="id.55.0.01.03">We can thus move the weight easily. </s><s id="id.55.0.01.04">Also, if we tie another pulley to the fixed place, pass the end of the rope we were holding through it, and then pull it through that pulley, we move the weight more easily. </s><s id="id.55.0.01.05">Again, if we tie another pulley to the weight to be carried, pass through it the end of the rope to be pulled and draw it, we can move the weight much more easily. </s><s id="id.55.0.01.06">Similarly also if we tie other pulleys to the fixed place and to the object to be carried and pass through them that end of the rope which we pull, we can move the weight more easily; the more the rope is bent on pulleys, the easier the drawing becomes. </s><s id="id.55.0.01.07">The easiest is when, instead of the ends which are tied to the fixed place, there is one piece of wood on which is placed an axle which moves, and it is called the manjanun; this manjanun is attached to the fixed place by another rope. </s><s id="id.55.0.01.08">Also the end tied to the weight is tied to another manjanun equal to the first and this manjanun is tied to the weight to be carried by another rope. </s><s id="id.55.0.01.09">It is necessary that the pulleys should be so placed in the manjana as not to touch one another as it would be difficult to move them. </s><s id="id.55.0.01.10">We snail explain later why more pulleys make for easier operation, and also why the other end of the rope is made fast to the fixed place. </s></p></chap> <chap n="56"><p type="head"><s id="id.56.0.00.01">Section 56 </s></p> <p><s id="id.56.0.01.01">The power following this is that produced by the wedge. </s><s id="id.56.0.01.02">It is very necessary in the pressing of perfumes and in the glueing processes necessary in carpentry when glueing has to be used to join things which are separate; but its greatest use is in uprooting rock from where it has been cut, and in separating it from what is in contact with its base. </s><s id="id.56.0.01.03">Of the rest of the powers none can do this either individually or in combination; the wedge alone operates by itself effectively in this, with minimal effort. </s><s id="id.56.0.01.04">When such craftsmen as use the wedge cease actively operating it, its effect does not cease. </s><s id="id.56.0.01.05">The sole contribution of the operators is the extent of the impetus; evidence of this is that often when there is no-one there striking in the sounds of cracks are heard in the rock from the action of the wedge. </s><s id="id.56.0.01.06">The smaller the wedge's angle the easier it is to work, I mean that light blows have an effect on it, as we shall later show. </s></p></chap> <chap n="57"><p type="head"><s id="id.57.0.00.01">Sect ion 57 </s></p> <p><s id="id.57.0.01.01">The tools we have mentioned so far are easy to operate, and are clear and self-explanatory. </s><s id="id.57.0.01.02">Their situation is very apparent when they are needed. </s><s id="id.57.0.01.03">In the case of the screw however there is some difficulty both in construction and operation, in that it is sometimes used on its own and sometimes in conjunction with some other power. </s><s id="id.57.0.01.04">It is no more than a wedge but it is one which is twisted and is not struck; it gets its impetus from something like the lever. </s><s id="id.57.0.01.05">We shall explain that in what we shall say later. </s><s id="id.57.0.01.06">Its construction is as I describe it: it is necessary to know that if the side of a cylinder moves on its surface, and if there is a point on the end of the side, which moves on the side together with the movement of the side, the side meanwhile turning round the cylinder once at some time and the point moving at the same time, so as to cut the whole length of the side, then the line which the point describes on the surface of the cylinder as it passes, is a twisted line. </s><s id="id.57.0.01.07">This line is called the line of the screw. </s><s id="id.57.0.01.08">It is drawn on the cylinder thus: in a plane we draw two lines which contain a right angle, one of which we make equal to the side of the cylinder, the other of which we make equal to the circle of the cylinder, which is the circumference of its base. </s><s id="id.57.0.01.09">Between the ends of these two lines we join a line which is the line which subtends the right angle. </s><s id="id.57.0.01.10">We place the line equal to the side of the cylinder on the side of the cylinder and we bend the other side, which contains the right angle, round the cylinder; the line subtending the right angle is twisted round the cylinder on the line we