<p>The primary concern of the questioner is whether and how the ancient Greeks, particularly Hipparchus, understood that the Sun is far away and that sunlight reaches us in nearly parallel rays, leading to the condition <span class="math-container">$\gamma \ll \delta$</span>. Aristarchus had estimated the distance to the Moon using the Sun-Earth-Moon angle at the half-moon phase. However, even without this estimation, it is straightforward to demonstrate the relationship.</p>
<p>At the distance of the Moon's orbit, the Earth's umbra is approximately 2.6 Moon diameters wide, and the Moon subtends an angle of about <span class="math-container">$0.5^\circ$</span>. Therefore, <span class="math-container">$\beta = 1.3 \times 0.5^\circ = 0.65^\circ$</span>. The Earth's radius must be more than its shadow's width at the moon, implying <span class="math-container">$\delta &gt; \beta$</span>. Moreover, because the Sun is larger than the Earth, <span class="math-container">$\gamma = R_\oplus / d_\odot &lt; R_\odot / d_\odot = \alpha = 0.25^\circ$</span> (half the angular diameter of the Sun). This establishes, without requiring knowledge of any absolute distances or sizes, that <span class="math-container">$0 &lt; \gamma &lt; 0.25^\circ$</span>, which is smaller than <span class="math-container">$\delta &gt; 0.65^\circ$</span> and allows setting reasonable limits on the size and distance to the Moon. Furthermore, since <span class="math-container">$0.25^\circ = 0.0044$</span> radians and the Sun is larger than the Earth, it follows that the Sun is more than 227 Earth radii away.</p>
<p>How did the ancient Greeks determine that the Sun is larger than the Earth? If the Sun were smaller, the Earth's umbra would expand outward instead of narrowing. This would result in much more frequent lunar eclipses—occurring nearly every month—and such eclipses would often be visible during the day, but they are never visible during the day.</p>
<p>Another way to conclude that sunlight travels nearly parallel is the observation that day and night are equal in length on the equinox. If the Sun were closer while still subtending an angle of <span class="math-container">$0.5^\circ$</span>, one would expect the night to last longer than the day. In ancient Greece, accurately measuring the relative lengths of day and night would have required highly consistent and reliable hourglasses which I presume they did have.</p>

CC BY-SA 4.0
2024-12-16T15:19:21.407Z
Thank you @eshaya for presenting this amazing analysis. Allow me to humbly disagree with the idea of "almost parallel sunlight". This works in many cases, but not here, since it would lead to $\beta\approx\delta$. The issue is that all the angles are tiny and none can be neglected. I've been studying Hyparchus method for months and I am almost convinced he didn't get $\gamma\approx 0^\circ$. From calculations I've been doing, considering information he could've had access to his time, I obtain $\gamma\approx 0.04^\circ$, along with $\beta\approx 0.682^\circ$ and $\alpha\approx 0.27^\circ$.
Brasil
CC BY-SA 4.0
2024-12-16T21:58:55.947Z
Notice that $\gamma\approx 0.04^\circ$ suggests that the Sun-Earth distance $D$ should be about $1,432R_\oplus$. This can't be achieved by the assumptions just raised in this answer. You may wonder how I got $\gamma\approx 0.04^\circ$. Started by assuming $D = 20d$, where $d$ is the Moon-Earth distance. This allows a first estimate of $d$ using parallax. For this, consider the solar eclipse you mentioned in your last comment in my question (On the Sizes and Distances): one can evaluate $d = 73R_\oplus$, which leads to $\gamma\approx 0.04^\circ$.
Brasil
CC BY-SA 4.0
2024-12-08T14:22:04.257Z
2024-12-12T22:03:33.087Z
2024-12-12T22:03:33.087Z
eshaya
<p>A quick search on web or books leads to Hipparchus evaluation of the distance to the moon from Earth, as sketched below, where left is the sun, middle is Earth, and right is the moon.</p>
<p><a href="https://i.sstatic.net/Y0D6GZx7.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Y0D6GZx7.png" alt="Eclipse of moon: sun to the left, Earth at the middle, and moon to the right." /></a></p>
<p>In the method, it is required to assume that the angular radius of Earth as seen from the sun, <span class="math-container">$\gamma$</span>, is negligible.
This is actually true, since <span class="math-container">$$\gamma = \arctan R/D \approx 0.0024^\circ,$$</span> where <span class="math-container">$R \approx 6.4 \times 10^3$</span> km is Earth's radius and <span class="math-container">$D \approx 1.5 \times 10^8$</span> km is the distance from Earth to sun.
Thus, the value of <span class="math-container">$\gamma$</span> is negligible in comparison, for instance, to the angles <span class="math-container">$\alpha$</span>, <span class="math-container">$\beta$</span>, and <span class="math-container">$\delta$</span>.</p>
<p>The value of <span class="math-container">$\alpha$</span> is easy to compute: it is the ratio between the lasting of a sunset or a sunrise to 24h (daily course time).
The angle <span class="math-container">$\beta$</span> is obtained from the measurement of the eclipse duration.
And <span class="math-container">$\delta$</span> is given from the relation between the angles of the triangle: <span class="math-container">$\delta + \gamma = \alpha + \beta$</span>.</p>
<p>Wherever this method is explained, all the author simply states that &quot;it is reasonable to assume <span class="math-container">$\gamma \approx 0$</span>&quot;.
Many of them don't provide a good reason for this to be accepted.
Some state that the Greeks were convinced of this because the sun rays come almost parallel to us.
However, this is not a good explanation, since the sun rays are responsible for the conic (and not cylindric!) shadow that causes the lunar eclipse.</p>
<p>So, my question is: <strong>which astrnomic event indicates that <span class="math-container">$\gamma \approx 0$</span> previously to the actual evaluation of the distance to the sun from Earth?</strong></p>

CC BY-SA 4.0
2024-11-28T17:59:02.780Z
The apparent size change of the sun from sunrise to noon.
eshaya
CC BY-SA 4.0
2024-11-28T19:41:56.153Z
What does it have to do with $\gamma$, @eshaya?
Brasil
CC BY-SA 4.0
2024-11-29T00:20:36.390Z
You’ll want to read Ptolemy’s Almagest, Book V, Chapter 17.
Pierre Paquette
CC BY-SA 4.0
2024-11-29T20:29:58.807Z
The Sun is an Earth radius closer at noon than at sunrise, hence one can determine the distance to the sun in Earth radii by measuring the angular size of the Sun.
eshaya
CC BY-SA 4.0
2024-11-30T02:21:56.523Z
You may probably be right, @eshaya, but how could they possibly have measured the angular size of Sun with such a precision to get the difference among noon and sunset/sunrise and discount the atmospheric refraction in ancient Greek (before Christ)?
Brasil
CC BY-SA 4.0
2024-11-30T18:49:05.503Z
The Chinese philosopher Mo Ti described an inverted image formed with a pinhole in the 5th century B.C. Aristotle wrote about his pinhole observations in the 4th century B.C.
eshaya
CC BY-SA 4.0
2024-12-03T00:00:14.743Z
@eshaya allow me analyze your hypothesis. Consider a cubic camera obscura 4 m in length (pretty large indeed). In this camera, the size of the sun's image would be of 3.77 cm at noon and would be about 2 $\mu$m smaller at set, while the size of moon's image would be about 3.62 cm at noon and would be about 0.6 mm smaller at set. The ability to measure the larger change (Moon) depends on the sharpness of the image and availability of instruments/technique. Considering these values, I believe they couldn't conclude anything about the comparative distances to Sun and to Moon by your hypothesis.
Brasil
CC BY-SA 4.0
2024-12-03T04:03:53.723Z
Could they show $\gamma$ << $\delta$ with your 4m camera?  Is the Sun bright enough for a longer camera?
eshaya
CC BY-SA 4.0
2024-12-03T15:23:26.373Z
I guess they couldn't. To answer precisely, I would have to repeat such experiment. But it is hard to construct a box 4m large. I guess that the small change in moon size is within any uncertainty due to both the measurement device (assuming they could have good apparatus with precision close to 1 mm) and image formation (edge sharpness).
Brasil
CC BY-SA 4.0
2024-12-07T04:40:46.333Z
I am not sure. But perhaps he was aware of Aristarchus earlier measurement which shows that the sun is ~20 times further than the moon.
Ryan Cole
CC BY-SA 4.0
2024-12-08T14:53:50.677Z
The methodology used by Aristarchus of Samos (c. 310 BC – c. 230 BC) is detailed in his only remaining work, ["On the Sizes and Distances"](https://en.wikipedia.org/wiki/On_the_Sizes_and_Distances_(Aristarchus)).
doubleunary
<the-moon><distances><parallax><hipparcos>
How Greeks proved that solar parallax is negligible?
Brasil
# How Greeks proved that solar parallax is negligible?

