Chapter 7.1, Problem 70AYU

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 $i°$ at a location $\theta °$ north latitude for any date between the vernal equinox and autumnal equinox. The declination of the Sun is defined as the angle $\theta$ between the equatorial plane and ray of light from the Sun. The latitude of a location is the angle 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, ${\cos }^{-1}\left(\tan i\tan \theta \right)$ must be expressed in radians.

Approximate the number of hours of daylight in New York, New York ( $40°45\text{'}$ north latitude), for the following dates:

Summer solstice $(i=23.5°)$

Vernal equinox $\left(i=0°\right)$

July 4 $\left(i=22°48\text{'}\right)$

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 New York , New york $\left(40°45\prime \text{north latitude}\right)$, for the following dates:

a. $\text{Summer solstice }i=23.5°$

Answer to Problem 70AYU

Solution:

a. $\text{12}\text{.0024 hours}$.

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. $\text{Summer solstice }i=23.5°$.

Formula used:

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

Calculation:

a. $\text{Summer solstice }i=23.5°;\Theta =40°45\prime \text{north latitude }=40.75°$.

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}23.5°\text{tan}40.75°\right)\right)/\pi \right]$

Convert degree into radians.

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0.410}^c\text{tan}{0.711}^c\right)\right)/\pi \right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0.0071\text{ X }0.0124\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0.0000880\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left(89.994\right)/\pi \right]$

$\text{D }=24\left[1–0.4999\right]$

$\text{D }=24\left[0.5000\right]$

$\text{D }=12.00\text{ hours}$

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 New York , New york $\left(40°45\prime \text{north latitude}\right)$, for the following dates:

b. $\text{Vernal equinox }i=0°$.

Answer to Problem 70AYU

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. $\text{Vernal equinox }i=0°;\text{ Θ }=40°45\prime \text{north latitude }=40.75°$

Formula used:

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

Calculation:

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}0°\text{tan}40.75°\right)\right)/\pi \right]$

Convert degree into radians.

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0}^c\text{tan}{0.711}^c\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0\text{ X }0.0124 \right)\right)/\text{π}\right]$

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

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

$\text{D }=24\left[1–0.0055\right]$

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

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

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 New York , New york $\left(40°45\prime \text{north latitude}\right)$, for the following dates:

c. $\text{July}4=22°48’$.

Answer to Problem 70AYU

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. $\text{July}4 i=22°48’=22.8°;\text{ Θ }=40°45\prime \text{north latitude }=40.75°$

Formula used :

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

Calculation:

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}22.8°\text{tan}40.75°\right)\right)/\text{π}\right]$

Convert degree into radians.

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0.3979}^c \text{tan}{0.711}^c\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0.0069\text{ X }0.0124 \right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left({\text{cos}}^{-1}\left(0.00008556\right)\right)/\text{π}\right]$

$\text{D }=24\left[1–\left(89.995°\right)/\text{π}\right]$

$\text{D }=24\left[1–0.4999\right]$

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

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