called "the twisted". </s><s id="id.57.0.01.11">We can divide the side of the cylinder into any number of parts we wish, and draw on every part a twisted line of the kind we have mentioned. </s><s id="id.57.0.01.12">We shall then have drawn on the cylinder many twisted lines; and we call the line which is twisted round once "one turn", and this is the line which joins the ends of the lines [containing] the right angle. </s></p></chap> <chap n="58"><p type="head"><s id="id.58.0.00.01">Section 58 </s></p> <p><s id="id.58.0.01.01">If we make a cut in the cylinder on this line so that it is possible to adjust to it a solid circle we have completed construction. </s><s id="id.58.0.01.02">We operate this screw in the way I describe. </s><s id="id.58.0.01.03">Its ends are made circular adjusted in two circular holes made in objects like poles so that it turns easily on them. </s><s id="id.58.0.01.04">Now let there be a piece of wood with a channel hollowed in the middle of it above the screw; let the second position of the screw be fitted in that channel so that one end of the second position of the screw is in the channel of the screw and the other in the channel which we have said is in the ruler. </s><s id="id.58.0.01.05">When people wish to move weight with this apparatus, they take a rope and tie one end of it to the weight and the other to the Tulos; they make holes in the head of the screw in which an Usqutali is placed. </s><s id="id.58.0.01.06">This is pulled downwards and the Tulos passes along the channel by means of a twist line; pulls the rope and so moves the weight; it is possible instead of the Usqutali for us to make a second handle at the end of the screw on the outside by one of the poles. </s><s id="id.58.0.01.07">If the screw is moved in this way the weight is drawn. </s><s id="id.58.0.01.08">Now the twisted cut in the screw can sometimes be made square and sometimes lentil-shaped. </s><s id="id.58.0.01.09">When the channel in it is cut at right angles it is square, and when an oblique cut is made and both cuts meet in one line it is lentil-shaped. </s><s id="id.58.0.01.10">The first is called "square", the second "lenticular". </s></p></chap> <chap n="59"><p type="head"><s id="id.59.0.00.01">Section 59 </s></p> <p><s id="id.59.0.01.01">Whenever the screw is used on its own, this is its description. </s><s id="id.59.0.01.02">But it is possible to add another power to it in which case its description is different; I mean that it is possible to add to it something like the wheel which I called "surrounding" which turns on an axle. </s><s id="id.59.0.01.03">We imagine a toothed wheel which turns on an axle. </s><s id="id.59.0.01.04">Let the screw be placed alongside the wheel, standing either at right angles to the plane of the ground or parallel to it. </s><s id="id.59.0.01.05">Let its groove be fitted to the teeth of the wheel, and let its ends revolve in circular holes in two uprights, resembling poles as we said earlier; let the end of the screw protrude outside the upright resembling a pole; round this end let there be either a handle or holes into which Usqutali are introduced, by which it can also be turned. </s><s id="id.59.0.01.06">Furthermore when we wrap the ends of the ropes which are tied to the weight round the axle on both sides of the wheel and turn the screw, the toothed wheel will turn and we shall move the weight. </s></p></chap> <chap n="60"><p type="head"><s id="id.60.0.00.01">Section 60 (H1) (Figs. 45, 46, 47) </s></p> <p><s id="id.60.0.01.01">We have now explained the composition and use of the five powers we mentioned earlier. </s><s id="id.60.0.01.02">The reason why each one of them moves a great weight by means of little power has been explained by Heron in his books on Mechanics. </s><s id="id.60.0.01.03">To follow this we now present the mechanical devices given in the third section of Heron's book which are suited to lightening the burden of work when we wish to move a great weight. </s><s id="id.60.0.01.04">He stated that objects which are to be drawn along the ground may be drawn on a "dabbaba"; the dabbaba is something composed of squared wood, the edges of which are firmly fixed and bent upwards; the weight is placed upon it. </s><s id="id.60.0.01.05">The ends of ropes, or the compound pulley, are tied to the ends of the squared wood and are then pulled either by hand or by means of what is called the Arghatus. </s><s id="id.60.0.01.06">When the Arghatus turns, it draws the dabbaba along the ground if Usqutali or planks are placed beneath it. </s><s id="id.60.0.01.07">If the weight is small Usqutali are used, if it