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In Aristotelian kinematics, it is necessary to distinguish the following in speaking of X traveling from A to B:
1. the distance or line AB which X travels over,
2. the time in which X travels AB,
3. the movement, which is the line AB understood as the line over which X travels in the given time.

Note that the movement is here conceived as a distance-traveled by X in an individual time, e.g. from Los Angeles to New York from 12:00 PM to 5:00 PM (Pacific time), 1 Sept. 2000. The distance-traveled by X from Los Angeles to Kaua'i one year later may be an equal distance-traveled in an equal time, but the distances-traveled and the times of travel are not the same. The distance traveled two years later from Los Angeles to New York may be the same distance, but it is not the same movement.

For convenience, the following translations will be used:

Greek literal translation translation used
is carried moves
carrying movement (a distance moved)
the carried object the moved
is changed changes
change change (a quantity of the change)
the changed object the changed

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Mechanica 1 858b1-10

First then the properties of the balance cause puzzlement: what is the reason why larger balances are more precise than smaller. And the principle of this cause puzzlement: why in the circle does the line which stands further from the center move faster than that one near to it which changes by nearly the same force, namely the smaller. For the faster is said in two ways. If it traverses an equal place in a smaller time, we say that it is faster, and if in an equal time it traverses more. The larger describes in an equal time a larger circle. For the outer is larger than the inner. And the cause of these things is that the line describing the circle moves with two movements.

Brief comments:
The defintion is of 'faster'. The notion of 'velocity' plays no role here, or anywhere else in the treatise.
Let TA be the time that A travels distance DA, where the movement is DA,TA.
Let TB be the time that B travels distance DB, where the movement is DB,TB.

The defintions of 'faster' are, given DA,TA and DB,TB:
1. A moves faster than B if DA = DB and TA < TB.
2. A moves faster than B if TA = TB and DA < DB.

In the discussion, the author only uses definition 2.

Mechanica 1 858b10-23

This proof is commonly described as 'the parallelogram of velocities' or even as 'the parallelogram of forces'. Both descriptions are very anachronistic and inappropriate. The composed lines are movements.

The diagram in modern editions and translations of the Mechanica involves a parallelogram. There is no reason to think, however, that the intended figure is not a rectangle. Whether a parallelogram is required depends on how one interprets the diagram at 858a2-6 and the relation between this problem and the related case of the rhombus at ch. 23. You may select which you prefer.

(diagram 1: moving picture or fixed picture )

And so whenever something moves in some ratio, it is necessary that the moved moves on a straight-line, and it becomes a diameter of the figure which the composed lines in this ratio make. For let the ratio which the moved moves be what AB has to AG; and let AG move towards B, and also let AB move down towards HG. (figure 2) Let A move towards D, and the line AB towards E. And so if the ratio for the movement is what AB has to AG, it is necessary that AD also have this ratio to AE. Therefore the small quadrilateral is similar to the larger, so the same diameter is also of them, and A will be at Z. In the same manner, it will be proved wherever the motion is marked off. For it will always be on the diameter.

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Mechanica 1 858b23-33 (converse of the previous)

And so it is obvious that what is moved on the diameter must move in two movements in the ratio of the sides. For if it moves in some other ratio, it will not move along the diameter. If it moves with two motions in no fixed ratio in any time, it is impossible the movement be straight. For let it be straight. And so when this is posited as diameter, and when the sides are filled out, it is necessary that the moved move with the ratio of the sides. For this was proved earlier. Therefore, what is moved in no ratio in no time will not produce a straight-line. For if it moves in some ratio in some time, it is necessary that there be a straight movement in this time for the mentioned reasons.



Mechanica 1 858b34-859a5 (application to circular movement)

How we interpret 858b10-23 depends partly on how we interpret this argument. (diagram 2) Modern translations make the argument general and so tend to make the arc BEG different from a quadrant. They then need to emend the second sentence of the text. Here we treat arc BEG as a quadrant. This makes the second sentence mean that when the point moving on a straight line, BD or BK, arrives at G, the straight line on which it rides is now moved to being a radius that is perpendicular again. One suspects that the radius KB was perpendicular to BD but later is rotated to KG perpendicular to DG, so that the line on which the point moves is BD. You may select which you prefer.

(basic diagram 1) Thus it becomes circular, when moving in two movements in no fixed ratio in any time. That the line describing the circle moves with two movements together is obvious from these things and the fact that what is moved along a straight line arrives onto the perpendicular, so that the line from the center is again perpendicular. (diagram 3: moving picture or still picture) Let there be a circle ABG, and let the end B move to D. Then it arrives sometime at G. And so if it had moved in the ratio which BD has to DG, it would have moved on the diameter BG. And now, since it moves in no fixed ratio, it moves on the circumference BEG.

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Mechanica 1 858a5-859a18 (the same force moves a larger circle more than a smaller one, in length of motion)

If of two things moved from the same force one is pushed back more and the other less, it is reasonable that the one pushed back more changes more slowly than the one pushed back less, which is thought to happen in the case of larger and smaller of lines from the center describing circles. For since the end point of the smaller line is nearer to the fixed point than the end point of the larger, just as held back in the opposite direction, the end point of the smaller moves to the center more slowly. And so this happens to every line describing a circle, and it moves with its natural motion along the circular-arc, but with its unnatural motion to the side and the center. But the smaller line always changes with its unnatural change always greater. For it is controlled more since it is nearer to the center which holds it back.


