Simple math, ancient Egypt:

Originally posted by @sonhouse
The bend at R=1 is what Einstein calculated, 1.75 arc seconds or about 8.5E-6 radians. I was just rounding out. R=2 is 1/2th of that, about 4.2E-6 radians or 0.875 arc seconds.

There are 1,296,000 arc seconds in a complete circle.


So if you have a telescope with resolution of 1 arc second it can split the circle into that many parts, 1.2 million. ...[text shortened]... the distance between Earth and sun, or about 93 million miles) AU stands for Astronomical Units.

It probably poops out because the further away the more the light rays are running true parallel. The longer
distance will yield fewer light rays, but you are losing light rays that are not correct. In other words, if viewed like a delta epsilon proof, the further the star, the smaller the delta which results. The smaller the delta, more bendable the light.

You can work that into your equation by limiting the range to the distance between the light source and the object bending the light.
F < distance between the two objects.

Basically what you are looking at is two opposing cones with the object doing the bending being the ice cream between. Each cone is identical.

Looking at it another way, if you view one ray of light, it would be the hypotenuse with the distance between the object being one leg and the radius of the ring is the other leg. The light is bent so that the angle of the light on the other side is the mirror image. It creates an isosceles triangle with the radius of the furthest ring being the height. This would only represent the furthest viewable ring. Funny how that one is similar to how light bounces off a mirror.


It makes sense that the larger the radius of the object doing the bending the more light at the focal point. The entire arc is being focused at one point. The smaller the arc length, the smaller the amount of light.

The larger the planetary body, the larger the lens.

As for the arc seconds, this would be calculated by 360*60*60. The two 60's are for minutes then seconds.

I get 4.848136811E-6 radians per arc second. You should use 2pi ÷1296000 in your formula. Never round off intermediate calculations. So 2pi ÷1296000÷n where n=ring radius÷planetary body radius.

Is there a way to calculate the maximum ring radius being bent?

Originally posted by @eladar
It probably poops out because the further away the more the light rays are running true parallel. The longer
distance will yield fewer light rays, but you are losing light rays that are not correct. In other words, if viewed like a delta epsilon proof, the further the star, the smaller the delta which results. The smaller the delta, more bendable the light ...[text shortened]... radius÷planetary body radius.

Is there a way to calculate the maximum ring radius being bent?

I have been corrosponding with Geoffry A Landis, a scientist and sci fi writer about the issue, he wrote a paper suggesting getting to the focal point, first focus would not be as helpful as people thought because of interference with the corona. If you had a telescope aimed at the sun and block out the suns light you could theoretically get many times ampilified views of some distant object using the sun as a giant lens. But he thinks you would have to go out way further than first focus say up to R=2, or 4 times the first focus distance or 200 odd billion miles out not 50.

As you go further out away from the sun and include higher and higher altitudes the amount of energy goes up that is focused but Landis says the energy density would be the same or go down due to the spot size being larger at the focus point.

One of my thoughts about these spikes of energy would be to try to use them as a free ride on solar sails, ride the spike as far as it will go but of course, if you were following Alpha Centauri, the ride poops out after 4 light years, but Sirius gives a longer 8 light year long ride.

I missed being first to make the formula for the general solution to the focus problem by about 2 years. Those folks wrote a major paper about the subject and used a version of the same formula as I did but they did not simplify it like I did, making 4G/C^2 into a constant, since 4 is a constant, G is a constant, C is a constant so C^2 is a constant so putting them all together as 3E-27 just means you don't have to deal with the whole fraction every time you do a solution. 3E-27 is rounded out, its really more like 2.97 or so but just rounding it out to 3 E-27 is close enough for government work🙂

So you do get the max energy focused just before the end where the diverging cone from the distant source compares to the converging cone of the light bending effect and at some point the diverging cone will be too great an angle for the converging cone to focus. I think that part of the work is original by me, a start of focus pointed out by Einstein and an end to the focus pointed out by me🙂 Of course my work is just a shadow of what big AL did....

Is there a way to calculate the maximum ring radius that is bent to the focal line?

If si I think I can help your equation using trig. Then if the focal length is linear you can use the beginning point and ending point to generalize with a limit built into the equation.

Originally posted by @eladar
Is there a way to calculate the maximum ring radius that is bent to the focal line?

If si I think I can help your equation using trig. Then if the focal length is linear you can use the beginning point and ending point to generalize with a limit built into the equation.

The way I did it was using trig on paper, charting the increased cone angle of light from a distance source to the decreasing cone angle of the bending and found it was roughly equal to the distance between the source and the sun. It might be modified by choosing stars like Betelgeuse which is a star the size of our solar system so that would throw off the angles somewhat. Not sure how much difference that makes.

Close in stars have a more extreme angle to the cone going by the sun than distant stars which would have a narrower cone so the light comes in tending to more straight on which is why it focuses further out than a close star like Alpha Centauri.

It seems like it might be a bit cumbersome to add trig to the simple R^2/ZM formula since I already know if we know the distance to a star we then know when the focus will end.

