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Where on Earth (75% of the time in the dark)

  1. Mar 27, 2015 #1
    ...would I have to be so I spend 75% of the time in the dark as the earth spins?

    Imagine the sun shines on the earth in such a way that it casts a shadow that is at 60 degrees to the axis of the spin of the earth (Assume the Earth's axis is 0 degrees) as shown in my diagram below.

    I am trying to work out what 'latitude' (shown as P) I have to be at so I spend 75% of my time in the dark and 25% of my time in sun light as the earth spins about its axis. I am not looking for the actual latitude, but how to calculate d2 (diameter) and h2 (height) from below.

    Can anyone help? I've been at this for ages!

    Fig_A.jpg
     
  2. jcsd
  3. Mar 27, 2015 #2

    Nugatory

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    Staff: Mentor

    How much trigonometry have you studied?
    (Knowing this will help us give you a better answer)
     
  4. Mar 27, 2015 #3
    I know the basics, SOH, CAH, TOA. So can work most simple trig problems out.
     
  5. Mar 27, 2015 #4
    If I look in the downward on the 'Earth' at the circle drawn out by the chord d2, then I know that 25% of the circumference of that circle needs to be in the light and 75% in the dark, which is 270 degrees and 90 degrees respectively. So I can make a segment that splits the circumference of a circle at 0 and 90 degrees, and thus work out the chord of that segment, and the arc which the shadow casts on the earth. But to do all that I would need to know d2 and I just can't figure out how to calculate that :(
     
    Last edited: Mar 27, 2015
  6. Mar 28, 2015 #5
    A lot of trial and error but I managed to solve this. Below are the un-simplified equations. If anyone could help simplify them that would be great :)

    From the diagram above:

    h2 = sin((cos(((1-(cos(θ/2)^2))*360)/2)+1)/2*θ)*(d/2)

    d2 = cos((cos(((1-(cos(θ/2)^2))*360)/2)+1)/2*θ)*d/2)*2

    Where θ is the angle of the shadow (60° in the example I gave) and d is the diameter of the earth (I called it d1 in the above diagram)
     
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