Sun Local Hour Angle and Latitude

In summary: It is asking you to solve for Lat, not just for sin(L). In summary, the problem involves finding the value of Lat in an equation involving H, Lat, and Dec. The solution involves simplifying the equation and using a trigonometric function to eliminate the variable. The final step is to solve for Lat using the simplified equation.
  • #1
moja_a
4
0

Homework Statement



if:
H = Local Hour Angle
Lat = Latitude.
Dec = Sun Declination.

cos(H) = -sin(a)-sin(Lat)*sin(Dec) / cos(Lat)*cos(Dec)


I wand to get The value of Lat .

The Attempt at a Solution


I Tried to make it simple By :
1 - multiply both sides by the denominator
1- square both sides
2- Rewrite cos^2(L) as 1-sin^2(L)
3- Rearrange into a quadratic equation in sin(L).
And this is what I got:

Cos^2(H).Cos^2(D).Cos^2(L) = -Sin^2(a) - 2Sin(a).Sin(D).Sin(L) + Sin^2(L).Sin^2(D)

Cos^2(H).Cos^2(D).(1-Sin^2(L) )= -Sin^2(a) - 2Sin(a).Sin(D).Sin(L) + Sin^2(L).Sin^2(D)

Cos^2(H).Cos^2(D) - Cos^2(H).Cos^2(D).Sin^2(L) = - Sin^2(a) - 2Sin(a).Sin(D).Sin(L) + Sin^2(L).Sin^2(D)

Cos^2(H).Cos^2(D) = -Sin^2(a)- 2Sin(a).Sin(D).Sin(L) + Sin^2(L).Sin^2(D) + Cos^2(H).Cos^2(D).Sin^2(L)

Cos^2(H).Cos^2(D) = -Sin^2(a)- 2Sin(a).Sin(D).Sin(L) + Sin^2(L) . ( Sin^2(D) + Cos^2(H).Cos^2(D) )

then I am Stuck can you help me please.
 
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  • #2
Welcome to Physics Forums.

So, is the problem asking you to solve the equation for Lat (or "L")?

As I read the equation, Lat appears in the form

sin(Lat) / cos(Lat)​

This can be replaced by a different, well-known trig function.
 
  • #3
Yes I used L = Lat
can you tell me about the function that replace it
 
Last edited:
  • #4
can you tell me about the function that replace it
 
  • #5
There are six standard, well-known trig functions. It is the one that is equivalent to

sin(θ) / cos(θ) = ___?​

Surely you know this?
 
  • #6
[tex][/tex]

may be I wrote the formula incorrectly

[tex]Cos H = \frac{sin(a)-sin(L)*Sin(D)}{Cos(L)*Cos(D)}[/tex]

I want to get the value of sin(L)
 
  • #7
[tex] cos(H) = \frac{sin(a)-sin(L)*sin(D)}{cos(L)*cos(D)} [/tex]

[tex] cos(L) = \frac{sin(a)-sin(L)*sin(D)}{cos(H)*cos(D)} [/tex]

[tex] cos(L) = \frac{sin(a)}{cos(H)*cos(D)} \; - \; \frac{sin(D)}{cos(H)*cos(D)}sin(L) [/tex]

define constants A and B appropriately and the equation becomes:

[tex] cos(L) = A - B sin(L) [/tex]

You could square both sides at this point, change cos2(L) to 1 - sin2(L) and solve the quadratic for sin(L).
 
  • #8
Or Sin(L)=[Sin(a)-Cos(L)*Cos(H)*Cos(D)]/Sin(D)
 
  • #9
RTW69 said:
Or Sin(L)=[Sin(a)-Cos(L)*Cos(H)*Cos(D)]/Sin(D)

How would that help? You've got a cos(L) on the RHS. You're trying to solve for L (eventually).
 
  • #10
You are correct, I should have read the problem statement a little closer.
 

What is Sun Local Hour Angle (LHA)?

Sun Local Hour Angle (LHA) is the angular distance between the observer's meridian and the hour angle of the sun at a given time. It is used to determine the position of the sun in the sky and is measured in degrees.

How is Sun LHA calculated?

Sun LHA is calculated by subtracting the observer's longitude from the hour angle of the sun. The hour angle of the sun can be calculated using the current time and the sun's right ascension, which is the celestial equivalent of longitude.

What is the significance of Sun LHA?

Sun LHA is important in celestial navigation as it helps determine the position of the sun relative to the observer. This information is essential for determining the observer's location on Earth.

How does Sun LHA relate to latitude?

The Sun LHA is directly related to the observer's latitude. When the sun's LHA is equal to the observer's longitude, the sun is directly overhead at the observer's location. When the LHA is equal to 90 degrees, the sun is on the horizon at the observer's due east or west, depending on the time of day.

What factors can affect Sun LHA?

The main factors that can affect Sun LHA are the observer's longitude, the time of day, and the time of year. The sun's LHA changes throughout the day as it moves across the sky, and it also changes throughout the year due to the Earth's tilt and its orbit around the sun.

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