Bobby I saw your comment at youtube and came here.

Here's three images to help you. (following this post, in separate posts) Take a look at these images, read carefully what I've annotated and the rationale. Here's some main points I want to impress upon you two:

1) my scale is quite accurate considering the atmospheric distortion, see person on balcony analysis.

2), angular scales based on objects closer to observer are less prone to refraction effects which accumulate over long distances. So I chose the one based on the Le Meridien Delphina in the vertical direction. (one can do a horizontal scale as well, see later comment)

3) the steeper the observation angle the less prone to refraction. peaks of mountains are the best target for observations, as I said in my video.

4) due to the nonlinear behavior of refraction, which changes with elevation angle, the image of distant mountains get's stretched and also it appears as if the "curvature" is hiding objects, but it's not the curvature its the non linear bending, because the peaks come out at the correct height !

5) measuring horizontal distances between buildings produce yet a third scale, which is more accurate, but I choose the vertical one based on the Le Meridien Delphina hotel because the propagation over water (approx 12 miles) introduces yet more compression to the image, and the horizontal scale will lead you astray!

6) I have yet a forth scale based on my image sensor size, pixel count, focal length, etc.. but that's reserved for my book

7) don't use the curvature calculator to figure out how much the hills in the foreground are hiding, Tom seems to be making that mistake as well as you Bobby. It's simple trigonometry. Now obviously the analysis is based on straight line propagation which we know apriory is not the case. But for the foreground scenery it is quite accurate. I apologize for my crude image in the video, I said the hills across the "bay" hide the mountain but the graphic was misleading. The tangential point is to the msl level at a distance of about 15 miles, just 3 more miles inland then the shoreline on the other side, so we can't say the water curvature hides any of the land mass in the foreground. Above this tangential line the mountain would still be visible if there were no obstructions above this line (red dotted line tangent to MSL), but obviously there's hills in the foreground! Now use simple trigonometry and see how much 0.01 rad hide at a distance of 117 miles. (since the angle is so small we don't need trigonometry, just multiply the angle in radians times the distance to get the arc length which is pretty damn close to the height calculated based on trig.)

**Bottom line folks, the mountain should not be visible, but not only do we see it, it's visible at the correct angular elevation! ** What are the chances refraction just happens to do that? none really, its peoples erroneous understanding of refraction that confuses them. The refraction downward assumed by almost everybody who defends the globe is

**ASTRONOMICAL** refraction, astronomers observe stars and planets SPEED up as they get closer to horizon and can calculate this accurately. At zero elevation it's about 0.5 degrees, the diameter of the sun. But

**ATMOSPHERIC** refraction is a whole new ball game. There's temperature gradients, various water vapor distributions, etc., not the same thing as astronomical refraction.

So atmospheric distortions get in the way of the flat earth phenomena, but when the atmosphere effects are understood and removed, we are left with the naked truth, that there is a damn strong flat earth phenomena! It is real, observable, and measurable! I encourage everybody to subscribe to my youtube channel, "JTolan Media1" I have some dynamite information coming, you don't want to miss it.

Anyway, hope this helps you in your quest for TRUTH!

v/r

-JT