[Bobby Shafto]

Certainly, the light from the mirror has bloomed due to the air (surface level moisture and haze), but that wouldn't defeat surface curvature.

Or can it?

See demo starting at the 6:40 mark of this follow-up video:



I still think lumping critique of this Monterey Bay experiment under the category of "flat earth can't science" is unfair, and this critique even gives the experimenters propers for their effort. I do think the apparent success of their experiment could give one pause about the shape of the earth, but as argued in the video above, simply busting the curvature calculator doesn't logically mean the earth is flat.

Now, if it's regularly repeatable and over ever-increasing distances, then a flat earth becomes more supportable. I just don't think it's going to happen. I'm not getting any of that super-refractive looming or "superior mirage" as they are calling it (but I disagree with that assessment) in my planned over-water observation points. So I predict I won't have the kind of success in Socal waters that they had over the Bay. And I also intend on going further than 13 miles. The real proof will be the 20 mile span from La Jolla to Carlsbad.

But the video above provides the closest thing I've found to a globe earth explanation for how that could happen. If they take the experiment to the Salton Sea, it should do away with some of the environmental issues (swell, sea spray, strong temperature inversion) that may have been at play in the Monterey observation.

Speaking of which, if the weather permits, my son and I are going to test out a signal mirror from two 800' spots 11.5 miles apart, just so I can see what an unobstructed reflected mirror light looks like (though I'm using a circular mirror of about 8" diameter and not the full-length mirror the flat earth researcher group used. I'm also curious to see if the FRS radios I have will cross that distance since I'd like to use them for coordinating in the surface-level over water test. I'll share a picture of the reflected light to see just how much the reflection blooms.

[Bobby Shafto]

Here's a short clip of today's mirror test. Note that this flash is the result of using a 7" diameter round mirror.

[stack]

Been thinking this topic is very much like what Navy Signalmen are trained to do, so I started looking into it.

In the US Navy Signalman Manual Earth’s curvature is mentioned several times in the manual. Specifically here:

"Geographic range: The maximum distance a light can be seen under conditions of perfect visibility, limited only by the curvature of Earth. It is expressed in nautical miles for the height of the observer's eye at sea level.”

http://www.hnsa.org/wp-content/uploads/2014/07/signalman.pdf

I found this cool calculator specific to ‘Geographic Range’ for light (signaling)

Signalman Geographic Range Calculator: http://www.csgnetwork.com/directnautrange.html

Might come in handy.

There’s also a ton in the manual about lookout procedures, ranging, vessel/plane identification, navigation on a globe, there’s even a whole chapter on various hull types of differing ships. Eg.: "HULL DOWN: Said of a vessel when, because of distance and curvature of Earth, only the superstructure is visible.”

There's also "The American Practical Navigator/Chapter 4” from the Navy. It includes tons of info about light identification, ranges, luminosity, etc. Pretty interesting stuff:

"A light’s geographic range depends upon the height of both the light and the observer. The sum of the observer’s distance to the visible horizon (based on his height of eye) plus the light’s distance to the horizon (based on its height) is its geographic range. See Figure 407c. This illustration uses a light 150 feet above the water. Table 12, Distance of the Horizon, yields a value of 14.3 nautical miles for a height of 150 feet. Within this range, the light, if powerful enough and atmospheric conditions permit, is visible regardless of the height of eye of the observer. Beyond 14.3 nautical miles, the geographic range depends upon the observer’s height of eye. Thus, by the Distance of the Horizon table mentioned above, an observer with height of eye of 5 feet can see the light on his horizon if he is 2.6 miles beyond the horizon of the light. The geographic range of the light is therefore 16.9 miles. For a height of 30 feet the distance is 14.3 + 6.4 = 20.7 miles. If the height of eye is 70 feet, the geographic range is 14.3 + 9.8 = 24.1 miles. A height of eye of 15 feet is often assumed when tabulating lights’ geographic ranges."



https://en.wikisource.org/wiki/The_American_Practical_Navigator/Chapter_4

[edby]

Been thinking this topic is very much like what Navy Signalmen are trained to do, so I started looking into it.

Presumably they do a lot of what they are trained to do, so how is it they didn't notice that the training did not match the reality, i.e. the reality that the earth is flat? That's what puzzles me.

[inquisitive]

Been thinking this topic is very much like what Navy Signalmen are trained to do, so I started looking into it.

Presumably they do a lot of what they are trained to do, so how is it they didn't notice that the training did not match the reality, i.e. the reality that the earth is flat? That's what puzzles me.

Which map shows the earth is flat?

[edby]

Been thinking this topic is very much like what Navy Signalmen are trained to do, so I started looking into it.

Presumably they do a lot of what they are trained to do, so how is it they didn't notice that the training did not match the reality, i.e. the reality that the earth is flat? That's what puzzles me.

Which map shows the earth is flat?

All maps are flat, except for globes. The map of the terrain is flat, therefore the terrain that it represents is flat.

[inquisitive]

Been thinking this topic is very much like what Navy Signalmen are trained to do, so I started looking into it.

Presumably they do a lot of what they are trained to do, so how is it they didn't notice that the training did not match the reality, i.e. the reality that the earth is flat? That's what puzzles me.

Which map shows the earth is flat?

All maps are flat, except for globes. The map of the terrain is flat, therefore the terrain that it represents is flat.

Many maps show the projection used, as you know.

[RonJ]

Light houses are built to put the light as high above the sea as is practical.  The higher the better under the round earth paradigm.  Seamen need these lighthouses for navigation.  If the earth were flat then it would be better to save the money used to build a tower and just build a more powerful & bigger light.  It wasn't done that way because the earth is round.

[Bobby Shafto]

Light houses are built to put the light as high above the sea as is practical.  The higher the better under the round earth paradigm.  Seamen need these lighthouses for navigation.  If the earth were flat then it would be better to save the money used to build a tower and just build a more powerful & bigger light.  It wasn't done that way because the earth is round.

The original lighthouse on Point Loma -- the beacon lighting the approaches to San Diego -- was built on the highest elevation of Cabrillo Point in 1855. During the 35 years it was in operation, it was the highest lighthouse in the US. But that proved to be it's bane, since the California coast is often visited by low ceiling marine layer and fog. In 1891, the old lighthouse was extinguished and the current lighthouse built on the lower 90' bluffs went into operation.

RonJ's point is valid though in that higher elevations are sought for communication antennas, lookouts, beacons and such to extend the line of sight otherwise limited by the (alleged) curvature of the earth. Heck, the mission of the aircraft I flew in for the Navy was predicated on the limitations of radar horizon facing surface-based long range surveillance radar systems. You can only put a radar antenna so high on a ship, but put one on an aircraft that can patrol at 20-30K ft and you increase the battle group's  detection distance exponentially. Limitations of surface radar range are not due perspective, or "compression" or optical opacity due to air density or waves. The radar horizon is akin to the visual horizon, in which line of sight (or radar propagation) is obscured by curvature of a globe surface. And that's the reason why distance to the horizon (visual or radar) changes with altitude.  That's the hallmark of a convex surface. That's how you "see" curvature.

[RonJ]

That's why the seamen used to carry land birds aboard their ship.  A lot of times you get a clue that land may be nearby when you spot a bunch of clouds that often appear above a land mass but are not above the open ocean. If the old time sailors thought that land was nearby both by seeing the clouds and by their primitive navigational methods they would release some land birds.  Those birds would naturally fly off the ship and would climb up in altitude for a better 'birds eye view'.  When the bird sees land then they would make a 'bee line' in that direction and give the sailors aboard the ship a good indication of the direction.  A crows nest on the mast of the ship was also manned when the approach to shore was expected.  The seamen in the crows nest could be counted on to be the first to see a good indication of the shore line.  These procedures all work because the earth is a globe and even the old sailors knew it. 

Don't think that these are just legends that are untrue.  I've personally witnessed what land birds do on modern day ships.  Usually when our ship would be at the dock doing cargo operations we would have many land birds hanging out & flying around on deck.  After leaving the dock and heading out towards our next port most of the birds would leave the ship and fly back towards shore. Sometimes there would be a few birds that would stay too long.  I've watched them take off, climb up, and then head off in the direction of land that we couldn't see with our own eyes at the level of the bridge.  We knew where land was because of our accurate maps and GPS.  Sometimes the birds would stay even longer.  They would leave the ship, climb to an altitude, fly in circles, not spot the land, and return to the ship.  Usually they would die after a day or two from lack of food & water.  Finding dead land birds on the deck of a ship is not an unusual thing.

[Bobby Shafto]

Here's a short clip of today's mirror test. Note that this flash is the result of using a 7" diameter round mirror.

Here's a quick capture from today's test from ~22 miles away, with significant surface haze. Amazing how a 7" diameter mirror still produces a detectable light flash:

[stack]

Here's a short clip of today's mirror test. Note that this flash is the result of using a 7" diameter round mirror.

Here's a quick capture from today's test from ~22 miles away, with significant surface haze. Amazing how a 7" diameter mirror still produces a detectable light flash:

And one that looks about 10’ tall.

[Bobby Shafto]

10' tall if at the distance of that apartment tower along the line of sight. But it's more than double that distance, so it might be more like 25'

The objective of this was to test out the use of a signal mirror for sighting between coastal points. It wasn't the point of this to analyze the shape of the earth, but given that there are a couple of identifiable landmarks along the line of sight, I invite anyone to work out the geometry and see if these elevations work out to support a flat plane or if they "dip" with distance.

I was at Cabrillo National Monument overlook at an elevation of 365' + 5' tripod for a total of 370' MSL

The 14-storey Sorrento Tower is ~8.7 miles away with a top height of 440'

The Pyramid on Miramar is ~15.16 miles away with a height of 574'

My signal mirror assistant was ~21.6 miles (113,804 ft) away at Hilltop Community Park at an elevation of 804' holding the mirror at eye level of 6' for a total of 810' MSL

Edit to add this graphic. Base of triangles is 370' "eye level" elevation. Values of vertical rise are elevations above that.



Do these figures validate a flat plane? Or are the angles less than what they should be? I propose using the utility housing on the top of the 14-story apartment building as a gauge. It's approximately 16-17' tall at a distance of 45,820' which puts that vertical 14 pixel height at about 0.02°. The tip of the pyramid and the signal mirror reflection are both lower than should be on a flat plane.

Is there a flaw in the maths or measurements? If not, why would there be "dip" in the angles?

[Bobby Shafto]

Amazing how a 7" diameter mirror still produces a detectable light flash:

And one that looks about 10’ tall.

10' tall if at the distance of that apartment tower along the line of sight. But it's more than double that distance, so it might be more like 25'

The diameter of that flash subtends about 0.02°, and at a range of 113,804' that makes it nearly 40'.

40' from this 7" mirror.

[Bobby Shafto]

Next step is to take this signal mirror to the beach and try to reproduce the result that in the OP video. The simple test is to first establish positive sighting by having my mirror assistant signal me from 70' parking lot overlook at Swami's (Encinitas) while I observe from the 25' view spot at La Jolla's Childrens' Pool, which is 12.9 miles away. I can see the top of the bluffs with a telescopic lens of my camera so it should be no problem.

Then, I'll have him descend down the stairs facing the cliff, stopping at intervals to confirm the flash is still visible and in line of sight. I've never been able to see below the roof of the life guard shack at the bottom of the stairs when viewing from that La Jolla perch 25' above the beach, so it will be interesting to see where (or if) the signal flash disappears. I'll have him go all the way to the water's edge there at Swami's.

Next, he'll climb the stairs back to a point where we should have visible contact, which I think should be at the landing of the life guard platform. And then I will move to the beach with camera (and/or telescope) and see if visible contact is retained or loss. Depending on the result, I'll have him either climb the stairs until signal sight is restored, or descend down to the beach to see if it can be seen at the water line across that span.

This is a rather simple procedure, and I may even YouTube stream the event so anyone here can watch. Planning on the first clear day during the week between Christmas and New Years. I'll want the sun to be behind me (in La Jolla), so anticipate it being 1200-1400 PST which will also be on falling tide toward the end of the week.

Not sure I want to complicate things, but if time and conditions permit, I might put my water level device on a tripod at the end of the sea wall protecting the Children's Pool area and line up the camera/telescope at a 15' view spot from the other side, in line with Swami's. Then have my signalman signal from the 15' level of the life guard tower.  That should be visible, but will it be level?



Regardless of the results, I may try to perform this a few times, just to see if varying atmospheric conditions alter the results.

I also hope to do this experiment across the span of La Jolla and the beach in Carlsbad in the sight line of the NRG power station, which is a distance of about 20 miles.

[stack]

Amazing how a 7" diameter mirror still produces a detectable light flash:

And one that looks about 10’ tall.

10' tall if at the distance of that apartment tower along the line of sight. But it's more than double that distance, so it might be more like 25'

The diameter of that flash subtends about 0.02°, and at a range of 113,804' that makes it nearly 40'.

40' from this 7" mirror.

Just a total approximation. The reflection from your mirror is around 5.7 times the size of the 7" mirror. The folks in the video were using what looks like your standard closet door wardrobe mirror, usually 16" x 48". So at 21.6 miles, perhaps the reflection from the wardrobe mirror would be roughly 5.7 times it's size, something like 91" x 273", or 7.5' x 22'. Which kind of makes it as big as a car, something someone mentioned in a counter video. I don't know how that translates back down to the 13 mile distance in question. But interesting nonetheless that the reflection is so much larger than the physical reflector. Probally the reason why Signalmen use the light to do their job at sea.

[Bobby Shafto]

The mirror is reflecting the sun, so if the sun were really emanating from the direction of the mirror, I'd expect it to flare just as it would if looking directly. The angular size of the mirror at these terrestrial distances is just a small fraction of the 0.55° diameter of the sun, so it's only able to reflect a portion. However, even that small portion is bright enough to "flare" beyond the 7" mirror boundaries.