is great then planks are used as it is easy to draw them. </s><s id="id.60.0.01.08">This is done as if the Usqutali twist they are unsafe because the weight will charge onwards. </s><s id="id.60.0.01.09">There are those who use neither Usqutali nor planks but set solid wheels in front of it and pull it. </s></p></chap> <chap n="61"><p type="head"><s id="id.61.0.00.01">Section 61 (H2) (Fig. 48) </s></p> <p><s id="id.61.0.01.01">He has asserted that among the devices by which weight can be lifted are those with two sections, those with three sections, and those with four sections. </s><s id="id.61.0.01.02">Those with one section are composed as I describe. </s><s id="id.61.0.01.03">A strong piece of wood is taken, taller than the height of the place to which we wish to lift the weight. </s><s id="id.61.0.01.04">[Even] if the wood itself is strong we put a rope round it, make it fast and wrap the rope round it as though it were woven, making the distances between each turn no more than four spans; in this way the wood will be stronger. </s><s id="id.61.0.01.05">Further more the rope which is bound round will serve as steps for the craftsmen who wish to climb to the top of the piece of wood, and their work will be easier. </s><s id="id.61.0.01.06">If however the wood is not strong the apparatus should be constructed from a number of pieces of wood. </s><s id="id.61.0.01.07">We have to assess the weight of the object to be carried and not let the wood be too weak to carry it. </s><s id="id.61.0.01.08">The section is then set up at right angles on some piece of wood. </s><s id="id.61.0.01.09">Three or four ropes are tied to the end of the section; these are stretched and tied in firmly fixed places so that the piece of wood does not incline when pulled, as the stretched ropes hold it. </s><s id="id.61.0.01.10">A compound pulley is attached to the top of it, and is tied to the weight to be carried, and it is drawn either by hand or with Arghatus until, when the weight to be carried has been raised and we want to put the stone on the wall, or wherever we wish, we untie one of the ropes which are on the opposite side from the weight and incline the section. </s><s id="id.61.0.01.11">We put an Usqutali under the object to be carried, on the side on which the sling is not wrapped round the stone. </s><s id="id.61.0.01.12">Then we slacken the compound pulley until the weight carried lies on that Usqutali. </s><s id="id.61.0.01.13">The sling is then loosened and the weight is moved with the lever so that it comes to the place we wish. </s><s id="id.61.0.01.14">Then people draw the piece of wood the section is on, with ropes, grasping them in their hands and they take it round to another part of the building after the stretched ropes have been loosened; then they tie them again and use it. </s></p></chap> <chap n="62"><p type="head"><s id="id.62.0.00.01">Section H3 (Fig. 49) </s></p> <p><s id="id.62.0.01.01">The apparatus in which there are two sections is composed as we describe. </s><s id="id.62.0.01.02">We make a square piece of wood which is called Udus and it is placed on the piece of wood on which is the section. </s><s id="id.62.0.01.03">We make its upper side incline slightly towards the interior of the Udus. </s><s id="id.62.0.01.04">We make two cuts in the Udus into which we put the ends of the section which are called Qula. </s><s id="id.62.0.01.05">On the upper side of the sections is placed that tool called Qalansuwa so as to join the two sections to one another from both sides. </s><s id="id.62.0.01.06">Another transversal is placed on them both from which is suspended the manjana of the compound pulleys. </s><s id="id.62.0.01.07">The other manjana is attached to the sling in which the stone is, and the ropes are either drawn by hand or by donkeys. </s><s id="id.62.0.01.08">The weight to be carried is then raised to where we consider fit. </s><s id="id.62.0.01.09">These sections should be tied with ropes and should be made reliable according to our earlier description. </s><s id="id.62.0.01.10">When the stone is put in its place the Udus is moved to another part of the building in which it is necessary to raise stones. </s></p></chap> <chap n="63"><p type="head"><s id="id.63.0.00.01">Section H4 </s></p> <p><s id="id.63.0.01.01">The apparatus in which there are three sections is as I describe. </s><s id="id.63.0.01.02">Three sections are made inclining to one another which interlock with one another and the manjana of the compound pulley is attached at the point where they interlock. </s><s id="id.63.0.01.03">The other manjana is tied to the weight to be