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Wednesday, November 15, 2017

Flat Earth Follies: O'ahu from Kaua'i

O'ahu seen from Kaua'i

There are a couple of these floating around the Flat Earth globe.

Terry Robinson has a video Flat Earth Proof: Oahu seen from Kauai more than 90 miles away

In this video he shows two different views

#1 is near Wailua River Over which seems to be very near 22.046347° -159.354560° (telephone pole and street sign match his video).  The claim is this is from 200' and Google Earth estimates 197' so this is also a good match.

Terry shows the islands being "108 miles center to center" -- however, this location is about 80 miles from the Kaena Point area (along Satellite Tracking Station Road), ~80.7 miles from a 1907' peak, and ~86.3 miles from Mount Ka'ala peak (4026') on Oahu.

So 108 miles, while accurate for center-to-center is irrelevant -- why give this value?

The Hidden height for these locations are 2157'-2631' / 2207'-2690' / 2622'-3185' giving the range as between 'Standard Refraction' and raw Hidden with no refraction.

The Tilt angle at this distance is a mere 1.249 degrees, undetectable at this distance under idea conditions -- much less when looking at sloping, uneven terrain.

So what's the problem here?  We're clearly NOT seeing the entirety of the island.

#2 is at ~150' elevation near Lihue airport, don't have exact location so haven't focused on this location yet. But it does show less of Oahu by the amount we would expect from a 50' drop in observer height.  So again, what is the problem here?

Flat Earth Follies: science says Moon Phases are caused by Earth's Shadow

I've yet to meet a Flat Earther who can correctly explain the actual model for how Moon Phases work.

Many of them will tell you some variation of "science says Moon Phases are caused by Earth's Shadow so how do we see the Moon and Sun in the Sky at the same time, LOL".


Here are a couple of videos that I think help to correctly explain the model that Moon Phases are caused, not by Earth's shadow on the Moon, but by the simple position of the Moon relative to the Earth and Sun.

Please note -- these demonstrations are not to scale -- they are only meant to show how the model works for the specific topic being discussed, not be perfect replications of every aspect of orbital mechanics.  If you want a better model then download Celestia and you can observe a model over thousands of years from every angle.

Before we get started with Moon Phases let's start with a very simple demonstration that you can go outside and replicate on any day where you see the Moon in the sky with the Sun:

This should explain anything you need to know about how Moon Phases work and this demonstration would be impossible on a Flat Earth with a nearby Sun because the angle of the light would be vastly different.

This only works with a very, VERY distant Sun.

But here is a more complete demonstration -- my main complaint with this demonstration is that they are far too close to the lamp and the lamp should be much larger, but the point is that you can very clearly see very similar patterns of light on the ball as we observe on the Moon.

Moon Phases Demonstrated

And this video tries to explain to you why we don't see an Eclipse every month (spoiler: it's because the orbit of the Mboon around the Earth isn't in the same plane as the orbit of the Earth around the Sun).

Why Don't We Have Eclipses Every Month?

Hopefully those will help Flat Earthers and non-Flat Earthers alike.

The simplest method to measure the geocentric lunar distance: a case of citizen science

Cool paper on measuring the distance to the moon by comparing the angular size:

The essence of this methodology is that the distance to the Moon changes over the course of the 6 hours between when the moon is near the horizon and near zenith sufficiently for us to measure this change in the angular size of the Moon.

From Figure 1 in the paper we're discussing the following geometry.  Please note that it is the observer that is suggested to be moving with the rotation of the Earth and not the Moon.

For ideal conditions (an observer near the Equator at the Equinox) the change in angular size would be expected to be simply:

Δϕ/ϕ ~ Rₑ/D ~ 1/60 ~ 1.7%

Therefore, if the Earth radius is ~3959 miles (6371 km) and rotating we would expect to observe a 1.7% change in the size of the Moon over this span.

The paper gives the more exact formula that also takes into account the observer latitude and the lunar orbital conditions, but would we expect this variation to be repeatable for all observers, at all times.

We do in fact see exactly this variation which is a variation that Flat Earthers are forced to appeal to magic to explain while the Heliocentric/Globe models explain it effortlessly.

It's a neat paper and I recommend giving it a shot if you have a decent camera and lens.

Navy Submarine RADAR History, Earth Curvature

In 'History of Airborne and Shipboard Periscope Detection Radar Design and Developments' we see the following in a discussion about radar horizon:
Not surprisingly, the radius of the earth enters into calculations of the geometrical range to the horizon; calculations of the range to the radar horizon account for refraction by assuming that the radius of the earth is 4/3 as great as its actual value. For sensor altitudes small in comparison with the radius of the earth, the distance to the radar horizon in nautical miles is 1.23 times the square root of the altitude in feet. Thus, for a typical airborne PDR altitude of 500 feet, the range to the radar horizon is 27.5 nmi, and for a nominal mast-mounted shipboard antenna height of 70 ft, it is about 10 nmi. Note that common surveillance radar heights for a U.S. cruiser or destroyer are 100-120 ft, and, for an aircraft carrier, 150-180 ft.
Just another nail in the coffin of claims that nobody ever takes curvature of the Earth into account.

