The curve of the sun?

[24Volts]

Hello,

In reference to the following link:

http://www.physicalgeography.net/fundamentals/6i.html

In reference to the graph on page 2 of 4 (Fig. 6i-2). What are they trying to tell me? If we look at the 50 degrees North line (The red one), It seems that approximately in mid June, the maximum hours of day length should be:

Sin(50) = 77% of a 24 hour day!

The maximum hours of the day should equal to = 0.77*24 = 18.4 hours

Why is the graph pointing more towards around 16 hours??

any help would be very appreciated!

Bob

 

[Arouse1973]

I take it as the sin function is used for the intensity of the sun not the length of the day. I might be wrong though.
Thanks
Adam

 

[Gryd3]

Hello,

In reference to the following link:

http://www.physicalgeography.net/fundamentals/6i.html

In reference to the graph on page 2 of 4 (Fig. 6i-2). What are they trying to tell me? If we look at the 50 degrees North line (The red one), It seems that approximately in mid June, the maximum hours of day length should be:

Sin(50) = 77% of a 24 hour day!

The maximum hours of the day should equal to = 0.77*24 = 18.4 hours

Why is the graph pointing more towards around 16 hours??

any help would be very appreciated!

Bob

You got two sections mixed up!
The sin(50) you are using is talking about light intensity based on the angle of incidence, and the graph you are looking at it indicating the length of day over time... The 50 degree line on our globe does not equal the angle of incidence...
I'm sure there is a proper formula for what you want to calculate... it will need to take into account the angle of our Earth's axis, the degree North or South of the equator, and our relative position from the sun. It's not quite as easy as using a trig function by itself

 

[(*steve*)]

In reference to the graph on page 2 of 4 (Fig. 6i-2). What are they trying to tell me? If we look at the 50 degrees North line (The red one), It seems that approximately in mid June, the maximum hours of day length should be:

Sin(50) = 77% of a 24 hour day!

It's more complex than that.. If it were that simple you would have to be at exactly 90 degrees of latitude to have 24 hours of daylight . As you can see, 70 degrees is sufficient to get 24 hour long days.

I suspect that 66.5 degrees would be sufficient to get at least one 24 hour day due to the Earth's (current) tilt of 23.5 degrees.

 

[24Volts]

Thanks all for your replies,

I would simply need to know where at a given latitude degree (ex: 45Deg North = Montreal), how much daylight is available every day throughout the year (365 days) based on the sunrises and sunsets. In other words for example, lets pick a date at random... suppose I want to know what time the sun comes up (Sunrise) and what time it goes down (Sun set) on September 25. I may may need this information for June 9th... or Dec 25 etc...

Also, I would need the same daylight between susnset and sunrise informations for different locations for example, Mexico (which is at 23Deg North).

Is there a simple formula to calculate this?

Thank you for al your help

 

[(*steve*)]

There is no simple formula, but the calculations are not magic.

The hardest part is determining the local time at a location because this is affected by politics as well as mathematics. Cannulating GMT time is easier.

Many web sites used for projecting solar panel performance can do these calculations. The length of the day is more important than the actual sunrise and sunset times.

Assuming a constant tilt in orbit and an integer number of days in a year simplifies calculations without much loss in accuracy as the difference from one day to the next is minimal.