# FECore Errors in LASER Level Experiments

Saturday, September 29, 2018 - 23:43 | Author: wabis | Topics: FlatEarth, Science
FECore's official conclusion from their LASER Level Experiments "no curve is detectable" and "FECORE measurements disprove the WGS84 model" (WGS84 is the official globe model) is based on errors in the calculations, using Snell's law wrong, assuming a flat earth and a wrong physical model of the atmosphere. In this article I show all errors made by FECore to come to the false conclusion.

## Introduction

FECore does not explain how they did their refraction calculations. The correct refraction calculation is the most essential part of the experiment. I figured out, how they did their refraction calculation. And I found fatal errors in FECore's document concerning refraction.

So their results and conclusions are all wrong.

## Data used for the Examples

I will use the values of the 4th measurement on the lake Ijssel as an example as published, but all their refraction calculations are wrong in the same way!

 Distance 28.68 km Laser height 2.85 m Laser visible height at target 0.85 m Temperature at 2.85 m height 11.8°C Temperature at 0.85 m height 7.6°C Refraction index n1 at 11.8°C 1.000283508 Refraction index n2 at 7.6°C 1.000287811

## Published Refraction calculation

This is an excerpt from the FECore document:

Angle of incidence (theta1): 0.0040°
Refractive index calculation
(based on Edlén Equation)

n1 = 1.000283508 (445 nm, 11.8°C, 74%)
n2 = 1.000287811 (445 nm, 7.6°C, 85%)

Angle of refraction is calculated with Snell's law:

sin theta2 = (n1 * sin theta1)/n2 = 0.003999983 degrees

Angle of deviation = 0.000000017°

We concluded the ambient conditions refracted the laser beam downward by maximum of 0.494 mm (0.0194 inches)

## FECore's Refraction Calculations

Here is what I figured out about the refraction calculation of FECore:

## FECore's Refraction Model

According to the refraction math that FECore applied, they assumed a flat earth and that the air consists of 2 flat layers with constant refractive indices each, derived from 2 atmospheric measurements at the LASER site and the target site respectively.

1. The lower layer with refractive index n2, calculated from atmospheric measurements at the Target, is assumed to go from ground to h1 = 2.85 m, which is just below the height of the laser. Above that, where the LASER is placed, they assumed a layer with refractive index n1, calculated from atrmospheric measurements at the LASER, see Calculating the Refraction Indexes of Air.
2. The angle of incidence θ1 was calculated from from h1, h2 and d as described at Calculating the Angle of Incidence.
3. They applied Snell's law to calculate the angle of refraction θ2 from the angle of incidence θ1 and the refraction indexes n1 and n2, see Applying Snell's law.
4. The difference between the angle of incidence θ1 and the angle of refraction θ2 from Snell's law gives their angle of deviation γ, see Calculating the Angle of Deviation.
5. Using the perspective equation they calculated the refraction correction c from d and γ, see Calculating the Refraction Correction.

Wrong: Inappropriate Refraction Model

The FECore refraction model is not an appropriate model for the atmosphere at all.

Refraction in the atmosphere is continuously, even if the earth is flat. It depends mainly on the temperature gradient (continuous change of the temperature with altitude), which results in a corresponding density gradient. In a density gradient a light ray is never straight but bent the whole way. If you have a density gradient, you cannot apply Snell's law like FECore did.

Wikipedia explains in simple terms how terrestrial refraction can be derived and used in practice:

Terrestrial Refraction; Wikipedia

## Calculating the Refraction Indexes of Air

FECore states they calculated the refraction indexes n1 and n2 based on Modified Edlén Equation.

