1. Mick West

    Mick West Administrator Staff Member

    Metabunk 2018-04-20 08-55-47.

    To use it you just put in the distance to the object in miles and your heigh above the surface in feet.

    The Metabunk curve calculator was written to provide a simple answer to the question: "how much of a distant object is hidden by the curve of the earth from a certain distance and view position". The following diagram is an exaggerated look at the numbers involved.
    Metabunk 2018-04-20 09-09-54.

    There's a version of this diagram on the calculator page that you can move around.

    Notice there are two sets of results given, "Results ignoring refraction" and "With Standard Refraction".

    The results ignoring refraction are the results if we ignore the effects of the air (the atmosphere) between the viewer (the eye position) and the target object.

    When light passes though air that varies in density then it gets bent towards the higher density air. Normally air gets less dense as you go higher (think of thin air at the top of a mountain). When the light is bent down then it makes things look higher than they are. It can also bend the light around the curve of the earth.

    This has the effect of (visually) flattening the Earth a bit, and we can use a hand rule of thumb (used by surveyors for 100+ years) of just doing the same calculation with a larger (7/6x) radius earth. That give you the second set of numbers.

    In reality the "ignoring refraction" numbers are almost always going to be too small. The atmosphere always exists, and it's always refracting. On average it's going to give you the "standard refraction" values. However there's going to be considerable variations. "Standard" here refers to the international standard atmosphere which is simply an average state of the atmosphere used as a baseline. It's like there's a "standard" temperature of 66°F/19°C. Again that's an average, over the whole year. On some days it's only going to be 66°F for a few minutes, some days (and/or locations) it's never going to be that value.

    So the "standard refraction" is just a typical value. Actual observations will vary.

    Once you've got the results, you can save them by clicking on "Permalink
    Metabunk 2018-04-20 09-23-06.

    then copy the URL from the browser for sharing. It will look something like:

    You don't have to click on "permalink" again, the URL will update automatically.

    There's also an "Advanced" version (click the "Advanced" checkbox just above the results). This gives you more options, and shows more results and the underlying math.

    This calculator arose from discussion in this thread:
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  2. Mick West

    Mick West Administrator Staff Member

    Today someone asked me about this formula, which I'd just extracted from here. I could not remember how I got it (four years ago), so I re-derived it from scratch. In doing this I realized it could be much simpler expressed (and explained) as

    r/cos( d/r - acos(r / (r+h) ) ) - r
  3. Mick West

    Mick West Administrator Staff Member

    I also realized that in my original diagram:
    Metabunk 2019-03-06 09-29-04.

    The line labeled "Distance" is the chord distance, but the calculator uses the line-of-sight distance (from "Eye" to the intersection with the target).

    I've updated the diagram to show the three ways the distance can be measured

    Metabunk 2019-03-06 09-37-44.

    Distance1 = Line of sight, from the eye/camera, grazing the horizon, to the target
    Distance2 = Arc distance or surface distance. From water level beneath the eye to water level beneath the target
    Distance3 = straight line between the Distance2 endpoints.

    I've clarified the text under "advanced" and added a calculation based on Distance3:

    Hidden3 = r/cos( 2*asin(d/(2*r)) - acos(r / (r+h) ) ) - r

    since d2 = 2 * r * asin(d3/2r)
  4. Mick West

    Mick West Administrator Staff Member

  5. Mick West

    Mick West Administrator Staff Member

    For reference.
    Metabunk 2019-03-06 17-26-54.

    Metabunk 2019-03-06 17-27-46.

    The second pages is a derivation of x directly from d3, which you can also get by substituting the previous equation of d2 in terms of d3.

    And here's the original derivation of x from d1 (and h and r, of course).

    I'd used the D1 method simply because I'm happier with pythagoras than with trig. But really r/cos( d/r - acos(r / (r+h) ) ) - r is such an elegant formula I wish I'd gone with that.

    Bottom line though, there's no real difference up to 100 miles. And it's not like one method is more accurate than another, it depends on what you are measuring as d.
    Last edited: Mar 6, 2019
  6. Rory

    Rory Senior Member

    Ay, I played with the different ways of calculating the distance and the variations in the ellipsoid and found it's just a few feet here and there - usually around the height of a tree, even over a hundred miles or so.

    I did add a step to calculate radius from latitude, which was quite nice. This is the equation for that for excel:



    Things I'd like to see on the calculator would be:
    • The addition of an obstruction calculator (yes/no tick box for "is there an obstruction?")
    • Predicted visible amount outputted, in addition to hidden
    • Variable refraction option (say plus or minus a percentage from 'standard')
    • The refracted results promoted above non-refracted (or non-refracted removed, or perhaps a simple check box for with/without refraction)
    • Outputs for predicted hidden/visible amounts for the flat earth 'notion' (generally only required when there's an obstruction, but perhaps always good to have it there in black and white: e.g., 'Predicted hidden amount: ZERO')
  7. Bobby Shafto

    Bobby Shafto New Member

    I just noticed this today. Was wondering if it'd been there all along and I had just not been observant.