(Asked by: username_0 on 2024-11-28)

A quick search on web or books leads to Hipparchus evaluation of the distance to the moon from Earth, as sketched below, where left is the sun, middle is Earth, and right is the moon.

[![Eclipse of moon: sun to the left, Earth at the middle, and moon to the right.](https://i.sstatic.net/Y0D6GZx7.png)](https://i.sstatic.net/Y0D6GZx7.png)

In the method, it is required to assume that the angular radius of Earth as seen from the sun, $\gamma$, is negligible.
This is actually true, since $$\gamma = \arctan R/D \approx 0.0024^\circ,$$ where $R \approx 6.4 \times 10^3$ km is Earth's radius and $D \approx 1.5 \times 10^8$ km is the distance from Earth to sun.
Thus, the value of $\gamma$ is negligible in comparison, for instance, to the angles $\alpha$, $\beta$, and $\delta$.

The value of $\alpha$ is easy to compute: it is the ratio between the lasting of a sunset or a sunrise to 24h (daily course time).
The angle $\beta$ is obtained from the measurement of the eclipse duration.
And $\delta$ is given from the relation between the angles of the triangle: $\delta + \gamma = \alpha + \beta$.

Wherever this method is explained, all the author simply states that "it is reasonable to assume $\gamma \approx 0$".
Many of them don't provide a good reason for this to be accepted.
Some state that the Greeks were convinced of this because the sun rays come almost parallel to us.
However, this is not a good explanation, since the sun rays are responsible for the conic (and not cylindric!) shadow that causes the lunar eclipse.

So, my question is: **which astrnomic event indicates that $\gamma \approx 0$ previously to the actual evaluation of the distance to the sun from Earth?**

**Tags:** the-moon, distances, parallax, hipparcos

> **Comments:**
> **username_1**: The apparent size change of the sun from sunrise to noon.  
> **username_0**: What does it have to do with $\gamma$, @username_1?  
> **username_2**: You’ll want to read Ptolemy’s Almagest, Book V, Chapter 17.  
> **username_1**: The Sun is an Earth radius closer at noon than at sunrise, hence one can determine the distance to the sun in Earth radii by measuring the angular size of the Sun.  
> **username_0**: You may probably be right, @username_1, but how could they possibly have measured the angular size of Sun with such a precision to get the difference among noon and sunset/sunrise and discount the atmospheric refraction in ancient Greek (before Christ)?  
> **username_1**: The Chinese philosopher Mo Ti described an inverted image formed with a pinhole in the 5th century B.C. Aristotle wrote about his pinhole observations in the 4th century B.C.  
> **username_0**: @username_1 allow me analyze your hypothesis. Consider a cubic camera obscura 4 m in length (pretty large indeed). In this camera, the size of the sun's image would be of 3.77 cm at noon and would be about 2 $\mu$m smaller at set, while the size of moon's image would be about 3.62 cm at noon and would be about 0.6 mm smaller at set. The ability to measure the larger change (Moon) depends on the sharpness of the image and availability of instruments/technique. Considering these values, I believe they couldn't conclude anything about the comparative distances to Sun and to Moon by your hypothesis.  
> **username_1**: Could they show $\gamma$ << $\delta$ with your 4m camera?  Is the Sun bright enough for a longer camera?  
> **username_0**: I guess they couldn't. To answer precisely, I would have to repeat such experiment. But it is hard to construct a box 4m large. I guess that the small change in moon size is within any uncertainty due to both the measurement device (assuming they could have good apparatus with precision close to 1 mm) and image formation (edge sharpness).  
> **username_3**: I am not sure. But perhaps he was aware of Aristarchus earlier measurement which shows that the sun is ~20 times further than the moon.  
> **username_4**: The methodology used by Aristarchus of Samos (c. 310 BC – c. 230 BC) is detailed in his only remaining work, ["On the Sizes and Distances"](https://en.wikipedia.org/wiki/On_the_Sizes_and_Distances_(Aristarchus)).  

