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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.

Wednesday, July 26, 2017

Quick Debunk: What do rockets 'push off of'?


What do rockets 'push off of''?

You gotta be kidding me right?  We learned this stuff in 10th grade.

The short answer is 'the same thing as everything else, mass'.

Flat Earthers will often cite an airplane as "pushing off the air" but this is simply not true.  They do PUSH THE AIR itself, but not OFF of it.  When the airplane propeller blades hit air molecules two forces are produced, one pushes the air backwards and the other pushes the airplane forward -- these are EQUAL and opposite forces.  It is the fact that the airplane has imparted a Force unto the air molecules in one direction that it is PUSHED in the other by this very action.

Neither do submarines "push off" the water, their propellers hit the water and there are two resulting forces -- one pushes the water back and the other pushes the submarine forward.

Let's talk about Newton's 3rd Law -- ALL PROPER FORCES occur in equal and opposite pairs.

And here is a simple experiment you can do that will demonstrate that it is NOT "pushing off the air" that is important but rather the velocity imparted to some other mass.

Sit in a rolling chair on a smooth, level surface.  Now wave your arms around wildly.  That is your control experiment.  The amount you move as you do this is your baseline.

Now throw a small but dense object (let's say a one foot wide, 50 lbs exercise ball) you will be pushed backwards MORE THAN if you throw a less massive but greater surface area object (say a 2' square board) -- this demonstrates that it is NOT the air that pushed you back.  In fact, try fanning the 2' square… did you even move backwards hardly at all compared to throwing the heavy ball?

If you cannot understand this as a thought experiment then ACTUALLY TRY IT.

If you study this relationship carefully you can find the exact amount which air resistance contributes (close to zero) and what is the contribution from the mass itself. And you'll find that F=ma (Force = Mass times Acceleration) is that contribution from the acceleration of the mass.

If we were pushing OFF OF the air only then we could push harder with a larger surface area but that very clearly is NOT what we observe happens.  The force with which we are pushed backwards has to do with the force we impart to the mass.  If you move the ball back and forth without letting go you'll move a lot more than you did waving just your arms but if you do the same with the 2' board you'll hardly move at all more than just waving your arms around caused -- and the amount you do move will be proportional to that extra mass plus a tiny bit from the surface area.

Indeed, you can do this in a vacuum and observe the equal and opposite reactions from ANY proper force.  A simple construct would be a rubber band device on wheels shooting some iron shot.  You should also notice a velocity difference - greater mass requires a greater force to go the same velocity. So the iron shot might go a little ways but the rubber band thingy might 'shoot off' like a rocket.

Ah ha! A CLUE!

In a rocket you have a fuel (like kerosene) *AND* an oxidizer (like liquid oxygen in a tank). These are combined and ignited in the rocket engine which releases enormous amounts of heat and produces very high kinetic energy in the exhaust gases. These exhausted gases expand and bump into the rocket nozzle which is cleverly designed to capture as much of that energy as possible by deflecting that gas OUT of the nozzle. The gases go out the back and, with equal and opposite force, the rocket is pushed the other direction. Side-to-side forces cancel out and reduce efficiency.

The F-1 engines Apollo used burned 1,738 pounds of RP-1 (refined kerosene) and 3,945 pounds of oxygen per second creating 1,500,000 lbf (6.7 MN) and 5 of these engines burned at once. With the exhaust gases at 5000 mph. The engine itself is 19 feet tall, these things are huge. The chamber pressure was 70 bar or 1015 pounds per square inch -- and all that was expelled out of the back end, pushing the rocket upwards.

The gas has a much smaller mass but rocket exhaust gases reach supersonic speeds. And with equal but opposite force it pushes against the engine nozzle as it is ejected out.

PS. Please NOTE that putting an object in front of a vacuum cleaner, which is creating a massive airflow, is NOT the same as testing IN a complete vacuum and thus invalidates the test.

PPS. YES, rockets CAN burn in a vacuum because they have their OWN oxidizer.

Flat Earth Self-Debunk: Danyang–Kunshan Grand Bridge meme shows Hangzhou Bay Bridge


Quick Debunk: The Flat Earth "Little Piggy" and "Cockermouth" balloon flights

Flat Earth'ers frequently post extracts from this 'Little Piggy' video series and claim #1 that the video isn't distorted, and #2 that it's proof the Earth horizon is 'flat' at 120,000 feet.  #CheckMate Ba'al worshippers!

and



Um...



Original uncut video is broken into many parts, starting with part 1 (playlist):



First of all, we can see very clearly this IS a fisheye lens:


Ok -- so that means all the images are distorted as per the curvilinear distortion field:


This is why the horizon bows down here -- it is below lens center (technically everything is squished in towards center to make room to "see more stuff").

ONLY along the horizontal lens center will the vertical distortion be minimized (as I covered in great detail here).

And yes, the lens IS what is warping the antenna.  We know this because we can see it in the later part of the video after the balloon pops.



So Flat Earthers are just lying when they selectively show segments of this video with lens distortion and then LIE to your face and tell you the antenna isn't straight.

Ok -- so other than Flat Earthers are uneducated, parroting, liars, what can we learn from this video?

Here is a frame from Pt 10 4:33 where the horizon comes very closer to lens center and the remainder is all BELOW lens center, which means the lens is bowing that section of the horizon UP -- working against seeing the horizon curvature and yet, we still can see very strong curvature:


If some Flat Earther can demonstrate to me (not merely assert, provide detailed evidence in an experiment that can be replicated) that shows that a GoPro camera in wide-angle mode would make something FLAT look exactly like this then I'll re-evaluate.

The evidence I have shows that a GoPro lens would FLATTEN this out

Until then -- Flat Earthers OWN videos show proof of the curvature of the Earth.

So thanks for the evidence (link to Horizon Calculations):



Another video I've seen less frequently referenced by Flat Earth (eg on reddit) is the "Cockermouth" video which Flat Earthers claim is a DIY balloon launch "using a non fisheye lens".  Unfortunately they also give any details about the camera and lens actually used so we really don't know what to expect here.  If it's a narrow field of view lens then we would expect to see only very slightly curvature of maybe a handful of pixels.

But what I do know is that Flat Earthers dishonestly claim that the footage that IS in that video shows the Earth being flat.  It simply doesn't and I'll show a technique here where you simply resize the width of an image to make the vertical changes stand out.

Here is the raw frame from the posted video where Flat Earthers CLAIM this proves the Earth is flat (sadly it is low-resolution):


Yeah -- that LOOKS flat.  So I can completely see how, if you were driven by confirmation bias and not doing research and actually seeking the truth, you might not bother to check it.

But a simple width compression (10x here) shows there absolutely is a curve in that image of several pixels (10ish) -- which gives us an estimated 30-40° camera FOV. If you can find any details I'll happily reevaluate and revise my data.



And here is another balloon with seemingly little or no fisheye, curve confirmed.