What the Middle-Aged Galileo Told the Elderly Galileo: Galileo’s Search for the Laws of Fall

Abstract

Recent historiographic results in Galilean studies disclose the use of proportions, graphical representation of the kinematic variables (distance, time, speed), and the medieval double distance rule in Galileo’s reasoning; these have been characterized as Galileo’s “tools for thinking.” We assess the import of these “tools” in Galileo’s reasoning leading to the laws of fall ($v^{2} \propto D$ and $v \propto t$). To this effect, a reconstruction of folio 152r shows that Galileo built proportions involving distance, time, and speed in uniform motions, and applied to them the double distance rule to obtain uniformly accelerated motions; the folio indicates that he tried to fit proportions in a graph. Analogously, an argument in Two New Sciences to the effect that an earlier proof of the law of fall started from an incorrect hypothesis (v ∝ D) can be recast in the language of proportions, using only the proof that v ∝ t and the hypothesis.

Appendices

Appendix A: The Proof of the Parabola $\varvec{v^{2} = 2gD}$ (Folio 152r)

In this appendix, the proof in folio 152r is reproduced,63 and recast in the notation in figure 5 and table 2.

First Paragraph

Let $ba$ be to $ad$ as $da$ to $ac$ and let $be$ be the gradus velocitatis in $b$.

$$\left[ {\frac{ba}{ad} = \frac{da}{ac}\;and\;be = v_{b} } \right]$$

In the paragraph, $ad$ is defined as the mean proportional between the distances $D_{ac} = ac$ and $D_{ab} = ab$:

$$ad \equiv D_{mp} = \sqrt {D_{1} D_{2} } \equiv \sqrt {ba \times ca}$$

That $be = v_{b}$ follows from the construction of the motions in table 1 or table 2, and the double distance rule.

Second Paragraph

And let $be$ be to $cf$ as $ba$ to $ad$, then $cf$ will be the gradus velocitatis in $c$:

$$\left[ {\frac{be}{cf} = \frac{ba}{ad}\;{\text{then}}\;cf = v_{c} } \right]$$

With the notation in table 2, this is:

$$\frac{{v_{b} }}{cf} = \frac{{D_{ab} }}{{D_{mp} }}\;{\text{then}}\;cf = v_{c}$$

(11)

The result follows from the construction in table 2, and the double distance rule.

Third Paragraph

Since therefore as $ca$ is to $ad$, $cf$is to $be$, also $\square ac$ will be to $\square ad$ as $\square cf$ to $\square be.$

$$\left[ {\frac{ca}{ad} = \frac{cf}{be}\;{\text{then}}\;\frac{{ac^{2} }}{{ad^{2} }} = \frac{{cf^{2} }}{{be^{2} }}} \right]$$

With the notation in table 2, this is:

$$\frac{{D_{ca} }}{{D_{mp} }} = \frac{{v_{c} }}{{v_{b} }} {\text{then}} \frac{{D_{ca}^{2} }}{{D_{mp}^{2} }} = \frac{{v_{c}^{2} }}{{v_{b}^{2} }}$$

(12)

The result follows, making $v_{c} = cf$ (equation 11).

Fourth Paragraph, Part 1

But as $\square ca$ to $\square ad$, $ca$ is to $ab$;

$$\left[ {\frac{{ca^{2} }}{{ad^{2} }} = \frac{ca}{ab}} \right] .$$

The result follows from the definition of mean proportional:

$$\frac{{D_{ca}^{2} }}{{D_{mp}^{2} }} = \frac{{D_{ac} }}{{D_{ab} }}$$

(13)

Fourth Paragraph, Part 2

But therefore as $ca$ to $ab$, $\square cf$ is to $\square be$. Therefore, the points $e$, $f$ are on a parabola

$$ \left[ {\frac{ca}{ab} = \frac{{cf^{2} }}{{be^{2} }}} \right] $$

From equation 12 and equation 13:

$$\frac{{D_{ca} }}{{D_{ab} }} = \frac{{v_{c}^{2} }}{{v_{b}^{2} }}$$

(14)

Appendix B: The Proof of the Straight Line v = gt (Folio 91v)

In this appendix, the proof in folio 91v is reproduced,64 and recast in the notation in figure 3 and table 1.

The folio starts with a proof for the composition of velocities. Then Galileo proposes a problem:

In motion from rest the moment of velocity and the time of this motion are intensified in the same ratio

$$\left[ {{\text{to prove that}}:\frac{{v_{c} }}{{v_{b} }} = \frac{{t_{ac} }}{{t_{ac} }}} \right]$$

For let there be a motion through ab from rest in a, and let an arbitrary point c be assumed; and let it be posited that ac is the time of fall through ac, and the moment of the acquired speed in c is also as ac,

$$\left[ {t_{ac} = ac\quad v_{c} \propto ac} \right]$$

[Explanation: in uniform motion, for fixed $v_{c}$, it is $d \propto t$. Then $d_{1} \equiv cd = 2D_{ac}$ ($v_{c}$ constant on $cd$), the time is $t_{ac} \propto d_{1} = 2ac$. This is the “conflation” observed by Palmerino.]

