Chapter 7.1, Problem 69AYU

In Problems 79-84, use the following discussion. The formula $D=24\left[1-\frac{{\cos }^{-1}\left(\tan i\tan \theta \right)}{\pi }\right]$ can be used to approximate the number of hours of daylight D when the declination of the Sun is $i^{\circ }$ at a location ${\theta }^{\circ }$ north latitude for any date between the vernal equinox and autumnal equinox. The declination of the Sun is defined as the angle i between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle $\theta$ between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth. See the figure. To use the formula, ${\cos }^{-1}(\tan i\tan \theta )$ must be expressed in radians.

Approximate the number of hours of daylight in Houston, Texas ( ${29}^{\circ }45\text{'}$ north latitude), for the following dates:

(a) Summer solstice $\left(i={23.5}^{\circ }\right)$

(b) Vernal equinox $\left(i=0^{\circ }\right)$

(c) July 4 $\left(i={22}^{\circ }48\text{'}\right)$

To determine

To calculate: The number of hours of daylight $\text{D}$ at the location $\text{θ}$ north latitude Approximate the number of hours of daylight in Houston, Texas $\left(29°45\prime \text{north latitude}\right)$, for the following dates:

a. Summer solstice $i=23.5°$.

Answer to Problem 69AYU

Solution:

a. $12.0024\text{hours}$.

Explanation of Solution

Given:

The declination of the Sun is defined as the angle $i$ between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle $\text{θ}$ between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth.

a. Summer solstice $i=23.5°\text{Θ }=29°45\prime \text{north latitude }=29.75°$

Formula used:

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}i\text{ tanθ }\right)\right)/\text{π}\right]$

Calculation:

a. Summer solstice $i=23.5°\text{Θ }=29°45\prime \text{north latitude }=29.75°$

$\text{D }=24\left[1–\left({\cos }^{-1}\left(\tan 23.5°\tan 29.75°\right)\right)/\pi \right]$

Convert degree into radians.

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0.410}^c\text{tan}{0.519}^c\right)\right)/\pi \right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0.0071\text{ X } 0.0090\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0.0000639\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left(89.996\right)/\text{π}\right]$

$\text{D }=24\left[1–0.4999\right]$

$\text{D }=24\left[0.5001\right]$

$\text{D }=12.0024\text{hours}$.

To determine

To calculate: The number of hours of daylight $\text{D}$ at the location $\text{θ}$ north latitude Approximate the number of hours of daylight in Houston, Texas $\left(29°45\prime \text{north latitude}\right)$, for the following dates:

b. Vernal equinox $i=0°\text{Θ }=29°45\prime \text{north latitude }=29.75°$.

Answer to Problem 69AYU

Solution:

b. $23.86 \text{hours}$.

Explanation of Solution

Given:

The declination of the Sun is defined as the angle $i$ between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle $\text{θ}$ between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth.

b. Vernal equinox $i=0°;\text{ Θ }=29°45\prime \text{north latitude }=29.75°$

Formula used:

$\text{D = 24}\left[\text{1 – }\left({\text{cos}}^{\text{-1}}\left(\tan i\text{ tanθ }\right)\right)\text{/π}\right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}0°\text{tan}29.75°\right)\right)/\pi \right]$

Calculation:

Convert degree into radians.

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0}^c\text{tan}{0.519}^c\right)\right)/\pi \right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0\text{ X }0.0090\right)\right)/\pi \right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0\right)\right)/\pi \right]$

$\text{D }=24\left[1–\left(1\right)/\pi \right]$

$\text{D }=24\left[1–0.0055\right]$

$\text{D }=24\left[0.9944\right]$

$\text{D }=23.86\text{hours}$

To determine

To calculate: The number of hours of daylight $\text{D}$ at the location $\text{θ}$ north latitude Approximate the number of hours of daylight in Houston, Texas $\left(29°45\prime \text{north latitude}\right)$, for the following dates:

c. $\text{July}4=22°48’$

Answer to Problem 69AYU

Solution:

c. $12.0024\text{ hours}$.

Explanation of Solution

Given:

The declination of the Sun is defined as the angle $i$ between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle $\text{θ}$ between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth.

c. $\text{July}4=22°48’\text{Θ }=29°45\prime \text{north latitude }=29.75°$.

Formula used:

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}22.8°\text{tan}29.75°\right)\right)/\text{π}\right]$

Calculation:

Convert degree into radians.

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0.3979}^c\text{tan}{0.519}^c\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0.0069\text{ X }0.0090\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0.0000621\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left(89.996°\right)/\text{π}\right]$

$\text{D }=24\left[1–0.4999\right]$

$\text{D }=24\left[0.5001\right]$

$\text{D }=12.0024\text{hours}$.