Earth shine algorithm

[SPASE_1976]

how can we find..

1) the f angle

2) the percentage of the Yellow area as viewing the SPHERE from the vessel ( not the 2d plane as seen in the picture )

 

[pilotpercy]

(1) R/d= sin(f) i think, its been a long time since ive used trig.

(2) as d tends to infinity, view of the sphere will tend to 50%, i think you will need to find a function for d for this

 

[SPASE_1976]

The Yellow area versus brown area percentage of the 2dPLANE is equal to
the SPHERE's Yellow area versus brown area percentage ?

i mean the 2d vs the 3d have the same Yellow %
do they have the same analogy ?

 

[pilotpercy]

yeah, from 2D you will see X% of the circle. in 3D you will see the same X% of the sphere

 

[Fizyk]

1. R/d = sin(f), so f = arcsin(R/d)
2. The percentage is a bit more tricky. I doubt there is the 2D-3D analogy.
First, we need to know k:
k = sqrt(d^2 - R^2)
z = R - (d - k*cos(f)) = R - d + k*cos(f) is a value that will be useful to calculate the area of the yellow part of the sphere, as it is:
A = 2*pi*R*z = 2*pi*R*(R-d+k*cos(f))

Percentage = A/(4*pi*R^2) = 0.5*(R-d+sqrt(d^2-R^2)*cos(f))/R
As sin(f) = R/d, cos(f) = sqrt(1-R^2/d^2) = sqrt(d^2-R^2)/d, hence:

Percentage = 0.5*(R-d + (d^2-R^2)/d)/R = 0.5*(1-d/R+d/R-R/d) = 0.5*(1-R/d)

Funny, much simpler than I thought it would be.

 

[Urwumpe]

The model is wrong...Earthshine does not stop despite of the sun. It is everywhere where a illuminated part of Earth is visible.

 

[SPASE_1976]

@Fizyk
can you simplify your equations ? i'm lost
is this for the 2d or the 3d ?
the 'z' is the yellow arc ?


heres what i figure out so far for the 2d version...


we know f so the <ab> angle = 180 - 90 - f
we divide it to the 360 degrees

FRAC=( 180 - 90 - f ) / 360 degrees

apply the above fraction to the total area of the circle

BIG_AREA = FRAC*( πR^2 )

the vertical red line VRL = k*sin(f)
we find the SIDE from the center of the planet to the visible limit (red line)

SIDE = SQR ( R^2 - VRL^2 )


the triangle [VRL,SIDE,R] has area...
( VRL*SIDE) /2

So the visible 2d area is

2*( BIG_AREA - VRL*SIDE /2 )

now we need to find the yellow area which ... i'm working on it

 

[tblaxland]

The model is wrong...Earthshine does not stop despite of the sun. It is everywhere where a illuminated part of Earth is visible.

It is half right, if I understand it correctly. He is trying to determine how much of the illuminated Earth is visible.

The part that is wrong is that how much (ie, what percentage) of the surface is illuminated and visible does not determine the Earth-shine but rather what percentage solid angle the illuminated part occupies when viewed from the vessel. The albedo of the Earth would also need to be taken into account (an average value could be taken, for a first approximation).

 

[SPASE_1976]

@tblaxland

The idea is to find the percentage of the illuminated surface to the vessel's visual 3d sphere.

we find the percentage of the bright side to the planet's visible circle THEN (not mentioned here) we find the percentage of the planet's visible circle to the vessel's visual 3d sphere and we have the fraction.


An addition would be to find if the sunligh reflects through the sea or land to the vessel -- must find the reflection's (SUN>EARTH>VESSEL) latitude/ longitude and sample a mini 'EARTH MASK' bitmap to see if theres land or sea at that point.

i posted this problem here as a math puzzle , i'm not writting an OVP


-- EDIT ---------------------------------------------


We must treat the vessel as a sphere
in this example the reflected light strikes mostly the vessel's 'bright side'

that doesn't make much difference to the human eye , we care about the vessel's 'dark side' which it's illumination is based entirely on the 'Earth-shine'

so we must find the GREEN area's percentage to the vessel's 'dark side' to get a coarse ambient level..

This model could be used for the planet's moon(s) too , but i don't recoment it , an easy way is just to declare a light source at the center of the planet.

 

[Fizyk]

@Fizyk
can you simplify your equations ? i'm lost
is this for the 2d or the 3d ?
the 'z' is the yellow arc ?

It's for 3D. 'z' is the... hmm... kind of "height" of the yellow part, it is the part of the (middle of the planet)-vessel line, that is between the red line and the planet surface.

I think the f=arcsin(R/d) part is pretty straightforward, so I'll focus on the percentage calculation.

I wrote the equations again in LaTeX to make them more readable. For some reason the text is right-aligned, sorry for that, but I think it is still easier to read.

A is the area of the yellow part on a sphere. 4 pi R^2 is the total area of the sphere.

 

[Urwumpe]

I would solve it with vectors and the angle between two direction vectors. The area is just simple again: The area is the integral of all meridians from 0 to the angle between the vectors. The length of a meridian on a unit sphere is pi. As we only look for the percentage, the answer is thus simple: