Question 1160776: 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(degree) and theta(degree) must be expressed in radians.
Approximate the number of hours of daylight in Quito, Ecuador for the following dates A) June 20 B) March 21st C) May 3rd
A = N23 26'
B = 0 00'
C = N15 30'
Answer by CPhill(2303)   (Show Source):
You can put this solution on YOUR website! ```python
# Let's inspect Quito's latitude in typical textbook problems, or calculate for theta = 0 degrees (0 radians).
# If theta = 0 (on the Equator):
# tan(theta) = tan(0) = 0
# tan(i) * tan(theta) = 0 for any finite declination i.
# cos^-1(0) = pi / 2
# D = 24 * [1 - (pi/2) / pi] = 24 * [1 - 0.5] = 12 hours!
import math
def calc_D(i_deg, theta_deg):
i_rad = math.radians(i_deg)
theta_rad = math.radians(theta_deg)
arg = math.tan(i_rad) * math.tan(theta_rad)
acos_val = math.acos(arg)
D = 24 * (1 - acos_val / math.pi)
return D
# Let's test with theta = 0
print("theta = 0:")
print("A (23° 26'):", calc_D(23 + 26/60, 0))
print("B (0°):", calc_D(0, 0))
print("C (15° 30'):", calc_D(15 + 30/60, 0))
```
```text
theta = 0:
A (23° 26'): 12.0
B (0°): 12.0
C (15° 30'): 12.0
```
```python
# Let's also check if theta could be 0° 14' S or something:
# If theta = -0.233°:
# But the problem states: "at a location of theta(degree) north latitude".
# In precalculus (e.g. Sullivan, Larson, or Stewart Precalculus), Quito is famous for being right on the equator (latitude 0°).
print("Quito is directly on the Equator (latitude theta = 0°).")
```
```text
Quito is directly on the Equator (latitude theta = 0°).
```
### Daylight Formula
The formula is given by:
$$D = 24 \left[ 1 - \frac{\cos^{-1}(\tan(i)\tan(\theta))}{\pi} \right]$$
where:
* $\theta$ is the latitude of the location (in radians).
* $i$ is the Sun's declination (in radians).
* $\cos^{-1}$ evaluates in radians.
---
### Latitude of Quito, Ecuador
Quito, Ecuador is situated directly on the Equator, so its geographic latitude in this standard mathematical model is:
$$\theta = 0^\circ = 0\text{ radians}$$
Because $\tan(0) = 0$, for **any** solar declination angle $i$:
$$\tan(i)\tan(\theta) = \tan(i) \times 0 = 0$$
$$\cos^{-1}(0) = \frac{\pi}{2}\text{ radians}$$
Substituting this into the daylight equation gives:
$$D = 24 \left[ 1 - \frac{\frac{\pi}{2}}{\pi} \right] = 24 \left[ 1 - \frac{1}{2} \right] = 24 \times 0.5 = \mathbf{12\text{ hours}}$$
---
### Solutions for the Given Dates
#### A) June 20 ($i = 23^\circ 26'$)
* **Declination:** $i = 23^\circ 26' = 23.433^\circ \approx 0.4090\text{ rad}$
* **Latitude:** $\theta = 0\text{ rad}$
$$D = 24 \left[ 1 - \frac{\cos^{-1}(\tan(0.4090)\times 0)}{\pi} \right] = 24 \left[ 1 - \frac{\frac{\pi}{2}}{\pi} \right] = \mathbf{12\text{ hours, } 0\text{ minutes}}$$
---
#### B) March 21st (Vernal Equinox, $i = 0^\circ 00'$)
* **Declination:** $i = 0^\circ = 0\text{ rad}$
* **Latitude:** $\theta = 0\text{ rad}$
$$D = 24 \left[ 1 - \frac{\cos^{-1}(0 \times 0)}{\pi} \right] = 24 \left[ 1 - \frac{\frac{\pi}{2}}{\pi} \right] = \mathbf{12\text{ hours, } 0\text{ minutes}}$$
---
#### C) May 3rd ($i = 15^\circ 30'$)
* **Declination:** $i = 15^\circ 30' = 15.5^\circ \approx 0.2705\text{ rad}$
* **Latitude:** $\theta = 0\text{ rad}$
$$D = 24 \left[ 1 - \frac{\cos^{-1}(\tan(0.2705)\times 0)}{\pi} \right] = 24 \left[ 1 - \frac{\frac{\pi}{2}}{\pi} \right] = \mathbf{12\text{ hours, } 0\text{ minutes}}$$
---
### Summary
For all three dates, Quito experiences **12 hours and 0 minutes** of daylight. Because Quito lies on the Equator ($\theta = 0^\circ$), the geometric horizon bisects the Sun's daily diurnal path equally throughout the entire year regardless of the Sun's declination.
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