To compute the refraction indexes n1 and n2, depending on pressure (standard 101,300 Pa used), temperature, humidity and LASER wavelength 445 nm, they seem to have used the calculator from the following website (I checked it for 3 measurements and got exactly the values they used):

Refractive Index of Air Calculator
Based on Modified Edlén Equation
https://emtoolbox.nist.gov/wavelength/Edlen.asp

Edlén published 1966 empirical equation for n of standard dry air and corrections for water vapour, based on experimental data. To get an impression of the Edlén formula see here:

## Calculating the Angle of Incidence

To compute the angle of refraction θ2 by Applying Snell's law you need the angle of incidence θ1:

Wrong: angle of incidence not measured

FECore did not measure the angle of incidence but calculated it by making the following assumptions:

They assumed a flat earth and calculated the dip angle = angle of incidence θ1 measured down from eye level at h1 = 2.85 m LASER height to h2 = 0.85 m target height in d = 28.68 km distance, which yields:

(1)
 \theta_1 = \arctan \left( { h_1 - h_2 \over d } \right) = 0{.}0039955216° \approx 0{.}00400°
where'
 \theta_1 ' =' 'angle of incidence = dip angle h_1 ' =' '2.85 m = LASER height, see Data used for the Examples h_2 ' =' '0.85 m = target height d ' =' '28,680 m = target distance

## Applying Snell's law

Even if we grant the 2 layer model, where Snell's law can be applied, FECore did it wrong. Snell's law says:

(2)
 { \sin\left( \theta_2^{\,\prime} \right) \over \sin\left( \theta_1^{\,\prime} \right) } = { n_1 \over n_2 }

Snell's law

where'
 \theta_1^{\,\prime} ' =' 'angle of incidence measured from the perpendicular to the boundary layer \theta_2^{\,\prime} ' =' 'angle of refraction measured from the perpendicular to the boundary layer n_1, n_2 ' =' 'refraction indexes of the two layers

Note: Both angles must be measured with respect to the perpendicular to the boundary layer.

FECore calculated the angle of refraction as follows:

(3)
\theta_2 = \arcsin\left( { n_1 \cdot \sin( \theta_1 ) \over n_2 } \right) = 0{.}00399998279°
where'
 \theta_2 ' =' 'angle of refraction \theta_1 ' =' '0.00400° = angle of incidence n_1 ' =' '1.000283508 = refraction index of top layer, see Calculating the Refraction Indexes of Air n_2 ' =' '1.000287811 = refraction index of bottom layer, see Calculating the Refraction Indexes of Air

Fatal Error: wrong angles used in applying Snell's law

FECore did not use the correct angle of incidence measured from the perpendicular to the boundary layer, but the angle measured up from the boundary layer. So they should have used \theta_1^{\,\prime} = 90° - 0{.}00400° instead, which yields a much bigger angle of refraction \theta_2 = 90° - \theta_2^{\,\prime}:

(4)
 \theta_2 = 90° - \arcsin\left( \sin( 90° - \theta_1 ) \cdot { n_1 \over n_2 } \right) = 0{.}168°

So with the right application of Snell's law the angle of refraction would be θ2 = 0.168°, which results in a angle of deviation of γ = 0.004° − 0.168° = −0.164°, and not only 0.000000017° as published!

Using the correct angle of deviation γ, the refraction correction would be c = −82.1 m, and not c = 0.494 mm. This is 166,000 times more than what they published.

A refraction correction of c = −82.1 m would mean, that the target is 81.3 m under the surface of the flat earth. This result proves that the refraction model FECore uses is wrong.

Sloppy: wrong direction of refraction stated

FECore stated: "We concluded the ambient conditions refracted the laser beam downward by maximum of 0.494 mm."

But if the angle of refraction θ2 = 0.00399998279° is smaller than the angle of incidence θ1 = 0.00400° then the laser beam is refracted upward.

Sloppy: invalid equation

The publised formula for Snell's law:

sin theta2 = (n1 * sin theta1) / n2 = 0.003999983 degrees

looks different than (4). The above published formula is not a valid equation, because sin theta2 (= sin 0.003999983°) is not equal 0.003999983 degrees. What they ment to write is the above formula (4).

