Friday, August 5, 2016

Old Bedford Level

The Old Bedford Level

In the year 1870, one John Hampden, Esq., of Swindon published a challenge in the Scientific Opinion, Jan 12th issue which read

He will acknowledge that he has forfeited his deposit if his opponent can exhibit to the satisfaction of any intelligent referee, a convex railway, river, canal or lake.

Hampden was a student of the Flat-Earther using the alias "Parallax", who was Flat-Earther Samuel Birley Rowbotham. Rowbotham had published a 16 page pamphlet in 1849 on his Flat Earth 'theory' and later expanded into a book 'Zetetic Astronomy: Earth Not a Globe' in 1865, and later into a 430 page book in 1881.

The "Old Bedford Level" is an fairly long, straight, and almost standing body of water found between Welney Bridge and Welches Dam in England (part of the Old Bedford river) which is 31688 feet long between these two points, almost exactly 6 miles long.

Welches Dam to Welney Bridge : 31688 feet = 6.0015151... miles



Rowbothham had previously used the "Old Bedford Level" stretch to try to measure the curvature of the Earth but failed due to ignoring the effects of refraction, and so wrongly concluded that this failed experiment proved his hypothesis. Now granted, the refraction would have been very extreme at that time but this is known to happen - especially in the evening over water when there is a thermal inversion created.

This can been observed very clearly in this time-lapse by Joshua Nowicki of Chicago.

It was none other than Alfred Russel Wallace who answered his charge, thinking to make an easy £500 (but having no idea how dishonest these people would be). His choice was Bala Lake but Hampden wanted to use the Bedford Level and Wallace agreed.

Now Wallace was an experienced surveyor who was knowledgeable about refraction and how Rowbotham could have been mislead by the effects of refraction since Rowbotham had chosen to put his telescope at just 8 inches over the water.

So Wallace designed a far superior experiment which, in his words (from Wallace, A. R. 1905. My life: A record of events and opinions. London: Chapman and Hall. Volume 2.):

The experiment finally agreed upon was as follows: The iron parapet of Welney bridge was thirteen feet three inches above the water of the canal. The Old Bedford bridge, about six miles off, was of brick and somewhat higher. On this bridge I fixed a large sheet of white calico, six feet long and three feet deep, with a thick black band along the centre, the lower edge of which was the same height from the water as the parapet of Welney bridge; so that the centre of it would be as high as the line of sight of the large six-inch telescope I had brought with me. At the centre point, about three miles from each bridge, I fixed up a long pole with two red discs on it, the upper one having its centre the same height above the water as the centre of the black band and of the telescope, while the second disc was four feet lower down. It is evident that if the surface of the water is a perfectly straight line for the six miles, then the three objects—the telescope, the top disc, and the black band—being all exactly the same height above the water, the disc would be seen in the telescope projected upon the black band; whereas, if the six-mile surface of the water is convexly curved, then the top disc would appear to be decidedly higher than the black band, the amount due to the known size of the earth being five feet eight inches, which amount will be reduced a little by refraction to perhaps about five feet.
Welney Bridge

In short, Wallace argued that a convexly curved Earth would match Figure 1 -- but a Flat Earth would match Figure 2.

Now, what was seen in the first 6" telescope, which both witnesses agreed upon, was that image appeared as


This was "sketched by Mr. Coulcher and signed by Mr. Carpenter as correct".  Now remember here, the lower target and the black band are supposed to have been in a line.

Mr. Carpenter then objected to this on the ground that "the telescope was not levelled, and because it had no cross-hair".  Which is irrelevant as neither the level or cross-hairs can change the geometry.  But Wallace borrowed a Troughton's level, which was acceptable to Carpenter.



The sketches then made in this inverting telescope were:

first sketch, from Welney Bridge; second, from the Old Bedford Bridge


All three sketches clearly show the highest marker as being below level (at 2) and the middle markers clearly extended well above the 'water level' marker (at 1).

