Need help with calculating vertical speed of ascending shadow at sunset

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.
 

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.
 

Bunkmeister

New 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.
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.
 

Bunkmeister

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

Mick West

Administrator
Staff member
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.

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.jpg

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.jpg
 

Bunkmeister

New 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.jpg

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.jpg

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?
 

Bunkmeister

New 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.jpg

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

Mick West

Administrator
Staff member
At some point towards the summit, the ascending shadow can only be from earth curve, right?

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.jpg

And here's a slider comparison of Google Earth's calculated view with and without terrain.


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

Again though, I doubt this will be of use.
 
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Bunkmeister

New 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.jpg

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

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

Again though, I doubt this will be of use.
Hi Mick, I appreciate your analysis!!
 

AtomPages

New Member
Digging this one back up. I happened to take a picture in my neighborhood that showed the sun just rising (I think it was at about 1.6 degrees altitude at this point), and it made me wonder if I could capture any photos of things casting shadows upward on one of the houses or fences in the neighborhood.
Sunrise_20220329_0704_small.jpg
I agree that if you already accept that the sun can appear to go below the horizon on a flat earth, you probably won't be convinced by shadows, but this got me wondering. One point FE's often make is that the sun casting shadows upward to clouds is actually a point in their favor, indicating a "local" sun that is below the clouds. Following that logic, if people started demonstrating pictures of the sun casting a shadow upward on a house, would that mean the sun was lower than the house?

My first attempt at capturing upward shadow pictures like this failed, since I haven't found a great spot where the sunrise light is bright enough to cast a noticeable shadow but low enough that the casting object is something I can easily measure the height of.
 

Mendel

Senior Member.
it made me wonder if I could capture any photos of things casting shadows upward on one of the houses or fences in the neighborhood.
This is difficult because at typical altitudes, the angles involved are rather small, so it requires some accurate leveling to show that the shadow goes up. And because the sun's diameter is 0.5⁰, you also get that much of a penumbra, aka a fuzzy shadow with no exact line between light and dark.

You may have more luck catching the sunrise below a clear plastic tube water level, but that, too, requires some altitude. Check out "refracted horizon dip" for the best angle you can achieve for a given altitude: https://www.metabunk.org/curve/
 

FatPhil

Senior Member.
Digging this one back up. I happened to take a picture in my neighborhood that showed the sun just rising (I think it was at about 1.6 degrees altitude at this point), and it made me wonder if I could capture any photos of things casting shadows upward on one of the houses or fences in the neighborhood.

Think big! Make a shadow of a mountain, and display it on a layer of clouds!

From https://flatearth.ws/shadow-on-clouds

Personally, I think the earth casting a shadow onto the underside of the cloud layer is good enough to disprove the FE thesis(*), a mountain is just to make it pretty too.

(* depending on quite how wacko their views are, some have no grasp on geometry at all.)
 

AmberRobot

Active Member
One point FE's often make is that the sun casting shadows upward to clouds is actually a point in their favor, indicating a "local" sun that is below the clouds.
The height of the Sun above the flat earth appears to be whatever the flat earther needs it to be to support the argument he is making at the time. When pressed for numbers no flat earther can give an answer. And whatever answer they do give can immediately be shown to be impossible through simple grade school geometry considerations.
 

Mendel

Senior Member.
Jeez, just how low does their sun go?
you could adress a lot of these concerns with "upward refraction" in the atmosphere, which unfortunately seems to be impossible to demonstrate, unless you're near the surface of a large body of water under favorable conditions, or simply photoshop downward refraction upside down (no joke, Dr John did that), because it goes against physical laws if you don't forget about the pressure gradient (again, Dr John).

Which is why I prefer challenging people on the sideways bending of light required to make the sun appear in the compass direction where it actually is, especially south of the equator in the summer.
 
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