    On a tangential topic, can I ask what value "the bulge" is to anything?

    I've been contemplating a presentation that deprecates the bulge (sagitta) in relation to calculations of earth curvature since I can't find any use for it. If anything, it fosters the notion that "the bulge" is what is obscuring, but that's only true when the horizon happens to coincide with it. Otherwise, I think it just contributes to the erroneous concept of the earth "rising up" like a wall or mound between observer and another point on the sphere.

    I've been wanting to address this "amount of curvature" terminology that is so rampant among globe busting, and the incessant conflation by those who think "the drop" and amount hidden are the same thing and that "the bulge" is what contributes to amount hidden. "Drop" is useful, but "bulge?"

    If I'm wrong, what am I missing?
  8. Mick West

    Mick West Administrator Staff Member

    I think it's perhaps useful to get a sense of scale. In the default, the "drop" is 10.67 feet, but the bulge is only 2.67 feet.

    I've not really seen anyone confused by it though.
  9. Bobby Shafto

    Bobby Shafto New Member

    I've repeatedly encountered globe "debunking" that uses height of the bulge as the obstruction and amount of "drop" what "should be hidden."

    Drop amount, at least, is useful for calculating amount of declination something in the distance will show compared to across a plane.

    Those with a propensity to misunderstand globe geometry misunderstand "the bulge" to be something that appears to rise before you, like a mound, hill or wall. Only if you were trying to sight along the chord would the bulge seem to rise, but who intuitively tries to use the chord as the sight line?

    It's just a campaign I'm on to wean people away from "the bulge" since it doesn't contribute to the geometry of "what should be seen or not" at all. The horizon is the obstruction. Not the bulge. Unlike "drop," I can't find any usefulness for the "bulge" other than for flat earthers to misuse it.

    Eventually, I'll publish my pitch (video) that we dispense with the "bulge." Just thought I'd voice it and see what anyone here thinks. Not a high priority thing, obviously. And yours and Walters calculators are so useful. I probably should pay you a subscription fee given the amount of times I rely on it.
    Last edited: Mar 7, 2019
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  10. Bobby Shafto

    Bobby Shafto New Member

    Yay. I see "bulge" has been replaced with "sagitta." May or may not be due to my input, but glad to see it nonetheless. Doubt it'll have any influence on the erroneous notion that the "bulge" (or sagitta) is responsible for obscuring line of sight, but it's a start.
  11. Mick West

    Mick West Administrator Staff Member

    It was you. I was doing a general revamp, still a work in progress. I'm also going to add an angular size version :)
  12. deirdre

    deirdre Moderator Staff Member

    hmm. noone knows what a sagitta is. i've never heard of that word in my life. you should at least put "bulge" in parenthesis next to it.
  13. Mick West

    Mick West Administrator Staff Member

    That's kind of the point though. The "bulge" sounds like it's an important number ("a 12-foot bulge of water") but it's not really relevant to any of the computations.

    Still, there's a potential for confusion. Maybe a note.
  14. DavidB66

    DavidB66 Member

    If there is a general revamp of the calculator I suggest there should be a look at the 'horizon curve' feature. At present this uses a default field-of-view of over 90 degrees, which is so wide as to be irrelevant for most purposes. It may be based on the maximum field of view of the (single) human eye, which is said to be about 95 degrees, but much of that is in peripheral vision, which the observer is only vaguely aware of. If it is applied to a camera, a shot with such a wide fov would probably show major lens distortion. A default of, say, 70 degrees might be more useful.
    The calculator also allows even wider fov's, up to 180 degrees and beyond. I notice that around 180 degrees the results seem to go mad. 'Horizon Curve Angle' (v1 or v2) suddenly jumps up to 90 degrees for an fov of 180, whereas Horizon Curve Fraction and Horizon Curve Pixels behave more sensibly. I don't even know what an fov of 180 degrees would mean for this purpose. Since the observer's viewpoint is always somewhere above the surface, the angle subtended at the viewpoint between two points on the horizon can never go up to 180 degrees. Unless there is a practical application for very high fov's, I suggest there might be a cutoff at, say, 100 degrees. I note that Walter Bislin's calculator/simulator seems to have a default fov of 65 degrees, and doesn't go beyond 92 degrees. (It is possible to enter higher values, but this doesn't seem to increase the visible curvature.)
  15. Mick West

    Mick West Administrator Staff Member

    Where are you getting that? The default FOV is 60:
    Metabunk 2019-03-19 13-55-58.
  16. DavidB66

