# The Rainy Lake Experiment

Saturday, July 20, 2019 - 00:50 | Author: wabis | Topics: FlatEarth, Knowlegde, Science, Experiment
The Rainy Lake Experiment was designed to show, how we can figure out the shape of the earth, Flat or a Globe, by observing and measuring a clever arrangement of targets over a distance of 10 km, taking terrestrial refraction into account and using modern equipment. The experiment is an advanced version of the Bedford Level experiment executed in 1838. The Experiment leads to the conclusion that the earth must be a Globe with a radius of 6371 km.

Rainy Lake Experiment Animation
walter.bislins.ch/RainyLakeAnimation

walter.bislins.ch/RainyLake

Short Version of the Report
In a Nutshell, e.g. for Presentations

## Overview

To come to a Conclusion the following requirements have been set:

## Execution

Although the whole Experiment was mainly George Hnatiuk's work, Jesse Kozlowski and Soundly joined him in early April 2018 to make GPS measurements (Jesse), images and videos (Soundly) of the Experiment. Later Walter Bislin created a Computer Model of the experiment, visualized the data in his GNSS Data Viewer and wrote this reported.

## Location

Rainy Lake is a relatively large freshwater lake (930 km2) straddling the border between the United States and Canada. Rainy Lake is part of an extremely large system of lakes forming the Hudson Bay drainage basin that stretches from west of Lake Superior north to the Arctic Ocean.

Img 1: Rainy Lake panorama on winter 2018 with Georges dog Khan at the snowmobile

During the winter months until May, the lake is covered by a meter-thick layer of ice. The lake was choosen for this experiment because George Hnatiuk lives right at the lake and is equipped with the tools and vehicles like a snowmobile necessary for the experiment. The lake provides an unobstructed path of about 10 km from George's home to a small island called Home Island, ideal for this experiment.

Img 2: Rainy Lake target locations. See Location Graph and Data for data assigned to the numbers.

## Choosing Target Heights

The target center heights were chosen in such a way, that on each earth model one of two rows of targets will be at eye level of an observer while the other row curves up (Flat Earth) or down (Globe). The lower targets are called Bedford targets, the upper targets are called Tangent targets.

Img 3: Rainy Lake Experiment Design: Flat Earth side-view

The lower Bedford targets consist of a row of 6 targets, all 1.85 m above water level. If the earth is flat, all this targets will appear at eye level for an observer at 1.85 m height. The upper Tangent targets will curve up for an observer at 3.91 m height.

Img 4: Rainy Lake Experiment Design: Globe-Earth side-view

The upper Tangent targets consist of a row of 4 targets, increasing in height with increasing distance. If the earth is a Globe, all Tangent targets will appear in line with eye level for an observer at 3.91 m height. The lower Bedford targets will curve down for an observer at 1.85 m height.

To see how the targets were planted and measured, see Planting the Targets and Measuring the Targets.

### Calculating Tangent Target Heights

The height h of the Tangent targets is the sum of the observer height ho and the curvature drop xi at the i-th target. On distances as small as 10 km the leaning can be neglegted. The target center height can be computed as follows:

(1)
source
where'
 $h_i$ ' =' 'center height of the i-th Tangent target $x_i$ ' =' 'drop at the i-th target $R'$ ' =' '7681.64 km = extended radius of the earth to account for Standard refraction k = 0.17.See Refraction Factor how this radius is obtained. $d_i$ ' =' 'distance to i-th target $h_\mathrm{o}$ ' =' '3.91 m = observer height

### Method used to Mount the Tangent Targets

The Tangent targets were first placed roughly at the pre-calculated target center heights. Then they were adjusted to the eye level height as seen through the auto level at the observer. Because the targets were adjusted at different times and days, refraction was slightly different for each target so the targets did finally not align perfectly. But the height deviations are within the calculated variations of common low refraction.

#### Calculated and Measured Tangent Target Heights

 Target (2) Target (4) Target (6) Target (7) Refraction Lift k = 0.17 0.063 m 0.254 m 0.551 m 1.194 m Lift Variation for k = ±0.1 ±0.037 m ±0.149 m ±0.324 m ±0.702 m Calculated Height (1) 4.218 m 5.151 m 6.602 m 9.740 m Measured Height 4.280 m 5.240 m 6.500 m 10.535 m Difference +0.062 m +0.089 m −0.102 m +0.795 m Refraction at Adjustment k ≈ 0 k = 0.11 k = 0.20 k = 0.06

The Calculated Heights are calculated from the curvature drop of a sphere with radius 6371 km with a standard refraction coefficient of k = 0.17 applied according to (1). The expected lift due to refraction for each target is shown in row Refraction Lift. The variation of the lift due to a refraction change of 0.1 is shown in row Lift Variation.

In row Measured Height is the measured target center height with respect to water level at the observer. I use the height with respect to water level at the observer rather than to water level at the target, because the targets were adjusted visually to the eye level at the observer. Measuring the Targets explains how the GPS Vectors to the target centers were measured.

Note: The water level elevation of the lake decreases about linearly with increasing distance from the observer. At the last target (7) the water level elevation is about 25 cm lower than at the observer with respect to the Reference Ellipsoid, due to variations in the gravitational field of the earth. This means that mean sea level is also 25 cm lower with respect to the Ellipsoid. The Geoid defines mean sea level with respect to the Reference Ellipsoid for each location on earth. See Obtaining Elevations for how the Geoid height is obtained.

The row Refraction at Adjustment shows the calculated prevailing refraction at the time each target was adjusted to eye level, using the measurements of the target center heights:

Refraction_at_Adjustment = 0.17 - Difference / (10 * Lift_Variation)

The different prevailing refractions at the adjustment times plus the current deviation from k = 0.17 at an observation results in the apparent height variations.

The Computer Model uses the measured GPS Vectors for the predictions after transforming them into a local coordinate system at the observer, see Target Positions and Sizes relative to the Observer. Therefore if the refraction in the Computer Model is set to the same refraction that was present when a photo was shot, the photo and the Computer Model image should match exactly.

## Size and Placement of the Targets

To optimize the visibility of the targets and to help better distinguish the targets, they are arranged in the following way:

Targets of the same size appear smaller with increasing distance (left image). To counter this perspective effect, the targets were built in such a manner, that their angular size at the observer is about the same. This way they appear the same size no matter how far away from the observer they are (center image). The target (7) is about 4 times larger than target (2), see Calculating Target Size. Target (1) and (6) are different for better recognition.

The targets are not placed in a straight row, so that they do not overlap each other as viewed from the observer (right image).

Img 5: Rainy Lake Experiment target design

The targets have two horizontal black or orange bars with a clearly visible gap between. A target is perfectly at eye level, if the crosshair of the theodolite or auto level is dead center in the gap.

Note: the slant of the orange target (6) is not intentional. It was damaged by strong winds.

### Calculating Target Size

To achieve that all targets appear the same size, they must have the same angular size $\theta$ at the observer. The target size s then depends on the distance d from the observer according to the following equation:

(2)
source
where'
 $s_i$ ' =' 'size of the i-th target at distance $d_i$ $\theta$ ' =' '1.6 m / 9459 m = 0.000169 = angular size of the targets in radian $d_i$ ' =' 'distance to the i-th target

See Target Positions and Sizes relative to the Observer for the calculated and choosen target sizes.

## Predictions for Bedford Targets

The Computer Model of the Experiment uses pre-calculated and measured target positions in ECEF coordinates. The following images are screenshots from this Model. For image 6 and 12 the pre-calculated data is used. For the comparison with the real images, the measured GPS Vectors in ECEF Cartesian coordinates are used.

Img 6: Predicted views of the Bedford targets from the 1.85 m observer location for Flat Earth and Globe → Model

Flat Earth: From an observer height of 1.85 m the Computer Model predicts that all Bedford targets appear exactly aligned with the horizon and eye level.

Globe: From an observer height of 1.85 m the Computer Model predicts that the Bedford targets appear all below eye level. The farther away from the observer, the more below eye level.

## Results for Bedford Targets

The Rainy Lake Experiment shows that from an observer height of 1.85 m all Bedford targets appear below eye level as predicted by the Globe model. The farther away a target is, the lower it appears. The horizon is predicted at 5.68 km, so the target (5) pole should appear just at the horizon.

Img 7: Globe prediction → Model
Img 8: Flat Earth prediction → Model
• Swap
• Globe Prediction
• Flat Earth Prediction
Img 9: Bedford target observation → Model
Img 10: Globe prediction for Bedford targets → Model
Img 11: Overlay observation with Globe prediction → Model
• Swap
• Real Picture
• Globe Prediction
• Overlay

The Flat Earth model predicts that all Bedford targets appear aligned with the horizon and eye level, which does not match the observation.

In this observation, refraction was about k = 0.27, see Measuring Refraction with the Computer Model. This is slightly more than standard refraction of k = 0.17. The highly zoomed in image is clear. This indicates that refraction is near standard and constant for the whole distance. Strong refraction would cause a distorted image, see Refraction Range of Clear Images.

The black vertical and horizontal lines in the image is the crosshair of the auto level. The horizontal crosshair indicates eye level of the observer at 1.85 m height. In the Computer Model, eye level is indicated by a magenta horizontal line labeled Eye-Level.

The Computer Model uses for this predictions the measured heights of the Bedford target centers rather than the pre-calculated target center heights and the observation matches the Globe model very well.

## Predictions for Tangent Targets

Img 12: Predicted views from the 3.91 m observer location of the Tangent targets for Flat Earth and Globe → Model

Flat Earth: The Computer Model predicts that the Tangent targets will appear above eye level from an observer height of 3.91 m. The farther away a target is, the higher it will appear.

Globe: The Tangent target center heights are computed in such a manner, that on the Globe they will all appear at eye level from an observer height of 3.91 m with Standard refraction k = 0.17 applied, see Calculating Tangent Target Heights.

## Results for Tangent Targets

The Rainy Lake Experiment shows that from an observer height of 3.91 m the Tangent targets are aligned more or less with eye level, which is predicted for the Globe-Earth. The deviation from eye level is the result of aligning the Tangent targets optically at different days with different refractions, rather than setting them to the calculated target center heights.

Img 13: Globe prediction for Tangent targets → Model
Img 14: Flat Earth prediction for Tangent targets → Model
• Swap
• Globe Prediction
• Flat Earth Prediction
Img 15: Tangent target observation → Model
Img 16: Globe prediction for Tangent targets → Model
Img 17: Overlay observation width Globe prediction → Model
• Swap
• Real Picture
• Globe Prediction
• Overlay

The Computer Model shows the predicted image using the measured target center heights, which match the observations very well. The Flat Earth prediction does not match the observation at all.

The image is clear over the distance of 10 km, so refraction can be assumed to be near standard, see Refraction Range of Clear Images. With the Computer Model a refraction of k = 0.27 was measured from the image, which is slightly more than standard refraction k = 0.17, see Measuring Refraction with the Computer Model.

The last target appears about 1 m too high, compared to the pre-calculated height. A refraction variation of k = ±0.14 results in a height variation of ±1 m at a distance of 9.5 km. So the divergence of +1 m is due to a lower prevailing refraction at the day and time this target was setup.

Note: This image was shot with a P900 camera, not through a theodolite or auto level. That's the reason why the crosshair is missing. The camera was placed beside the auto level at the same height of 3.9 m.

### Another Day

The following observation was made earlier than the observation above, when the Tangent target (6) was not yet damaged by strong winds.

Img 18: Globe prediction for Tangent targets → Model
Img 19: Flat Earth prediction for Tangent targets → Model
• Swap
• Globe Prediction
• Flat Earth Prediction
Img 20: Tangent target observation → Model
Img 21: Globe prediction for Tangent targets → Model
Img 22: Overlay observation width Globe prediction → Model
• Swap
• Real Picture
• Globe Prediction
• Overlay

This image was shot through an auto level. Note that the magenta eye level line matches the horizontal eye level crosshair exactly.

The picture is clear with a small layer of inferior mirage above the surface. At the Tangent target heights refraction can be assumed to be near standard, see Refraction Range of Clear Images. With the Computer Model a refraction of k = 0.187 was measured from the image, which is slightly more than standard refraction k = 0.17. The targets in this observation are better aligned with eye level as the other observation above, which is due to less refraction.

## GPS Vectors

Throughout this report I call the locations directly measured using survey grade GNSS receivers and expressed in ECEF Cartesian coordinates GPS Vectors, to emphasize the fact, that this vectors have no relation to the shape of the earth.

### Coordinate Systems

A position in 3D space can be represented by a vector with 3 components, called the coordinates of the vector. There are many coordinate systems to represent the same 3D vector. All have in common, that a certain vector points from an origin to the same location in space. The length of the vector is the same in every coordinate system. Likewise is the distance from one vector to another the same in every coordinate system. Only the coordinate values are different in each coordinate system. To uniquely define a vector, you have to state in which coordinate system the coordinates of the vector are expressed.

The choice of the coordinate system used depends on the application. Common coordinate systems are:

• Cartesian Coordinate Systems like the Earth Centered Earth Fixed (ECEF) (x,y,z) coordinates used internally by GPS/GNSS for all calculations
• Spherical Coordinate Systems used in spherically symmetric applications where the vector components are represented with 2 angles and a radius (φ,θ,r)
• Geodetic (Ellipsoidal) Coordinate Systems used in geodesy where the vector components are represented as latitude, longitude and ellipsoid height (φ,λ,h)
• Cylindrical Coordinate Systems used in cylindrically symmetric applications like the Flat Earth where the vector components are represented with 2 angles and a height or latitude, longitude and height above the zero plane (φ,λ,h)

Note: although Geodetic and Cylindrical Coordinates both have a height component h, their value for a certain GPS Vector are not the same, because in Geodetic coordinates height is measured from the surface of an ellipsoid, while in Cylindrical (Flat Earth) coordinates height is measured from the base plane through the origin.

It is possible to transform a vector from one coordinate system into another system without changing the location the vector is pointing to and without changing the length of the vector or the distance between different vectors. See WGS84 Coordinate System for the transformation between ECEF Cartesian coordinates and Geodetic Ellipsoidal coordinates.

Projections on the other hand change the vectors, their length and the distances between them. If you have a vector, say in the Geodetic coordinate system of the Globe model, and simply use the same coordinate values in the Cylindrical coordinate system of the Flat Earth model, you are making a projection which changes the length of all vectors and the distances between them. So the Flat Earther claim, that Geodetic coordinates work the same on the Cylindrical coordinate system of the Flat Earth, is not true. That's the reason why all distances on the Flat Earth model are not the same as on the Globe model. The real distances are only correct on the Globe model.

