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Tuesday, September 26, 2017

Celestia Recreation: Apollo 11 Earth AS11-36-5355

I happen to own an original 8"x8" print of this image so this was fun.  This is the NASA high-resolution scan of the image designated AS11-36-5355.

Apollo 11 Earth AS11-36-5355
A little research showed that this was taken on 1969 July 16th around 1211Z from 98,000 nautical miles - here is the result in Celestia:

Celestia recreation of AS11-36-5355

This is the power of the scientific model we have built, based on thousands of years of detailed observations and study of physics.  Almost all of the planetary model that enables this exquisite level of detail to be exactly recreated is based on a SINGLE observation -- that mass mutually attracts proportional to GmM/r²

From this and the basic physics of motion we get the planetary motions that predict and recall the motions of the planets over thousands of years -- and we also find that we must add Relativity to get the motions down to the next level of accuracy.  No measurement made thus far have proven this model to be wrong.

Monday, September 25, 2017

Flat Earth Claim: Flight from Bali diverted to Alaska proves the Flat Earth?

Flat Earth Claim



Globe Earth Reality


In true Flat Earth-style they "researched" this thoroughly (right?).  So I guess that means they are lying?  Or maybe they just can't research, you be the judge...

That flight went from Bali to Taiwan, and THEN to Los Angeles -- it did NOT cross the Pacific from Bali going to LA as claimed.

"Gr8Believer", in true Flat Earther fashion, refuses to accept this and doubles down:


So I dug around a bit and found the actual flight data from Flight Aware on flight CAL8 from 2015 Oct 7th:

http://flightaware.com/live/flight/CAL8/history/20151007/1550Z/RCTP/KLAX


Once you look at the Taiwan -> LAX flight path (ON A GLOBE) and take into account the distances, flight speeds, arrival times, etc -- you can see this matches the Globe model and it makes perfect sense.  There really isn't another good option - they flew about 400-500 miles out of the way total and had to refuel with a total delay time of a little over 3 hours.

There is absolutely zero evidence this flight was Bali to LAX, there are no China Air flights from Bali to LAX, the flight goes Bali to Taiwan to LAX, at best he latched on to some slightly vague statements in the press articles.

The Mercator projection used for most "flattened maps" (google maps, flight aware, many others use this projection-style) makes the Great Circle "direct path on a sphere" look like a big arc -- but that's because the map is a distortion of the Globe (it has to be, Gauss's Theorema Egregium is (indirectly) a mathematical proof that there is no way to flatten a sphere into a flat map without distortions and tears).  Tighten a string over an actual Globe and it will match the flight path I show below.

I put all the data into a nice (?) graphic, for ease of refuting in certain online discussions. Here I include the Globe Great Circle path (red) and the yellow line is approximately the diversion path so I could estimate the distance, and thus the travel time.


and the saved tweet in case it is deleted:


Thursday, September 21, 2017

Flat Earth Consequences

Just a few of the consequences of believing in Flat Earth...

(Thanks to JW3HH for the idea)

You must believe that every astronomer and most of the scientists are either liars/traitors/in on a VAST conspiracy or are utterly incompetent.  Because everyone using an Equatorial mount can see it track objects in the sky all night and track the Sun all day but somehow the Sun is magically taking a sharp turn at turn so it isn't REALLY going on the other side of the Earth DESPITE this being exactly where the EQ mount is tracking it and the way the EQ mount works is it simply tracks a circle in a fixed plane.  So the sun is moving EXACTLY in this plane all day but magically after sunset it would have zoom around never ACTUALLY going below the observers ground-level only to reappear at dawn in the same plane.

You have to believe that Gravity doesn't exist despite a vast array of evidence that no Flat Earther has ever actually defeated other than to blindly call scientists liars.  Almost every college-level physics student repeats the classic Cavendish experiment but dozens of different methods all agree (see Compton's method).  They even do experiments where they move a very large mass above a fountain of cold atoms and observe the resulting decrease in the rate at which the atoms fall.  Furthermore, the undeniable acceleration of gravity is unlike ANY OTHER type of acceleration known because it does not register on an accelerometer -- rather you register 1G at rest on Earth's surface and that drops to ZERO when the object is in Free Fall.



That, despite the fact you can easily walk to the front of the Concorde going 1354 miles per hour, somehow you should be tossed around like a rag doll because you are moving along with the Earth's surface at 1040 mph.

That the horizon rises to "eye-level" despite the very clear evidence from surveyors and even phone theodolite apps that it does NOT.  The angle measured here (by ME) matches the prediction of Globe.


