1. Trailspotter

    Trailspotter Senior Member

    • Agree Agree x 1
  2. Henk001

    Henk001 Active Member

    Neatly matches the curvature calculator with a little bit more refraction though (1.22 in stead of 1.16, nothing unusual)
  3. Trailblazer

    Trailblazer Moderator Staff Member

    I was looking into this so-called "Bishop experiment". It seems pretty clear that he was looking in the wrong direction.

    This post on the FE Society forum sums it up (although the references to "west" should read "east")



    This is the beach that Tom Bishop refers to seeing with the naked eye, and being able to see people splashing in the water through a telescope:


    Clearly this photo is taken looking eastwards towards the prominent dunes at Sand City about 4 miles away:



    This piece of "evidence" still seems to be circulating, but it doesn't pass the most basic fact checking.
    Last edited: Jun 1, 2017
    • Like Like x 1
  4. Clouds Givemethewillies

    Clouds Givemethewillies Active Member

    "Drop is the amount the curve of the earth drops away from level, it can be calculated in various ways
    which are all about the same for under 100 miles distance
    Drop as r-sqrt(r^2-d^2); drop = 784.85 meters (perpendicular to level line)
    Drop calculated as 8 inches per mile squared = 8 * d*d / 12; = 6.67 km (6666.67 m)
    Drop calculated with trig = r/cos(d/r) - r; = 784.89 meters (perpendicular to target surface)"I've

    Something went a bit wrong here for d=100 km., metric units.
    Also it would be good if the dip calculation included refraction.

    ps. I would be tempted to make the planet, and it radius a constant, to make it more foolproof. I've been caught out!
    Last edited: Jun 1, 2017
  5. Mick West

    Mick West Administrator Staff Member

    The "8 inches per mile squared" equation was being used without converting to metric. The metric versions is the somewhat less handy "0.07845590756046217 meters per kilometer squared". I've fixed that. I suppose you could use a different approximations of "about 78mm per km squared" or even in a pinch "about 8cm per km squared"
  6. Clouds Givemethewillies

    Clouds Givemethewillies Active Member

    Thanks. One would not want metric FEers getting the wrong answer.
  7. Mick West

    Mick West Administrator Staff Member

    I don't have a "refracted bulge", as I'm not sure it really makes sense. Maybe I should just move "Bulge" into the advanced section?
  8. Clouds Givemethewillies

    Clouds Givemethewillies Active Member

    Seems that LadderCam got a bit lucky!
    For viewer height of 51 metres
    MickCalc - Refracted Dip = 0.212 Degrees, (0.0037 Radians)
    LadderCam - Measured dip 11/2940*360/(2*Pi) = 0.214 degrees
    • Winner Winner x 1
  9. drhex

    drhex New Member

    Mick West, I find the curve calculator very useful when arguing with flat earthers, thank you.:)
    If I may ask for improvements, it would be to
    * Use consistent terms. x/Hidden/Obscured seems to mean the same thing
    * Augment the diagram at the bottom so that in includes lines showing what is meant by "drop"
  10. Bas Koning

    Bas Koning New Member

    This. I was just going to say the same.
    The 'a', 'b', 'd', 'x' never come back in the summary at the top.
    The user has to guess what 'obscured' is.
    if 'hidden' is obscured, then what is 'drop'.
    These things are not clear.

    Also, display the math used. Don't just show the answer, show how you got to that answer. (EDIT: ah, 'advanced' shows them, title is not clear)

    Also, and I know I am pushing it now, perhaps, it would be super cool if you could show the answer... in the diagram. Some people are really visual. They barely read numbers. Showing the actual angles and sizes could help. A lot. For visually understanding what is going on. 'Just' draw the answer out, zoom in on the relevant part. Scale misconceptions are one of the things that keep flat earth going. (Keep the old diagram too, for when the data makes a very very thin triangle, people can still see which parameters are where.) To make things more visible when needed you could have some 'stretch/shrink horizontal/vertical 10times' buttons. Add current magnification on the axis in the diagram. It is worth the effort. Many people use this calculator, and the impact would be much larger if the answers where given in a visual observable way.

    Make the length of the curved line between observer and object known.

    Allow for other combinations of data input: for example, if 2m is obscured, with eyes at 1.4 meter, give all the other numbers. Or: Of the bulge is 20cm, and eyes are at 3m, give all the other numbers. Etc.
    Only doing 'distance' and 'eye' as input is a bit very restrictive.

