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

  1. Mar 27, 2015 #1


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    ...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!

  2. jcsd
  3. Mar 27, 2015 #2


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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


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    I know the basics, SOH, CAH, TOA. So can work most simple trig problems out.
  5. Mar 27, 2015 #4


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    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


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    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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