And even videos using a water level




You have to believe that EVERY Satellite is a fake -- despite the fact that millions of Satellite TV dishes ONLY work when pointed very high in the sky (especially in Southern US) and even the SLIGHTEST misadjustment makes it stop working.

That every image of a globe is either Faked or is the result of a "Fish Eye Lens" -- EVEN videos done by Flat Earthers that do NOT use a Fish Eye Lens show curvature when they are sharp enough, the horizon clearly visible, and have a sufficient Field of View.  Himawari 8 & 9, GOES-16, DISCOVR/EPIC all deliver high-resolution full frame images of Earth at a high rate.

That the thousands of people who work in Antarctica lie about seeing 24 hours of sunlight during the northern winter months.

That you CAN see Polaris from sea-level more than a few degrees south of Equator despite nobody being able to produce evidence of such.

That the altitude (the angle above the Horizon) of Polaris in the Northern latitudes equaling your Latitude is some kind of Jedi Mind Trick by "NASA".  On a Flat Plane this would be impossible.  And yet, the angles to Polaris somehow manage to EXACTLY work on a Globe (see also Sun angles).



That southern star trails do not rotate around an area in the Southern Sky exactly opposite where the center of rotation is in the Northern Sky (which BTW is *NOT* Polaris -- Polaris is 20 arc minutes off center and is actually 2 stars itself (very close together).  Like this one taken from the Chimborazo volcano in Ecuador taken in extreme wide angle view (the circular border is actually the entire horizon).



That star trails somehow magically form a Circle for all observers (also why EQ mounts work) when ANY attempt to recreate this shows that a Dome of stars would give elliptical star trails except at the North Pole.

That mountains magically shrink with distance.





I bet you can think of more...

Sunday, July 30, 2017

Globe Impossible! Coriolis Force and Airplanes

Does the Coriolis Force on airplanes flying North prove the Globe is impossible?

This is almost a shaggy dog story so I'll spoil the ending here for the tl;dr -- it acts like something between gentle breeze to a stiff wind, is corrected the same way, and there is pretty much no net effect as it is constantly countered by holding the heading steady.

What is the Coriolis Force?


Very simply stated, it is an apparent force (not a Proper Force that we measure as an acceleration on a mass) caused by inertia in a rotating reference frame.

The simple example is you are on a merry-go-round and throw a ball towards the center, it appears to be deflected. But from the point of view of someone NOT on the merry-go-round the ball goes in a straight-line once you release it.  So that's why it's not a Proper Force.

Translating that to a globe, if you are on the Equator you have an Eastward velocity of about 464.9 m/s (or 1040 mph).  As you move North in an airplane you retain this initial Eastward velocity but the Earth rotational speed here is slightly less - so you appear to be moving East slightly faster than the ground at this point.

Simple right?

There are actually several ways that this manifests on a rotating sphere like the Earth -- the main effect is the horizontal component resulting from horizontal motion - but there is also the Eötvös effect which causes an object moving Westward to be deflected towards Earth center (making it heavier on a scale) and an objecting moving Eastward would be deflected upwards (making it lighter on a scale).

Wolfie6020 has produced several videos showing this effect as well as the centrifugal change by latitude:



We're just going to look at the horizontal dimension here - a plane flying North and the apparent Eastward deflection.

The whole equation is a vector multiplication where

Coriolis acceleration = -2 Ω × 𝑣

The sign here really doesn't matter -- it just tells you which way on the axis, we just want magnitude anyway.  But let's strip it down to just the north vector (so our east/west and vertical motion is zero) so we end up with:

acceleration = 2 ω sin(φ) 𝑣ₙ

Omega (ω) is simply our rotational rate which is the angle of rotation divided by the time - I'm going to ignore the altitude component of the airplane and just use 1 sidereal day (86164.09054 seconds) over the full circle so that is 2π/86164.09054s.

For our velocity north we will use 𝑣ₙ = 250 m/s (~559.2 mph).

The final component here is φ which is our latitude, the sin(0°) = 0 and sin(90°) = 1 so this simply scales the whole thing by latitude.  Since the maximum effect is at 90° let's look at that value:

2×(250m/s)×(2π/86164.09054s)×sin(90°) ~ 0.03646 m/s² [wolfram|alpha]

Gravity is 9.8 m/s² so this is 1/269th the effect of gravity (and we will show below, equal to no more than a stiff crosswind and far below typical winds aloft).

