You are way off. **The Reunion Tower is only 561 feet tall. **

561 ft

**above the ground it stands on**. So you need to add that ground level to the height of the tower for the tower's height above MSL. Mean Sea Level (or a similar reference level). Then consider the height of the observer above that level, and the height of the intervening hills, also

**measured from the same level**. If all these are set out, on a flat plane, it can be seen that the hills inbetween cannot intrude on the observer's sightline to the Reunion Tower, such that only the ball of the tower can be seen. In fact, the whole tower should be visible if all this is set out on a flat plane... (EDIT correction; MOST OF the tower. The top 361 feet or so)

Now, look at it this way:

Observer = ~800-850 ft

Colleyville terrain obstruction = ~600-650 ft

Dallas = ~450 ft

If you consider the elevation of Dallas, ~450 feet, as simply 0 feet, then consider the point of the observer to be ~400 feet above that and the elevation in between, the Colleyville area, to be ~200 feet above that zero point of elevation, then the photo looks **"EXACTLY"** as it should look.

No, it does not, for the photographer states that most of the 561 feet of the Reunion Tower is HIDDEN BY THE HILL, and it is

561 ft ABOVE your zero reference level for Dallas.... so the photographer would be looking up from 450 to 561, with a 200 hill between - a hill which cannot intrude on his sightline

You are purposely ignoring the roughly 100 feet of trees and buidlings above the elevation of the Colleyville 600-650 foot elevation.

Just using the figures you quoted to start with. Now you want to change them? Makes no difference, anyway

And for those who keep saying the observer should be looking UP, that is just a bunch more nonsense. You are looking at a city that is FORTY FIVE MILES AWAY! The tallest building is 915 feet

So if the observer is at 800, and he's looking at something 915, he's looking up, isn't he? Makes no difference how far away, if it's all on a flat plane, and the city hasn't shrunk into the ground .....

He would be looking level if he looked at something of 800, and down if he looked at something of 700. No?

The horizon is at eye-level. That is where the buildings are.

Eye level is the observer's height. At least one of the buildings is 561 feet high, so the whole of it cannot be AT eye level ...

I'd like to issue you a challenge.

Fire up your CAD diagram machine and create a diagram of your ROUND EARTH depiction of that photo, from the observer to the target city--using ACCURATE dimensions, showing ACCURATE curvature/drop for that distance (1350 feet), with the observer on the left and the target on the right - just like you've been showing with a FLAT ground line and make this one with a ROUND ground line. Wow us with your results.

Don't have CAD, but ... why should I? I've shown that it's totally inconsistent with a flat plane. You tried to redefine the reference level for the plane, such that Dallas = zero level, but even that does not fit with the photo.

What IS the reference level for all the heights you have quoted for the land? Are they consistently defined from one level? If so, what is that reference level?