Not sure I need anything besides that but it would be interesting to see nonetheless.

Originally posted by @sonhouse
The way I did it was using trig on paper, charting the increased cone angle of light from a distance source to the decreasing cone angle of the bending and found it was roughly equal to the distance between the source and the sun. It might be modified by choosing stars like Betelgeuse which is a star the size of our solar system so that would throw off the ...[text shortened]... ll end.

Not sure I need anything besides that but it would be interesting to see nonetheless.

I was just trying to come up with an equation which self limits.

Originally posted by @eladar
I was just trying to come up with an equation which self limits.

It would be interesting to see. How would you go about that? Comparing an incoming angle (light from source) to outgoing angle (bending inwards of gravity lens) which at some point is equal which would indicate the end of focus.

So it would have to incorporate star distance and end of focus.

So it would have to include something like the line connecting center to center of the source star with the sun as one leg of a skinny triangle.

I can see it as more of a list, since the bending angle for each R number is a constant for a given mass, like the sun's 2E30 Kg, will always produce the same bending angle for a given R.

So we have the line connecting center to center and then the angle of the cone as a variable, where the first angle of interest is the angle of the cone that skims the surface of the sun, a very skinny cone indeed. That would be the starting point and the end point would be where the cone divergence angle = the bending inwards angle.

Originally posted by @sonhouse
It would be interesting to see.

The maximum ring radius would limit the domain. By limiting the domain you naturally limit the range.

Originally posted by @eladar
The maximum ring radius would limit the domain. By limiting the domain you naturally limit the range.

Yes but that domain is still a variable since distance is a variable which means a narrower cone angle for a greater distance between source and the sun.

Originally posted by @sonhouse
Yes but that domain is still a variable since distance is a variable which means a narrower cone angle for a greater distance between source and the sun.

Of course it would not be universal since the length of the focal line depends on the distance between the objects.

Originally posted by @eladar
Of course it would not be universal since the length of the focal line depends on the distance between the objects.

You are getting the idea. But how to put it all together? It seems clear light from a star 1000 light years away will make a focal line 1000 light years long and no more and a star 10 light years away makes a focal line 10 light years long, so perhaps all we need is to prove that point for some number say between 1 and 1 million light years then forget needing the formula to do it for a given star, just knowing the distance tells us the length of the focal line.

Originally posted by @sonhouse
You are getting the idea. But how to put it all together? It seems clear light from a star 1000 light years away will make a focal line 1000 light years long and no more and a star 10 light years away makes a focal line 10 light years long, so perhaps all we need is to prove that point for some number say between 1 and 1 million light years then forget nee ...[text shortened]... mula to do it for a given star, just knowing the distance tells us the length of the focal line.

If the number of radii is constant, then at least one leg is known for each object. I would imagine the bend depends on the density and radius of the object.

Does the 1R bend remain constant for all planetary bodied no matter how dense?

Originally posted by @eladar
If the number of radii is constant, then at least one leg is known for each object. I would imagine the bend depends on the density and radius of the object.

Does the 1R bend remain constant for all plentary bodies no matter how dense?

Total mass is the key, if it is a planet of pure uranium it will mass much more than Earth but if it is the same size as Earth then the bending angle will be greater by the degree of difference between total mass of the body compared to something else, the sun, Earth, Jupiter, whatever. It's the total mass that counts and if you know the radius you have the solution in hand.

Nice thing about orbiting bodies, if they know the orbital mechanics they can work out the mass. Size is a bit trickier.

Originally posted by @sonhouse
It would be interesting to see. How would you go about that? Comparing an incoming angle (light from source) to outgoing angle (bending inwards of gravity lens) which at some point is equal which would indicate the end of focus.

So it would have to incorporate star distance and end of focus.

So it would have to include something like the line connec ...[text shortened]... ng point and the end point would be where the cone divergence angle = the bending inwards angle.

I was assuming the rate of change along the focal line is linear. This would mean one could find a ratio with the only variable is the distance away from the star.

Is this the length of the focal line the distance from the star minus the 1R focal point distance?

Originally posted by @eladar
I was assuming the rate of change along the focal line is linear. This would mean one could find a ratio with the only variable is the distance away from the star.

Is this the length of the focal line the distance from the star minus the 1R focal point distance?

Yes, that's correct, but it is not a huge difference considering the whole thing goes light years long. Compare 50 billion miles to 5800 billion miles (one light year) so less than a 1% difference even for one light year so 4 light years, 1/4th of one percent change and for Sirius, 1/8th of one percent difference.

Originally posted by @sonhouse
Yes, that's correct, but it is not a huge difference considering the whole thing goes light years long. Compare 50 billion miles to 5800 billion miles (one light year) so less than a 1% difference even for one light year so 4 light years, 1/4th of one percent change and for Sirius, 1/8th of one percent difference.

When you do your equations, what it is that you are trying to calculate? Is it each particular focal length for each individual R? Or are you trying to find the R based on the focal length?

I don't remember seeing the answer to this question, does the strength of the gravitational field increase the bend at 1R?