Let's assume a loaded 439,985 kg 747

That means the maximum force from the above Coriolis acceleration is about 16 kN [wolfram|alpha] from Force = Mass × Acceleration.  For comparison, the force of gravity is 4312.7 kN.

Let's also consider how fast the ground is moving given our latitude:

Tangential speed = distance / time = 2π × R × cos(φ) / 86164.09054s

φ = 0° = 2π × 6378137m × cos(0°) / 86164.09054s ~ 465.1 m/s ~ 1040 mph [wolfram|alpha]
φ = 1° = 2π × 6378137m × cos(1°) / 86164.09054s ~ 465.0 m/s ~ 1040 mph [wolfram|alpha]
φ = 45° = 2π × 6378137m × cos(45°) / 86164.09054s ~ 328.9 m/s ~ 736 mph [wolfram|alpha]
φ = 89° = 2π × 6378137m × cos(89°) / 86164.09054s ~ 8.1 m/s ~ 18 mph [wolfram|alpha]

So our first 1° North is ~69.1 miles and we've only needed to reduce our Eastward velocity by 0.1 m/s (0.22 mph) over about 7.5 minutes of flight time.  This is down in the noise and doesn't even register as a gentle breeze.  The real winds, thermal, and pressure differences will overwhelm any such noise.

The maximum effect is when we're at the Pole -- consider we're about 1° or about 69.1 miles from the North Pole the Earth (and our plane) are both moving East now at just 8.1 m/s (18 mph) and we've flown 6150 miles slowly and continuously reducing our Eastward velocity down.  By the time we fly over the North Pole we reduce it to Zero, and then as we proceed to fly due South from the North Pole we need to (conversely) slowly start increasing our Westward speed to remain on course.  And since we're already crabbed slightly Westward we just begin the slow and continuous reduction of the crab angle.

Now let's consider the effect of a crosswind on the plane.  The specific values will vary of course but we're just look at ONE example in order to help us think about the problem in more general terms.

Air density at 10km (~35,000 feet) (p) is 0.4135kg/m³ [via Engineers Toolbox]

I estimated the profile area of the 747-400 by pixel count at 553

Coefficient of drag I placed between a circular tube and a square tube at 0.6

Wind Speed of 15.3 m/s (about 34 mph, or 30 knots - actual winds aloft can hit over 200 knots)

So we can plug these into our equation for

force of air resistance = ½×p×C×A×v²

and we find the force is about 16 kN

0.5×(0.4135kg/m³)×(0.6)×(553m²)×(15.3m/s)² = 16 kN [wolfram|alpha]

By comparison, a 120 knot crosswind (not uncommon at 35,000 feet) would be on the order of 261 kN of force [wolfram|alpha]

So, the maximum Coriolis force on the plane is about like a stiff wind but far less than typical winds aloft.

In fact, we can treat this exactly like a shift in the wind angle taking us off course by just subtracting the Coriolis vector from the Wind vector -- and indeed the plane cannot tell any difference between the two.  Pilots and autopilots know how to correct for this drift.  If we're drifting Eastward then we turn the plane a bit Westward until our true heading is corrected.  This turning of the plane Westward to counteract this drift would, in turn, automatically cancel out our Eastward excess velocity as we fly.  In addition to that, you are constantly fighting air resistence to keep the plane moving forward -- any velocity changes show up as airspeed changes which are going to cancel out very quickly by just keeping the plane on the correct heading and at the desired altitude and airspeed.

So the cumulative effect is pretty much nil at the end and the instantaneous effect just feels like wind and is easily corrected for.

For East/West motions the deflection is up or down and we handle this by keeping actual vertical speed near zero -- this automatically cancels out the Eötvös effect.  Again, this is a tiny fraction of the gravitational force.

So no issue with Coriolis forces are found for flying airplanes.

Saturday, July 29, 2017

Quick Debunk: YouTube: Flat Earth 6.95 mile curvature zoom test

Another COMPLETELY dishonest Flat Earther video -- call me shocked.



Once again, they completely IGNORE the height of the observer.

QUOTE:

curvature drop (H) = 8 inches * d * d = 8 * 6.95 * 6,95 = 386.85 inches = 32 feet (9.5 meters)

Where did they demonstrate an accurate height of the camera above the water?

For that to be accurate that camera lens has to be half-way under the water so the center of the lens is EXACTLY at zero elevation.  Any higher than that and they are lying to you.

Oh there it is -- the camera is well above the water...  I estimate about 4 feet above the water level.


So what DO we end up seeing?  Do we see the shore of the 6.95 mile away beach?

