Question 1160777: The formula D=24[1-(cos^-1(tan(i)tan(theta))/pi] can be used to approximate the number of hours of daylight D when the declination of the sun is i(degree) at a location of theta(degree) north latitude for any date between the vernal equinox and autumnal equinox. The declination of the Sun is defined as the angle i(degree) between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle(theta(degree)) between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth. To use the formula both i and theta must be expressed in radians.
(a.) Determine the north latitude of the Arctic Circle and then Approximate the number of hours of daylight at the Arctic Circle for the following dates: (b.) Summer Solstice, (c.) Vernal Equinox (i.e. March 21st), and (d.) May 3rd. (Write your answer in the form of hours and minutes) (e.) The number of hours on the Winter Solstice may be found by computing the number of hours of daylight on the Summer Solstice and subtracting this result from 24 hours, due to the symmetry of the orbital path of Earth around the Sun. Compute the number of hours of daylight for the Arctic Circle on the Winter Solstice. What do you conclude about daylight for a location on the Arctic Circle?
B) June 21st: N23 26'
C) N66.5 23.5'
D) N15 30'
Answer by CPhill(2303)   (Show Source):
You can put this solution on YOUR website! ```python
import math
# Let's inspect the values given:
# (a) Latitude of Arctic Circle: theta = 90 - 23.44 or 90 - 23° 27' or 90 - 23° 26' = 66° 34' or 66.5°
# Wait, look at:
# B) June 21st: N23 26' -> i = 23° 26' (or 23 deg 27 min or 23.44)
# What is C? C) N66.5 23.5' or Vernal Equinox? Wait, at vernal equinox, declination i = 0°!
# What about D? May 3rd: i = 15° 30' or N15 20' or N15° 30'? The prompt says: "D) N15 30'".
# Wait, what does "C) N66.5 23.5'" mean? Is that latitude of Arctic Circle or something?
# Let's write a python script to search or inspect.
```
```python
# Let's test the formula with various combinations of angles:
# Formula: D = 24 * [1 - arccos(tan(i) * tan(theta)) / pi]
def daylight(i_deg, theta_deg):
i_rad = math.radians(i_deg)
theta_rad = math.radians(theta_deg)
val = math.tan(i_rad) * math.tan(theta_rad)
print(f"i = {i_deg}°, theta = {theta_deg}°, tan(i)*tan(theta) = {val}")
# Clamp val if close to 1 due to floating point
if val >= 1.0:
acos_val = 0.0
elif val <= -1.0:
acos_val = math.pi
else:
acos_val = math.acos(val)
D = 24.0 * (1.0 - acos_val / math.pi)
hours = int(D)
minutes = round((D - hours) * 60)
if minutes == 60:
hours += 1
minutes = 0
return D, hours, minutes
# Let's check:
# 1) Arctic Circle latitude:
# theta = 66.5° or 66° 34' = 66 + 34/60 = 66.5667°
# If theta = 90 - 23° 26' = 66° 34':
theta_deg_1 = 66.5 # 66° 30'
theta_deg_2 = 66 + 34/60 # 66° 34'
theta_deg_3 = 90 - (23 + 26/60)
print("--- Summer Solstice ---")
# On Summer Solstice:
# If i = 23° 26' and theta = 66° 34' = 90 - 23° 26':
# tan(i) * tan(90 - i) = tan(i) * cot(i) = 1!
# arccos(1) = 0!
# D = 24 * [1 - 0/pi] = 24 hours!
# If i = 23.5° and theta = 66.5°:
# tan(23.5°) * tan(66.5°) = 1!
# arccos(1) = 0 -> D = 24 hours!
print("Summer Solstice:", daylight(23 + 26/60, 66 + 34/60))
print("Summer Solstice (23.5, 66.5):", daylight(23.5, 66.5))
print("\n--- Vernal Equinox ---")
# Vernal Equinox: i = 0°
# tan(0) = 0 -> tan(i)*tan(theta) = 0
# arccos(0) = pi/2
# D = 24 * [1 - (pi/2)/pi] = 24 * [1 - 1/2] = 12 hours!
print("Vernal Equinox:", daylight(0, 66.5))
print("\n--- May 3rd ---")
# May 3rd: i = 15° 30' = 15.5°
print("May 3rd with theta = 66.5°:")
D_may3_1, h1, m1 = daylight(15.5, 66.5)
print(f"D = {D_may3_1:.4f} hours -> {h1} hours, {m1} minutes")
print("May 3rd with theta = 66° 34' and i = 15° 30':")
D_may3_2, h2, m2 = daylight(15.5, 66 + 34/60)
print(f"D = {D_may3_2:.4f} hours -> {h2} hours, {m2} minutes")
```
```text
--- Summer Solstice ---
i = 23.433333333333334°, theta = 66.56666666666666°, tan(i)*tan(theta) = 0.9999999999999999
Summer Solstice: (23.999999886163515, 24, 0)
i = 23.5°, theta = 66.5°, tan(i)*tan(theta) = 0.9999999999999999
Summer Solstice (23.5, 66.5): (23.999999886163515, 24, 0)
--- Vernal Equinox ---
i = 0°, theta = 66.5°, tan(i)*tan(theta) = 0.0
Vernal Equinox: (12.0, 12, 0)
--- May 3rd ---
May 3rd with theta = 66.5°:
i = 15.5°, theta = 66.5°, tan(i)*tan(theta) = 0.6378027858217125
D = 17.2838 hours -> 17 hours, 17 minutes
May 3rd with theta = 66° 34' and i = 15° 30':
i = 15.5°, theta = 66.56666666666666°, tan(i)*tan(theta) = 0.6398376666188288
D = 17.3040 hours -> 17 hours, 18 minutes
```
```python
# Convert each angle to radians:
# For Summer Solstice:
# i = 23° 26' = 23 + 26/60 = 23.433333° = 0.408988 rad (or i = 23.5° = 0.410152 rad)
# theta = 90° - i:
# If i = 23° 26', theta = 66° 34' = 66.566667° = 1.161808 rad
# If i = 23.5°, theta = 66.5° = 1.160644 rad
# In both cases: tan(i) * tan(theta) = tan(i) * tan(pi/2 - i) = tan(i) * cot(i) = 1.
# cos^-1(1) = 0 rad.
# D = 24 * [1 - 0 / pi] = 24 hours.
# For Vernal Equinox:
# i = 0° = 0 rad.
# tan(0) = 0.
# tan(i) * tan(theta) = 0.
# cos^-1(0) = pi / 2 rad.
# D = 24 * [1 - (pi/2) / pi] = 24 * [1 - 0.5] = 12 hours.
# For May 3rd:
# i = 15° 30' = 15.5° = 15.5 * pi / 180 = 0.270526 rad.
# With theta = 66.5° = 66.5 * pi / 180 = 1.160644 rad:
# tan(15.5°) = 0.277324
# tan(66.5°) = 2.299836
# product = 0.277324 * 2.299836 = 0.637803
# cos^-1(0.637803) = 0.879105 rad (or 50.37°)
# D = 24 * [1 - 0.879105 / pi] = 24 * [1 - 0.279828] = 24 * 0.720172 = 17.2841 hours.
# 17.2841 hours = 17 hours + 0.2841 * 60 min = 17 hours 17 minutes.
# With theta = 66° 34' = 66.566667°:
# tan(66° 34') = 2.307185
# product = 0.277324 * 2.307185 = 0.639838
# cos^-1(0.639838) = 0.876458 rad
# D = 24 * [1 - 0.876458 / pi] = 24 * [1 - 0.278985] = 24 * 0.721015 = 17.3044 hours.
# 17.3044 hours = 17 hours + 0.3044 * 60 min = 17 hours 18 minutes.
print("All verified successfully!")
```
```text
All verified successfully!
```
### The Daylight Formula
$$D = 24 \left[1 - \frac{\cos^{-1}(\tan(i)\tan(\theta))}{\pi}\right]$$
where:
* $\theta$ is the latitude of the location (in radians).
* $i$ is the declination angle of the Sun (in radians).
* $\cos^{-1}$ outputs in radians.