    DavidB66 Member

    That's odd. I have the calculator bookmarked here: https://www.metabunk.org/curve/?d=1&h=100000&r=3959&u=i&a=a&fd=94.4&fp=1920

    which shows a default FoV of 94.4 degrees. But several of the other defaults are also different. Maybe I just bookmarked an obsolete version that is somehow still accessible? [Added: or the settings are just some that I once searched for and unintentionally got 'frozen into' the bookmarked version. But I can't remember or imagine using that particular set of settings. For example, to the best of my memory I have never even touched the 'image width' setting. ]
    Last edited: Mar 19, 2019 at 3:25 PM
  17. Mick West

    Mick West Administrator Staff Member

    Yeah, you just want:

    Otherwise, it will read in the parameters from the URL. You get that URL when you click on "Permalink", you can reset them with "Reset"
    Metabunk 2019-03-19 16-42-01.
  18. Rory

    Rory Senior Member

    "Sagitta" is good. Like you say, "bulge" isn't really relevant, and makes people think there's a hill of water. If they can't understand the word "sagitta", but want to, they can always take five seconds to google it, and then they'll learn something, and maybe even go on from there to learn something else.

    I like how the "standard refraction" comes first now, and is highlighted. And the "no refraction" has been termed "geometric".
  19. deirdre

    deirdre Moderator Staff Member

    the curve of the ball vs a flat plane isnt relevant to computations?
  20. Mick West

    Mick West Administrator Staff Member

    The sagitta isn't used in the computations. It's a side effect.

    The curve of the ball is defined by the radius. So we use the radius in the calculation.

    The bulge height is the height in the middle, but this height is not the height of the horizon. It's not a height that obscures anything
    Metabunk 2019-03-20 06-55-33.

    See where the horizon is, above? The Sagitta is "related" in the sense that it's also calculated using the radius. But it's not really useful in calculating the hidden amount.
  21. deirdre

    deirdre Moderator Staff Member

    i know because you have a big green X marked "horizon", which is quickly showing me the horizon is not the same as the bulge/sagitta. In your OP pic i see the horizon is past the bulge.

    But if the sagitta is irrelevant to the calculations, then why even have that marked? (although unmarked, or marked sagitta, pretty sure all the non mathematicians of the world are still going to call it the bulge.. so if you guys like sagitta better that's cool)
  22. deirdre

    deirdre Moderator Staff Member

    ps. the dots on all my browsers are red
  23. Rory

    Rory Senior Member

    Ditto. I'm guessing they used to be blue?

    Did you notice the one that says "= Draggable" is also draggable. That's cool. :)

    I've got a little j under my green line (line of sight to horizon). You?
  24. Mick West

    Mick West Administrator Staff Member

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  25. Mick West

    Mick West Administrator Staff Member

    Spurred on by @Rory's angular shenanigans, I've added a purely angular-size based calculation of the hidden amount.

    Metabunk 2019-03-20 16-16-57.
    This, of course, is in part to answer some critics of curve calculators in the FE community who claim the curve calculator ignores angular size. So I worked out equations that ignore things like calculating the drop or the horizon distance, and instead "simply" calculate the angular size of the portion of the target that is hidden.

    It's somewhat pointless, as the end result is the same. It also might not be understandable if you didn't understand the old way. However, that's the math, and it can be verified.

    Like before, there's three important numbers:

    r = radius of the Earth
    h = viewer height
    d = distance to target

    d here is Distance2 on the calculator diagram, i.e. the distance along the surface of the earth.

    The dip to the horizon is just a very simple acos(r/(r+h))

    The dip to the base of the target is more complicated, as we don't have a right angled triangle, so we have to use the Sine Rule and the Cosine Rule. Fun high school math
    Metabunk 2019-03-20 21-48-40.

    So we just use the law of cosines to find the length of the line from the viewer to the base of the target

    EyeToBase = sqrt(r^2 + (r+h)^2 - 2*(r+h)*r*Cos(d/r))

    Then the law of sines to get the dip angle of the base

    BaseDip = d/r + Asin((r+h)*Sin(d/r)/EyeToBase) - PI/2

    In code
    var r = 3959*5280; // radius of the earth in feet
    var HorizonDip = Math.acos(r/(r+h))
    var EyeToBase = Math.sqrt(r**2 + (r+h)**2 -2*r*(r+h)*Math.cos(d/r))
    var L = Math.asin((r+h)*Math.sin(d/r)/EyeToBase)
    var BaseDip = (d/r) + L - Math.PI/2;
    var AngularHidden = BaseDip - HorizonDip
    (Note: all angles are in radians. If degrees are used then d/r should be d/r*180/PI)
    Last edited: Mar 21, 2019 at 6:23 AM
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