<p><em><a href="https://en.wikipedia.org/wiki/Araucaria_columnaris" rel="nofollow noreferrer">Auracaria columnaris</a></em> (Cooks pine) on average lean towards the equator.  Most of the pines in the Northern Hemisphere lean southward and most of the pines in the Southern Hemisphere lean northward.</p>
<p><a href="https://i.sstatic.net/8fCvH.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8fCvH.jpg" alt="enter image description here" /></a></p>
<p>The above picture is from <a href="https://www.researchgate.net/publication/317275781_Worldwide_hemisphere-dependent_lean_in_Cook_pines" rel="nofollow noreferrer">Johns et al. 2017</a>, who performed a study of 256 Cooks pines across 18 regions on 5 continents.  Though they don't publish their data set in the paper, we can see that the leaning effect is more pronounced for the higher latitudes.</p>
<p><a href="https://i.sstatic.net/1zLKe.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1zLKe.png" alt="enter image description here" /></a></p>
<p>At higher latitudes, the apparent location of the Sun in the sky becomes more biased in the opposite direction.  For example, in far Northern Alaska, the Sun mostly appears in the Southern half of the sky.  Could <a href="https://en.wikipedia.org/wiki/Phototropism" rel="nofollow noreferrer">phototropism</a> be an explanation for the Cooks pine lean? Does the North/South component of the Sun's average incident angle by latitude match the above table?</p>
<p><strong>Question:</strong> Consider a perfect blackbody disk elevated slightly from the ground.  For a given latitude <span class="math-container">$L$</span> what is angle <span class="math-container">$\theta$</span> between a fixed disk and the local tangent plane that maximizes yearly absorbed wattage from the Sun (assume no horizon obscura).</p>

CC BY-SA 4.0
2021-03-01T19:53:54.290Z
There is arguable more biology here than astronomy. The average postion of the sun at 10 degrees North is... about 10 degrees south, and at 20 degrees North it is 20 degrees south.  As for whether this is conventional phototropism or something specific to Cooks pine, that would seem to requier a botanist to answer.
James K
CC BY-SA 4.0
2021-03-01T20:37:41.793Z
@JamesK But at 90 degrees North, the average angle isn't 90 degrees South.  One has to take into account only when the Sun is above the horizon, right?
Connor Garcia
CC BY-SA 4.0
2021-03-01T20:44:24.453Z
Perhaps, you're right, things will get weird at very high latitudes, but for the low latitudes at which these trees grow things should be simpler.
James K
CC BY-SA 4.0
2021-03-01T22:59:37.897Z
@JamesK Based on your comments and thinking a bit more about this problem, I tried to make the question more explicit and clearly astronomy based in an edit.
Connor Garcia
CC BY-SA 4.0
2021-03-01T23:04:52.493Z
I'm pretty sure, (but only by looking at the local solar farms and guessing) that to maximise wattage you would want the trees at the equator to be completly horizontal, and the trees in the northern hemisphere to be inclince by nearly 90 degrees to the *North*. Angling towards the South would tend to reduce the cross section that was exposed to the powerful midday sun.
James K
CC BY-SA 4.0
2021-03-01T23:10:41.367Z
@JamesK Uhoh! I think you may be right, let me try one more edit.
Connor Garcia
CC BY-SA 4.0
2021-03-02T03:28:26.250Z
@JamesK I think that depends on how much front-back shadowing there is vs. how much top-down shadowing there is.   In the horizontal-at-the-equator scenario, one side of the tree gets all the light and the other side gets none.  If it were vertical, I think there could be more needle sunlight exposure overall, especially given the somewhat conical shape of the tree.
Eric Jensen
<the-sun><coordinate>
How is the incidence angle of the Sun related to the lean angle of the Cooks pines?
Connor Garcia
# How is the incidence angle of the Sun related to the lean angle of the Cooks pines?

(Asked by: username_0 on 2021-03-01)

*[Auracaria columnaris](https://en.wikipedia.org/wiki/Araucaria_columnaris)* (Cooks pine) on average lean towards the equator. Most of the pines in the Northern Hemisphere lean southward and most of the pines in the Southern Hemisphere lean northward.

[![enter image description here](https://i.sstatic.net/8fCvH.jpg)](https://i.sstatic.net/8fCvH.jpg)

The above picture is from [Johns et al. 2017](https://www.researchgate.net/publication/317275781_Worldwide_hemisphere-dependent_lean_in_Cook_pines), who performed a study of 256 Cooks pines across 18 regions on 5 continents. Though they don't publish their data set in the paper, we can see that the leaning effect is more pronounced for the higher latitudes.

[![enter image description here](https://i.sstatic.net/1zLKe.png)](https://i.sstatic.net/1zLKe.png)

At higher latitudes, the apparent location of the Sun in the sky becomes more biased in the opposite direction. For example, in far Northern Alaska, the Sun mostly appears in the Southern half of the sky. Could [phototropism](https://en.wikipedia.org/wiki/Phototropism) be an explanation for the Cooks pine lean? Does the North/South component of the Sun's average incident angle by latitude match the above table?

**Question:** Consider a perfect blackbody disk elevated slightly from the ground. For a given latitude $L$ what is angle $\theta$ between a fixed disk and the local tangent plane that maximizes yearly absorbed wattage from the Sun (assume no horizon obscura).

**Tags:** the-sun, coordinate

> **Comments:**
> **username_1**: There is arguable more biology here than astronomy. The average postion of the sun at 10 degrees North is... about 10 degrees south, and at 20 degrees North it is 20 degrees south.  As for whether this is conventional phototropism or something specific to Cooks pine, that would seem to requier a botanist to answer.  
> **username_0**: @JamesK But at 90 degrees North, the average angle isn't 90 degrees South.  One has to take into account only when the Sun is above the horizon, right?  
> **username_1**: Perhaps, you're right, things will get weird at very high latitudes, but for the low latitudes at which these trees grow things should be simpler.  
> **username_0**: @JamesK Based on your comments and thinking a bit more about this problem, I tried to make the question more explicit and clearly astronomy based in an edit.  
> **username_1**: I'm pretty sure, (but only by looking at the local solar farms and guessing) that to maximise wattage you would want the trees at the equator to be completly horizontal, and the trees in the northern hemisphere to be inclince by nearly 90 degrees to the *North*. Angling towards the South would tend to reduce the cross section that was exposed to the powerful midday sun.  
> **username_0**: @JamesK Uhoh! I think you may be right, let me try one more edit.  
> **username_2**: @JamesK I think that depends on how much front-back shadowing there is vs. how much top-down shadowing there is.   In the horizontal-at-the-equator scenario, one side of the tree gets all the light and the other side gets none.  If it were vertical, I think there could be more needle sunlight exposure overall, especially given the somewhat conical shape of the tree.  

---

## 0 answers

