1. Bunkmeister

    Bunkmeister New Member

    I am entering the world of making flat earth debunking videos on YouTube.

    I would like to focus on the concept of shadows ascending up a mountain at sunset.

    I plan to make a few table top demonstrations showing that as long as a light source is higher in elevation than an object, that object cannot cast a shadow whose vertical height exceeds its own. I will then show that when we lower the light below the height of the object, the shadow can then cast upward due to the negative angle of the light.

    The culmination of my video will be this time lapse of Mount Everest: https://www.istockphoto.com/my/video/mt-everest-at-sunset-gm539252432-98158951

    Since nothing is taller or equal height to Everest, the ascending shadow must be caused by curvature. That is the gist of it.

    However, I am also interested in demonstrating that the vertical speed of the shadow ascent up the mountain (feet or meters per minute) is consistent with a 15 degree per hour sun movement. Is there any way I can correlate that?

    I am guessing there would be enough topography data for me to call out specific elevation points in the video. Just not sure how I would calculate this speed, if it is even calculable.
     
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  2. Tedsson

    Tedsson Member

    Wouldn’t you need to know the height above mean sea level the video was taken from.

    Or take a reference point which you can clearly establish from the topography.

    Subtract that from the known height of Everest.

    Know the actual duration of the shadow casting.

    That should enable you to calculate how rapidly the shadow creeps up the mountain in FPS or mph.

    You might have to make a few assumptions such as Everest is effectively vertical.
     
  3. Rory

    Rory Senior Member

    Doesn't it being a time lapse video kind of mess with that?
     
  4. Bunkmeister

    Bunkmeister New Member

    I have identified the photographer and am seeing if he can supply me with the original footage or verify the duration shown in the video.
     
  5. Bunkmeister

    Bunkmeister New Member

    I have identified the photographer and am going to see if he can provide location detail. I am imagining the elevations of the various camps on the mountain are well known.
     
  6. Bunkmeister

    Bunkmeister New Member

    Anyhow, your variables laid out make sense. I will report back if I can get these critical details.
     
  7. Mick West

    Mick West Administrator Staff Member

    You can calculate the rough angular speed of a shadow moving over an object of height h cast by an object at distance d as atan(h/d)/time. If you are only calculating angular speed then the distance to the sun (S) is irrelevant.

    Metabunk 2018-12-01 10-14-11.

    There's a variety of problems here - what height is the shadow actually moving over? How far away is the peak or horizon that's casting the shadow?

    But the deeper problem is that this isn't really going to convince anyone. It should be blindingly obvious that if the sun remains above the horizontal plane (as the FE folk believe) then it can never go below the horizon. The details of how the shadows move, and the speed at which they move just details. If you think the sun is always 3,000 miles above the flat earth then you've basically rejected the conventional (straight line) model of light and optics (and maybe even geometry). So a demonstration that involved the conventional model is unlikely to sway them.

    Show them a picture of the sun casting shadows upwards, clouds illuminated from below. You should not need math to understand this means the sun is lower than the clouds.
    Untitled_Panorama2a.
     
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  8. Bunkmeister

    Bunkmeister New Member

    Mick, thanks for your feedback. I will ditch the vertical feet math. However, I think I need to identify the tallest nearby peaks that are casting shadows, and then determine the portion of visible mountain which is above that elevation. At some point towards the summit, the ascending shadow can only be from earth curve, right?
     
  9. Bunkmeister

    Bunkmeister New Member

    Also, cool picture! What are wee seeing there? A contrail?
     
  10. deirdre

    deirdre Moderator Staff Member

    FEers have bendy light.
     
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  11. Mick West

    Mick West Administrator Staff Member

    It's from whatever is on the horizon in the direction of the sunset on that day. If you can see the ocean, or it's a flat plain, then you could describe that as the curve of the earth. But this is the view due west from Everest.
    Metabunk 2018-12-01 12-53-51.

    And here's a slider comparison of Google Earth's calculated view with and without terrain.
    Metabunk 2018-12-01 12-56-26. Metabunk 2018-12-01 12-55-38.


    So basically it's some other mountains casting the shadow.

    Again though, I doubt this will be of use.
     
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  12. Mick West

    Mick West Administrator Staff Member

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  13. Bunkmeister

    Bunkmeister New Member

    Hi Mick, I appreciate your analysis!!
     
  14. Rory

    Rory Senior Member