---

## 1 answers

### Answer by username_1 on 2024-12-08. Score: 1

The primary concern of the questioner is whether and how the ancient Greeks, particularly Hipparchus, understood that the Sun is far away and that sunlight reaches us in nearly parallel rays, leading to the condition $\gamma \ll \delta$. Aristarchus had estimated the distance to the Moon using the Sun-Earth-Moon angle at the half-moon phase. However, even without this estimation, it is straightforward to demonstrate the relationship.

At the distance of the Moon's orbit, the Earth's umbra is approximately 2.6 Moon diameters wide, and the Moon subtends an angle of about $0.5^\circ$. Therefore, $\beta = 1.3 \times 0.5^\circ = 0.65^\circ$. The Earth's radius must be more than its shadow's width at the moon, implying $\delta > \beta$. Moreover, because the Sun is larger than the Earth, $\gamma = R\_\oplus / d\_\odot < R\_\odot / d\_\odot = \alpha = 0.25^\circ$ (half the angular diameter of the Sun). This establishes, without requiring knowledge of any absolute distances or sizes, that $0 < \gamma < 0.25^\circ$, which is smaller than $\delta > 0.65^\circ$ and allows setting reasonable limits on the size and distance to the Moon. Furthermore, since $0.25^\circ = 0.0044$ radians and the Sun is larger than the Earth, it follows that the Sun is more than 227 Earth radii away.

How did the ancient Greeks determine that the Sun is larger than the Earth? If the Sun were smaller, the Earth's umbra would expand outward instead of narrowing. This would result in much more frequent lunar eclipses—occurring nearly every month—and such eclipses would often be visible during the day, but they are never visible during the day.

Another way to conclude that sunlight travels nearly parallel is the observation that day and night are equal in length on the equinox. If the Sun were closer while still subtending an angle of $0.5^\circ$, one would expect the night to last longer than the day. In ancient Greece, accurately measuring the relative lengths of day and night would have required highly consistent and reliable hourglasses which I presume they did have.

> **Comments:**
> **username_0**: Thank you @username_1 for presenting this amazing analysis. Allow me to humbly disagree with the idea of "almost parallel sunlight". This works in many cases, but not here, since it would lead to $\beta\approx\delta$. The issue is that all the angles are tiny and none can be neglected. I've been studying Hyparchus method for months and I am almost convinced he didn't get $\gamma\approx 0^\circ$. From calculations I've been doing, considering information he could've had access to his time, I obtain $\gamma\approx 0.04^\circ$, along with $\beta\approx 0.682^\circ$ and $\alpha\approx 0.27^\circ$.  
> **username_0**: Notice that $\gamma\approx 0.04^\circ$ suggests that the Sun-Earth distance $D$ should be about $1,432R_\oplus$. This can't be achieved by the assumptions just raised in this answer. You may wonder how I got $\gamma\approx 0.04^\circ$. Started by assuming $D = 20d$, where $d$ is the Moon-Earth distance. This allows a first estimate of $d$ using parallax. For this, consider the solar eclipse you mentioned in your last comment in my question (On the Sizes and Distances): one can evaluate $d = 73R_\oplus$, which leads to $\gamma\approx 0.04^\circ$.  

---