This is demonstrated by Wallace's diagram where the top line from A represents level.


Of course, no amount of clear evidence would ever stop a Flat Earther so Carpenter bizarrely declared the Earth Flat on this evidence. But Walsh "decided without any hesitation that I had proved what I undertook to prove" and proceeded to publish the results and grant Wallace the winnings (which ultimately turned out to be bad due to English law who much later would void the bet).

This began a long campaign by Hampden of about fifteen years against Wallace - including Hampden threatening Wallace's life, mailing violent threats to Wallace's wife, and Hampden making numerous libelous claims against Wallace. Hampden at one point was placed in prison and had several judgments made against him.

MRS. WALLACE,
Madam—If your infernal thief of a husband is brought home some day on a hurdle, with every bone in his head smashed to pulp, you will know the reason. Do you tell him from me he is a lying infernal thief, and as sure as his name is Wallace he never dies in his bed.
You must be a miserable wretch to be obliged to live with a convicted felon. Do not think or let him think I have done with him.
JOHN HAMPDEN.
Hampden was later forced to publish retractions and repeatedly serve prison terms, but would then continue with his outright lies and liable against Wallace until his eventual death.


Mathematically speaking, over those 31688 feet we would expect an object which is at our same elevation to be about 24.0182 feet below a strictly straight-line tangent extended out from the original position..  Of course, optical sight-lines are NOT straight lines, they are warped and distorted by small changes in the refractive index of the air.  Wallace observed about 22 feet of drop below the exact level point - which left about 2 feet which was likely due to a combination of refraction and measurement error.

But even this doesn't deter Flat Earthers from designing experiments which continue to ignore refraction and continue to give false results and reports as a result.

Perhaps they will never learn...

Tuesday, August 2, 2016

Flat Earth Follies: buildings would crumble due to walls not parallel!

The first problem with the idea that walls for big buildings would 'crumble' if the Earth was round is that, whatever the actual angle between the walls, gravity would be pulling STRAIGHT DOWN so long as the walls are plumb.  Now, no wall is PERFECTLY plumb anyway and what I intend to show is that the angle is SO SLIGHT in practice between the walls that even this point is irrelevant.

So just how out of parallel would plumb walls be in a really big building.

The Willis Tower is 195 feet wide at the widest point and 1,729 feet tall.  Technically the walls aren't that tall but we will just assume they are.

Let's figure out how parallel those walls would be on a round Earth.

Earth's radius is 20890566 feet.

So what we need to know is, given two lines from the center of the Earth out through the surface which are 195 feet apart, what distance are they at when they are 1729 feet further out.

Our good friend the Right Triangle comes into play here.  We just need to figure out the angle the lines make at 195 feet apart from one Earth radius away, and then extend that same angle out an additional 1,729 feet.

We will use this Angle calculator to help us - select 13 decimal places option.

First we calculate the angle:

g = 195 feet
r = 20890566 feet

gives us an angle of α = 0.0005348192579° - a VERY TINY ANGLE.  This is a clue.

Now we just need figure out what g would be at the top, when we add 1729 to r = 20892295

r = 20892295 feet
α = 0.0005348192579°

Which gives us g of 195.016 feet.  So it's a MERE 0.016 feet wider at the top over 1729 feet.  That is WELL BELOW the margin of error for measuring how plumb a wall of that size is.

TWO TENTHS OF AN INCH from bottom to top.  A surveyor couldn't even detect that - you would need specialized equipment to get a reading that accurate over half a kilometer.

So this is an absurd concern.  No contractor would be expected or required to factor in something so slight into a building design.

The Earth is BIG should be your take-away from this.