    Distance is almost never the thing you actually have available to put in: people put in the curved line between the two points as 'distance' mostly now I guess. If I walk 2 km besides a canal, I am not walking distance (d) I am walking the curved line...

    Make distance between top of both objects known. This way people can check things like bridges and fences being longer at the top then at the bottom.

    One thing that really bugs me at the moment is tiny distances. The granularity of the numbers is just a bit too large, IMHO. Take for instance (metric) distance 1.6km, eye height 0.1m, obscured: 0.02 meter (aka 2 cm). Plus or minus 1 cm. But I need 0.020 meter, plus or minus 1 mm for the calculations of my experiment:

    The smallest scale visible curvature. To do this take a small brightly colored object (like a matchbox), and measure its side as exact as possible. Say it is 2.00 cm, plus or minus half a millimeter.
    Calculate how much water would be needed to hide the object behind the curvature of the earth. (eyes at 10 cm up, +/-0.5cm would be closest to the ground that is practically): About 1.6km.

    Find any odd shaped lake where the water does not move (or at least not much). Or canal.

    On google maps, find two pieces of beach on the lake that are 1.6km away.
    This lake for example would do just fine https://www.google.nl/maps/dir/51.7...m0!1m3!2m2!1d5.8232206!2d51.7460769!3e2?hl=nl

    Put matchbox on the waterline. Go the other side. Put eyes at 10 cm. Matchbox gone. Stand up. Matchbox appears. Use goggles or telescope or naked eye: same thing.

    So to be on the sure side I can round everything up roughly to the safe side. But I want to know the smallest possible scale, pretty exact. Lakes with 1.6 km are far more common then lakes with 2.6 km open water. The smaller the lake, the easier^2 it is to repeat this experiment.

    (Would mirages superior be able to mess with things significantly on that scale? Perhaps?!)
    Last edited: Jul 8, 2017
  11. Bass In Your Face

    Bass In Your Face Active Member

    Mick posted this simulator a while back, although it is not integrated with the meta calculator.. (but I swear there was a slightly different one that existed on this forum linked in another spot // EDIT: There was but I cant find it.. you could drag the grid around, [not just the earth], and you could also access a menu by right clicking, enabling the grid, labels, other geo points, zooming in, etc //).. anyways:


    Last edited: Jul 8, 2017
  12. Bas Koning

    Bas Koning New Member

    Yeah that is very good. It still needs something more.

    We need to opensource the code somehow.

    -The distance camera target should be on it.
    -Zooming with mouse scroll wheel in and out.
    -It already can position: camera, target and earth by dragging, but it also needs a way to input two numbers by hand then push button: calculate the rest.
    -Dimensions: metric? Meter? Dunno. Has to be on it.
    -switch metric/imperial
    -'show me the math' button
    -Remove bug: only count as 'hit' if target line crosses camera line after horizon. Before horizon, 'Obscured' should be zero:

    Then we have to think about implementing refraction I guess... hmmmm..
    Last edited: Jul 8, 2017
  13. Elan

    Elan New Member

    Hi Mick,

    I didn't know how to post in this general thread without replying to a specific posting so I am just doing it here. I have a question regarding the image at the bottom of the page with the calculator that you built. https://www.metabunk.org/curve/
    Is the secant line labeled "distance" just drawn from the point where the person is standing to some arbitrary point on the circle or is there some significance to that other point?
  14. Mick West

    Mick West Administrator Staff Member

    The other point is where the object being viewed is. Like, for example an island, a distant shore, a ship or a lighthouse.
  15. SquillagusNiggle

    SquillagusNiggle New Member

    Apologies if this has been covered, but something strikes me as a fundamental missed assumption of the calculator: the distance is not the linear distance between the two points. It is actually the length of the ARC between the two points. Maps are drawn by measuring the Earth's surface, the geoid, and will give the distance between two points as the length of the Earth's arc between them. Your calculator will consistently underestimate the blocked-off amount of the object as the linear distance is always less than the arc length, and this will get very much worse as distance increases.
  16. Mick West

    Mick West Administrator Staff Member

    For under 200 miles it's a negligible difference. And if you click on "advanced" it will give you both numbers.

    e.g. 200 miles, from 20 feet elevation: