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. :)
     

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  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.
     

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