How could Horrocks have measured the AU?

Question

I have always understood that the great historical significance of the transits of Venus, and the reason for the expeditions mounted to observe it, were that, by observing it simultaneously from two distant locations, the absolute distance to the Sun could be measured. But I read in several sources, including some of my texts and two articles in Wikipedia, that Jeremiah Horrocks's observations of the 1639 transit of Venus allowed him to

make an estimate of the distance between the Earth and the Sun, now known as the astronomical unit (AU)

Horrocks's made his observations from a single location, and thus could not have been using parallax (as was done 130 years later) to arrive at his estimate.

Without knowledge or the sizes of both the Sun and Venus, how could he have performed the necessary calculations? Was he simply using the " well-informed guess as to the size of Venus" mentioned in the Wikipedia article. If so, what made it "well-informed"

Answer 1

I guess there is an explanation at http://en.wikipedia.org/wiki/Transit_of_Venus,_1639#Results - "Horrocks tentatively... proposed a law which stated that all planets (with the exception of Mars) would be the same angular size when viewed from the Sun, this being 28 arc seconds. This ... led Horrocks to the false conclusion that the distance between each planet and the Sun was about 15,000 times its radius. Thus he estimated the distance from the Earth to the Sun to be approximately 60 million miles (97 million km)"

Answer 2

Horrocks did not, in fact directly use his Venus transit observations to measure the AU, but to confirm a relationship that he postulated between planetary radii and orbital distance, using which the AU could calculated simply from an estimate of the radius of Earth.

During the early 17th century it was widely believed that the size of a planet was related in some way to its distance from the Sun. As expressed by Kepler (1618):

Nothing is more in concord with nature than that the order of magnitude should be the the same as the order of the spheres, so that among the six primary planets, Mercury should ... obtain the most narrow sphere; that next to Mercury should be Venus, which is larger, but still smaller than Earth's ...

The significance of Horrocks's (1639) Venus transit observation, combined with earlier observations of Mercury and the contemporary estimates of the relative sizes of the orbits of the Mercury, Venus and Earth, is that it allowed him to be more explicit¹:

All planets (with the exception of Mars) are the same angular size when seen from the Sun, this size being 28 seconds of arc.

From this "law"² (sometimes called "Horrocks Law") arriving at an estimate of the AU is a matter of simple trigonometry, involving only the Earth's radius,³ $R_E$, since if the Earth subtends an angle of $28''$ when viewed from the Sun

$$R_E / 1 \text{AU} = \tan (28''/2)$$

so that

$$1 \text{AU} = (\tan 14'')^{-1}R_E \approx 14,733 R_E$$

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(1) Horricks observed an angular size for Venus of $76''$ and used contemporary estimates of the relative Sun-Venus and Sun-Earth distances, taking orbital eccentricities into account, to compute the angular size of Venus as seen from the Sun:

$$76'' \cdot ( 0.98409 \text{AU} - 0.7200 \text{AU}) / 0.7200 \text{AU} \approx 27.876''$$

A similar calculation based on Gassendi's 1632 observation of Mercury (using $20''$ and Kepler's values for Sun-Mecury and Sun-Earth distances) also produced an angular size of about $28''$. This, along with the general beliefs of the time, were apparently enough to convince Horrocks that the angular size from the Sun was $28''$ for all planets — notably Earth, which made his calculation of the AU in terms of $R_E$ possible.

(2) The term is modern. Horricks was more careful: "I do not put forward this conjecture as a certain demonstration, but as a probability".

(3) For which many contemporary estimates were available.