Mechanica 1 859a18-b19 (the same force moves a larger circle more than a smaller one, in length of motion, geometrical/mechanical argument)

(general diagram)

And it is clear from these things that the smaller of the lines from the center describing circles changes with regard to the unnatural magnitude more than the larger.

(diagram 1) Let there be a circle BGDE, and another, a smaller, in this, CNMX, about the same center A. And let the diameters be extended, GD and BE in the larger, with MC, NX in the smaller . And let the oblong (i.e. rectangle) be filled out, DYRG. (diagram 2: moving or fixed) If, in fact, AB describing a circle will have come to the same point from where it set out for AE, it is clear that it moves to itself. Similarly AC will have come to AC as well. But AC moves slower than AB, just as was said, since the pushing back is greater and AC is held back more. (diagram 3 = general diagram) Let AQH be drawn, and from Q to AB let a perpendicular QZ be drawn in the circle. And again from Q let QW be drawn parallel to AB, and WU to perpendicular AB,* as well as HK. In fact lines WU and QZ are equal. Therefore, BU is smaller than CZ. For equal straight-lines when placed in unequal circles at right angles to the diameter cut off smaller segments than the they do in larger circles (see brief proof below), but WU is equal to QZ. In fact in as much time in which AQ moved CQ, in so much time the end point of BA has moved in the larger circle a distance larger than BW. For the natural movement is equal, but the unnatural is smaller: BU is smaller than ZC. But they ought to be proportional, as the natural magnitude is to the natural, so is the unnatural magnitude to the unnatural. Therefore, it has traversed a circular-arc, HB, larger than WB. But it is necessary that it traverse HB in this time. For it will be here whenever the unnatural magnitude happens to be pairwise proportional to the natural magnitude. If in fact the natural magnitude is larger in the larger circle, then the unnatural magnitude might instead occur here in only one way, in order for B to move with movement BH in the time in which point C moved CQ.** For here magnitude KH*** is natural for point B (for the line KH is perpendicular from H), but it is unnatural to KB. But as HK is to KB, QZ is to ZC. (diagram 4) But it is obvious if lines are joined from B, C to H, Q (since triangles BKH and CZQ are similar). (diagram 3 = general diagram) But if the movement which B makes is smaller or larger than HB, they will not be similar and the natural to the unnatural will not be pairwise proportional either.

*The manuscripts have: . The text may have been: (let WU be drawn perpendicular to AB).

**Bekker's text reads , which everyone translates as above. Nonetheless, this meaning is clearly forced but correct, so that the text needs some fixing, though perhaps not to the extent of Heath's, .

***Here too, everyone translated the received text in this way, although it is very forced. I read instead of the manuscripts (the center).

General note: What are the natural and the unnatural movements? The Oxford translation's diagram makes the natural motion sideways, but it should be downwards (as Heath or Blancanus). Hence, this diagram (after Blancanus) has the natural motiions be KH and ZQ. The unnatural motions are BK and CZ towards the center, i.e. towards line GD or NX. These are proportional: KH : ZQ = BK : CZ.

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Mechanica 1 859b19- 860a2 (why larger balances are more precise than smaller balances)

The reason why the point more distant from the center moves faster from the same force is clear through what was said, while the reason by larger balances are more precise than smaller is obvious from these things. For the rope becomes the center (since this stays put), while the lines from the center are on each side of the scale. And so from the same weight it is necessary that the end point of the scale move faster by as much as it is distant form the rope, and some weights placed on small balances are not clear in regard to perception, while in large ones they are clear. For nothing prevents it moving a magnitude smaller than would be obvious to sight. In the case of a large scale the same weight made a visible magnitude. And some are clear in both cases, but much more in the case of the larger ones due to the fact that the magnitude of the downward push from the same weight is much larger in the larger ones. And for this reason the dealers in purple use trickery in setting them up them for skimming by not placing the rope in the middle and by infusing lead into one side of the beam or by making the balance of wood towards the root where they tends to push down or if it has a knot. For it is heavier where the root of the wood is, but a knot is a sort of root.

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Lemma for Mechanica 1 859a18-b19: BU < CZ (proof after Heath)

(diagram 1) Given UW = ZQ, and UW and ZQ perpendicular to diameters BE and CM. (diagram 2) First extend UW to P and ZQ to L, to the respective circles. Thus PW intersects BE in circle BGED and QL intersects CM in circle CNMX. (diagram 3: alternating or single) By Euclid 3.35:
WU * UP = BU * UE
QZ * ZL = CZ * ZM

Since WU = UP = QZ = ZL,
BU * UE = CZ * ZM.

Hence, BU : CZ = ZM : UE.

If ZM < UE then BU < CZ.

We need to show that ZM < UE.
We do this by showing that AZ < AU.
(diagram 4) Join AQ and AW to form right triangles AZQ and AUW, where LZ = UW.

AW2 - AU2 = UW2 = LZ2 =AL2 - AZ2

Hence, AW2 + AZ2 = AL2 + AU2

Since AW2 > AL2, it follows that AZ2 < AU2,
and AZ < AU.

Since AM < AE and AZ < AU, then MA + AZ < UA + AE.

Hence, BU < CZ, so that BU < CZ.

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