Though I will admit I'm not 100% convinced the Bishop variation actually works. If light is pushed up more the further away an object is, that seems like it would do the opposite of make distant objects vanish - it'd be *harder* for lower, nearer objects to blot out a more distant one.

Maybe he means the electromagnetic accelerator below the Earth attracts it and pulls light down? Which would keep the more distant light from reaching you, but the direction it'd been seen in would be wrong. I swear you need light to curve in a u shape. Hm. There might need to be two variables.

It's actually the opposite.

With something accelerating light upwards, it will curve up. i.e. initially light going down at some angle will go to a shallower and shallower angle until eventually it starts going upwards.

This means the light from the object will hit your eye from a lower angle than the line directly to the object.

This will make it appear lower than it is.

And importantly, that means eventually the light path would need to pass through Earth to reach you.

Conversely, if light was accelerated downwards, then it curves down, then the opposite happens, so the object appears higher.

This allows objects that should be blocked by an object to still appear as the light can start going up and over the object before curving down.

Hypothetically can be done with a ship going over the horizon too. At least nets a ballpark figure, even if one might not know both x and y in perfect detail.

The part which makes it problematic (which I'm sure they know) is that it is claimed as an approximation.

So for something near Earth where the distance covered is small, it wouldn't work.

But ignoring that, we can try with something like Polaris.

We can note that differentiating the function we get y'=(xb/c^2)^(1/3)

(note: sign is due to how I was doing it, focusing on Polaris at x-0, and looking at locations to the right)

So for a latitude of 45 degrees, where y'=1, that gives us:

-1=(xb/c^2)^1/3

-1=xb/c^2

x=-c^2/b.

So if Earth didn't block the way, the light would be horizontal a distance of -c^2/b away horizontally.

This means it must be horizontal at a distance of:

(3/4)*(x ^4*b/c^2)^(1/3)

=(3/4)*((-c^2/b)^4*b/c^2)^(1/3)

=(3/4)*((c^8/b^4)*b/c^2)^(1/3)

=(3/4)*((c^6/b^3))^(1/3)

=(3/4)*(c^2/b)

away vertically (below).

This means it will follow a formula of:

y-(3/4)*(c^2/b) = (3/4)*((x-c^2/b)^4*b/c^2)^(1/3)

Or for more a more general case, if you need light to be travelling at an angle of

*a* as it cuts through the ground, this means:

y'=-tan(a)=(xb/c^2)^1/3

-tan(a)=(xb/c^2)^1/3

-tan(a)^3=xb/c^2

x=-tan(a)^3*c^2/b

y=(3/4)*((-tan(a)^3*c^2/b)^4*b/c^2)^(1/3)

=(3/4)*((tan(a)^12*c^8/b^4)*b/c^2)^(1/3)

=(3/4)*((tan(a)^12*c^6/b^3))^(1/3)

=(3/4)*tan(a)^4*c^2/b

Anyway, doing that I get this:

https://www.desmos.com/calculator/lfhky8uh1sAnd while altering the constant will change where they meet, they will never all line up.