Technikatörténeti szemle 10. (1978)

A MÉRÉS ÉS A MÉRTÉKEK AZ EMBER MŰVELŐDÉSÉBEN című konferencián Budapesten, 1976. április 27–30-án elhangzott előadások II. - Maula, E.: A szögmérés kezdetei

But how will the desired position of the instrument be found ? In other words, what is the heuristic insight succouring the proof? Speaking about the connection between Menaechmus's solution and the so-called Platonic solution by means of a „machine" consisting of two right-angled rulers, Heath (op. cit., i, p. 257) notes: „That it is possible for the machine to take up the desired position is clear from the figure of Menaechmus, ...., although to get it into the required position is perhaps not quite easy". We can, however, suggest even two procedures leading to the required position, and at least one of them explains the alleged confusion of a discrete proportion with a continuous one, while both explain the use of ,,curved lines" occurring in the heuristic part but disappearing in the proof. It may be added that these methods were found in a ,.practical" way, i.e. by the use of the reconstructed instrument. The analogy between these heuristic ideas and the striking disappearance of the auxiliary parameter n in dealing with the combinations of spherical motions in the theory of homocentric spheres (by means of the methods of analysis and synthesis) is conspicuous, however. The Spider's Dance In the first method the measuring unit is released from the pivot, and the plumbstring (but not the auxiliary trunk) of thearachne is needed. We illustrate the steps of the method by simplified diagrams. 1° One makes the initiative (first) guess (x

1

), i.e. one makes the arms of the instrument go through the points A, B and keeps the apex M

1 of the right angle AMN on the axis of the enoptron. Now the apex of the other right angle, N

x

, does not meet the other axis (Fig. 11). A perpendicular is dropped (by means of the plumb-string) from N

x to MjB. From similar triangles AO:M

x

O = MjOiJXjQj. = N^BO, or, in other words, AO:x

x = (

Xl

—e^y^ = y^BO + e^, where e

t = 00

1 is the first error term. Thus x

x is too great by e

1; as compared with the final, required solution. 2°The second guess is x

2 == x

1

—e

1

. The situation is shown in the figure (Fig. 12). From similar triangles AO :M

2

0 = M

2

0

2 :N

2

0

2 = N

2

0

2 :B0

2 or, in other words, AO:x

2 = (x

2 + e

2

):y

2 = y

2

:(BO—e

2

), where e

2 < e

x is the second error term. Thus x

2 is too small by e

2

. 3° The third guess is x

3 = x

2-r-e

a

. The situation is shown in the figure (Fig. 13). From similar triangles AO:M

3

0 = M

3

0

3

:N

3

0

3 = N

3

0

3

:B0

3 or, in other words, AO:x

3 = (x

3

—e

3

):y

3 = y

3

:(BO + e

3

), where e

3 < e

2 < e

l is the third error term. 4° The next guess is x

4 = x

3

—e

3

. The same procedure will, in principle, be repeated ad infinitum, until e„ = 0 and the so—called Platonic solution is achieved. These are the main points to be observed. First, the same rationale is discernible at every step of the procedure (the use of similar triangles), and it is the same in the final proof—situation also. Second, the method converges whatever the initial guess. Third, the procedure can indeed be indicated by means of a „eurved line", on which the points N

n are situated. Noting that if, for instance, AO = 1 and OB = 2, then l:OM = (x + OM):y = y:(2—x), whence OM = (y

2

—2x+x

2

):(2—x), the equation of this ,,curved line" (of the fourth degree) can readily be obtained. Third, at each step of the procedure, when the proportionality of the sides of similar triangles