I have another larger circular mirror I will turn into a signal mirror (with reflective tape and a sighting hole) and hope to try that too to compare. I imagine it will produce an even larger flare. The rectangular mirror the F.R.E.E. group used would be larger still, but I don't think it's necessary. The objective is to make positive identification of the location of the signaler. If using an ever-larger mirror is required, then that must mean something is happening optically that is distinguishable from the earth being flat. I also don't want to be toting that large of a mirror around, and I also want a mirror that I can positively target and not just wiggle around.

If the size of the mirror turns out to matter, then that can be part of the analysis.

I'm not sure how much the intervening atmosphere contributes to the bloom in size of the light. One of the globe earth responding videos suggested that the moisture in the air near the surface of the water helped make the light appear above the physical curved horizon, as if the aerosols along the light path were retransmitting the light. I don't know if I buy that, but they do have an interesting demonstration of how a light source hidden behind a cylinder has it's light become visible when aerosol particulates are created in the intervening space. I linked to that video earlier in this thread, but I really don't think that's what's happening in the Monterey video. We're not just seeing a diffused light but you can make out the actual rectangular shape of the mirror.

My feeling is that it's a strongly refractive mass of air close to the surface, almost like a surface duct. I'm in the uncomfortable spot of maybe taking Tom's "compression" explanation and using it to explain how, maybe, 20-30 feet from the water surface up is being compressed into a thin band, not producing a mirage (superior mirage) but to cause a looming effect. The air near the surface of the water would have to be quite dense, meaning it would be quite cooler than the rarer air just tens or maybe a hundred feet above. I don't know what the sea and air temp were that day or what the humidity/dew point was, so it's hard to tell. All I can do is speculate.

Which is why I think it's important to see if their results are repeatable. And not just once, but on different days. And maybe at different locations. I mentioned before hoping to participate in their return to the Salton Sea in February where they plan on using this technique there, and over greater distances. I think with both globe and flat earthers working together, the exercise will be less likely to be conducted in a way to confirm a bias. We'll both be able to balance each others tendencies to try to produce results we might want.

I've rambled off the point, so getting back to the mirror type/size, for this first round, I intend to stick with my mirror configuration for both the 12.9 mile and then the 20+ mile tests. And I will be paying close attention to and recording sea and air temperature and other meteorological conditions, including the ozone/air quality indices since this pesky "haze" is unpredictable and may be a difference factor between my results and those from Monterey.

I'm still hoping for an opportunity to get to Monterey myself sometime after the New Year, though my focus then will be on the Bishop Experiment recreation. However, this mirror technique might come in handy for that too. I just would need to enlist a volunteer to drive around to Santa Cruz. I may not have the time for that and will just have to rely on visual identification of Santa Cruz shoreline landmarks.

[Tumeni]

How can a level, near sea level, low lying landmass extending inward for many miles, adjoin a coastal water that's part of spherical Earth (with water covering 72% of earth) without being flooded to the point the circles curve is complete and the rest of the land mass is sticking up above the curve? 

Example; Draw a horizontal line on the inside of a circle from one point to the other. Then ask yourself what is keeping that line from being flooded with water because it is below the oceans curve.

Southern Louisiana cannot exist if Earth is a ball. Do you see my point?

By drawing a line, do you mean like this, drawing line c under the arc s?

https://en.wikipedia.org/wiki/Circular_segment

[Bobby Shafto]

Do these figures validate a flat plane? Or are the angles less than what they should be? I propose using the utility housing on the top of the 14-story apartment building as a gauge. It's approximately 16-17' tall at a distance of 45,820' which puts that vertical 14 pixel height at about 0.02°. The tip of the pyramid and the signal mirror reflection are both lower than should be on a flat plane.

I'll finish this tangential observation with the calculation to see if applying curvature can reconcile the discrepancy in vertical angles between those landmarks.

As illustrated in the diagram above, if the earth was flat, the vertical angle between the apartment tower and the pyramid would be 0.14°-0.09°=0.05° and the vertical angle between the pyramid and the signal flash would be 0.22°-0.14°=0.08°.

But in analyzing the photo and gauging angles by the apartment tower we can see they're not. The former angle is 0.02° and the latter 0.04°.



Solve for a curved earth and applying the curvature "drop" figures, the angles are depressed by the following amounts:



Vertical angle between Pyramid and Apartment Tower is 0.05°-0.03°=0.02°
Vertical angle between the Signal Flash and the Pyramid is 0.09°-0.05°=0.04°

[Max_Almond]

Handily, I've already got a spreadsheet that does these calculations from my mountain ranges tests so I was able to enter them and do a check pretty quickly. Results of viewing angles from observer:

To Sorrento Tower: flat 0.087°, sphere 0.0243°
To pyramid: flat 0.146°, sphere 0.0363°
To mirror: flat 0.221°, sphere 0.0647°

So we agree on the flat results, but not on the sphere. Not sure why.