If i wrote something in the sky what would the radius of visibility on the ground? If i wrote something in the sky 8500-10,000 ft high 40 ft tall letters what would be the visibility radius on the ground? Or what formula would i use to come up with the answer?
 A: Let's assume that you want to write your letters in a square large enough for a little phrase and spacing. Call $\theta$ the angle of the effective field of view (FOV) of the eye (most of our visible area is a little blurred, we have almost a 180° view but we can see clearly only a small angle). here's a little picture for a better explanation 
that's why we can see little object near us but only big objects at greater distances.
Speaking of formulas "reading" a square with a side lenght L at an altitude H requires a minimum FOV of
\begin{equation}
\theta_{min} = 360°\frac{L}{2\pi H}
\end{equation}
where the differences between the square and the sperical sector are assumed negligible and the observer is "under" the centre of the square (if you want a more precise formula we can discuss later, I also don't know the precise value of $\theta_{min}$, I assume it's about $\approx$ 10° for be able to read a text, but more research/googling is needed).
this image should be helpful to visualize the geometry.

if $\theta_{min}$ is too low you must decrease the altitude H o make the text bigger (increase L).
If the text is closer or bigger enough to be read, we can start moving out from the area below the square (that is also the nearest). Given H, L and $\theta_{min}$ we can calculate the radius by first calculating the max distance $D_{max}$ from the square to have it readeable inverting the formula that i wrote before (where H is swapped with $D_{max}$ and the fact that we are viewing the square a little to the side is considered negligible)
\begin{equation}
D_{max}= 360°\frac{L}{2\pi \theta_{min}}
\end{equation}
and finally we can find the radius of visibility on the ground $R_{max}$ using the pythagoras theorem.
\begin{equation}
 R_{max} = \sqrt{D_{max}^2 - H^2}
\end{equation}
Post Scriptum
The calculations in this explanation are super semplifed for a better and easier understanding, but you should still achieve a good result. If you have any question we can discuss in the comments.
The formulas should be correct, it's my first time here and I hope I didn't make any mistake.
Sorry for the bad english
A: The size of the letters and their distance will be proportional to size and distance where they are to be calculated.  This is arrived by calculating angle those letters subtend at eye.
Let say if their visibility radius is to be calculated at 1 ft distance then:
$$\frac{sz}{1 ft} = \frac{40 ft}{10000 ft}$$
=> $$sz = .0004 ft$$
A: 40 ft at 10,000 feet is 15 sec. It would easily focus onto a single retinal cone cell which is about 30 sec across.