raised, and the stone is raised as we have described. </s><s id="id.63.0.01.04">The base of this device with three sections is more reliable than others, only it is not suitable in many places, and is only suitable when the weight to be carried has to be lifted in the middle of the apparatus. </s></p></chap> <chap n="64"><p type="head"><s id="id.64.0.00.01">Section H5 (Fig. 50) </s></p> <p><s id="id.64.0.01.01">The apparatus which has four sections is suitable for very great weight, for in it four sections are set up on the ground, and the sides of its base are parallel, and its size is such that it is possible for the stone to be manipulated easily within it. </s><s id="id.64.0.01.02">At the ends of the four sections are placed transversals which join them. </s><s id="id.64.0.01.03">Then other transversals are placed above the highest and the lowest transversals, and by means of these the apparatus is made reliable. </s><s id="id.64.0.01.04">The same is done in the whole apparatus. </s><s id="id.64.0.01.05">The compound pulley is tied in the middle and the weight is lifted by it. </s><s id="id.64.0.01.06">It is necessary that nails, pegs, and holes should be avoided in the apparatus, and especially in the places on which weight is carried. </s><s id="id.64.0.01.07">It is essential that the wood of the apparatus be bound only with ropes. </s></p></chap> <chap n="65"><p type="head"><s id="id.65.0.00.01">Section H6 (Fig. 53) </s></p> <p><s id="id.65.0.01.01">Because the sling in which the stone is carried is underneath it, and this makes it difficult to put the stone on the building easily, the thing called the Alaq is made. </s><s id="id.65.0.01.02">We make the base of the stone the figure ABGD and we draw a figure HZHT. </s><s id="id.65.0.01.03">Let line HZ be parallel to line HT and line KL to line MN. </s><s id="id.65.0.01.04">Let both HZ HT be greater than KL MN; let HH be equal to KM and let a place corresponding to this figure we have mentioned be hollowed out and let the extent to which it is hollowed out be in accordance with the thickness of the stone. </s><s id="id.65.0.01.05">Let the figure HZ HT be at right angles to all the sides. </s><s id="id.65.0.01.06">Let the figure KLMN be parallel at the lower area so as to resemble the shape of a female Balaqinus. </s><s id="id.65.0.01.07">Let the narrow side of it be equal to KLMN and the broad side be equal to HZHT. </s><s id="id.65.0.01.08">Let there also be a male Balaqinus of iron fitted to the female one, and on its upper side which is the narrow one let there be a ring in it, either attached to it or formed from its own substance. </s><s id="id.65.0.01.09">It is lowered on HZHT so as to settle firmly in the deep place. </s><s id="id.65.0.01.10">Then it is brought up to the female Balaqinus so that it cannot be drawn so as to be pulled out upwards. </s><s id="id.65.0.01.11">A chock of wood is made and is placed at HZHT which prevents the male Balaqinus from coming out. </s><s id="id.65.0.01.12">We make the Balaqiniat several. </s><s id="id.65.0.01.13">Then the manjana of the compound pulleys are attached to the rings as they were attached to the sling by which the stone was lifted. </s><s id="id.65.0.01.14">The weight is raised so that we can put it in the necessary place without hindrance. </s><s id="id.65.0.01.15">Then the chocks are lifted so the Balaqinu come out, and are fitted on to another stone. </s></p></chap> <chap n="66"><p type="head"><s id="id.66.0.00.01">Section H7 </s></p> <p><s id="id.66.0.01.01">The stone is also lifted by crabs having three sections and having four sections. </s><s id="id.66.0.01.02">Things resembling chocks are put on the feet of the crab and their ends are fitted in cuts in the sides of the stone. </s><s id="id.66.0.01.03">Isfinat are put above the head of the crab in the firm parts of the sections on their upper sides where they have depth. </s><s id="id.66.0.01.04">If then the weight pushes the chocks outwards because of its weight, the isfin will provide contrary pressure and the weight will be firmly fixed on it. </s><s id="id.66.0.01.05">The compound pulley is then attached to the bar which passes through the section(s). </s><s id="id.66.0.01.06">Manjan one is attached above the upper part of the compound pulley and the stone is drawn in this way. </s></p></chap> <chap n="67"><p type="head"><s id="id.67.0.00.01">Section H8 (Fig. 51) </s></p> <p><s id="id.67.0.01.01">This can be done in another way which is easier than the one which we mentioned, and more reliable. </s><s id="id.67.0.01.02">It is this. </s><s id="id.67.0.01.03">We make the base of the stone