### Vectors in ECEF Cartesian Coordinates

GPS Vectors: The locations calculated by GNSS receivers and error corrected with any of the methods mentioned at Improving GPS accuracy with Differential GPS are vectors in ECEF Cartesian (x,y,z) coordinates, not in geodetic coordinates latitude, longitude and ellipsoid height or geoid elevation.

GNSS receivers calculate the ECEF Cartesian coordinates (GPS Vectors) from the measured distances to multiple satellites with exactly known positions in space (in about 20,200 km altitude for the Navstar GPS system) using Multilateration. GNSS receivers can export the measured distances to the satellites and the vectors calculated from them in the RINEX data format for later processing on a PC.

A GPS Vector has its origin at the mass center of the earth and points to a certain location in space which may lie on the surface of the earth or anywhere above it. The ECEF Cartesian coordinate system is fixed to the earth. As the earth rotates, the coordinate system rotates with it.

So each measured target GPS Vector points to a position in 3D space, independent of the shape of the earth. Some satellites and the ISS, use the same coordinates gathered with GNSS receivers on board to calculate their positions in space [1]. They don't use Geodetic coordinates for their trajectory calculations.

For example, the water level GPS Vector of the observer location has the following (x,y,z) coordinates in the ECEF Cartesian coordinate system: (-243772.154, -4217822.457, 4762585.891). The length of this vector can be calculated using Pythagoras: l = √ x2 + y2 + z2 = 6,366,449.207 m. This is approximately the mean radius of the earth.

Note: because the location is not at sea level and the earth is not a perfect sphere but in the first approximation an Ellipsoid and more accurate a Geoid, the radius of curvature at any location on the surface of the earth diverges slightly from the length of the vector from the center of the earth to that location.

### Geodetic Coordinate System

The measured GPS Vectors in ECEF Cartesian coordinates are only transformed into Geodetic Ellipsoidal coordinates (latitude,longitude,elevation) for geodesy, mapping and navigation. Most calculations like GPS/GNSS internal calculations, trajectory calculations or air and space navigation calculations are done in ECEF Cartesian coordinates, because vector calculations are especially easy to carry out.

Note: Transforming a vector into another coordinate system is not a projection onto a Globe, Flat Earth or map. Coordinate System Transformations are not projections. Coordinate System Transformations retain the vector pointing to a certain location in space. Coordinate System Transformations retain the length of vectors and the distance between vectors. Projection Transformations do not retain the vectors. Distances between vectors do change. That's the reason why distances on the flat earth AE map and other projections of the whole earth onto a plane map are completely wrong and why distances on flat maps of small parts of the earth are only approximately correct.

### GPS Vectors and the Computer Model

All positions that are used in the Computer Model of the Rainy Lake Experiment are derived from the measured GPS Vectors in ECEF Cartesian coordinates, see Measuring the Targets.

Walter Bislin converted the GPS Vectors into local coordinates required by his Computer Model using his WGS84 Calculator. The WGS84 Calculator can also transform the GPS Vectors from the ECEF Cartesian coordinate system into the Geodetic coordinate system and vice versa. The transformations from the WGS84 Calculator into latitude, longitude and ellipsoid heights confirm the same data also provided by Jesse Kozlowski.

## Visualizing GPS Vectors

All target locations were measured using Differential GPS (DGPS). DGPS are enhancements to the Global Positioning System (GPS) which provide improved location accuracy from the 15-meter nominal GPS accuracy to about 1-3 cm in case of the best implementations. A Base Station or network of Base Stations calculates differential corrections for its own location and time and this correction signal is then transmitted to the mobile GNSS receiver which corrects its measurements accordingly, see Improving GPS accuracy with Differential GPS.

Jesse Kozlowski measured the GPS Vectors to the locations of each target and the observer locations with his Differential GPS Equipment as described at Measuring the Targets. All data is provided in ECEF Cartesian coordinates (GPS Vectors) plus Elevation (see Obtaining Elevations).

So we have a set of vectors from the center of the earth to the locations of all targets and observation locations in a coordinate system that has no relation to the shape of the earth and is not affected by refraction or perspective.

We can now import the GPS Vectors into a 3D software like the GNSS Data Viewer, written by Walter Bislin. The Viewer displays the GPS Vectors as white markers and allows to make calculations between the data points. The software allows to import and overlay an image onto the display to see how observations match the measured GPS Vectors.

Img 23: Overlay GPS Vectors with image of Bedford targets → App
Img 24: Overlay GPS Vectors with image of Tangent targets → App

If we look along the points at a shallow angle we can recognize that the Bedford target, ice and water level GPS Vectors do not lie on a plane but curve down. The radius of the curvature can be calculated with the software, see Measuring the Radius of the Earth.

We can also recognize that the targets on the images appear as higher than the GPS Vectors, as farther away they are from the observer. This is due to refraction and the images are consistent with the calculations of Apparent Lift due to Refraction. We can calculate the refraction from the deviation of the target images from the corresponding GPS Vectors, see Measuring Refraction.

## Measuring the Radius of the Earth

In the GNSS Data Viewer we can select 3 Bedford target markers of the same height or 3 water level markers and calculate the radius of the earth from the arc that this markers span. The ECEF Cartesian coordinates of all data points are listed in the Table at the bottom of the GNSS Data Viewer.

Img 25: Measuring the radius of the earth from 3 data points → App

In this example, the calculated radius of the earth from 3 selected markers is R3pt = 6025 km. That is only 363 km or 5.4% too less.

The distance between the data points is too small to get an accurate value for the radius of the earth with this method. A height deviation of only 1 cm results in a change of the calculated radius of Rerr = 71.1 km/cm. The height variation of the selected points in this example is dHel = 18 cm which results in an uncertainty for the radius of about ±9 · 71.1 km = ±640 km.

The exact radius of the earth depends on the location and the measuring direction, because the earth's surface is in a first approximation an Ellipsoid. The exact radius of the Ellipsoid at this location in the direction of the targets is calculated as Re12 = 6388 km.

We also have to consider that sea level of the earth forms a Geoid. The surface of the Rainy Lake is curved more than the reference Ellipsoid of the earth. The Ellipsoid height of the last water level position is about 24 cm less than the Ellipsoid height of the water level at the observers, see GNSS Ellipsoid and Height Data, due to variations in the gravitational field of the earth. So we can expect to measure a too small radius of the earth from the data points.

The GNSS Data Viewer calculates the radius as described in Method to calculate the Radius of the Earth from 3 Points.

Why are all data points sitting near each other in the Data Viewer when they are more than 1 km appart?

This is a highly zoomed in view almost perfectly along the targets. So due to perspective compression we can clearly see that the water level points are curving over a horizon. Water and ice level of the last 2 targets are beyond and below the horizon, indicated by brown horizontal ground lines. Targets nearer than the horizon are indicated by green ground lines.

P1 is the observation point for the Bedford targets. The thick glowing red lines lie on the eye level plane of the reference point Pref. Due to the curvature of the earth the points P1 and P2 lie below this plane. Remember, all data points are Vectors in ECEF Cartesian Coordinates that are independent of the shape of the earth. The points of equal height above the surface show a curve.

This can only be if the surface of the earth is curved. If the earth were flat then all points with the same height would lie on a plane.

For more examples of perspective compression see Comparison of Globe and Flat-Earth Model Predictions with Reality.

## Refraction

Refraction is a major factor when observing distant objects. Refraction changes the apparent vertical position with respect to a reference line like eye level. Because the Rainy Lake Experiment is about the relative vertical positions of targets, refraction has to be taken into account.