and assume again any point $b$: I say that the time of fall through $ab$ to the time through $ac$ will be as the moment of velocity in $b$ to the moment in $c$

$$\left[ {\frac{{t_{ab} }}{{t_{ac} }} = \frac{{v_{b} }}{{v_{c} }}} \right]$$

Let $as$ be the mean [proportional] between $ba$ and $ac$;

$$\left[ {as = \sqrt {ba \times ac} } \right]$$

and since the time of fall through $ac$ was set to be $ac$, as will be the time through $ab$: it thus has to be shown that the moment of speed in $c$ to the moment of speed in $b$ is as $ac$ to $as.$

[Explanation: because it is $\frac{{t_{ac} }}{{t_{ab} }} = \frac{{d_{1} }}{{d_{2} }}$, it has to be proved that $\frac{ac}{as} = \sqrt {\frac{ac}{ab}} = \sqrt {\frac{{d_{1} }}{{d_{2} }}} = \frac{{v_{b} }}{{v_{c} }}.$]

The Proof

First Paragraph

Assume the horizontal [line] $cd$ to be double $ca$, but $be$ to be double $ba$: it follows from what has been shown, that the [body] falling through $ac$, deflected into the horizontal $cd$, will traverse $cd$ in uniform motion in an equal time as it also traversed $ac$ in naturally accelerated motion; and, similarly, it follows that $be$ is traversed in the same time as $ab$: but the time of $ab$ itself is $as$: therefore, the horizontal [line] $be$ is traversed in the time $as$

$$\left[ {cd = 2ca,be = 2ba} \right]$$

Second Paragraph

Let $eb$ be to $bl$ as the time $sa$ is to the time $ac$; and since the motion through $be$ is uniform, the space $bl$ will be traversed in the time $ac$ according to the moment of speed in $b$: but according to the moment of speed in $c$, in the same time $ac$ the space $cd$ will be traversed; but the moments of speed are to one another as the spaces, which according to these moments are traversed in the same time: therefore the moment of speed in $c$ is to the moment of speed in $b$ as $dc$ to $bl$.

The text states relations among the uniform motions that follow from table 7:

$$\frac{cd}{bl} = \frac{{v_{c} }}{{v_{b} }}\;{\text{and}}\;\frac{bl}{be} = \frac{{t_{ac} }}{{t_{ab} }}.$$

(15)

Table 7 Double distance rule in the accelerated motion in figure 3

When Galileo refers to ac and as also as time, he is using what Palmerino calls “conflation” of coordinates. This shows that Galileo requires a separated vertical line to represent time. In the Two New Sciences, he uses the coordinates speed versus time, and an independent line for distance.

Third Paragraph, Part 1

But as $dc$ to $be$, so are their halves, i.e., $ca$ to $ab:$

$$\left[ {\frac{dc}{be} = \frac{2ca}{2ab}} \right].$$

The uniform motions are associated with uniformly accelerated motions, using the double distance rule:

$$\frac{{D_{ac} }}{{D_{ab} }} = \frac{ac}{ab} = \frac{dc}{be}$$

(16)

Third Paragraph, Part 2

But as $eb$ to $bl$, so $ba$ to $as$:

$$\left[ {\frac{eb}{bl} = \frac{ba}{as} {\text{that is }}\frac{{t_{ab} }}{{t_{ac} }} = \frac{ba}{as} } \right].$$

This result uses the double distance rule: $\frac{{d_{2} }}{{d_{mp} }} = \frac{{2D_{ab} }}{{2D_{mp} }}$; from equation 16 and table 7

$$\frac{{D_{ba} }}{{D_{mp} }} \equiv \sqrt {\frac{{D_{ba} }}{{D_{ca} }}} = \frac{{t_{ab} }}{{t_{ac} }}.$$

(17)

Third Paragraph, Part 3

Therefore, by the same [ex aequali], as $dc$ to $bl$, so $ca$ to $as$: that is, as the moment of speed in $c$ to the moment of speed in $b$, so $ca$ to $as$,

$$\left[ { \frac{dc}{bl} = \frac{ca}{as} {\text{that is}} \frac{{v_{c} }}{{v_{b } }} = \frac{ca}{as}} \right].$$

This is analogous to paragraph 3, part 2: from the double distance rule: $\frac{{d_{1} }}{{d_{mp} }} = \frac{{2D_{1} }}{{2D_{mp} }}$; from equation 16 and table 7

$$\frac{{D_{ca} }}{{D_{mp} }} \equiv \sqrt {\frac{{D_{ba} }}{{D_{ca} }}} = \frac{{v_{c} }}{{v_{b} }}.$$

(18)

Third Paragraph, Part 4

That is, the time through $ca$ to the time through $ab$. Quod erat demonstrandum

$$\left[ {\frac{{v_{c} }}{{v_{b} }} = \frac{{t_{c} }}{{t_{b} }}} \right].$$

This follows from equation 17 and equation 18.