I still for the life of me can't work out what you are on about but lets fill in some maths numbers and see if it helps you describe what you are trying to solve since you are actually communicating.
No idea what this means in a stand alone I am assuming something to do with the sun brightness so lets answer that.
The sun output has been studied by climate science for the last 20 years or so so we know the actual direct solar irradiance at the top of the atmosphere fluctuates by about 6.9% during a year (from 1412 Watts per square meter in early January to 1321 Watts per square meter in early July) due to the Earth's varying distance from the Sun.
No way of doing this as a ground based observation because clouds and even the temperature of atmosphere change everything at the ground where you and I live.
I think that is what you are suggesting you are going to try and measure but it can't be done at ground level because of the atmosphere issue. So if that is what you are suggesting it won't work.
The solar irradiation at the actual sun surface is remarkably stable even the appearance or not of sunspots changes the value little and I will ignore that for now because I don't think that is what you are on about.
Ok I am assuming from this you realize there is a flight time of 8 minutes +- a little bit depending on time of year and distance we are from sun.
The photons of light fly in a direct straight line so that means the photon that is leaving the sun right now will not hit earth because earth will have moved in the 8 minutes of flight time.
The only photons that will hit earth are the ones that leave the surface of the sun slightly forward of our current position and the earth and the photons sort of collide in a 3D collision.
This is exactly the same when we fire lasers at the moon you have to fire forward of the moon position you are aiming at to allow for flight time. There are actually many reflection targets on the moon science uses
http://www.space.com/20865-soviet-moon-rover-lunokhod-laser.htmlhttp://www.princeton.edu/~achaney/tmve/wiki100k/docs/Lunar_Laser_Ranging_Experiment.html
As the last link explains it is almost the same as "a rifle to hit a moving dime (US coin) 3 kilometers (two miles) away."
So I am not understanding what part of this you are struggling with the photons get emitted from the suns surface in a straight line like a bullet. The ones that leave slightly ahead of a straight line between earth and the sun currently we will collide with in 8 min time.
Sun was in p1 if I see P1 where is the sun Where I'm ?
how far from p1
This makes no sense to me lets turn it back to a rifle, bullet and target because that is exactly what we have.
My bullet from my gun take 8 min to reach the target but the target is moving in that 8 min. If I aim directly at the target my bullet will miss and fall behind it. What I have to do is aim slightly in front of the motion of the target to hit it.
Lets say I am not a good sniper so instead of a rifle I get myself a machine gun. What I do is spray a pile of bullets ahead of the target increasing my chances of hitting it. Anti-aircraft guns work on that principle they lay down a field of fire ahead of the current position of the aircraft.
The sun is like the ultimate machine gun it lays down a massive amount of photons in 360 degree the exact number can be calculated to be 1 x 10E45 photons per second
Here is the calculation by Dr Brian Cox.
http://harveyjohnson.wordpress.com/2013/...un-in-a-second/Earth will encounter only a small fraction of those 1x1045 photons emitted every second being the ones emitted pointing slightly forward of earths current position.
You can do a quick calculation of what percentage of the suns light we hit easily
1360 Watts per square meter is a rough average sunlight per square meter on earth which equals approximately 3x10E21 photons.
The cross sectional area of earth is
pi*6378000^2= 127,796,483,130,631 square meters.
I will round that to 1.3x10E14 square meters
So earth gets hit by 3x10E21 x 1.3x10E14 = 3.9x10E35
So the sun emits 1x1045 photons per second of which earth get hit by 3.9x10E35 photons per second so that is 0.0000000000039%.
So as you can see only a tiny percentage (0.0000000000039%) of photons emitting out from the sun per second ever hit the earth and those that do are the photons that were pointing at our current position 8 minutes ago just like the bullet and target example.
Now I have no idea if any of that helps you but at least you have numbers and mathematics to start.