So lets talk about how refraction works and can be calculated:

Light travels at a slower speed in transparent media than in vacuum. [2] The amount of the slow-down is expressed with the index of refraction n = c/v, where v is the speed of light in the medium and c is the speed of light in vacuum. If light encounters a change in the index of refraction and the boundary of the change is not perpendicular to the direction of the light ray, the direction of the light ray changes at the boundary, always in the direction of the higher index of refraction. We can use Snell's law to calculate the angles of the light ray with respect to the perpendicular to the boundary layer.

The density of the atmosphere generally decreases exponentially with increasing altitude. The index of refraction is proportional to density. Because the density change is not abrupt but continuous, the light rays do not suddenly change direction, but are bent in an arc towards the denser part of the atmosphere, which is usually downwards. This is even the case for light rays that start perpendicular to the density gradient, i.e. horizontal in the atmosphere. This means that objects in the distance appear higher than with a straight line of sight, because the bent light rays of objects reach the observer from more above.

Refraction is not a constant phenomenon. It depends strongly on the current atmospheric conditions along the light path and therefore fluctuates on the way to the observer. Since it is difficult to measure the actual refraction from the object to the observer, an average value is obtained which can be calculated from the atmospheric conditions at the observer's location, at least for shorter distances of only some km. But these values can be used for longer distances too, if there are similar conditions along the light path. The average value corresponds to a light beam following an arc of constant radius RR.

Atmospheric refraction can be expressed with different values. Some are independent on the distance of the observed object, like the Refraction Coefficient and the Refraction Factor. Others, like the Refraction Angle and the Apparent Lift due to Refraction, depend on the distance to the object.

### Refraction Coefficient

The refraction Coefficient k, often simply called refraction, is defined as the ratio of the mean radius of the earth R to the curvature radius of the bent light rays RR:

(3)

per Definition

where'
 $k$ ' =' 'refraction Coefficient $R$ ' =' '6371 km = mean radius of the earth $R_\mathrm{R}$ ' =' 'radius of the curved light ray from the object to the observer in km

If the light rays are not curved, their radius RR is infinite. This means that for non-curved light rays the refraction Coefficient is k = 0. If the light rays follow the earth's curvature, which is quite possible, then k1. In this case the earth appears flat or even concave, see Strong Refraction.

Note: The occurence of the radius of the earth in the refraction Coefficient definition has no influence on refraction itself. It is simply a convenient definition and works the same for any shape of the earth, because the only variable that depends on refraction is the radius of curvature of the light ray, not the shape of the earth.

Standard Refraction: On average the atmosphere has a certain pressure, temperature and density gradient. This average is called International Standard Atmosphere. On standard atmospheric conditions refraction is called Standard Refraction.

For Standard Refraction a value of k = 0.13 is often used in survey. Another frequently used value assumes a radius of curvature for light rays of RR = 7 · R, which corresponds to a refraction Coefficient of k = 0.14 or a Refraction Factor a = 7/6. The equation below gives a refraction Coefficient of k = 0.17 at sea level, decreasing with altitude.

The difference in the apparent lift of the observed object is small between the different values for Standard refraction between k = 0.13 and k = 0.17: for an altitude measurement of a target in a distance of 1000 m the difference in apparent lift is only about 3 mm while the drop due to earth's curvature is 78 mm. The apparent lift (looming) of a target at 1000 m at k = 0.17 is 13 mm, which is 1/6 of the curvature drop. Because the target appears 1/6 of the real drop loomed upwards to 65 mm apparent drop, we have to correct the measurement by multiplying it with 6/5 = 1.2 to get the real drop.

The refraction Coefficient can be calculated from air pressure, temperature and temperature gradient as follows:

(4)
source
where'
 $k$ ' =' 'refraction Coefficient $P$ ' =' 'air pressure at the observer in mbar or hPa or 1/100 Pa, Standard = 1013.25 mbar $T$ ' =' 'temperature at the observer in Kelvin, Standard = 288.15 K = 15°C $\mathrm{d} T/\mathrm{d} h$ ' =' 'temperature gradient at the observer in K/m or °C/m, Standard = −0.0065°C/m

The refraction Coefficient decreases with increasing altitude due to a decrease in air pressure. Refraction is very sensitive to temperature gradients, i.e. small changes in temperature in the height range of the light ray. Near the surface refraction can change considerably if the surface temperature is different than the air above, which is almost always the case.

Above a surface that is cooler than the air (ice, water), you can get refraction Coefficients greater than k = 1 easily, which means that light rays follow the curvature of the earth for hundreds of km. Laser tests over water to prove the earth is flat are therefore flawed, because strong refraction causes the laser to bend along the surface to the observer, see Strong Refraction at Bedford Targets.

To get accurate measurements you have to assure that the line of sight is always some meters above the surface, the temperature of the surface and the air is about the same (no visible distortions, clear image) and the distance to the object is not too far, because the apparent lift due to refraction increases with the square of the distance.

Further usefull readings with explanations and simulations of refractions:

### Refraction Factor

In order to be able to use the same equations e.g. for the calculations of the obscuration of objects by the curvature of the earth and taking refraction into account, there is a trick: simply replace the radius of the earth R by an increased apparent radius of the earth R', which can be calculated from the refraction Coefficient k, and assume the light ray as straight. This does not change the geometry between the earth and the light ray.

I denote the multiplier as Refraction Factor a:

(5)

per Definition

(6)

where'
 $R'$ ' =' 'apparent extended radius of the earth due to refraction $R$ ' =' '6371 km = real radius of the earth $a$ ' =' 'refraction Factor $k$ ' =' '

A value of a = 7/6 corresponds to Standard refraction k = 0.14.

Note: In the literature the refraction Factor is often denoted as K (big K). To avoid confusion with the refraction Coefficient k I use a instead of K.

The refraction Factor can be calculated from the refraction index gradient or refractivity gradient as follows:

(7)
source
where'
 $a$ ' =' 'Refraction Factor $R$ ' =' '6371 km = Radius of the earth $\mathrm{d} N/\mathrm{d} h$ ' =' 'Rrefractivity gradient in 1/km, standard is −22.4/km for k = 0.143 or a = 7/6 $\mathrm{d} n/\mathrm{d} h$ ' =' 'Refraction index gradient in 1/km $N$ ' =' '(n − 1) · 106 = Refractivity $n$ ' =' 'Refraction index $h$ ' =' 'Altitude

### Apparent Lift due to Refraction

Due to the fact that the density of the atmosphere decreases with increasing altitude, light rays get bent towards the surface of the earth. So light rays from a target reach an observer from more above than without refraction. The target appears higher than it is in reality. Refraction can vary a lot near the surface. If this is the case, images appear distorted and unsteady. If an image is calm and clear, then refraction of light rays, that do not pass near the surface, can be assumed to be bent according to standard of about k = 0.13..0.17.

If we know the mean refraction Coefficient k we can calculate how much each target appears to be lifted as measured at the target with the following equation:

(8)
(9)
(10)
source
where'
 $l$ ' =' 'apparent lift of target due to refraction with respect to a fixed reference line like eye level of the observer $\rho$ ' =' 'refraction angle $k$ ' =' 'refraction coefficient, standard refraction is k = 0.17 $k/R$ ' =' '1/RR = curvature of light ray $R_\mathrm{R}$ ' =' 'radius of curvature of light ray $d$ ' =' 'distance of target from observer $R$ ' =' '6371 km = radius of the earth

Note: This equations can be used for any shape of the earth. The factor R is only present because the refraction Coefficient k is defined as the curvature of the light ray 1/RR times R. The curvature of the light ray itself is independent of the shape of the earth.

The following table lists the expected lifting for each target depending on some refraction coefficients.

 Target (1) Target (2) Target (3) Target (4) Target (5) Target (6) Target (7) values in m 1095 2169 3234 4363 5434 6429 9459 1.82 1.844.27 1.81 1.845.24 1.73 1.676.50 10.54 0.55 0.370.41 0.55 0.740.86 0.92 0.650.65 1.6 0.01 0.04 0.08 0.15 0.23 0.32 0.70 0.02 0.06 0.14 0.25 0.39 0.55 1.19 0.03 0.10 0.22 0.40 0.63 0.88 1.90 0.04 0.15 0.34 0.61 0.95 1.33 2.88

As we can see the farther away a target is, the more it appears lifted by refraction. The last Tangent target (7) can easily appear lifted higher than its size even by moderate refraction changes, as shown at Strong Refraction at Tangent Targets.

Note: The Bedford target centers were set exactly 1.85 m above the lake water level. In the table above are the target center heights listed as used by the Computer Model . Due to variations in the gravitational field of the earth, expressed by the Earth Gravitational Model (EGM96 Geoid), see Obtaining Elevations, the Ellipsoid height of the lake surface at the last target (7) is about 25 cm lower than the Ellipsoid height of the lake surface at the observer. The Computer Model does not take the Geoid variations into account. To simulate the exact geometrical 3D positions of the targets, all target center heights were adjusted for a perfect sphere by using the Ellipsoid heights of the targets with respect to the Ellipsoid height at the observer. This results in the reduced target center heights shown in the table above. Using this adjustments the target images of the Computer Model match the observations perfectly, while the horizon appears about 24 cm too height on the Computer Model, because it does not model the surface drop of 25.1 cm of the Geoid.

### Horizon above Eye Level

If refraction causes the horizon to appear above eye level (concave earth), the horizon will be as far away as we can see, if there are no obstructions. The ground will fade slowly into the sky and therefore the horizon is not a distinct line. Note, there will always be a horizon, because there are always light rays that never reach the ground. At refractions k < 1, including Standard refraction, the Globe has a distinct horizon at a certain distance, see equation (23). The Flat Earth on the other hand only has a distinct horizon when light is bent upwards due to negative refraction.

Below are 2 images from a Refraction Simulation that show how the horizon will appear at Standard refraction for the Globe and the Flat Earth model. The simulation predicts that on the Flat Earth there is no distinct horizon line like on the Globe, if refraction is greater than 0. To clearly see this effect in reality you have to strongly zoom in on the horizon.

Distinct horizon on the Globe model
No distinct horizon on the Flat Earth model

The targets are 3 m high and 6 m wide.

## Measuring Refraction

There are two methods used to measure refraction in the Rainy Lake Experiment:

Both methods require that we can measure or calculate the 3D vectors in space to the target centers. Method 1 uses the GPS Vectors directly measured using Survey grade GNSS receivers. Method 2 uses the calculated vectors predicted by the Globe and Flat Earth Computer Model .

A third method would be Measuring Refraction from Observations with a theodolite. This was not done in this Experiment, but I will describe how it could be done anyway.

### Measuring Refraction from GPS Vectors

The GPS Vectors to the target centers were directly measured using Survey grade GNSS receivers. They represent the real geometrical positions in space, not dependent on refraction, perspective or the model of the earth.

The following screenshots show the overlay of photos of the Bedford and Tangent targets with the corresponding GPS Vectors in the GNSS Data Viewer. The photos are taken through the auto level. The virtual camera of the Viewer was set to the observer locations and the zoom and viewing direction was adjusted to match the vertical positions of the nearest target centers and the horizontal positions of all targets.

Img 23: Overlay GPS Vectors with image of Bedford targets → App
Img 24: Overlay GPS Vectors with image of Tangent targets → App

The white markers show the positions the GPS Vectors are pointing to and can be used as a reference. The apparent lift of the targets is due to refraction and can be measured from the image.

Note: The equations for the refraction Coefficient (11) and (12) can be used for any shape of the earth. The factor R is only present because the refraction Coefficient k is defined as the curvature of the light rays times R. The curvature of a light ray itself is independent of the shape of the earth.

Note: Use positive values for l if the image appears higher than the GPS Vectors and negative values if the image appears lower than the GPS Vectors.

Bedford Target Refraction: The apparent lift of the Bedford target (5) at 5.4 km distance is about 0.5 m, estimated from the 0.9 m target plate vertical size. Applying formula (10) results in a refraction of:

(11)

Tangent Target Refraction: The apparent lift of the Tangent target (7) at 9.5 km distance is about 1 m, estimated from the 1.6 m target plate vertical size. Applying formula (10) results in a refraction of:

(12)

The measured refraction at the Bedford targets from 1.85 m observer height was slightly higher than at the Tangent targets from 3.91 m observer height, which is consistent with what atmospheric refraction predicts.

The markers in the Viewer were aligned exactly to the vertical positions of the first target centers, but in reality this targets were also affected by refraction and should appear slightly above this markers. Therefore this refraction calculations results in slightly too small values: 0.05 smaller than compared with Measuring Refraction with the Computer Model.

The GPS Vectors clearly show the earth is not flat.

### Measuring Refraction with the Computer Model

We can use the Computer Model to measure refraction from an image as follows:

We import an image of the Bedford or Tangent targets into the Computer Model, set the observer height accordingly, set the zoom to match the zoom of the image and pan the model camera into the direction the real camera was aimed. We have to use the simulation of the Globe Model to measure refraction, because refraction is not simulated for the Flat Earth Model. But we can calculate refraction for the Flat Earth Model from the values obtained with the Globe Model.

With zero refraction set in the Computer Model, all targets on the images will appear too high and can not be matched with the image by tilting the model camera. But when we apply refraction with the Refraction slider, we can make the computer graphics match the images perfectly. We can then read off the refraction Coefficient k from the Computer Model:

#### Flat Earth Refraction from Globe Refraction

The corresponding refraction Coefficient of the Flat Earth Model can be calculated from the refraction value k of the Globe Model as follows:

(13)

Explanation: If refraction k = 0 is set on the Globe Model, the earth appears like a globe with radius 6371 km. In reality the Flat Earth would appear as such a globe if light is bent upwards with a radius of curvature of 6371 km. This light ray curvature corresponds to a refraction Coefficient of −1: so kFE = k1 = 01.

On the other hand, if refraction k = 1 is set on the Globe Model, the earth appears flat, because at refraction k = 1 light rays get bent down with a radius of curvature of 6371 km. In reality the Flat Earth would appear flat if refraction is zero: so kFE = k1 = 11.

This correlation between kFE and k is only valid for k < 1 respectively kFE < 0, where we have a distinct horizon. In other circumstances we have to execute the Calculations without a Horizon.

So the measured Flat Earth refraction Coefficients from the Computer Model are:

• Result for Bedford Targets: kFE = 0.2701 = −0.730
• Result for Tangent Targets: kFE = 0.1871 = −0.813

As explained at Refraction Range of Clear Images such high negative refractions would cause very distorted images. We don't see distortions in the used images and this values are not consistent with the values obtained at Measuring Refraction from GPS Vectors.

This results show, the earth can not be flat.