## Calculating the Angle of Deviation

Their "angle of deviation" is simply:

(5)
\gamma = \theta_1 - \theta_2 = 0{.}00000001721°
where'
 \gamma ' =' 'wrong angle of deviation \theta_1 ' =' '0.00400° = angle of incidence \theta_2 ' =' '0.00399998279° = wrong angle of refraction

Using the correct angles we get:

(6)
 \gamma = \theta_1 - \theta_2 = -0{.}164°
where'
 \gamma ' =' 'correct angle of deviation \theta_1 ' =' '0.00400° = angle of incidence \theta_2 ' =' '0.168° = correct angle of refraction

This would be the correct angle of deviation if the FECore model would be correct, which it is not!

## Calculating the Refraction Correction

The refraction correction c is simply the deviation of the laser at the target due to refraction compared with the unrefracted beam. An upward bended beam causes a positive refraction correction value. FECore computed the refraction correction using the following equation, which I have reconstructed from the published values. This formula would be correct if the FECore's refraction model would be correct:

(7)
c = d \cdot \gamma = 0{.}494\ \mathrm{mm}
where'
 c ' =' 'wrong refraction correction d ' =' '28,680 m = distance between LASER and target \gamma ' =' '0.00000001721° = wrong angle of deviation, see Calculating the Angle of Deviation

Error: the angle of deviation γ must be given in radian units in formula (7), not in degrees!

So without this error the refraction correction would be:

(8)
c = d \cdot \gamma \cdot ( \pi / 180° ) = 0{.}00861\ \mathrm{mm}
where'
 c ' =' 'still wrong refraction correction d ' =' '28,680 m = distance between LASER and target \gamma ' =' '0.00000001721° = wrong angle of deviation, see Calculating the Angle of Deviation

So the refraction correction should even be about 57 times smaller. Using the correct angle of deviation, obtained by applying Snell's law with the correct angles, we would get a refraction correction:

(9)
 c = d \cdot \gamma \cdot ( \pi / 180° ) = -82{.}1\ \mathrm{m}
where'
 c ' =' 'correct refraction correction d ' =' '28,680 m = distance between LASER and target \gamma ' =' '−0.164° = correct angle of deviation, see Calculating the Angle of Deviation

Note: this refraction is 7.5 times bigger than the refraction correction c = −11.0 m using the mainstream refraction model at standard refraction.

## Summary

FECore used the difference between the (wrong) angle of incidence θ1 and the (hence also wrong) angle of refraction θ2 as the angle of deviation γ. In their model refraction takes place only once directly at the laser.

In reality the density gradient of the air acts on the whole way to the target and bends the light to a curve. Such a curve can be approximated by an arc of a certain radius for practical purposes. Over long distances the angle of deviation γ is considerably bigger than what FECore has calculated. In the mainstream refraction model this angle is called refraction angle ρ = −γ. A positive refraction angle means the target appears lifted. The greater the distance, the greater the refraction angle ρ.

In FECore's refraction model the angle of deviation γ does not depend on distance d and the refraction correction c increases linear with distance. In the accepted mainstream refraction model the refraction angle ρ grows linear with distance and the refraction correction c (the negative of the value called Lift absolute in the Advanced Earth Curvature Calculator) grows with the square of the distance d. This is the case for flat earth and globe earth.

Lets recap what we have so far:

Angle of Deviation
published angle of deviation using FECore's refraction model with errors γ = 0.00000001721°
angle of deviation using FECore's refraction model correct γ = −0.164°
angle of deviation (refraction angle) using the mainstream refraction model, standard refraction, at 28.68 km γ = −ρ = −0.022°
Refraction Correction
Published refraction correction using FECore's refraction model with errors c = 0.000494 m
Refraction correction using FECore's refraction model with using radian units but Snell's law error c = 0.00000861 m
Refraction correction using FECore's refraction model without errors c = −82.1 m (not 0.000494 m)
Refraction correction using the mainstream refraction model1), with standard refraction, at 28.68 km c = −11.0 m

Using the wrong refraction angle, a not working refraction model (target would be 81.3 m below the surface) and calculation errors, FECore concluded, refraction can be neglegted. And therefor the WGS84 model is proven wrong.