Prediction: Denver Shadows, Sept 22, 2016 Equinox

Prediction: Denver Shadows, Sept 22, 2016 Equinox

For 39° 44' 21.119" N / 104° 59' 25.08" W in Denver Colorado on Sept 22, 2016 (the autumnal equinox for 2016 at which point the sun will be directly over the equator) at noon local time (19:00Z) the shadow cast by a plumb 5 meter tall pole should be ~4.12858 meters long (not that we can measure that accurately but should be close).   This calculation assumes a Spherical Earth with the sun 93 million miles away.

This location is ~4401481 meters from the equator (found using Google Earth Pro which uses the actual Geo data rather than assuming a simple sphere) which means that, on a Flat Earth, with the Sun 4000 mile away, you would have a shadow length of just 3.4186972 meters.


For that to match a Flat Earth model the sun would have to be just 5330.51 km (or just 3312 miles) high.

But then that measurement cannot match the shadow lengths at other latitudes.

Every latitude will get a different value for the 'height' of the sun to match the shadow length.


Children around the world will all be getting shadow length data on this date as part of the Noon Day Project.  You can use their data to show that the shadow lengths simply do match a Flat Earth model.


What exact latitude & longitude are you at?  What shadow length predictions do you get?

Flat Earth Follies: How to derive 8" per mile squared and why it's wrong


From the Pythagorean theorem we have the relationship: a² + b² = c²

In our case we will let c = R+δ; b=R; a=d; where R is the radius of the Earth (we will use 3959 miles), d is our distance, and δ which is our unknown quantity (the drop height) so we can put these into the equation as follows:

(R+δ)² = R² + d²

We first want to solve for δ which gives us:

δ = √ [R² + d²] - R

Even this formula assumes a perfect sphere, so it is also an approximation, but if you plug in the right values for R you'll get a pretty good answer for the drop height at some distance and it remains close to the actual value over the full range of values; and we will show that [8"×d²] does not.

Below is the geometry for this equation. To understand this, imagine that you are at loc₁ and you go straight out for 3959 miles where you are at loc₀, and then, from THAT point, you draw a line back (from where ever you are, doesn't matter how far) to the center of the Earth - the length of that line, minus 3959 miles is your "Drop Height".




You can understand from this that there is NO distance (the length of 'a') from which you could NOT draw a line back to the center of the Earth. What happens is the further, and further out you go - the length of (a) VERY SLOWLY gets closer and closer to R+δ but no matter how big (a) becomes, it will always be just a hair smaller.

Some people seem to think that you can give this equation a distance value that is "too big" (bigger than one radius usually), this is clearly wrong. ANY VALUE for distance is perfectly fine, you just need to understand WHAT the equation is giving you back. Drop height CAN be far far greater than the radius of the Earth once you understand what you are measuring. I'll deal with why people THINK this is the case below and I'll show that they are just completely wrong.

From here, what the originator of [8"×d²] apparently did was to take our fairly accurate formula (which, as a reminder, is)

δ = √ [R² + d²] - R

and approximate it using a short-cut by using a truncated version of the Taylor series expansion assuming d is small, which gives you:

(√ R² - R) is zero so they ignored that, they kept the second term, and then ignored all terms after that, making this approximately equal to:

δ = [1/(2×R)] d²

This is why [8"×d²] is not accurate for longer distances, they must have assumed distance would be small to get a simple formula.  NOTE: I was incorrect in which mathematical approach was used, see Addendum below for the similar method Rowbotham documented, but I doubt his was the original method.  I'll let this stand as it is the same fundamental approach and formula both ways and this is a nice way to let Wolfram|Alpha do the work for us so we can see how this term comes out of the mathematics.

So now we just need to find 1/(2×R) and then apply a scaling factor to convert miles to inches and since there are 63360 inches per mile we just multiply [1/(2×R)] by 63360, which gives us:

δ = [1/(2×R) × 63360] d²

Now we can plug in various values as R and see what we get:

For measurements along Earth's Equatorial radius (3965.1906 miles) you get: 7.989527…
For measurements along Earth's Common radius (3959.0000 miles) you get: 8.00202...
For measurements along Earth's Average radius (3956.5467 miles) you get: 8.006982…
For measurements along Earth's Polar radius (3949.9028 miles) you get: 8.020451…

You can just see someone thinking.. "meh, 8 inches is in there somewhere".