Pretty much every book on RADAR discusses both the effect of Earth curvature and refraction on RADAR.

Monday, November 13, 2017

Flat Earth Follies: All Moon Phases Occur At The Same Time!

This Flat Earther was telling me that I don't understand the Heliocentric/Globe model because we should see all phases of the Moon at the same time.  Say what!?
LOL -- okay.  After some discussion (and research on my part) I realized his misunderstanding comes from the FAA FAR/AIM manual (which is also copy & pasted on other sites) which reads:

For practical purposes, phases of the Moon and the percent of the Moon illuminated are independent of the location on the Earth from where the Moon is observed. That is, all the phases occur at the same time regardless of the observer's position.

And ok, I guess if you don't understand science at all you could misread the "phases occur at the same time" but since it explicitly says "phases of the Moon and the percent of the Moon illuminated are independent of the location on the Earth from where the Moon is observed" trying to read that as ALSO saying everyone sees a different phase of the moon is just dishonest.

When it says "all the phases" it clearly means "any/each phase" -- meaning there aren't any phases which do not follow this rule that everyone will see the same phase on the same day "For practical purposes".

That "For practical purposes" phrase is also important because, in actuality, observers far apart actually see very slightly different sides of the moon -- but because it's about 238,900 miles away even a few thousand miles apart is only a small shift.  And this small shift is exactly what we observe in reality.

Sunday, November 12, 2017

Flat Earth Follies: White Cliffs of Dover from Boulogne

Another extremely clever Flat Earth 'proof' the Earth is Flat...


/Google for source image/

Original by Dave Cawkwell on Blogspot
Confirms it is from Boulogne, France -- Dave Cawkwell says he is about 10 meters up, I think a little higher but I'll go with 10 meters (32 feet).

Based on the position of the wall we appear to be somewhere very near the location marked below, which puts us 29.5 miles away -- not 31 miles away -- and 2 feet higher than the Meme.

What else?  Oh yeah, those are the Cliffs West of Dover which puts them up as high as 437 feet, not just 350.

And once we account for Refraction, completely ignored in the original.

So I find that we expect about 252 feet of the Cliffs to be hidden leaving about 185 exposed.

I also see a small band of an inferior mirage line running along the bottoms of the cliff so the true horizon is likely obscured by the mirage effect by some amount (few pixels worth here).

We don't have the exact camera and lens details to get Field of View but we can estimate the angular size of the Cliffs by finding their angular size at 29.5 miles:

2*arctan(437/2/(29.5*5280)) = 0.1607°

And looking at this and this we get a good estimate of the height of the walls to around about 20' high (from the walkway where the people are fishing to the top) and we know the end is about 1.3 miles away from the camera position - which gives us an angular size of about 0.1669° -- very close to the angular size of the whole Cliffs of Dover at 437' high!

Very good match.

And putting all this together, here is a Meme version you can use for debunking.

Friday, November 10, 2017

Celestia Recreation: Unmanned Apollo 4 Image

NASA posted this today:

I liked to grab the original scans and what raw data is available from EOL which we can find here for frame AS04-1-580 - so NASA cropped, cleaned up the image, and rotated it for presentation.

Image Caption: Coastal Brazil, Atlantic Ocean, West Africa, and Antarctica, looking west, as seen from the earth-orbital Apollo 4 (Spacecraft 017/Saturn 501) unmanned space mission. This picture was taken when the Spacecraft 017 and the Saturn S-IVB (third) stage was orbiting the earth at an altitude of 9,544 nautical miles.

That's ok - we can do the same thing very easily.  All I've done here is one 90 degree rotation, lowered the brightness, slight increase in saturation, and adjusted the white point so the clouds are whiter.  It's the exact same image, just adjusted for what a human eye would see.  In fact, how this appears on my display will be DIFFERENT from how it appears on your display -- there is no "true" image.  I could get the same visual effect by adjusting my monitor settings. But most people would see the background as washed out on their monitor if we don't adjust it down from the original scan.

Here is the Celestia CEL data for recreating this view:


And here is the recreation of this view in Celestia -- We know from the caption we're supposed to be "looking west" and see the coast of Africa along the terminator and then South America near the western edge and that is exactly what this image shows at the date and time indicated.

Celestia Recreation

There isn't a lot of land mass in view where I can line things up better but this is a pretty good match with the solar glare in about the right spot

So I guess consider this a prediction that the Apollo 4 was likely at roughly Lat: 23.58559, Long: 10.03758, Alt: 17676km, Date/Time: 1967 Nov 09 1811Z

Here is a partial weather map for the Northern Hemisphere from that date from the NOAA library, looks like a significant low-pressure system over North Africa but we can't really see much between Africa and South America.