Problem with earth curvature calculator - Hope for revised version ?

Kriss3d

New Member
Ok aparently i didnt do a great job at explaining so ill try again.

Flat earhers very often use the earth curvature calculator here and point out a vast drop that they then ask "wheres the missing drop".

When using the metabunks curvature calculator they enter a distance. But this distance you enter isnt the distance as google maps would present it.
Basically your curvature calculator shows quite a drop that isnt whats matching up with reality because they enter the curvature line where it should be the line of sight - simply because they can look up the distance along the curve but not the direct line.

Would it be possible to get the calculator revised so you enter the distance along the curvature - the google maps distance, and the hight of the viewer and it calculates the right amount of hidden ? As it is now its being used wrong thus they get the wrong result as the line of sight cant be looked up.

Im afraid im not very good with these kind of things.
 

Mick West

Administrator
Staff member
Basically your curvature calculator shows quite a drop that isnt whats matching up with reality because they enter the curvature line where it should be the line of sight - simply because they can look up the distance along the curve but not the direct line.

Would it be possible to get the calculator revised so you enter the distance along the curvature - the google maps distance

For anything under 200 miles these distances are essentially the same. But if you click on "Advanced" it will calculate it using the curve (great circle) distance.
https://www.metabunk.org/curve/?d=200&h=1000&r=3959&u=i&a=a&fd=60&fp=3264
Distance d = 200 Miles (1056000 Feet)
Radius of Earth r = 3959 Miles (20903520 Feet)
Distance to horizon, a = sqrt((r+h)*(r+h) - r*r), a = 38.73 Miles (204470.14 Feet)
Amount obscured x = sqrt(a*a - 2*a*d + d*d + r*r)-r, x = 3.28 Miles (17336.85 Feet)

The following is the amount obscured calculated using slightly more complex trig
And assuming the 'distance' is the distance across the surface of the earth
This is more accurate for large distances and heights
But essentially the same for distances under 100 miles
True hidden = r/cos( d/r - asin(sqrt(h*(2*r+h)) / (r+h) ) ) - r = 3.29 Miles (17356.31 Feet)

'Bulge' b = r-sqrt(4*r*r - d*d)/2, b = 1.26 Miles (6669.41 Feet)
Bulge is the amount of rise of the earth's curve from a straight line connecting two points on the surface
Content from External Source
The difference there, with 200 miles distance viewed from 1,000 feet up, is just 3.28 miles vs. 3.29 miles. So there's no great benefit to using the great circle distance.
 

Nik Kinze

New Member
The problem when using these formulas to calculate the apparent curvature works, but ONLY regarding a full circumference path. If your position between points A and B lie on an oblique angle, (one not parallel to the diameter), the curvature can be less than 2" per mile squared.
 

Mick West

Administrator
Staff member
The problem when using these formulas to calculate the apparent curvature works, but ONLY regarding a full circumference path. If your position between points A and B lie on an oblique angle, (one not parallel to the diameter), the curvature can be less than 2" per mile squared.

Your objection makes no sense. The calculator figures the distance of a line between A and B that grazes the surface at the horizon. There's no such line as "the diameter".

If you'd like to explain, please draw a diagram.
 

Robin McDonald

New Member
The flat earthers are having a field day this week with your addition of “observations made over water near the horizon are inaccurate. They are practically dancing in the street. If you have ever explained it I am having an impossible time finding your explanation in this forum. It would be useful if you would provide On the calculator page a brief paragraph have what you mean. I can only assume putting your eyeline just above the water brings the horizon near. But since I really can’t tell for sure what you intended this statement has essentially made the calculator useless as a proofing tool for Flat earth arguments unless you explain how low to water is too low to be accurate.
 

Mick West

Administrator
Staff member
But since I really can’t tell for sure what you intended this statement has essentially made the calculator useless as a proofing tool for Flat earth arguments unless you explain how low to water is too low to be accurate.
I originally said "horizon grazing", but they didn't understand that. The problem is that for all viewer heights, there's lines of sight that pass very close to the water, but that's not one of the inputs.

It's essentially discussed here:
https://www.metabunk.org/why-flat-earth-laser-tests-are-misleading.t10625/
 
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