What Does the 4a Angle Mean in Light Trails?

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The discussion revolves around understanding the "4a angle" in light trails, particularly how it is calculated and its significance. The angle is derived from the relationship between the observer's eye position and the light rays reflecting from water waves at varying angles. It is explained that if water were a flat mirror, only one spot of light would be visible, but waves create multiple reflections, leading to the 4a angle. The total deviation from a flat surface is twice the angle of inclination, which contributes to the observed angle. The conversation highlights the importance of geometry in visualizing this phenomenon and confirms that the calculations align with the expected outcome.
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Hi

I've read about "light trails" in http://www.islandnet.com/~see/weather/elements/glitter.htm

but I didn't understand how they have calculated the 4a angle, and what really is the meaning of it.
 
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Your eye is located at a fixed spot. The waves are at many different angles at many different locations. Consider waves with maximum inclination "a" at many different locations. If the location is too far or too near, the wave with maximum inclination "a" will not reflect light to your eye position. The angle "4a" is formed by the position of your eye, and the light rays that reach your eye position from the closest and farthest wave locations able to reflect light to it.
 
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Thank you, but how did they get it was exactly 4a?
 
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I cheated and looked in the section on "Glitter" in Lynch and Livingston's book. If the water was a perfectly flat mirror, you would see one spot of light. With waves at many angles going from "-a" to "+a", you will see spots corresponding to mirrors at all those angles. The crucial figure is not available for preview. My guess is they assumed that the sun's rays are incident exactly perpendicular to the horizontal, so the angle of incidence and angle of reflection are both zero from a perfectly flat surface. When the surface is inclined at some angle, the total deviation from the zero of the flat surface is twice the angle of inclination because you add the angle of incidence and the angle of reflection. So the total range of angles seen is twice the total range of angles in the water waves. http://books.google.com/books?id=4Abp5FdhskAC&printsec=frontcover#PPA83,M1
 
Hey thank you again
but why the total deviation is not 90-a?
 
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Do you mean like that:

(I've added in blue the 2a angles)
 

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Yes, that's what I thought. But when I worked it out, I got this. The sum of angles at the eye seems to still work out to 4a. I flunked the last geometry test I took about 15 years ago - I wonder what the right answer is. :confused: (The attached picture is my alteration of a picture from Joseph Shaw's http://www.etl.noaa.gov/about/eo/science/glitter/glitter.html )
 

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Thank you, I see that with a bit of geometry it does works!
 
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Thanks - it's nice to have someone check my geometry too, and it explains a really pretty phenomenon!
 
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