Why 8" per mile squared is wrong


Now we can compare our better formula with this [8"×d²] estimation formula, I'll compare for Earth radius of 3959 miles:

Earth Radius
3959


Distance (miles)
Actual Drop (miles)
Estimated Drop (miles)
ERROR (miles)
1
0.000126
0.000126
0.000000
10
0.012629
0.012626
0.000003
100
1.262744
1.262626
0.000118
1000
124.341891
126.262626
-1.920735
2000
476.502339
505.050505
-28.548166
3000
1008.260915
1136.363636
-128.102721
3959
1639.871493
1979.000126
-339.128633
4000
1668.937544
2020.202020
-351.264476

From this is it easy to see that [8"×d²] is somewhat accurate for distances up to about 100 miles and then it is absolutely terrible after that.  And now we know why - they threw out all the little corrections in the Taylor series that keep it accurate in order to make a simple "rule" that was "good enough" for what they needed at the time.

So my advice is to gently inform those using it that it's only fairly accurate to about 100 miles but don't fret a few inches here and there when discussing the 'curvature' of the Earth.

The more important issue to address is...

Why It's Even More Wrong Than That


But more importantly, this [8"×d²] formula only gives you this somewhat wrong value for Drop Height, when what we actually care about, almost universally, is the height of a distant object that would be obscured for an observer at some elevation (h₀). I derive the correct formula to use for (h₁) height hidden by curvature at some distance in my other blog post but here is the short version:




Additional Notes


There are TWO other ways to potentially measure Drop Height.

The first one (which is the most common mistake) would be to imagine that we go directly out some distance along our tangent and then drop down perpendicular to our original plumb line. Like this:



But, if this were the case, when we go out a distance of one Earth radius (3959 miles) then the drop would be exactly 3959 miles also. This is CLEARLY nowhere even CLOSE to [8"×d²] which gives us just 1979 miles. Impossibly wrong. This geometry is the one where you couldn't have a distance greater than the radius - but I have NEVER seen anyone actually write this equation and it would be fairly complex because you have to find the intersection of a perpendicular line on a circle. This would be absurd on a spheroid anyway because nobody ever cares about A-B.

So we can clearly eliminate this as the intended equation modeled by [8"×d²] - it doesn't make geometric sense and it's not even in the BALLPARK mathematically.


The final one is what most people seem to actually do without intending to. They measure the distance using something like Google Earth which is giving you the Great Circle distance along the curve. This is NOT the same as the straight-line distance! And then they input this distance in the WRONG equation [8"×d²], which just compounds the error. You can see the difference between 'a' and curveDistance in this image below - however, for distances under 50 miles it doesn't make much difference at all.  I see absolutely no use in giving a derivation for computing a drop height based off this value so I'll leave it here.


Addendum

Since originally writing this derivation, I found the text below in Rowbotham's book 'Zetetic' (the Flat Earthers Bible).

Rowbotham apparently cited a more geometric method from Britannica here, to the same end result. I hereby admit my error in the method although it is based on the same reasoning and the end result is identical and I find both approaches yield insights so I'm leaving the original for now.

More importantly my research was intended to confirm my geometrical assumptions (that CD would not be parallel to AB but rather form ADC with D being the intersecting point on the sphere) about the meaning of the distance (segment BC) and this 100% confirms it.


I also found a fun additional tidbit just following this part in which Rowbotham very clearly is aware of refraction -- which makes his 'Bedford Level Experiment' a complete sham. He also could not possibly have not understood that the viewers elevation would have mattered in how far one could see objects past the viewers horizon point.


Monday, August 1, 2016

Flat Earth Follies: TEH CONSPIRACY!