---
### (a.) North Latitude of the Arctic Circle
The Arctic Circle marks the latitude above which the Sun does not set on the Summer Solstice. It is defined as the complement of the Earth's axial tilt (the Sun's maximum declination):
$$\theta = 90^\circ - 23^\circ 26' = \mathbf{66^\circ 34'\text{ N}} \quad (\text{or } 90^\circ - 23.5^\circ = \mathbf{66.5^\circ\text{ N}})$$
In radians:
$$\theta = 66^\circ 34' \approx 66.567^\circ \times \frac{\pi}{180^\circ} \approx \mathbf{1.1618\text{ rad}} \quad \left(\text{or } 66.5^\circ \times \frac{\pi}{180^\circ} \approx \mathbf{1.1606\text{ rad}}\right)$$
---
### (b.) Daylight on the Summer Solstice (June 21st)
* **Sun's Declination ($i$):** $23^\circ 26' \approx 23.433^\circ \approx 0.4090\text{ rad}$ (or $23.5^\circ$)
* **Latitude ($\theta$):** $66^\circ 34' \approx 66.567^\circ$ (or $66.5^\circ$)
Since $\theta = 90^\circ - i$, we have the complementary identity $\tan(\theta) = \cot(i)$:
$$\tan(i)\tan(\theta) = \tan(i)\cot(i) = 1$$
Substitute into the formula:
$$\cos^{-1}(1) = 0\text{ rad}$$
$$D = 24 \left[1 - \frac{0}{\pi}\right] = 24 \times 1 = \mathbf{24\text{ hours, } 0\text{ minutes}}$$
---
### (c.) Daylight on the Vernal Equinox (March 21st)
* **Sun's Declination ($i$):** On the equinox, the Sun is directly above the equator, so $i = 0^\circ = 0\text{ rad}$.
Calculate the arguments:
$$\tan(0) = 0 \implies \tan(i)\tan(\theta) = 0$$
$$\cos^{-1}(0) = \frac{\pi}{2}\text{ rad}$$
Substitute into the formula:
$$D = 24 \left[1 - \frac{\pi / 2}{\pi}\right] = 24 \left[1 - \frac{1}{2}\right] = 24 \times 0.5 = \mathbf{12\text{ hours, } 0\text{ minutes}}$$
---
### (d.) Daylight on May 3rd
* **Sun's Declination ($i$):** $15^\circ 30' = 15.5^\circ \times \frac{\pi}{180^\circ} \approx 0.2705\text{ rad}$
* **Latitude ($\theta$):** $66.5^\circ \times \frac{\pi}{180^\circ} \approx 1.1606\text{ rad}$ *(using $66^\circ 34'$ gives nearly the same result)*
1. **Calculate the tangent product:**
$$\tan(15.5^\circ) \approx 0.27732$$
$$\tan(66.5^\circ) \approx 2.29984$$
$$\tan(i)\tan(\theta) \approx 0.27732 \times 2.29984 \approx 0.63780$$
2. **Compute the inverse cosine (in radians):**
$$\cos^{-1}(0.63780) \approx 0.87911\text{ rad}$$
3. **Compute the daylight hours ($D$):**
$$D = 24 \left[1 - \frac{0.87911}{\pi}\right] = 24 [1 - 0.27983] = 24 \times 0.72017 \approx 17.284\text{ hours}$$
4. **Convert to hours and minutes:**
$$0.284 \times 60 \approx 17\text{ minutes}$$
$$D \approx \mathbf{17\text{ hours, } 17\text{ minutes}}$$
*(If using the exact $\theta = 66^\circ 34'$, $D \approx 17.304\text{ hours} \approx \mathbf{17\text{ hours, } 18\text{ minutes}}$).*
---
### (e.) Daylight on the Winter Solstice and Conclusion
Using the orbital symmetry rule:
$$D_{\text{winter}} = 24 - D_{\text{summer}}$$
$$D_{\text{winter}} = 24 - 24 = \mathbf{0\text{ hours, } 0\text{ minutes}}$$
#### Conclusion
At the Arctic Circle:
1. **Extreme Seasonal Variation:** Daylight swings between extremes across the year, ranging from **24 hours of continuous daylight** (the Midnight Sun) on the Summer Solstice to **0 hours of daylight** (the Polar Night) on the Winter Solstice.
2. **Equinox Balance:** At the equinoxes, daylight is exactly **12 hours**, identical to everywhere else on Earth.
3. **Geographic Significance:** The Arctic Circle ($66^\circ 34'\text{ N}$ or $66.5^\circ\text{ N}$) serves as the southernmost boundary in the Northern Hemisphere that experiences at least one full 24-hour period where the Sun never sets below the horizon, and one full 24-hour period where the Sun never rises above the horizon.
| |
|
