Expert: Scotto - 4/10/2011

Question
Hello,

Sorry for bothering you, but I asked you a question regarding the distance between the centres of sun and moon when the sun has a 10% eclipse. Your advice was quite helpful and I think I've cracked it. Can you please see if it's okay:
Let A be angle between two radii forming a sector.
Since both have the same radius, A= 1/2r^2.A
Total area = pir^2 . However 10% of sun's area is covered by 10% of moon's so 0.1+0.1= 0.2 and 0.2pi.r^2= 1/2r^2.A which gives A=0.4pi. Let y=distance between two centres. so Cos A=y/r, which then gives y=1.62r.

Terribly sorry for taking up your time.
Thanks
May

Answer
It's OK, since I enjoy answering them.
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Now for 10% of the area to be covered, the area in the eclipse is 0.1S.

If y = distance between the two centers, then we know that (y/2)/r = cos(A/2) since y/2 is the
length of the base, r is the hypotenuse, and A/2 is the angle.  This gives y = 2r*cos(A/2).

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Back to my old idea, I think I've figured it out.

If 10% of the sun's area is covered by 10% of the moons area,
Let S be the total surface area.

That means the total will area be S + S - 0.1S = 1.9S, since 10% of the area is overlapping.

It is known that S = πr² and the angle of intersection is A.

The amount of space inside a circle of angle A of a circle is (A/(2π))S = Ar²/2.

The height of half of this is h/2 = r*sin(A/2) and the width still in the sun is w = r*cos(A/2).
Now the area of the triangle inside is wh/4, but there are two triangles, so it is wh/2.
That is r²sin(A/2)cos(A/2)/2.  It is known that sin(2x) = 2sin(x)cos(x),
so sin(2x) = 2*sin(x)*cos(x).  Given x = A/2, this turns r²sin(A/2)cos(A/2)/2 into r²sin(A)/4.

Thus, the total area of half of the amount of area in the eclipse is Ar²/2 - r²sin(A)/4.
That is (Ar²/2)(1 - sin(A)/2).  This needs to be 0.05S, which is 0.05πr².

Solving gives 0.05r² = (Ar²/2)(1 - sin(A)/2).
Dividing both sides by r² gives 0.05 = (A/2)(1 - sin(A)/2).

Using Newton's method, this gives
x   f   f'
0.400000000 -0.111058166  -0.308060484
0.039492335  0.030643642  -0.480261691
0.103298470  0.001013667  -0.448491295
0.105558640  0.000001166  -0.447370631
0.105561247 -0.0000000001  -0.447369339
0.105561247  0
You might check this out and see if it is OK.