I redid the graphic I had in my video, to clearly show the foreground hills blocking the view of the mountain, in case that wasn't clear in my video. I did say the hills across the water would be blocking it.

And this updated graphic: how did you derive the 10 milliradian angle that the foreground hills would produce above the line tangent to a curved earth surface. It looks like that is based off of your angular gauge as well, measured from where you placed its origin: at the shoreline of Santa Monica beach. Is that correct? If so, then it hinges on the accuracy of the gauge based on the Le Meridien Delfina as well. Yes?

I think the answer is "yes." The reason it seems why you claim none of Mount San Jacinto should be visible from Malibu Bluffs on a spherical earth is because you've concluded that those foreground hills form an obstructing angle of 10 mrad (0.57°); a value which it appears you derived from gauging angular dimensions in the imagery of hotel Le Meridien Delfina in San Monica.

Is that a sensible basis on which to form such a presupposition?

If we were to try to assess whether San Jacinto was visible from Malibu Bluffs before you ever set out to record it, how would we do? I would say we'd using topological data and geometry to predict whether or not the summit of San Jacinto should be visible over, or hidden by, that ridge in the foreground, ignoring atmospheric predictions in the interim. I question whether your measured/calculated for theta Θ in your diagram is accurate:

First, some flat and spherical earth geometry diagrams, just to make sure we're on the same sheet of music. Here is a diagram of flat earth lines and angles:

P = Position of observer

O = Obstructing object

T = Target

Straight line G is the flat earth ground surface (sea level)

Straight red line EL is the level eye sight from P, parallel to G.

Straight red line SL is the sight line, angled from P to the top of O.

Θ is the measure of that angle.

Compare this with the geometry of a spherical earth:

P, O and T are all defined the same but they no longer parallel. They are tilted away from each other.

Earth ground surface G is no longer a straight line but is curved. It still represents sea level.

Red Line EL is still straight and perpendicular to a vertical line at P, but it is not parallel to G as it was for flat earth.

There is a new essential line and point that are not applicable for flat earth.

Straight blue line TL drawn from P that is tangent to G at point H (horizon).

This forms an angle that's not part of the flat earth diagram.

Alpha (α) is the "dip" angle between tangent line TL and eye level line EL. On a flat earth, there is no "dip" angle and what is perceived as a horizon is believed to be coincident with eye level.

Straight yellow line SL is defined as it was for flat earth, but the angle theta (Θ) is not formed by it and Eye level line EL but rather with tangent line TL.

All clear?

Assuming there are no other limiting obstacles and ignoring for now atmospheric effects, we can plug in known values for elevations and distances to calculate the angles and the amount of T that is obscured by O in both flat and spherical earth configurations.

Flat Earth GeometryP elevation = 150 ft

O elevation = 450 ft

Elevation delta between P and O = 300 ft

P-O distance = 105,072 ft (19.9 miles)

Flat Earth Θ = arctan(300/105072) = 0.164° (2.86 milliradians)

P-T distance = 617,760 ft (117 miles)

T elevation = 10,800 ft

Flat Earth T elevation obscured = 617,760*sin(0.164°) = 1768 ft

Flat Earth T elevation visible = 10,800-1768= 9032 ft

Spherical Earth GeometryEarth radius ≈ 20,903,500 ft (3959 miles)

P elevation = 150 ft

H distance = 79,190 ft (15 miles)

Horizon "Dip" angle α = 0.217° (3.8 milliradians)

P-O Distance = 105,072 ft (19.9 miles)

0 elevation = 450 ft

Spherical Earth O elevation obscured = 16 ft

Spherical Earth O elevation visible = 434 ft

Spherical Earth Θ = arctan(434/105072) = 0.237° (4.1 milliradians)

P-T distance = 617,760 ft (117 miles)

T elevation = 10,800 ft

Spherical Earth T elevation hidden by curve (below TL) = 6937 ft

Spherical Earth T elevation obscured by O (above TL) = 617760*sin(0.237°)= 2555 ft

Spherical Earth T total elevation obscured = 9492 ft

Spherical Earth T elevation visible = 10,800-9492= 1308 ft

In order to present a 10 milliradian obstructing angle above the spherical earth tangent line, the hill elevation at 19.9 miles would have to be 1051' feet above that tangent line (1067' - 16' hidden). But they are not:

Your angular gauge needs to be reconciled with the vertical elevations of other features (Baldwin Hills, Pacific Plaza) before it can be used to declare what should or shouldn't be visible over a spherical earth topography.

When I did this previously, I came up with 1450' of San Jacinto that should be visible even without atmospheric refraction. Doing the math now that number is reduced to 1308'. I don't know why the difference. If I find my earlier notes, I'll try to find the source. I may have simply used different values for some of the parameters or rounded differently. Or maybe I've made an error here. Whatever the reason, the Baldwin Hills do not rise high enough to obstruct the view of San Jacinto from Malibu if the earth is curved.

In fact, if the earth is flat, much more of San Jacinto should be visible than the 2500-2600' that appears in the captured images. The curved earth geometric calculation is closer to what is observed than the flat earth calculation.

Of course, atmosphere (atmoplane) plays a significant role in that. The question is, does it make more of the mountain range appear or less? If light is being refracted downward, then it should be more, as is the case if the earth is spherical. If light is being refracted upward as you say, then it should be less.