1Everett B Anderson 10/29/2018 | 13:19

Sly Sparkane did an analysis of FECORE's experiment and showed that they failed to properly level their laser, also invalidating all their results.

Of course, FECORE ignored his analysis, just like they are going to ignore your analysis, and continues to boast about their experiment.

I do not expect that FECore takes care of the errors. But at least they are listed here for reference.

3indio007 11/4/2018 | 17:58

I find it humorous that your own advanced calculator shows that hitting the target was not possible even with a refraction coefficient of .937.

The 2 layer model is a strawman BTW
You might convince a bunch of high school students that don't know anything, but anyone that knows anything about the "science" of atmospheric refraction can see it's garbage.

Snells law is an ideal case ot really applicable to this.

Like Ive said before. Where's the empirical test data ?
Where is the test data showing light bends down into an optically denser medium ESPECIALLY at small angles of incidence?

You guys have a bunch of words and no actual empirical data.

#### Refraction exceeding k = 1

indio007: I find it humorous that your own advanced calculator shows that hitting the target was not possible even with a refraction coefficient of .937.

As I noted in my App, it is limited to refraction coefficients of max k = 1. This is a limitation of the 3D model used, not a physical limitation. Refraction can easily become much more than ±1. I am currently working on a refraction simulation that correctly traces each light ray through the whole atmosphere taking refraction into account at each point on the way from the target to the observer, where the refraction at each point is calculated from atmospheric values given by multiple temperture measurements at observer and target. The simulation so far shows exactly the images we can observe in reality: loomin, sinking, towering, inferier and superier mirages etc.

A temperature increase of some degrees per 100 m with altitude is enough to get refraction greater than 1. In this case you can see all the way to the ground no matter how far the shore is away given the visual conditions are good enough. So LASER experiments over cool or even frozen water, where such temperature gradients are easily achieved, are flawed. It is expected to be able to see the LASER in such conditions in the real world.

#### Refraction is scientifically well understood

indio007: but anyone that knows anything about the "science" of atmospheric refraction can see it's garbage.

Refraction in generall and atmospheric refraction in particular are well known physical effects. Refraction is taken into account in astronomical obserevations and survey since ever. Surveyors know that measurements over cool water at low altitude always are subject to strong refraction. They never ever make measurements in such conditions.

Did you know that Refraction is measured and taken into account in measurements of large structures as wings of airplanes in montage halls, eg. by Boeing. Temperature gradient in hall can be big, cold floor, hot a roof. dT/dh = 0.5°C/m, d = 30 m, -> deflection = 0.4 mm.

#### Fermat's Prinicple and Atmospheric Refraction

indio007: Snells law is an ideal case not really applicable to this.

Snells law is universal. It is a physical law. But it is applied only to cases where the touching medias are isotropic, like in calculating optical lenses.

For the atmosphere with a desnity gradient, resulting in a gradient of refractivity, you have to apply Fermat's principle. Snell's law can be derinved from Fermat's principle as a special case for two touching isotropic media.

The refraction formula is derived by using the principle that light travels on the path that takes the least time to traverse (Fermat's principle). The formula contains the refractive index along the light path, which in turn depends on atmospheric pressure, temperature, temperature gradient and humidity at each point of the path. There exist empirical formulas for the refractive index in air derived from real measurements.

This formulas are tested and confirmed by multiple independent experiments again and again. The formulas are rather complex. But you can derive simplified versions from it that can be used as approximations in many cases. The formula I used in my App is such an approximation which is also used by surveyors. The deviation from reality is only some percent but it depends on the current weather conditions.

#### Empirical Data for Refraction

indio007: Like Ive said before. Where's the empirical test data?

#### Experiments showing Refractoin in Medium with variing Refractivity

indio007: Where is the test data showing light bends down into an optically denser medium ESPECIALLY at small angles of incidence?

You are using the wrong "angle of incidence" like FECore did. The angle of incidence is measured from the perpendicular to the boundary layer, not vertically.

The vertical angle is 90° - "angle of incidence" and is called elevation or altitude angle. In fact refraction at small elevation angles is the strongest. Think about: if you look 90° up or down, you look perpendicular to the atmospheric layers and light gets not bent then. Refraction decreases with the cos(θ), where θ is the elevation angle.

You can even see the LASER bending over the water in some FECore footage!

indio007: You guys have a bunch of words and no actual empirical data.

I think it's the other way around. Who makes videos for hours, only talking about something they have no clou about, having no models, know nothing about math and physics and do not know how to measure anything properly? Who is only providing word salad and wild claims without backing up anything?

The documents from FECore are full of errors, missconcenptions and lies. These people have no basic understanding of math and physics but claim they can debunk established science with applications in multi billion dollar businesses. What has any flat earth ever produced worth a single penny?

You have access to the internet and the search enginge google. If you look outside YouTube you can find all the requested data, experiments and peer reviewed papers archived since decades or even centuries. You only have to look for it, instead of watching silly YouTube videos.

5indio007 11/10/2018 | 03:02

You are full of hot air. Your 2 layer model is garbage and you know it.
You said this
"For Standard-Atmosphere this results in a maximum value of approx. k = 0.17 which decreases continuously with increasing altitude of the observer and is practically zero at an altitude greater than 40 km."

So please tell us under what condition the earthly refraction will exceed .937. BTW I know the coefficient is mathematically used to increase the radius and then fed into a calculation using the new larger radius.
.937 means you are calculating a sphere of 7668 miles radius.

there there is the the formula for deriving K (which you got from wikipedia) in the first place. It assumes a spherical earth!
You should have actually READ the citation ,

Monitoring of the refraction coefficient in the lower atmosphere using a controlled setup of simultaneous reciprocal vertical angle measurements", Journal of Geophysical Research, American Geophysical Union

here's a choice quotes
"The local refraction coefficient χ is essentially a function
of the temperature gradient ∂T/∂z, and depends only slightly on pressure p and temperature T."

The temperature gradient is a consequence of an assumed sphere. There is no gradient with a line of sight with a constant altitude like on a flat surface.

It also says that is ONE method used in the literature and there are others. The paper doesn't say the formula is correct at all. In fact they say the standard coefficient used is actually incorrect at 30 meters or below.

"The Gaussian refraction coefficient, though often suitable to describe refraction effects well above the ground, is
not representative for the lower atmosphere Brocks 1950a, 1950b; Bomford, 1980. This region involves
the surface layers to a height of about 30 m 1984 and, importantly, includes the near-ground domain in
which optical geodetic measurements are often carried out (e.g., geometric levelling in general and
trigonometrical heighting in flatter regions). Here, the temperature of the air strata, particularly its vertical
temperature gradient, is strongly influenced by daily variations in the surface temperature. This may result in
either negative and positive values of the refraction coefficient k near the ground with differences to the
Gaussian value as large as two orders of magnitude Brocks, 1950a, 1950b; Hübner, 1977; Eschelbach,
2009
. "

I might as well pile on eh?

Practical Formulas for the Refraction Coefficient
https://ascelibrary.org/doi/10.1061/%28ASCE%29SU.1943-5428.0000124
" If temperature vertical profiles are unknown then the refraction coefficient cannot be reliably determined."

As far as the rest , like the papers you link. I don't think you actually read them. You certainly didn't read the first two. The second one especially destroys any refraction current model.

Funny thing is
I read most them ,well over a year ago. Some of those have no empirical test data and are models.

As you usual you are just another one that think that they can bluff there way through things.

indio007: Your 2 layer model is garbage and you know it.