Flat Earth requires these ALL to be fakes!

https://www.youtube.com/shared?ci=7pUM6cPesow
https://www.youtube.com/shared?ci=21kwJWJNoQE
https://www.youtube.com/shared?ci=z72VFZCCZmI
https://www.youtube.com/shared?ci=EH9gytTA758
https://www.youtube.com/shared?ci=_zwTpuF8-Qo
https://www.youtube.com/shared?ci=CSDRQa_r9Uk
https://www.youtube.com/shared?ci=FFLclXbsN5M
https://www.youtube.com/shared?ci=-Luf1LKRkuc
https://www.youtube.com/shared?ci=V63JcShpqTg
https://www.youtube.com/shared?ci=EdN9Y2QoyWE
https://www.youtube.com/shared?ci=fGpfc8TxKrU
https://www.youtube.com/shared?ci=ev8B14swprA
https://www.youtube.com/shared?ci=3LsYD37YhEY
https://www.youtube.com/shared?ci=J_rA5eXZ_D0
https://www.youtube.com/shared?ci=Wm6pLXfqJSU
https://www.youtube.com/shared?ci=g-pvefbvGEk
https://www.youtube.com/shared?ci=7WRMItzrxO4
https://www.youtube.com/shared?ci=pZW3hrtYoUE
https://www.youtube.com/shared?ci=wGnT1mQEbo8
https://www.youtube.com/shared?ci=zy6G1-Qn_j8
https://www.youtube.com/shared?ci=4kOGRBcG6sM
https://www.youtube.com/shared?ci=VXKe9FCzjw4
https://www.youtube.com/shared?ci=VExmGcJknTI
https://www.youtube.com/shared?ci=jWwZI9oxTHY
https://www.youtube.com/shared?ci=XQXXe5_a4s4
https://www.youtube.com/shared?ci=mvNapBIoaBk

And THOUSANDS more… people all over the world, at many different times…

All have to be IN on the conspiracy.

Flat Earth Follies: Erroneous parallax measurement of the Sun

I often see Flat Earth folks trying to "measure how far away the sun is" by using altitude (angle of the sun in the sky from that location) values from two far away cities.

The first obvious Folly here is that they invariably IGNORE that on a curved Earth you cannot simply compare two altitude values -- each location has 0° on THEIR horizon, so they are rotated about the Earth's center.  So before we can USE this altitude information we have to already know the shape of the Earth (which we do, it's a Spheroid).

Imagine a viewer at the North Pole viewing the Sun which is in Equinox over the Equator.  The person at the Equator sees the Sun at 90° altitude, directly overhead.  But the person at the North Pole would see the Sun on the Horizon at 0° -- that is an IMPOSSIBILITY on a Flat Earth.

We also know that a nearby Sun is IMPOSSIBLE because of the actual parallax that would be visible!

Imagine a Sun that is only 4000 miles up and directly overhead observer A, an observer B a mere 1000km away would see the Sun at an angle of 8.8° lower!  That is HUGE.  If this were true ANYONE could trivially measure this enormous shift just by having two people 1000 km away from each other.  THIS NEVER HAPPENS.  This conclusively shows that the Sun *cannot* be 4000 miles up.  You can test this out using this calculator.  But don't just test only ONE distance -- if you are going to try to validate your model you need to test at multiple distances between actual observers and see if you observe the amount of parallax expected.



The actual parallax for two viewers on the Earth 1000km distance from each other is just 0.0001981409° - this is WELL below the ability to measure without fairly sophisticated equipment.  This is why you cannot simply use altitude reports which only have a few decimal points of precision, you need about 9 decimal points to get in the right ballpark.




Flat Earth Follies: Satellite sizes

Some Flat Earthers ask why don't SEE the satellites.

Well, we CAN see the ISS because it is, compared to most satellites, very large at 240'x356' and much closer than most satellites at just 249 miles.  While satellites are usually under 40 feet and are much further out.

