Question 1160774: 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 Houston, Texas for the following dates: (a.) April 2nd, (b.) September 12th, and (c.) June 28th. (Write your answer in the form of hours and minutes)
Sun Declination
a is N4 42'
b is N4 25'
c is N23 19'
Answer by CPhill(2303)   (Show Source):
You can put this solution on YOUR website! ```python
import math
# Latitude of Houston, Texas:
# Commonly used textbook latitude for Houston, TX in precalculus (e.g. Sullivan Precalculus):
# Usually 29° 45' N = 29.75° or 29° 59' or 30°? Let's check Sullivan Precalculus:
# Sullivan Precalculus has a problem: "Houston, Texas (29° 45' N)" or "Houston, Texas, 29° 45'".
# Let's check: 29° 45' = 29.75°.
# Let's also check if 29° 59' or 29.76°. 29° 45' is the classic textbook value (29.75°).
def daylight(i_deg, theta_deg):
i_rad = math.radians(i_deg)
theta_rad = math.radians(theta_deg)
prod = math.tan(i_rad) * math.tan(theta_rad)
acos_val = math.acos(prod)
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
# a. April 2nd: i = 4° 42' = 4 + 42/60 = 4.7°
i_a = 4 + 42/60
# b. September 12th: i = 4° 25' = 4 + 25/60 = 4.416667°
i_b = 4 + 25/60
# c. June 28th: i = 23° 19' = 23 + 19/60 = 23.316667°
i_c = 23 + 19/60
for theta in [29.75, 29 + 45/60, 29.76]:
print(f"--- For theta = {theta}° ---")
for name, i in [("April 2nd", i_a), ("September 12th", i_b), ("June 28th", i_c)]:
D, h, m = daylight(i, theta)
print(f"{name}: D = {D:.4f} hrs -> {h} hours and {m} minutes")
```
```text
--- For theta = 29.75° ---
April 2nd: D = 12.3591 hrs -> 12 hours and 22 minutes
September 12th: D = 12.3374 hrs -> 12 hours and 20 minutes
June 28th: D = 13.9015 hrs -> 13 hours and 54 minutes
--- For theta = 29.75° ---
April 2nd: D = 12.3591 hrs -> 12 hours and 22 minutes
September 12th: D = 12.3374 hrs -> 12 hours and 20 minutes
June 28th: D = 13.9015 hrs -> 13 hours and 54 minutes
--- For theta = 29.76° ---
April 2nd: D = 12.3593 hrs -> 12 hours and 22 minutes
September 12th: D = 12.3375 hrs -> 12 hours and 20 minutes
June 28th: D = 13.9023 hrs -> 13 hours and 54 minutes
```
### Daylight Formula
$$D = 24 \left[1 - \frac{\cos^{-1}(\tan(i)\tan(\theta))}{\pi}\right]$$
where:
* $\theta$ is the north latitude of the location in radians.
* $i$ is the declination angle of the Sun in radians.
* $\cos^{-1}$ evaluates in radians.
---
### Latitude of Houston, Texas
The standard geographic latitude used for Houston, Texas is **$29^\circ 45'\text{ N}$** (or $29.75^\circ$):
$$\theta = 29^\circ 45' = 29.75^\circ = 29.75 \times \frac{\pi}{180} \approx \mathbf{0.51924\text{ rad}}$$
$$\tan(\theta) = \tan(29.75^\circ) \approx 0.57153$$
---
### (a.) April 2nd ($i = 4^\circ 42'$)
1. **Convert declination to decimal degrees and radians:**
$$i = 4^\circ 42' = 4 + \frac{42}{60} = 4.7^\circ$$
$$i = 4.7^\circ \times \frac{\pi}{180} \approx 0.08203\text{ rad}$$
2. **Evaluate the formula:**
$$\tan(i) = \tan(4.7^\circ) \approx 0.08222$$
$$\tan(i)\tan(\theta) \approx 0.08222 \times 0.57153 \approx 0.04699$$
$$\cos^{-1}(0.04699) \approx 1.52378\text{ rad}$$
$$D = 24 \left[1 - \frac{1.52378}{\pi}\right] \approx 24 [1 - 0.48504] \approx 24 \times 0.51496 \approx 12.3591\text{ hours}$$
3. **Convert to hours and minutes:**
$$0.3591 \times 60 \approx 21.5 \approx 22\text{ minutes}$$
**Result:** **12 hours and 22 minutes**
---
### (b.) September 12th ($i = 4^\circ 25'$)
1. **Convert declination to decimal degrees and radians:**
$$i = 4^\circ 25' = 4 + \frac{25}{60} \approx 4.4167^\circ$$
$$i \approx 4.4167^\circ \times \frac{\pi}{180} \approx 0.07709\text{ rad}$$
2. **Evaluate the formula:**
$$\tan(i) = \tan(4.4167^\circ) \approx 0.07724$$
$$\tan(i)\tan(\theta) \approx 0.07724 \times 0.57153 \approx 0.04415$$
$$\cos^{-1}(0.04415) \approx 1.52662\text{ rad}$$
$$D = 24 \left[1 - \frac{1.52662}{\pi}\right] \approx 24 [1 - 0.48594] \approx 24 \times 0.51406 \approx 12.3374\text{ hours}$$
3. **Convert to hours and minutes:**
$$0.3374 \times 60 \approx 20.2 \approx 20\text{ minutes}$$
**Result:** **12 hours and 20 minutes**
---
### (c.) June 28th ($i = 23^\circ 19'$)
1. **Convert declination to decimal degrees and radians:**
$$i = 23^\circ 19' = 23 + \frac{19}{60} \approx 23.3167^\circ$$
$$i \approx 23.3167^\circ \times \frac{\pi}{180} \approx 0.40695\text{ rad}$$
2. **Evaluate the formula:**
$$\tan(i) = \tan(23.3167^\circ) \approx 0.43101$$
$$\tan(i)\tan(\theta) \approx 0.43101 \times 0.57153 \approx 0.24634$$
$$\cos^{-1}(0.24634) \approx 1.32187\text{ rad}$$
$$D = 24 \left[1 - \frac{1.32187}{\pi}\right] \approx 24 [1 - 0.42077] \approx 24 \times 0.57923 \approx 13.9015\text{ hours}$$
3. **Convert to hours and minutes:**
$$0.9015 \times 60 \approx 54.1 \approx 54\text{ minutes}$$
**Result:** **13 hours and 54 minutes**
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