No, we some typical blurry as all hell P900 footage where we can *maybe* barely make out the top of the rock wall.  Why do Flat Earthers refuse to buy a good camera?


So Let's see what that view looks like up closer.  What I did was mark a 3D line in Google Earth Pro from their claimed observer location to the "dome" on that middle building.

That gives me their line-of-sight to that building.


If I put my view down closer to the water then I cannot see the archways on that building.  This means the observer line-of-sight is well above the water at this point.  This is why I put the observer at 4 feet.

That puts the estimated hidden height at 13.5 feet.


This matches very well with what we actually observe in the video.

We CANNOT see the shoreline, we do not see a very substantial portion of the rock wall.

That wall is about 11 feet high, and a couple more feet down to the water line puts us right at about 13 feet hidden.


Even at just 2 feet above the water we would only have 15 feet hidden if you apply standard surveyor refraction of 14%


What no Flat Earther will honestly address here is why don't we CLEARLY see the bottom of the rock wall?  Where did it go?  Why is it hiding?

They will appeal to "perspective" but I've shown again and again that perspective CANNOT hide just the bottom of an object -- it makes the WHOLE object smaller in proportion.  If it was too small to see the bottom it would be too small to see the top.

Thursday, July 27, 2017

Quick Debunk: Another Flat Earth fraud claiming to find "photoshop" evidence in NASA image

Many Flat Earthers will post this meme:



Which is, of course, a complete and utter fraud and lie on the part of Flat Earthers.

The meme features a squashed and cropped version of this image of Buzz Aldrin coming down the ladder of Apollo 11 in frame AS11-40-5868:


Flat Earth liars have purposefully extended the frame and added in the picture of the Earth.


Now, WHO is lying to you?


The ever astute @FlatSlugBrains also pointed out that it would be impossible for the Earth to appear that low in the sky from the Apollo landing sites (any of them, much less Apollo 11).  This wasn't obvious to me at first but it is a logical consequence of the Moon being tidally locked to the Earth, the position of the Earth in the sky doesn't change much on the Moon!  It wobbles around a bit (lookup Lunar Libration) but stays in that area of the sky.

The Apollo 11 landing site is located at 0.6875 N by 23.4333 E

Here is the resulting view -- the Earth is VERY high in the sky -- I had to expand the Field of View to about 80° to capture it -- so that's about 70° above the horizon.


If you download Celestia and want to recreate this here are the steps:

Navigation > Goto Object (Enter: Moon, 0.6875 N, 23.4333 E, 1 (km))
Navigation > Select Object (Enter: Earth)
hit 'C' to center the Earth
Shift-Left Drag (up and down) to change the field of view
Left Drag to position Earth
Left/Right arrow keys to rotate the view

Observant readers will notice my time is way in the future.  You can use 'L' to speed up time ten x per keypress -- leave Earth centered in that case to see how it wobbles around.  The backslash key will return time to normal speed '\' (don't confuse with forward slash '/').


In case it is deleted, here is the meme in question:



Do "NASA" full frame images of the Earth show a perfect circle or an ellipsoid?

To answer this  I grabbed one of the DISCOVR/EPIC frames epic_1b_20170726023808_02 and I found that the Earth was approximately 1506 pixels wide -- this is the semi-major (a) axis:


Next, we need to calculate how tall we would expect the Earth Ellipsoid to be IF it wasn't a perfect circle.  To do this we get the model value from what is called WGS-84 (World Geodetic System from 1984), which is the current measured values for Earth.  It defines the measured semi-major axis (a) and something called 'flattening', using this you can calculate the semi-minor axis (b) using a simple equation that defines this relationship for an ellipse:

ƒ = (a − b)/a
Therefore, b = a - aƒ

From WGS-84 we have:

1/ƒ = 298.257223563
Therefore, ƒ = 1/298.257223563

So given our semi-major axis (a) is 1506 pixels wide we would expect our semi-minor axis (b) to be approximately:

b = 1506 - (1506 × (1/298.257223563)) ~ 1501 pixels

And that is what we find, to within 1 pixel, in the image from DISCOVR.  Since the edges are fuzzy and the atmosphere isn't helping so we are within our margin of error.  But this also clearly is not a perfect sphere, we're 4 pixels from it being a perfect sphere.

We can also estimate that since the Earth equatorial radius is 6378137.0 meters radius that we're seeing about (6378137.0*2)/1506 = 8470.3 meters PER pixel here (or 5 1/4 miles per pixel).

So this image comports extremely well with what we expect from an ellipsoidal Earth.