1. Rory

    Rory Senior Member

    I've been working on a calculator that outputs the predicted visible amount of a distant landmark when there's an obstruction above the horizon, such as a view over terrain with a ridge or intervening mountain. It seems to be working pretty well, and I'd say it's just about finished, but I have one more modification I'd like to make.

    First of all, here's the diagram of what it's calculating:

    IMG_20190125_113610004.

    I've gone about it three different ways, firstly two using the drop below eye level - one factoring in for tilt, one more simple without - and then another using angular size, and they all return more or less the same results (for example, for the recent Pikes Peak shot: 1760, 1765, and 1761 feet respectively).

    One sticking point I had was that I needed different equations for whether a target was above or below eye level, but the angular size method seems to have dealt with that, and so that's the one I'll probably go with.

    Last thing I want to do, though, is take into account where the line of sight actually 'hits' a mountain. For the San Jacinto pictures, for example, while the peak was 122.6 miles away, the part of the mountain seen at the bottom of the visible portion was around 117 miles away. I believe this makes a difference to what the predicted value should be.

    It doesn't seem terribly difficult to do this, but I seem to have got myself stuck, and figured the quickest way out of this would be to post this here and see what y'all say.

    The latest version of the calculator is attached. First tab is for San Jacinto (simplified drop trig only); second tab is for Pikes Peak (both drop trigs); and third tab is using angular size, with the 'mountainside' equations missing.

    Hope that all makes sense. :)
     

    Attached Files:

  2. Rory

    Rory Senior Member

    Here's a diagram showing 'the simplified trig':

    sphere diagram copy.

    This doesn't factor in the tilt, nor the difference between 'base to base' and 'peak to peak' distances.

    It's accurate enough, though, and useful as a verifier to show that the more complex methods achieve pretty much the same result.
     
  3. Mick West

    Mick West Administrator Staff Member

    It does not make a difference to the angular amount that is hidden. The angle (y, in your first diagram) remains the same.

    It's kind of a semantic question. What are you calculating, and from what? The important amount, it seems to me, is how much the peak of the mountain is visible above the peak of the obstruction. If you measure that in feet then you have to ask "at what distance", and the only distance that makes sense is at the distance of the peak — you treat the mountain as a silhouette at the peak's distance.

    But from an observation POV the point labeled "mountainside" IS actually lower, so you might look up that height on in Google Earth, and say that's what the real-world "hidden height" is. But that's the hidden height at that distance. For it to be meaningful you've then got to scale it up to the distance of the mountain.

    Example, if you visually identify the height of the "mountainside" point as being at 1,500 feet and 117 miles away, but the peak is 122.6 miles away, then (ignoring curve) you need to multiply 1,500 by 122.6/117 (i.e. 1,572 feet)

    If you want to get more complicated and fully account for curve, you can simply reuse your existing math. Treat the "mountainside" point as the obstruction, and see what that obscures at the peak distance
     
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  4. Rory

    Rory Senior Member

    Finally got around to 'finishing off' the obstruction calculator. I decided the best way to approach the 'mountainside equation' would be to have it output the predicted visible elevation above sea level at a known distance. This returns the following (using the recent San Jacinto shot as an example):

    upload_2019-2-9_13-10-4.

    I've tidied up all the equations and included them in step-by-step form, so that they're easier to change and verify. I've also added a feature that calculates the radius of the Earth from the observer's latitude, as well as the ability to more easily vary refraction (let me know if the range I've posted above for 'normal' is okay).

    It's all on a single page now, and doesn't require different equations depending on whether the target peak is above or below eye level. Do give it a try and let me know what you think. Spreadsheet attached.
     

    Attached Files:

  5. Rory

    Rory Senior Member

    Can somebody help me with this please? I'm not quite getting my head around how and where refraction should be included.

    The spreadsheet attached doesn't factor in for refraction, and the results seem right.

    Relevant cells are C14 (refraction coefficient), C32 (Earth's radius, used in many of the equations), and C34 (Earth's radius multiplied by refraction coefficient, which I believe I should be using in place of C32, but which doesn't currently feature in any of the equations).

    Cheers.
     

    Attached Files:

    Last edited: Feb 20, 2019
  6. Mick West

    Mick West Administrator Staff Member

    Seem right how?

    In my curve calculator, I use the adjusted radius. If you do that in yours you get:

    Metabunk 2019-02-20 06-53-13.
    I'm not sure this is something you can simply use the refraction coefficient for though.

    I'm going to do a visual calculator preset in my refraction simulator, which should give us another data point. Sometime this morning.
     
  7. Rory

    Rory Senior Member

    I guess what we know from reality is that, for the above example, the lowest visible point is at around 7000 feet amsl, and the amount of the peak visible is around 3750 feet.

    The results without factoring for refraction are pretty much bang on. I just don't know why.
     
  8. Mick West

    Mick West Administrator Staff Member

    Okay, new preset:

    https://www.metabunk.org/refraction/?~(p~'Visual*20Obstruction*20Calculator)_
    Metabunk 2019-02-20 08-08-52.

    This consists of two targets, which you edit here:
    Metabunk 2019-02-20 08-09-35.

    I've set them in the preset to:
    Obstruction: 19.9 miles, 180 feet
    Target: 117 miles, 10834 feet:

    The target is marked with percentage lines, so you can read off the highest point obscured by the horizon and/or the obstruction, and multiply that by the height. Here it is without refraction
    Metabunk 2019-02-20 08-12-15.

    Horizon = 64.5% * 10834 = 6987 feet. (compare with Curve Calc hidden value, 7006 feet, same)
    Obscuration top - 73.2% * 10834 = 7930 feet (compare with spreadsheet, 6850, different!)


    Now WITH standard atmosphere refraction
    Metabunk 2019-02-20 08-18-17.

    Horizon - 52.5% * 10834 = 5688 feet, compare with CC 5870 feet, similar enough for the rough rule of thumb 7/6
    Ob Top - 61.8% * 10834 = 6695, which is similar to your spreadsheet value of 6850

    I'm not sure exactly what accounts for the differences. However here the targets is 2D, and I've set the distance to 117 miles (the distance to the visible point) as opposed to 122.6 for the actual peak, so I would suspect that has an effect, but I'm not sure which way.
    Metabunk 2019-02-20 08-25-55.
     
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  9. Mick West

    Mick West Administrator Staff Member

    So that means in my simulator the results WITH refraction are bang on :)

    It's not making a HUGE difference in the question of globe vs. flat, as in all situations more than half the mountain is below the ocean horizon.
     
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  10. Rory

    Rory Senior Member

    Okay, I found the oversight. There was one cell in each step that had factored refraction, I just wasn't spotting where it was: the formulae included C32*C14, which I've now changed to C34.

    It makes no difference, but it answers the question and tidies it up a bit.

    A couple of other minor tweaks done, but it's basically the same. Latest - and hopefully last! - version attached.
     

    Attached Files: