#### DarkStar

##### Active Member

It is time to revisit the ellipsoid question as mentioned by @Rory here and Mick's thoughts here. And Rory linked to this awesome page with Earth ellipsoid and geoid radius calculator integrated and explains the math behind it: SummitPost

So here is my brain dump in the hopes that we can collaborate on this and at least all understand the situation better, even if we don't update the calculator itself.

History:

So based on the now Infamous Balaton Lake Debacle (IBLD) a discussion thread ensued in another forum and one of the posters again mentioned the Ellipsoid and posted some calculations which showed ~1.67 meters *per* kilometer over this Lake - which is greater than the curvature. I never thought it would matter that much so I hadn't really focused on this specific problem before in my calculations. But it does matter and it can matter a whole lot especially as the direction of observation goes more North/South (the earth radius is increasing as you move South, and at some angle away from South you get some fraction of that effect).

Ok, so here is the scenario I looked at...

One of my first forays into the Flat Earth world was looking at the view from Olcott, NY toward Toronto and the famous CN tower. It was obvious that observer height mattered so I first derived the equations for this calculation (Mick's Metabunk calculator was more barebones at that time but I DID find his early GeoGebra calculator, which also got me interested in GeoGebra & Metabunk

So I got the distances, estimated the observers height, checked the elevation profile from Google Earth, AND I even used the rechneronline calculator to get the approximate ELLIPSOIDAL Earth radius at this point. I plugged in my numbers for no refraction and 7% refraction and got the typical, fairly reasonable results.

Which, frankly, I think is usually sufficient to show that the views we see are compatible with a curved Earth.

HOWEVER, upon revisting this due the oddness of the IBLD results (because the FEW observations we can make from that mess really did NOT fit the spherical model - they didn't fit a flat 'model' *cough* either. Refraction didn't really explain it either) I started seeing some significant impact on the results -- they fit much better once you accounted for the significant delta in Earth radius along that 6 kilometer run (for example, instead of the laser having a slight downward slope it actually had a slight upward slope, but "MSL" was also rising under you).

So I revisted the Toronto view and here is what I found:

Elevation, Radius(includes Geoid), & Geoid in meters, lat/long in decimal (neg = S/W)

CN Tower: 43.641729 -79.384076 / E=81 / R=6367960.569 / G=-35.626

Olcott Pier: 43.339961 -78.717597 / E=75 / R=6368073.305 / G=-35.312

Olcott Pier Geoid is

All of sudden your expected hidden height is then maybe significantly less than the expected ~255 meters at 63.6km. But I'm not sure yet how to actually calculate the view. Can I count it as a 110m high observer? And if I then plug that into a spherical calculator I don't think that is going to give me a solid answer.

Mick suggested in his post that you can just scale the results... I'm not sure. So what I propose is that we tackle nailing this down so that all of our predictions can be more accurate. I do think these findings

My next approach is to the take the equation for a circle and derive the horizon equation from it - and then do the same for the ellipse, that is -- get a an equation I can plug the observation elevation into and get the horizon distance back out that fits to the ellipse. That should be pretty easy -- I just need to find the time (instead I thought I would share my findings so far).

When you have an elevation (especially from Google Earth) that is giving you the Orthometric height as shown below, so to really compare two Orthometric heights you need the Geoid, and the Geoid is a delta from the Ellipsoid.

Here is how you get the radius at some latitude from the WGS-84 ellipsoid:

f=1/298.25722356 << given by the model ('flattening')

a=6378137 << given by the model

b=a-(a*f) << computed

θ = geocentric latitude (tan θ = (b²/a²) tan ϕ)

ϕ = geographic latitude

R(θ) = √(((a²×cos(θ))² + (b²×sin(θ))²) / ((a×cos(θ))² + (b×sin(θ))²))

Latitude ϕ also needs to be the correct 'kind' of latitude, I assume Google Earth gives you geographic latitude so you have to rescale it, (b²/a²) ~ 0.9933 -- so it's a slight correction to the slope but 42° becomes 41.81°

From there the Geoid height is hard to get (you have to download the database and do some complex computations based on spherical harmonic expansions to get the height for any given point).

However, they do have a 'Geoid Height' file you can download and just do some interpolation (which should be more than accurate enough for what we're doing). I think if we can JUST get the ellipsoid model working we will be in much better shape. That's my next step anyway.

Handy publication notes/reference:

http://www.ngs.noaa.gov/GEOID/PRESENTATIONS/2007_02_24_CCPS/Roman_B_PLSC2007notes.pdf

http://www.esri.com/news/arcuser/0703/geoid1of3.html

http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/egm08_wgs84.html

http://earth-info.nga.mil/GandG/wgs84/gravitymod/

UPDATE: I did the same thing for the famous 'Bedford level' area between the bridge at Welney and the bridge near the Welches dam. Because this is England I did it in feet (actually feet are slightly more fine-grained).

Bedford level: 31,688' distance (WGS84 curvature distance)

Welney bridge: 52.519628° 0.250938° / E=12' / R=20881755.992' / G=154.577'

Welches dam: 52.451411° 0.162850° / E=11' / R=20881837.357' / G=155.071'

delta R = 81.365'

So this means that level ground across that 6 mile difference is 81' further from the center of Earth mass at one end than the other. That does NOT mean there is an 81' slope - this is the equipotential gravity level line. The elevation is harder to tell from this data but appears to be approx 1 foot but that could just be the ground around the water, not necessarily the water level itself. The Geoid shift seems to be about 6 inches (again, higher at Welches which is further south).

This page seems to have some of the math we need (start at 'Radius of Curvature Derivation - General Case'):

http://www.cohp.org/local_curvature.html

So here is my brain dump in the hopes that we can collaborate on this and at least all understand the situation better, even if we don't update the calculator itself.

History:

So based on the now Infamous Balaton Lake Debacle (IBLD) a discussion thread ensued in another forum and one of the posters again mentioned the Ellipsoid and posted some calculations which showed ~1.67 meters *per* kilometer over this Lake - which is greater than the curvature. I never thought it would matter that much so I hadn't really focused on this specific problem before in my calculations. But it does matter and it can matter a whole lot especially as the direction of observation goes more North/South (the earth radius is increasing as you move South, and at some angle away from South you get some fraction of that effect).

*And I am *NOT* talking about E/W observations following a line of Latitude which get smaller and smaller as you move North -- Latitude lines are NOT great circle lines - so they are NOT the path under a line of sight (except at the equator . You have to get that image out of your head. When we talk about point A to point B you have to intersect a plane with the center of the Earth and follow the surface of that intersection which doesn't follow lines of latitude (again, except at the equator).*Ok, so here is the scenario I looked at...

One of my first forays into the Flat Earth world was looking at the view from Olcott, NY toward Toronto and the famous CN tower. It was obvious that observer height mattered so I first derived the equations for this calculation (Mick's Metabunk calculator was more barebones at that time but I DID find his early GeoGebra calculator, which also got me interested in GeoGebra & Metabunk

So I got the distances, estimated the observers height, checked the elevation profile from Google Earth, AND I even used the rechneronline calculator to get the approximate ELLIPSOIDAL Earth radius at this point. I plugged in my numbers for no refraction and 7% refraction and got the typical, fairly reasonable results.

Which, frankly, I think is usually sufficient to show that the views we see are compatible with a curved Earth.

HOWEVER, upon revisting this due the oddness of the IBLD results (because the FEW observations we can make from that mess really did NOT fit the spherical model - they didn't fit a flat 'model' *cough* either. Refraction didn't really explain it either) I started seeing some significant impact on the results -- they fit much better once you accounted for the significant delta in Earth radius along that 6 kilometer run (for example, instead of the laser having a slight downward slope it actually had a slight upward slope, but "MSL" was also rising under you).

So I revisted the Toronto view and here is what I found:

Elevation, Radius(includes Geoid), & Geoid in meters, lat/long in decimal (neg = S/W)

CN Tower: 43.641729 -79.384076 / E=81 / R=6367960.569 / G=-35.626

Olcott Pier: 43.339961 -78.717597 / E=75 / R=6368073.305 / G=-35.312

Olcott Pier Geoid is

**112.736m HIGHER**than CN Tower - 6m elevation difference + 3.5m viewer = ~110m higher!All of sudden your expected hidden height is then maybe significantly less than the expected ~255 meters at 63.6km. But I'm not sure yet how to actually calculate the view. Can I count it as a 110m high observer? And if I then plug that into a spherical calculator I don't think that is going to give me a solid answer.

Mick suggested in his post that you can just scale the results... I'm not sure. So what I propose is that we tackle nailing this down so that all of our predictions can be more accurate. I do think these findings

**that the ellipsoidal difference will be significant in some measurements (especially for observations at those mid-latitudes running more N/S).***suggest*My next approach is to the take the equation for a circle and derive the horizon equation from it - and then do the same for the ellipse, that is -- get a an equation I can plug the observation elevation into and get the horizon distance back out that fits to the ellipse. That should be pretty easy -- I just need to find the time (instead I thought I would share my findings so far).

When you have an elevation (especially from Google Earth) that is giving you the Orthometric height as shown below, so to really compare two Orthometric heights you need the Geoid, and the Geoid is a delta from the Ellipsoid.

Here is how you get the radius at some latitude from the WGS-84 ellipsoid:

f=1/298.25722356 << given by the model ('flattening')

a=6378137 << given by the model

b=a-(a*f) << computed

θ = geocentric latitude (tan θ = (b²/a²) tan ϕ)

ϕ = geographic latitude

R(θ) = √(((a²×cos(θ))² + (b²×sin(θ))²) / ((a×cos(θ))² + (b×sin(θ))²))

Latitude ϕ also needs to be the correct 'kind' of latitude, I assume Google Earth gives you geographic latitude so you have to rescale it, (b²/a²) ~ 0.9933 -- so it's a slight correction to the slope but 42° becomes 41.81°

From there the Geoid height is hard to get (you have to download the database and do some complex computations based on spherical harmonic expansions to get the height for any given point).

However, they do have a 'Geoid Height' file you can download and just do some interpolation (which should be more than accurate enough for what we're doing). I think if we can JUST get the ellipsoid model working we will be in much better shape. That's my next step anyway.

Handy publication notes/reference:

http://www.ngs.noaa.gov/GEOID/PRESENTATIONS/2007_02_24_CCPS/Roman_B_PLSC2007notes.pdf

http://www.esri.com/news/arcuser/0703/geoid1of3.html

http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/egm08_wgs84.html

http://earth-info.nga.mil/GandG/wgs84/gravitymod/

UPDATE: I did the same thing for the famous 'Bedford level' area between the bridge at Welney and the bridge near the Welches dam. Because this is England I did it in feet (actually feet are slightly more fine-grained).

Bedford level: 31,688' distance (WGS84 curvature distance)

Welney bridge: 52.519628° 0.250938° / E=12' / R=20881755.992' / G=154.577'

Welches dam: 52.451411° 0.162850° / E=11' / R=20881837.357' / G=155.071'

delta R = 81.365'

So this means that level ground across that 6 mile difference is 81' further from the center of Earth mass at one end than the other. That does NOT mean there is an 81' slope - this is the equipotential gravity level line. The elevation is harder to tell from this data but appears to be approx 1 foot but that could just be the ground around the water, not necessarily the water level itself. The Geoid shift seems to be about 6 inches (again, higher at Welches which is further south).

This page seems to have some of the math we need (start at 'Radius of Curvature Derivation - General Case'):

http://www.cohp.org/local_curvature.html

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