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Scott, MIT Graduate
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How do you use a clinometer to calculate sunrise delays ...

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How do you use a clinometer to calculate sunrise delays from published sunrise times for specific sites in a given area? I use other methods for calculating the sunrise delay but need a method which allows me to calculate the sunrise delay as I stand on specific sites. We use the delays to determine which sites are mist suitable for erecting buildings. I need to be able to say precisely how many minutes the sun is visually delayed from published times for any site. For example, if I am standing on a site in an area that the published data says the sun will rise at 6:13 AM and a hill to the East is obstructing my vision of the sunrise until some minutes later, I want to be able to calculate that delay with the clinometer without having to be there in the morning at the time of the sunrise. I am assuming there is a way to do this after establishing what angle the top of the obstruction is. Since I have never used a clinometer I have no idea how to do this and need to buy one.
Submitted: 7 years ago.
Category: General
Expert:  Scott replied 7 years ago.

Hi Dan, and welcome to JA!

There are a few things that you need to consider.

First of all, on different days of the year, the sun will rise from slightly different locations (further south in the winter, and further north in the summer). So to do the calculations generically, we need to assume that the hill to the east is constant in height (or else the adjustment from the published data will vary per day). The next assumption that I'll make is that the earth is flat. That may seem drastic, but on the small scale that you're dealing with, it's a fair assumption.

Next, the clinometer... You can measure the exact angle to the top of the hill from your location, in degrees (your clinometer should come with specific instructions on how to do that, since they are all slightly different). Call that number of degrees x.

Now, remembering our "flat earth" assumption, the sun has to travel 180 degrees during the day (this number will also vary drastically depending on the time of year). So to compute the adjustment to the length of the day, take x/180 away from the day length.

An example will help. Say your tables indicate a sunrise of 6am and a sunset of 6pm. The day is 12 hours long (that's 720 minutes). In your measurement, you see that the hill gives an angle of 2 degrees. Therefore, the proportion you need to take away is 2/180, or 1/90. 1/90th of 720 minutes is 8 minutes. The sunrise at your location will be about 6:08am.

One more example. Let's say during the summer, at the same place, the sunrise is 4am, and the sunset is 8pm. Now the day is 16 hours long (or 960 minutes). Using the same proportion (1/90), we see that the sunrise is nearly 11 minutes later than indicated, or about 4:11am.

In summary, we had to make a number of assumptions, the biggest ones being:

  • The hill doesn't vary in height, so that even though the sun rises in slightly different places each day, we don't have to repeat the calculation.
  • The sun travels a complete 180 degree path each day. This is not true, but allows the calculations to be much, much simpler.

If you don't want to perform these calculations yourself, I have found a software package ($99) that can do it for you (with hills, etc.). The software would also allow you to see how the path of the sun changes throughout the year.

http://www.wide-screen.com/sunPATH/index.shtml

Let me know if you have any questions about any part this. Good luck!

-Scott

Scott, MIT Graduate
Category: General
Satisfied Customers: 17710
Experience: MIT Graduate, United States Traveler, Information Finder
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