<p>If you are reducing everything to plain geometry, you can compute the fraction of the "sunny" arc of the circle of latitude for a given location. (Hint: draw a diagram showing the tropic of Cancer, the circle of latitude, the terminator and the axis, all as straight lines.) The height of this segment of a circle of latitude (at $\lambda$) is</p>

<p>$$h = \frac{\cos(\lambda) + \sin(\lambda)\times \tan(\epsilon)}{\cos(\lambda)}$$</p>

<p>The angle of this segment is</p>

<p>$$a = 2\times \arccos(1 - h)$$</p>

<p>I get </p>

<ul>
<li><p>New York: 14.9297092807291</p></li>
<li><p>London: 16.4195095586004</p></li>
</ul>

<p>which is fairly close to the "expected" numbers you have provided. The difference is very likely due to Mathematica using a geoid rather than a sphere and including atmospheric refraction.</p>

CC BY-SA 3.0
2017-03-07T22:00:04.290Z
2017-03-07T22:29:34.823Z
2017-03-07T22:29:34.823Z
laune
<p>The difference is due to two things. First, using geometry calculates the time from the center of the sun crossing the horizon in the east to the center of the sun crossing it in the west. But sunrise is when the edge of the sun's disk crosses the horizon and sunset when the other edge crosses it, so your day length is low by the time it takes the sun to "move" one solar diameter, or about 2 minutes (half a degree apparent diameter). At higher latitudes it would be a bit longer because the sun is "moving" at an angle.</p>

<p>The other difference is the effect of refraction, which makes the sun appear higher than it really is when it's near the horizon; and because the angle at which the sun rises above the horizon or sinks below it increases with latitude, so does the length of time it takes the sun to overcome the effect of refraction, so the real length of the day at the equinox varies from about 15h 06m at the equator to over 15 1/2 hours inside the polar circles (in fact, at the poles the sun, or most of its disk, would be slightly above the horizon for the entire 24h day). This effect is more pronounced at the solstice, so the length of the day at the summer solstice can be half an hour or even more than an hour longer (again at higher latitudes) than the length of the night at the winter solstice.</p>

CC BY-SA 3.0
2017-12-09T14:16:58.147Z
2017-12-09T14:16:58.147Z
Michael
<p>I want to be able to predict how long the summer solstice lasts at any part of world given the maximum tilt angle $φ = 0.4101524 \; \mathrm{rad}$ and the latitude $θ \; \mathrm{rad}$. I specifically want to do this using a geometric approach. The problem is, I never had experience with spherical trig before, and the values I got are somewhat off from their real values. So I want to know if that's because I didn't do the math correctly, and/or because of physical reasons (e.g. the Earth isn't perfectly spherical, the tilt changes slightly etc.).
Here is how I did it:</p>

<p>Let the Earth be represented as a perfect sphere of radius $R$ moving at a constant angular velocity. Let the Earth's rotational axis be in the z direction, and the axis of the circle forming the dark/light boundary be tilted by angle $φ$ relative to the Earth's axis. During the summer solstice, our location of interest (say, in the northern hemisphere) will trace out a circular path of radius $r &lt; R$ perpendicular to the z-axis. These three circles, $C_{\text{rot}}$, $C_{\text{bound}}$ and $C_{\text{path}}$ create a spherical triangle. Using the arc length $s$ of this triangle along $C_{\text{path}}$, we can calculate the length of the day in hours using: $$24 \times \frac{πr + 2s}{2πr}$$</p>

<p>The point where $C_{\text{rot}}$ and $C_{\text{path}}$ meet is located at $$P(\text{rot/path}) = (R\cosθ, 0, R\sinθ)$$</p>

<p>To find the point where $C_{\text{bound}}$ and $C_{\text{path}}$ meet, we first observe that $$y = \sqrt{R^2 - x^2}\sinφ$$ and $$z = \sqrt{R^2 - x^2}\cosφ$$
Which we can solve for $x$ by setting $z$ equal to the $z$ from the previous point
$$x = R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}}$$
Plugging this back into $y$ and $z$, we finally get
$$P(\text{bound/path}) = \left (R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}},R\sinθ\tanφ,R\sinθ \right )$$
Now these two points are on the same plane in the z-axis, which is the plane of $P_{\text{path}}$. The distance $d$ between these points is 
$$d = \sqrt{\left (R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}} - R\cosθ \right)^2 + R^2\sinθ^2\tan^2φ}$$
From this we can get $s$ by using the arc length formula:
$$s = 2\arcsin \left (\frac{d}{2r} \right )r$$
I used Mathematica to compare the calculated and expected values in 2017:</p>

<p>New York: $θ = 0.7105724077$; length (calculated) = 14 hours 56 minutes; length (expected) = 15 hours, 5 minutes</p>

<p>London: $θ = 0.8989737191$; length (calculated) = 14 hours 25 minutes; length (expected) = 16 hours, 38 minutes</p>

<p>As you can see, there is a noticeable difference (10+ minutes).</p>

CC BY-SA 3.0
2017-03-07T21:55:08.743Z
A circular path of radius r<R perpendicular to the z-axis cannot be one side of a spherical triangle - only great circles can be used as sides. I don't think it matters as you are not relying on one of the formulas for solving spherical triangles.
laune
CC BY-SA 3.0
2017-03-08T00:34:01.770Z
Oh, I wasn't aware of that. Does the triangle have a special name then?
Denn
CC BY-SA 3.0
2017-03-08T06:50:38.667Z
No - it isn't a triangle. In analogy to plane geometry you could have sectors (e.g. two meridians, one circle of latitude) or segments (e.g. the ecliptic and a circle of latitude near the equator). Applying the law of sine or cosine is, of course, restricted to triangles made from geodesics.
laune
<earth><rotation>
Calculating length of summer solstice day using tilt and latitude
Denn
# Calculating length of summer solstice day using tilt and latitude

(Asked by: username_0 on 2017-03-06)

I want to be able to predict how long the summer solstice lasts at any part of world given the maximum tilt angle $φ = 0.4101524 \; \mathrm{rad}$ and the latitude $θ \; \mathrm{rad}$. I specifically want to do this using a geometric approach. The problem is, I never had experience with spherical trig before, and the values I got are somewhat off from their real values. So I want to know if that's because I didn't do the math correctly, and/or because of physical reasons (e.g. the Earth isn't perfectly spherical, the tilt changes slightly etc.).
Here is how I did it:

Let the Earth be represented as a perfect sphere of radius $R$ moving at a constant angular velocity. Let the Earth's rotational axis be in the z direction, and the axis of the circle forming the dark/light boundary be tilted by angle $φ$ relative to the Earth's axis. During the summer solstice, our location of interest (say, in the northern hemisphere) will trace out a circular path of radius $r < R$ perpendicular to the z-axis. These three circles, $C\_{\text{rot}}$, $C\_{\text{bound}}$ and $C\_{\text{path}}$ create a spherical triangle. Using the arc length $s$ of this triangle along $C\_{\text{path}}$, we can calculate the length of the day in hours using: $$24 \times \frac{πr + 2s}{2πr}$$

The point where $C\_{\text{rot}}$ and $C\_{\text{path}}$ meet is located at $$P(\text{rot/path}) = (R\cosθ, 0, R\sinθ)$$

To find the point where $C\_{\text{bound}}$ and $C\_{\text{path}}$ meet, we first observe that $$y = \sqrt{R^2 - x^2}\sinφ$$ and $$z = \sqrt{R^2 - x^2}\cosφ$$
Which we can solve for $x$ by setting $z$ equal to the $z$ from the previous point
$$x = R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}}$$
Plugging this back into $y$ and $z$, we finally get
$$P(\text{bound/path}) = \left (R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}},R\sinθ\tanφ,R\sinθ \right )$$
Now these two points are on the same plane in the z-axis, which is the plane of $P\_{\text{path}}$. The distance $d$ between these points is
$$d = \sqrt{\left (R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}} - R\cosθ \right)^2 + R^2\sinθ^2\tan^2φ}$$
From this we can get $s$ by using the arc length formula:
$$s = 2\arcsin \left (\frac{d}{2r} \right )r$$
I used Mathematica to compare the calculated and expected values in 2017:

New York: $θ = 0.7105724077$; length (calculated) = 14 hours 56 minutes; length (expected) = 15 hours, 5 minutes

London: $θ = 0.8989737191$; length (calculated) = 14 hours 25 minutes; length (expected) = 16 hours, 38 minutes

As you can see, there is a noticeable difference (10+ minutes).

**Tags:** earth, rotation

> **Comments:**
> **username_1**: A circular path of radius r<R perpendicular to the z-axis cannot be one side of a spherical triangle - only great circles can be used as sides. I don't think it matters as you are not relying on one of the formulas for solving spherical triangles.  
> **username_0**: Oh, I wasn't aware of that. Does the triangle have a special name then?  
> **username_1**: No - it isn't a triangle. In analogy to plane geometry you could have sectors (e.g. two meridians, one circle of latitude) or segments (e.g. the ecliptic and a circle of latitude near the equator). Applying the law of sine or cosine is, of course, restricted to triangles made from geodesics.  

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## 2 answers

### Answer by username_1 on 2017-03-07. Score: 3

If you are reducing everything to plain geometry, you can compute the fraction of the "sunny" arc of the circle of latitude for a given location. (Hint: draw a diagram showing the tropic of Cancer, the circle of latitude, the terminator and the axis, all as straight lines.) The height of this segment of a circle of latitude (at $\lambda$) is

$$h = \frac{\cos(\lambda) + \sin(\lambda)\times \tan(\epsilon)}{\cos(\lambda)}$$

The angle of this segment is

$$a = 2\times \arccos(1 - h)$$

I get

* New York: 14.9297092807291
* London: 16.4195095586004

which is fairly close to the "expected" numbers you have provided. The difference is very likely due to Mathematica using a geoid rather than a sphere and including atmospheric refraction.


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### Answer by username_2 on 2017-12-09. Score: 2

The difference is due to two things. First, using geometry calculates the time from the center of the sun crossing the horizon in the east to the center of the sun crossing it in the west. But sunrise is when the edge of the sun's disk crosses the horizon and sunset when the other edge crosses it, so your day length is low by the time it takes the sun to "move" one solar diameter, or about 2 minutes (half a degree apparent diameter). At higher latitudes it would be a bit longer because the sun is "moving" at an angle.

The other difference is the effect of refraction, which makes the sun appear higher than it really is when it's near the horizon; and because the angle at which the sun rises above the horizon or sinks below it increases with latitude, so does the length of time it takes the sun to overcome the effect of refraction, so the real length of the day at the equinox varies from about 15h 06m at the equator to over 15 1/2 hours inside the polar circles (in fact, at the poles the sun, or most of its disk, would be slightly above the horizon for the entire 24h day). This effect is more pronounced at the solstice, so the length of the day at the summer solstice can be half an hour or even more than an hour longer (again at higher latitudes) than the length of the night at the winter solstice.


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