ABGD. </s><s id="id.67.0.01.04">Let there be in it a cut HZHT so as to be like a plane with parallel sides, and let it be of moderate depth. </s><s id="id.67.0.01.05">In the places in which are sides HZ HT are made re-entrant depths. </s><s id="id.67.0.01.06">Let there be the solid portion above them. </s><s id="id.67.0.01.07">Let there be two levers of iron resembling a right angle K L, and let there be at the top of them a ring or hole so that when they are put together and combined it is possible to introduce into the hole(s) a strong bar which will hold these two levers firm. </s><s id="id.67.0.01.08">Between these two levers is placed a third lever in the upper part of which there is also a hole, and the bar which we have mentioned passes through it and the other two holes. </s><s id="id.67.0.01.09">Let the bar have a tip on both the sides, so that when the bar is withdrawn and removed, the two levers which resemble a right angle descend. </s><s id="id.67.0.01.10">Their two hooked ends enter the deep places which were hollowed out underneath. </s><s id="id.67.0.01.11">The third lever is placed between these two so that the depth of HZHT is filled by the three of them and there are no gaps and no looseness in it. </s><s id="id.67.0.01.12">Furthermore the bar enters the three holes and the manjanun of the compound pulley is attached to it. </s><s id="id.67.0.01.13">The other manjanun is attached to a higher apparatus and when we do that, the stone is raised. </s><s id="id.67.0.01.14">It is made reliable by what it grasps. </s><s id="id.67.0.01.15">For the lever placed between the levers resembling a right angle, prevents them from coming out. </s><s id="id.67.0.01.16">When the stone has been raised in this way, and then lowered, so that it is possible for the builder to remove the bar and the middle lever, then the two hooked levers are removed and the apparatus is moved to another place. </s><s id="id.67.0.01.17">The iron used in this apparatus should not be very dry in order that cracks {do} not appear in it, and it should not be very soft so as not to fold because of the weight of the stone. </s><s id="id.67.0.01.18">But it is necessary that it be between the two states mentioned and that in it there should be no soldering nor cracks, when it is in operation. </s><s id="id.67.0.01.19">For if that is the case one cannot be sure that the stone will not fall and destroy the craftsmen. </s></p></chap> <chap n="68"><p type="head"><s id="id.68.0.00.01">Section H9 </s></p> <p><s id="id.68.0.01.01">That is the information we have on the lifting of weight. </s><s id="id.68.0.01.02">It has been mentioned that all the devices must be within the limitations of place and time and such necessary materials as are present. </s><s id="id.68.0.01.03">An example of this is the practice of some people in situations where they employ devices when they wish to lower a great weight from the top of a lofty mountain which has a steep cliff. </s><s id="id.68.0.01.04">The ancients used to lower that on beasts only which is dangerous insofar as the weight often tilted thus destroying the beasts and breaking the carts with the force of its descent. </s><s id="id.68.0.01.05">Some people devised a method in which they made two smooth tracks from the position from which they wished to lower the weight until they reached the bottom of the hill, and they made two carts on each of which were four wheels. </s><s id="id.68.0.01.06">They put one of the carts, the one which carries the weight and lowers it and is propelled from above, on one track, and they put the other cart on the other track at the bottom. </s><s id="id.68.0.01.07">They attached pulleys in a fixed place between the two tracks at the top, tied ropes to the cart which carries the weight, and threaded them through the pulleys. </s><s id="id.68.0.01.08">Then they led them back to the lower cart. </s><s id="id.68.0.01.09">They put in the lower cart stone of moderate size such that its combined weight is slightly less than the weight which they need to lower. </s><s id="id.68.0.01.10">They pull the cart the weight is in with beasts. </s><s id="id.68.0.01.11">The lower cart then by its correspondence to the other cart lowers the weight slowly. </s></p></chap> <chap n="69"><p type="head"><s id="id.69.0.00.01">Section H 10 (Fig. 52) </s></p> <p><s id="id.69.0.01.01">It is asserted [that] some people wanted to set up very great cylinders on their bases by this device, and in the same way cylinders with defined tops: </s><s id="id.69.0.01.02">They used to tie ropes to the end of the cylinder and attach them on pulleys fixed to