### Measuring Refraction from Observations

We can calculate refraction from real world observations. First we calculate the expected drop of the horizon or a target without refraction for the Globe and Flat Earth model. Then we measure the effective drop with a theodolite. From the difference of the expectation and the measurement we can calculate the refraction Coefficients for both models.

To calculate the expected horizon drop x and drop angle φ without refraction, depending on the observer height h, for the Globe we can use the following good approximations. Note, we use positive signs for drops below eye level:

(14)
source
where'
 $x_\mathrm{globe}$ ' =' 'expected drop of the horizon from eye level for the Globe at k = 0 $\varphi_\mathrm{globe}$ ' =' 'expected drop angle from eye level in radian for the Globe at k = 0 $h$ ' =' 'observer height $R$ ' =' '6,371,000 m = radius of the earth

The expected horizon drop and drop angle of the Flat Earth model is simply 0.

(15)
where'
 $x_\mathrm{FE}$ ' =' 'expected drop of the horizon from eye level for the Flat Earth at k = 0 $\varphi_\mathrm{FE}$ ' =' 'expected drop angle from eye level in radian for the Flat Earth at k = 0

Now we measure the real horizon drop xreal or drop angle φreal with a theodolite. If we measure the drop angle φreal we have to convert it into the drop distance xreal using the following equation. Note, we use positive signs for drops below eye level:

(16)

source

(17)

good approximation

source

where'
 $x_\mathrm{real}$ ' =' 'measured drop from measured drop angle φreal $d$ ' =' 'distance to the horizon as a reference (for the globe model without refraction) $h$ ' =' 'observer altitude $R$ ' =' '6,371,000 m = radius of the earth

Now we calculate the apparent lift from the differences between expectation and measurement. Note, we use positive signs for drops below eye level:

(18)
where'
 $\Delta l_\mathrm{globe}, \Delta l_\mathrm{FE}$ ' =' 'apparent lift due to refraction

Now we can calculate the refraction Coefficients for Flat Earth and Globe model using the same equation for both models, beacuse the curvature of the light rays over distances not much more than 10 km is identical for all practical purposes for both models, see Refracted Light Rays on Globe and Flat Earth:

(19)
where'
 $k_\mathrm{globe}$ ' =' 'refraction Coefficient for Globe model $k_\mathrm{FE}$ ' =' 'refraction Coefficient for Flat Earth model $\Delta l_\mathrm{globe}$ ' =' 'apparent the lift due to refraction on the Globe model $\Delta l_\mathrm{FE}$ ' =' 'apparent the lift due to refraction on the Flat Earth model $d$ ' =' 'distance to the horizon, see (17) $R$ ' =' '6371 km = radius of the earth

This method only works if there is an unobstructed and distinct horizon, which is not always the case, see Horizon above Eye Level. If there is no distinct horizon, you have to use the Calculations without a Horizon.

If the calculated refraction value lies within the Refraction Range of Clear Images, the image will look clear. Outside this range it will look distorted. If we have a clear image and the calculated refraction Coefficient for one of the models lies outside the range for clear images, then this model does not fulfill the prediction and is falsified.

#### Refracted Light Rays on Globe and Flat Earth

In this Simulation of Atmospheric Refraction we can see that the light rays bend exactly the same amount on Globe and Flat Earth model. The only difference is that we see the horizon at different vertical angles with respect to eye level. The apparent lift of the horizon or any object is practically identical in both models. The fact that the air layers are curved on the Globe but flat on the Flat Earth has no effect on distances much smaller than the radius of the earth.

Note: This is not true if we measure in layers near the ground and have strong refractions in this layers causing mirages and compressions. [3] In this case Globe and Flat Earth do not look the same at the same atmospheric conditions. We can use the Simulation of Atmospheric Refraction to investigate such behaviour. But if we observe objects above such layers from above such layers the difference is negligible.

#### Calculations without a Horizon

If we don't have a distinct horizon, we have to use another target with a known height and distance to calculate its position with respect to eye level for zero refraction. For the Globe model we can use the following equation. We use positive values for drops below eye level:

(20)
source
where'
 $x_\mathrm{globe}$ ' =' 'drop of the target from eye level, negative for targets above eye level $\varphi_\mathrm{globe}$ ' =' 'drop angle from eye level in radian, negative for targets above eye level $d$ ' =' 'distance to target (line of sight or along surface is approx the same) $h_\mathrm{O}$ ' =' 'observer height $h_\mathrm{T}$ ' =' 'target height $R$ ' =' '6,371,000 m = radius of the earth

For the Flat Earth Model we can use:

(21)

We can then measure the raise or drop of the target with respect to eye level with a theodolite. The remaining calculations are the same as described at Measuring Refraction from Observations with a distinct horizon.

## Strong Refraction

Lets look whether we can determine the shape of the earth from images with strong refraction. To do so we first have to investigate how strong refraction influences an image.

The biggest influence on atmospheric refraction has the temperature gradient dT/dh (change in temperature in a certain change of altitude). On standard atmospheric conditions below 11 km altitude the temperature gradient according to the International Standard Atmosphere is constant −0.0065°C/m, which results in a refraction Coefficient at sea level of k = 0.17, called Standard Refraction.

Distorted images are caused by turbulent air. Turbulent air is caused by temperature gradients that diverge from standard. So distorted images are always a sign of strong refraction. In fact, there are devices that measure refraction by analysing the turbulence of the air in images [4] [5]. On the other hand, a clear image is a sign of a temperature gradient near standard which causes low refraction. But note, low refraction is not zero refraction. Because the air density is always decreasing with increasing altitude, even on calm air conditions there is always some standard refraction.

### Refraction Range of Clear Images

The following table lists some refraction values and tells how the earth (flat and globe) is expected to look like according to the Accepted Model of Atmospheric Refraction:

Refraction dT/dh [°C/m] Image Flat Earth appearance Globe appearance
k = −1 −0.20 distorted globe with R = 6371 km
horizon below eye level
globe with R = 3185 km
horizon below eye level
k = −0.83..0 −0.17..−0.034 distorted globe R > 7682 km to flat
horizon below eye level
globe R = 3481..6371 km
horizon below eye level
k = 0 −0.034 clear flat
horizon at eye level
globe R = 6371 km
horizon below eye level
k = 0.17
standard
−0.0065
standard
clear concave
horizon above eye level
globe R = 7682 km
horizon below eye level
k = 0..0.3 −0.034..0.015 clear flat..concave
horizon above eye level
globe R = 6371..9101 km
horizon below eye level
k = 0.3..1 0.015..0.13 distorted concave
horizon way above eye level
globe..flat R > 9101 km
horizon below eye level
k = 1 0.13 distorted concave
horizon way above eye level
flat
horizon at eye level
k = 1..2 0.13..0.29 distorted concave
horizon way above eye level
flat..concave
horizon above eye level
k > 2 > 0.29 distorted concave
horizon way, way above eye level
concave
horizon way above eye level

The following graph shows the correlation between the refraction Coefficient and the predicted horizon position with respect to eye level for the Flat Earth and Globe model.

This graph is not accurate, but shows the point for the following discussion. The real correlation between refraction and horizon drop is not linear. The slope of the horizon lines depends on the observer height, but the Globe Horizon must cross eye level at k = 1 and the Flat Earth Horizon must cross eye level at k = 0 and the lines are parallel.

The refraction range producing clear images is shown with a blue background. This range depends partially on the observed distance. Near objects appear less distorted than far objects. Here is the approximate range shown for the 10 km observing distance of the Rainy Lake Experiment.

We can see that for a certain horizon drop the refraction Coefficient of the Flat Earth model is kFE = kGlobe1, at least for k values less than 1, as explained at Flat Earth Refraction from Globe Refraction.

If refraction is 0 then the Flat Eearth horizon is at eye level, while the drop for the Globe corresponds to a sphere with radius 6371 km. If refraction is 1, the Globe horizon is at eye level, the Globe looks flat, while the Flat Earth will appear concave. For a Flat Earth to appear as a globe with radius 6371 km, refraction has to be −1. Such conditions can only occur over hot ground and are confined to low altitudes over ground. The corresponding images will look very distorted, because the temperature gradient for such refractions causes turbulent air.

### Strong Refraction at Bedford Targets

If refraction gets greater than 1, even on the Globe the horizon raises above eye level and the earth appears concave shaped. An observation of refraction greater than 1 is shown in image 28. You can tell that there is strong refraction because the image looks very distorted and the corresponding video shows a very instable scene.

Img 27: Clear image, low refraction, watch video
Img 28: Distorted image, strong refraction, watch video

Note: Eye level on both images is where the horizontal crosshair is, not at the center of the image! Both imagages are taken from 1.85 m observer height through an auto level.

Globe Model: According to Refraction Range of Clear Images refraction in image 27 is low, about standard, because the horizon is below eye level and the image is clear. Refraction on image 28 is strong because the image is distorted. The horizon is above eye level, which means refraction is greater than 1. Both images are consistent with the predictions of atmospheric refraction on the Globe model.

Flat Earth Model: Measuring Refraction from Observations for the Flat Earth gives for image 28 a refraction coefficient greater than 1. According to Refraction Range of Clear Images it is expected that such an image on the Flat Earth will look very distroted. This is the case, so this image supports the Flat Earth model. But in image 27 the refraction calculated for the Flat Earth model is strong negative, because the horizon appears way below eye level. According to Refraction Range of Clear Images this image should look very distorted, but it looks clear. This means the assumption of a Flat Earth is not consistent with this observation.

Images of the Bedford targets with strong refraction are not consistent with the Flat Earth model.

Calculations

Although the apparent lifting of the horizon may appear unbelievable strong here, keep in mind, we are looking through an auto level with high magnification. The refraction angle on standard refraction to the horizon at the farthest target at 9.5 km distance is only ρ = 0.0072°, see formula (9). The lift of the horizon on image 28 compared to image 27 is about 15 m, which corresponds to a refraction Coefficient of k = 2.1, see formula (10). The refraction angle at this refraction is still less than ρ = 0.09°. But this is enough to raise the horizon 9 m way above eye level! The earth appears concave.

Laser Tests

Refraction always fades to standard with increasing altitude. When you look over a surface, light gets influenced the most as it approaches the surface. At least parts of the light rays from objects behind the horizon always streak the horizon where refraction is the strongest. In looming conditions, where refraction is greater than standard, images appear increasingly compressed near the surface, because refraction there is strongest and lifts up the lower parts of objects more than the upper parts.

If refraction directly over the surface is greater than 1, which is very often the case on looming conditions, light rays streaking the surface can follow the curvature of the earth for hundreds of km. Shooting a laser great distances over water or ice is a bad idea to measure curvature, because there are always light rays streaking the surface, which then can bee seen in any distance until they get absorbed by the atmosphere.

It is obvious that in conditions like in image 28 a Laser placed on the ground at 10 km distance or more would be visible from the ground at the observer. Image 28 was taken before dawn, when the air was still warm but the ground was cold. This can cause strong refractions of k > 1.

After sunset, when the ground is cooling faster than the air above, near the ground you can easily get refraction coefficients way beyond 1. [3] Flat earthers alway conduct their laser test very near the ground after sunset over cool water or ice. In such conditions it is expected to see a laser over any distance. Such experiments to prove the earth is flat are flawed, as the Rainy Lake Experiment shows.

### Strong Refraction at Tangent Targets

Here are some images from the Time lapse of Refraction that show that strong refraction at higher elevation is not as strong as on lower elevation, but still can raise targets in the distance considerably.

• Time lapse at 13 s
• at 16 s
• at 20 s
• at 28 s
• at 38 s
• at 39 s

We can recognize strong refraction in images, because images look distorted on strong refraction. So on the first images refraction is low, while on the last images it is high. The effect is even better visible on a Time lapse of Refraction during a day.

Lets analyse refraction from the first and last image for the Flat Earth and Globe model.

Img 31: Clear image, low refraction
Img 32: Distorted image, strong refraction

Globe Model: According to Refraction Range of Clear Images refraction in image 31 is low, about standard, because the horizon is below eye level and the image is clear. Refraction on image 32 is strong because the image is distorted. On both images the horizon is below eye level, which means refraction is less than 1 on both images. Both images are consistent with the predictions of atmospheric refraction for the Globe model.

Flat Earth Model: Because the horizon in both images is below eye level, according to the graph at Refraction Range of Clear Images, the corresponding refraction Coefficients in both images must be negative for the Flat Earth model. Refraction in image 31 must be less negative than in image 32, because the horizon is lower. As image 32 with less negative refraction than image 31 already looks very distorted, image 31 should look even more distorted. But image 31 looks very clear. This means the assumption of a Flat Earth is not consistent with this observation.

From the distortions of the images can be concluded, that refraction must be in the range from standard refraction to refraction k < 1. In this refraction range on the Globe model the horizon will appear always below eye level. This is consistent with observation. On the Flat Earth model the horizon in this refraction range would always appear above eye level, which is inconsistent with observation.

Images of the Tangent targets with strong refraction are not consistent with the Flat Earth model.

Calculations

The apparent lift of the Tangent target (7) between the first and last image is estimated from the target size of 1.6 m to about 2.9 m. This corresponds according to formula (10) to a refraction difference of k = 0.23. The measured refraction Coefficient for the Globe model of the first image is k = 0.18. So the refraction Coefficient of the last image is k = 0.18 + 0.23 = 0.41.

The strongest observed refraction at the Bedford targets with an observer height of 1.85 m (image 28) was about k = 2.1. The strongest observed refraction at the Tangent targets with an observer height of 3.91 m (image 32) was about k = 0.41. This is consistent with the prediction that refraction near the surface is usually stronger than higher up.

### Time lapse of Refraction

The camera was at a height of about 4 m above water level next to the auto level, which is at the level of the Tangent targets. Note how refraction gets stronger and stronger towards the evening as the ground cools down faster than the air above, creating a strong negative density gradient directly above the surface, which bends light down and looms up the horizon and the targets. You can see how the horizon and the targets get lifted more and more and the images get more distorted. The horizon appears lifted about 2.9 m at the end.

## Night Observations

Img 33: Lantern through T2 theodolite
Img 34: Lantern through TOPCON theodolite

Part of the Rainy Lake Experiment was to show the Bedford targets at night. For this purpose lantern were mounted above the Bedford targets and observed through the two theodolites from a location more to the side of the line of targets. The heights of the theodolites were set exactly the same height over water level as the lantern.

Img 33: Magnified view through T2 theodolite shows the lantern drop with distance

It could be observed, that the lantern lights were all below the horizontal eye level crosshair. The farther away the target, the lower the lights appeared, which is expected on the Globe Earth.

Note: The drop of the targets shown here appears much less than in the images taken at daylight. But keep in mind that this images show a view more from the side with much less zoom, while the daylight images show a view along the targets with strong zoom. Looking along the targets causes the drop to appear much more pronounced due to zoom perspective compression.

Another reason may be refraction. Over cool water or ice at the evening refraction is commonly stronger than standard and lets the earth appear flatter than it is. But we can not derive any quantitative conclusions from this images.

## Drone Observations

• Next
• h 1
• h 2
• h 3
• h 4
• h 5
• h 6
• h 7
• h 8

The image sequence above was filmed from a Voyager 4 Drone from ground to about 100 m altitude. The drone uses a gymbal stabilized camera which keeps it pointing at eye level during the whole ascent. The yellow lines mark the horizon at the ground and at the highest altitude.

We can observe that the horizon drops with increasing altitude and that more and more of the landscape comes into view from behind the horizon. The Globe model predicts a horizon drop of 0.29° for an altitude of 100 m. The rate at which the horizon drops as a function of altitude is such, that it first drops fast and then slower and slower according to the following equation:

(22)
source
where'
 $\alpha$ ' =' 'drop angle in radian; multiply with 180°/π to get degrees $R$ ' =' '6371 km = radius of the earth. Use a radius of 7670 km to acount for standard refraction k = 0.17 $h$ ' =' 'observer altitude

The Flat Earth model predicts that the horizon will stay at the same position in the frame during the whole ascent.

In the center of the image is a corridor of 32 km length of ice. So the trees in the background of the coridor are 32 km far away. The distance to the Globe horizon from 100 m altitude with standard refraction is calculated as 39.2 km.

(23)
source
where'
 $d$ ' =' 'distance to the horizon $h$ ' =' 'observer altitude $R$ ' =' '6371 km = radius of the earth. Use a radius of 7670 km to acount for standard refraction k = 0.17

The Drone observation of a horizon drop matches the Globe model, but not the Flat Earth model.

## Data and Tools

Follow the links below for all data gathered and processed and the tools used:

## Conclusion

All Observations match the Globe Model predictions very well, but contradicts the Flat Earth Model.

Refraction can have a great influence on observations. But distorted images tell when there is strong refraction. Indications of low, standard refraction are clear and steady images. Under those conditions the observations match the Globe Model extremely well.

To produce the same observations on a Flat Earth, refraction would have to be very strong and negative, bending light upward all the time and to any altitude. Such refractions requires temperature gradients of −15°C per 100 m all the way up to high altitudes, see Simulation of Atmospheric Refraction. This is physically impossible. Negative refractions can only occur in small layers above the surface with temperature inversions and produce inferior mirages or visible compression and distorted images.

The overall refraction of the atmosphere is always positive, bending light down, see Refraction Coefficient as a Function of Altitude. Positive refraction lets the earth appear flatter than it is and can bring objects into view that are expected to be hidden by the curvature, if we assume no refraction. The comparison of the observations with the GPS Vectors gathered using Differential GPS shows that all targets are lifted consistent with the calculations of standard to medium refraction, see Visualizing GPS Vectors.

The physics of atmospheric refraction can not produce the observations of the Rainy Lake Experiment for a Flat Earth model.

There is only one conclusion left: The earth can't be flat. The data is consistent with a Globe with a radius of about 6371 km. The Rainy Lake Experiment shows this even better than the Bedford Level Experiment, because we also have GPS measurements, which are not affected by refraction or perspective.

## Videos

See the following video documentations for the planing, execution and discussion of the Rainy Lake Experiment:

[1]
Rainy Lake Survey Progress Report
George Hnatiuk, Soundly, and Jesse Kozlowski get together for a hangout live stream to discuss the data and observations collected on Rainy Lake and plans for completing the work.
[2]
Survey Overview Rainy Lake - is the earth curved?
George Hnatiuk: This is the first part of a series of videos that will follow an ongoing set of measurements on a frozen lake in northern Minnesota to determine the contour of the earth's surface.
[3]
Laser with corner reflector prism
George Hnatiuk: This was a quick test to see how difficult it is to aim my laser to a corner reflector placed 1.36 miles away. The corner reflector effectively reflected the laser light back to me. The laser did not have a beam collimator attached.
[4]
Survey Update 1 --- Rainy Lake Minneosta
George Hnatiuk: This is the first of weekly updates for the survey on Rainy Lake in northern Minnesota which discusses the point of observation and placement of the reference poles.
[5]
Survey Update 2 --- Rainy Lake Minnesota
George Hnatiuk: This update details the deployment of marker targets and reflectors for a laser to shoot and return light back.
[6]
Survey Update 3 - Rainy Lake Minnesota
George Hnatiuk: This update shows the deployment of the six mile target and discusses why a laser is not the correct tool for aligning targets separated by long distances.
[7]
Survey Update 4 --- Rainy Lake Minnesota
George Hnatiuk: This video details the deployment of target poles and targets at 4 miles and 6 miles from the observation station which are used to survey the contour of the water surface of Rainy Lake in northern Minnesota.
[8]
Survey Update 5 -- Rainy Lake Minnesota
George Hnatiuk: On a previous day, the target heights were adjusted to be centered about a tangent line set by an auto level. Some preliminary measurements are made on the 1.36 mile target (time 5:02) and Fransen Island target at 2.72 mile.
[9]
Survey Update 6 -- Rainy Lake Minnesota
George Hnatiuk: Setting the lower Bedford targets for a Bedford level type of test. All Bedford targets are set to 73" above the water surface, a constant height with distance from the spotting scope.
[10]
George, Soundly & Jesse Talk About The Work Planned On Rainy Lake
Soundly and Jesse Kozlowski are traveling on 3/30 to meet George Hnatiuk. In this live hangout, the three will discuss their plans for the work to be performed conducting and recording surveying observations on the frozen Rainy Lake.
[11]
GPS Signal Test
George Hnatiuk: This video is a simple demonstration that GPS receivers capture signals that originate from the sky above rather than ground based towers.

## References

Does GPS work on the ISS?; Robert Frost, Specialist on GPS use for space navigation
https://www.quora.com/Does-GPS-work-on-the-ISS
Refractive index
The refractive index or index of refraction of a material is a dimensionless number that describes how fast light propagates through the material.
https://en.wikipedia.org/wiki/Refractive%5Findex
[3]
Terrestrische Refraktion
Variation of the Refraktion Coefficient near the Ground; Terrestrial Refraktion in Astronomic Navigation
https://de.wikipedia.org/wiki/Terrestrische_Refraktion
Concepts and Solutions to Overcome the Refraction Problem in Terrestrial Precision Measurement; Prof. Dr. Hilmar Ingensand
Keywords: Refraction, dispersometry, scintillometry, temperature gradient
https://www.fig.net/resources/proceedings/fig_proceedings/fig_2002/Js28/JS28_ingensand.pdf
Modeling Atmospheric Refraction Influences by Optical Turbulences Using an Image-Assisted Total Station; Dipl.-Ing. Dr. techn. Alexander Reiterer (2012)
Study presents a method for the determination of the influence of refraction on the basis of optical measurements.
https://geodaesie.info/sites/default/files/privat/zfv_2012_3_Reiterer.pdf
[6]
Monitoring of the refraction coefficient in the lower atmosphere using a controlled setup of simultaneous reciprocal vertical angle measurements
This paper shows that near the surface refraction can vary from -3 to 16 during the day. As the surface is heated by the sun, refraction is negative. As the surface cools refraction is positive, much greater than 1 after sunset.
http://onlinelibrary.wiley.com/doi/10.1029/2010JD014067/abstract;jsessionid=929EF8B10D42832FBFFA1D197CBAC3CB.f02t01
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