Here is Martin Lewis' image:



We can use this Angular Size Calculator to figure out how big something would appear in our field of view.

So the ISS comes in at 62.5 arcseconds and we can just barely see it with the naked eye (little more than a dot).  Being large it also reflections a LOT more light, 85440 square feet (well, it's not solid so maybe 1/4 that) -- so that's maybe 21360 sq feet versus 1600 sq. feet -- or about 7% as much light IF it were at the same distance -- but they aren't.  Most satellites are 5 times or more further away.  So that makes them even dimmer (to the square of the distance)!

Your 40' satellite at 2000 km is only 1.26 arcseconds -- a tiny tiny fraction of the ISS in total area.

Too small, too faint.  Here is what a powerful 14" telescope can see of a normal satellite when it is close to Earth, this one was about to reenter the atmosphere (Thierry Legault Emmanual Rietsch):





A $400 Orion 8945 SkyQuest XT8 Classic Dobsonian (8") Telescope -- which gives you about 400x useful magnification max is going to just barely be able to see that and only for a VERY short time as it flits across the viewfinder.  You just aren't likely to see most satellites from amateur ground-based telescopes with the exception of the ISS (because it is large and fairly close) and a few others.

You need very expensive equipment to capture these.

If you want to try to catch one here is a good resource listing the brightest.

The Math


This friendly Flat Earther wanted to share his 'math' on why we should expect to see so many satellites...


Here is the image offered:



So let's see how our friend did on his math...

Earth diameter: 7917.5 miles
Earth diameter pixels in image (measured from image): 281
Miles PER Pixel in that image: 7917.5/281 = 28.1762
# Pixels for Big Satellites in your image (measured from image): 14
A Satellite in your image would be 14*7917.5/281 = 394.4662 MILES LONG

Actual Big Satellite ~ Bus (see below): 35 feet

35 feet is 35/5280 = 0.0066 of a mile

So even a 35 foot SQUARE satellite would only take up:

(35^2)/((7917.5/281)*5280)^2)*100 = 0.0000055% of a pixel

http://talkingpointsmemo.com/idealab/satellites-earth-orbit
"Size varies. Communication satellites can be as big as a small school bus and weigh up to 6 tons, the Federal Communications Commission says. Most weigh a few tons or less. Some that are used briefly are 4 inch cubes and weigh about 2 pounds."
https://www.reference.com/vehicles/long-school-bus-feet-3c674c9adc10c1bd
"An average 64-passenger school bus is 35 feet long"
And just to show you how bad it gets, let's consider the VOLUME of that space.

Satellites mostly orbit between 99 miles and 22237 miles.

So we take the volume of a sphere 1 Earth radius (3958.75 miles) + 22237 miles and subtract from that the smaller sphere of 3958.75+99 miles and we get:

Volume of sphere = (4/3)πr³
[volume sphere r=(3958.75+22237)] - [volume sphere r=(3958.75+99)]
[7.52976×10^13] - [2.79862×10^11] = 75017738000000 CUBIC MILES

For our satellite we will assume it is a full 35' cube = 0.00000029127 cubic miles or

0.00000029127 / 75017738000000 * 100 = 0.000000000000000000388 PERCENT of the space.

Something that small is reflecting that much less light, plus the light falls off with the SQUARE of the distance -- and you are pitting it against vastly closer and thus brighter objects but your camera can't even capture the highlights in the clouds when taking a picture of your friend in the shade.

To be fair, some regions, such as the geosynchronous orbits, are more crowded but those are the ones that are about 22237 miles away.   The angular size of a distant object is given by α = 2*arctan(g/r/2) so that means that even a 100' satellite, 22237 miles away would be a mere 0.176 arcseconds in size - good luck resolving that.

So yikes, off by about 20 orders of magnitude which is just comically wrong. That is SOME SERIOUS rounding function you got there Flat Earth!