a place higher than the weight like a tower. </s><s id="id.69.0.01.03">They would attach at the other end things large enough for stones to be put in, and things like boxes would hold them. </s><s id="id.69.0.01.04">Then they put in a greater quantity of stone than the weight which is to be lifted so as to overcome it. </s><s id="id.69.0.01.05">Then they put the stone in the boxes gently and so set up the cylinder after they had tied the base to wood or with ropes. </s><s id="id.69.0.01.06">They used to pull these to the two sides in order that the cylinder should stand on its own base. </s></p></chap> <chap n="70"><p type="head"><s id="id.70.0.00.01">Section H 11 </s></p> <p><s id="id.70.0.01.01">Some people, from what is asserted, wished to draw heavy things on the sea by this device; they made a raft of square wood which they made reliable with nails and pegs. </s><s id="id.70.0.01.02">They put walls on it and lowered it to the place of anchorage where they wished to load the weight. </s><s id="id.70.0.01.03">Before that they would attach beneath it sacks filled with sand and the mouths of the sacks were tied; and they would bring two ships up to the raft. </s><s id="id.70.0.01.04">Then they used to bind the raft to the two ships by its walls which were on the two sides, with ropes. </s><s id="id.70.0.01.05">Then they used to put the weight on the raft and untie the mouths of the sacks so that the sand went out of them. </s><s id="id.70.0.01.06">Then they set the raft in motion with the two ships, with the load on it. </s></p></chap> <chap n="71"><p type="head"><s id="id.71.0.00.01">Section H 11a </s></p> <p><s id="id.71.0.01.01">It is asserted that some people lifted stone from the depths of the sea in this way: they tied loops in the ends of ropes and placed them round the stone. </s><s id="id.71.0.01.02">They put bars of iron in the loops and tied them with other ropes so that the loops should not twist. </s><s id="id.71.0.01.03">Then they brought two ships near to one another after that, as we said where we described the raft, and they tied them with ropes and filled them with water up to the place of the covered sides which are in the ship. </s><s id="id.71.0.01.04">Then they set up on those tied ropes the apparatus to which the compound pulley is attached, and they raised the stone with it. </s><s id="id.71.0.01.05">Then they emptied the water which was in ships which rise and the stone is raised. </s><s id="id.71.0.01.06">Then they drew up the stone with the compound pulley and put it on the mats which are in the two ships. </s><s id="id.71.0.01.07">Then they drew the two ships to the land. </s><s id="id.71.0.01.08">They extracted stones from the depth of the water on something which they placed beneath it. </s></p></chap> <chap n="72"><p type="head"><s id="id.72.0.00.01">Section H12 </s></p> <p><s id="id.72.0.01.01">It is asserted that some people set up straight walls which had inclined because of earthquakes in that they dug a hole in the ground in the area towards which the wall was leaning and made it run the whole length of the wall. </s><s id="id.72.0.01.02">In it they put square pieces of wood a short distance away from the wall. </s><s id="id.72.0.01.03">Then they set up pieces of wood between that wood already placed there and the wall. </s><s id="id.72.0.01.04">They attached compound pulleys to the ends of the upright pieces of wood, put their ropes on an Arghatus and pulled them. </s><s id="id.72.0.01.05">It used to turn out that these matters triumphed by their power. </s><s id="id.72.0.01.06">The movement used not to be fast but the wall would return little by little as the power was applied bit by bit, and would become upright. </s><s id="id.72.0.01.07">When it was upright they would leave it for a few days so that the stones could settle on one another. </s><s id="id.72.0.01.08">Then they would untie the bits of apparatus from it when that had happened, and leave the wall. </s><s id="id.72.0.01.09">The End. </s></p> <p><s id="id.72.0.02.01">Pappus's introduction to Mechanics finishes. </s><s id="id.72.0.02.02">Much praise be to God and may God bless Muhammad and his pure family. </s><s id="id.72.0.02.03">This copy was annotated from a copy after it had been written. </s><s id="id.72.0.02.04">Ahmad ibn Mohammad ibn 'Abd al-Jalil wrote from a copy of the Banu Musa in Rabi al-akhar of year 352 and this writer is the Sheikh as-Sijzi. </s><s id="id.72.0.02.05">The completion of this copy was in the middle of Jumada l'ula of the lunar year 688 at Maragha in the Yaqawiyya school. </s></p></chap></body> <back></back> </text> </archimedes>