What 2 layer model? My model is based on an atmosphere with a continuous temperature/pressure/density gradient. No layers.

indio007: So please tell us under what condition the earthly refraction will exceed .937

Simple math. Use the refraction formula and solve for dT/dh, because this is the value with the most effect. You can use any suitable value for T and P, eg. from standard atmosphere. It does not change the outcome much. So the formula to calculate the tempertaure gradient that is needed to get a certain refraction you can use:

(10)
 { \mathrm{d} T \over \mathrm{d} h } = { T^2 \over 503 \cdot P } \cdot k - 0{.}0343
where'
 \mathrm{d} T / \mathrm{d} h ' =' 'temperature gradient in °C/m T ' =' 'temperature in Kelvin, eg. 288.15 K for standard atmosphere P ' =' 'pressure in mBar, eg. 1013.25 mBar for standard atmosphere

So to get a refraction k = 0.937 you need a temperature gradient of:

(11)
 { \mathrm{d} T \over \mathrm{d} h } = { 288{.}15^2 \over 503 \cdot 1013{.}25 } \cdot 0{.}937 - 0{.}0343 = 0{.}118°\mathrm{C}/\mathrm{m}

So a temperature increase of only 0.12°C per meter is enough to get refraction of about k = 1. This is easy the case over cool water or ice. In effect, in such conditions directly over water refraction can get much more than 1. But remember, this is only a relatively small layer. The gradient gets less and less with altitude and so does refraction. But you have to go some meters above the ground to avoid strong refraction values.

indio007: there is the the formula for deriving K (which you got from wikipedia) in the first place. It assumes a spherical earth!

This formula can be found in many surveyor books. It's not only in Wikipedia. It does NOT assume a spherical earth. I think you mean this because of the definition of k = Rearth / Rray. They could have choosen any value for Rearth. It would only change the value of k by some arbitrary factor. If you can choose any value, why not the radius of the earth so you can compare the radius of the light ray with something meaningfull, like say the alleged radius of the earth?

indio007: The temperature gradient is a consequence of an assumed sphere. There is no gradient with a line of sight with a constant altitude like on a flat surface.

This is an wrong assumption you make. The temperature gradient has nothing to do with the shape of the earth. You can have a temperature gradient in a hall between the ground and the roof, no curvature involved. Fermat's principle states that light gets bent in a vertical refractivity gradient due to a density/temperature gradient even if the light travels initially horizontally and the gradient is constant with constant altitude. If this were not the case, optical fibers would not work. Think about: there must be a physical difference for light between traveling through a non-gradient horizontal layer and a horizontal layer with a vertical gradient, even if the light ray is horizontal. This can be mathematically proven using calculus, but you can also make a simple experiment to show this:

indio007: It also says that is ONE method used in the literature and there are others. The paper doesn't say the formula is correct at all. In fact they say the standard coefficient used is actually incorrect at 30 meters or below.

Yes there are other methods to calculate refraction, depending on what atmospheric data you have. And yes, below 30 meters or so refraction can get really unpredictable. And yes all formulas are approximations from empirical measurements. So what?

Surveyors know about refraction and its influence to measurements. They know in which conditions refraction is such that the needed accuracy of a measurement can not be achieved. They either avoid to measure in such conditions or change the way they measure. If you half the distance between two points you half the error due to refraction. So by dividing a distance into smaller parts they can achieve any precision they need. Comparing the measurements between 2 points, done many times with different distances they can calculate how much refraction is goeing on. And then there are devices that are able to measure refraction directly along the line of sight so the measurements are corrected accordingly automatically by the device.

indio007: As you usual you are just another one that think that they can bluff there way through things.

Do you really think all the scientists and surveyors since centuries are that stupid that they can not manage refraction? They have studied refraction in all details. They have checked their measurements using different methods, different devices and different distances again and again. You can get your exact position by astronomical measurements and cross check your terrestrial measurements against that. Today we use GPS. And guess what, all measurements are in agreement with each other to the stated precision.

So no, this is not a bluff. There are literally hundreds of thousands surveyors involved to measure the whole globe since centuries. The fact that all maps are correct proves that they master their job.

I wonder what image you have from other professions. There are not only idiots at work. Perhaps you should begin your research by first going to an university and learn math and physics and the language used in technical papers, so you are able to really understand the papers you are reading. It is not enough to read them, you have to understand them. You have to know exactly the context they are written in, else you get only missunderstandings.

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