Is Max Eberle as big as a goof as I now think?

Get_A_Grip said:
Watch the video. He is basically at the same height as the peak of the mountain. 5400 feet. The curvature of the earth is 8 inches per mile squared. The video includes the calculation of how much of the top of the mountain would be hidden by dropping beneath the horizon. (~770 ft). I see no problems with his calculations.
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I've watched the video. If you don't know why you can see the two mountains despite the 770 ft "drop" according to the calculations, you'd really like a magic trick I used to do with my kids when I used to steal their noses from their face. Hint: it was my thumb....

Please tell me which company you work for as a scientist. I'm looking for my next "short".
 
MuchoBurrito said:
Okay, I watched (parts of) the video related to this point. If you don't see the problem with his calculation it's because you aren't thinking clearly.

at 27:40 he sets the "eye height" to ZERO, as though his eye height were at sea level, when he's actually at 5400 feet! According to him, whether you're at exactly sea level, or at the top of a mountain, your "eye height" is always ZERO. He thinks that the "eye height" is relative to the thing you are observing. It is not. That is simply not how this calculation works. Your eye height is relative to ground or sea level. Fact. he completely misunderstands how this works.

Go to https://dizzib.github.io/earth/curve-calc/?d0=34&h0=5400&unit=imperial yourself. He's at 5400 feet of altitude, which means his "eye height" in order to determine his horizon distance is 5400 feet. Because the horizon distance is a function of your altitude relative to ground / sea level. At a height of 5400 feet, and a distance of 34 miles from the thing he's observing, you will see that there are 0 feet hidden by curvature, since the horizon at 5400 feet is 90 miles.

But don't bother because math is a conspiracy.
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He is doing this because he's only interested in seeing the top of the mountain (the peak) and where the peak should drop over the horizon due to the curvature.

If you watch the other video I linked related to seeing buildings in Chicago over Lake Michigan, it might make you happier, because he is viewing the buildings from sea level. Same concept.
 
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Get_A_Grip said:
He is doing this because he's only interested in seeing the top of the mountain (the peak) and where the peak should drop over the horizon due to the curvature.

If you watch the other video I linked related to seeing buildings in Chicago over Lake Michigan, it might make you happier, because he is viewing the buildings from sea level. Same concept.
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No, here is where you (and he) are fundamentally misunderstanding this phenomenon and this calculation. This is not a matter of opinion, it's math. And he's doing it incorrectly, he would fail his high school math quiz doing it like this, because it's incorrect.

The horizon as it appears to you, is a function of your eye level relative to ground/sea level. Which is why the horizon changes as you gain altitude. Go to the website yourself, and look at the diagram below the calculator, which explains what the variables mean.

The diagram on the website makes it perfectly clear. "h0" is your eye height from ground level. Which is 5400 feet, yet he's putting it in a 0 feet, because he's an idiot. It's not a matter of opinion. He doesn't understand how to do math.
 
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MuchoBurrito said:
No, here is where you (and he) are fundamentally misunderstanding this phenomenon and this calculation. This is not a matter of opinion, it's math. And he's doing it incorrectly, he would fail his high school math quiz doing it like this, because it's incorrect.

The horizon as it appears to you, is a function of your eye level relative to ground/sea level. Which is why the horizon changes as you gain altitude. Go to the website yourself, and look at the diagram below the calculator, which explains what the variables mean.

The diagram on the website makes it perfectly clear. "h0" is your eye height from ground level. Which is 5400 feet, yet he's putting it in a 0 feet, because he's an idiot. It's not a matter of opinion. He doesn't understand how to do math.
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Because he's using those equations to predict the amount of drop from the peak due to the curvature. He's never going back to view from sea level. He's computing the drop and then showing on the picture where the peak should descend to starting at 5400 feet and that the peak wouldn't drop beneath the horizon at sea level, but it would drop below the terrain out in front of him. Like I said, if you want to see a similar example with the same calculation where buildings are viewed from sea level, look at the other vid.
 
Shawn Armstrong said:
I've watched the video. If you don't know why you can see the two mountains despite the 770 ft "drop" according to the calculations, you'd really like a magic trick I used to do with my kids when I used to steal their noses from their face. Hint: it was my thumb....

Please tell me which company you work for as a scientist. I'm looking for my next "short".
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Go view the other vid if you prefer a viewing from sea level. You obviously can't understand that the PEAK of the mountain is what should drop 770 feet and should be obscured by the terrain in front of him. The peak won't drop 5400 feet and be invisible.
 
Get_A_Grip said:
Because he's using those equations to predict the amount of drop from the peak due to the curvature. He's never going back to view from sea level. He's computing the drop and then showing on the picture where the peak should descend to starting at 5400 feet and that the peak wouldn't drop beneath the horizon at sea level, but it would drop below the terrain out in front of him. ...
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So you admit that he's doing it completely wrong. Thanks for that, at least.
 
Bob Jewett said:
So you admit that he's doing it completely wrong. Thanks for that, at least.
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You guys crack me up. No, he's not doing "it" wrong. He set up an experiment from viewing at ~5400 feet and calculating the drop due to the curvature. Each point from sea level to 5400 feet obviously drops the same about. You guys need to get_a_grip.
 
Get_A_Grip said:
You guys crack me up. No, he's not doing "it" wrong. He set up an experiment from viewing at ~5400 feet and calculating the drop due to the curvature. Each point from sea level to 5400 feet obviously drops the same about. You guys need to get_a_grip.
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Yes, the peak of the opposite mountain will drop 770 feet. And so if you're standing at sea level you won't see it.

But he's not standing at sea level, he's standing at 5400 feet! So he will have a perfectly good view! At 5400 feet, his horizon is 90 miles, and the mountain is only 34 miles away! Do the calculation at the website. In order to figure out at what distance the horizon will be to you, simply input your eye height (5400 feet).

There's no more sense in discussing this. You are literally failing an easy math quiz right now, and arguing with your textbook.

Go to the website he used (I posted it earlier). See the instructions / diagram. You will quickly see why he's making a mistake.
 
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You guys are serious?

Nice.

30 pages and 447 posts about Max's topic.

His name has been recited...how many times?

Think Max is smiling?

Yup.

Chuckling a bit, perhaps.

NO publicity is bad publicity. Ask Max if he believes this.

Yup.

Robin Snyder
 
MuchoBurrito said:
Yes, the peak of the opposite mountain will drop 770 feet. And so if you're standing at sea level you won't see it.

But he's not standing at sea level, he's standing at 5400 feet! So he will have a perfectly good view! At 5400 feet, his horizon is 90 miles, and the mountain is only 34 miles away! Do the calculation at the website. In order to figure out at what distance the horizon will be to you, simply input your eye height (5400 feet).

There's no more sense in discussing this. You are literally failing an easy math quiz right now, and arguing with your textbook.

Go to the website he used (I posted it earlier). See the instructions / diagram. You will quickly see why he's making a mistake.
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He'll actually be able to see a peak height of 5385 ft above sea level for a distance of about 178 miles, if he's standing on a lookout point of 5400 ft in height above sea level.

The power of math.....
 
MuchoBurrito said:
Yes, the peak of the opposite mountain will drop 770 feet. And so if you're standing at sea level you won't see it.

But he's not standing at sea level, he's standing at 5400 feet! So he will have a perfectly good view! At 5400 feet, his horizon is 90 miles, and the mountain is only 34 miles away! Do the calculation at the website. In order to figure out at what distance the horizon will be to you, simply input your eye height (5400 feet).

There's no more sense in discussing this. You are literally failing an easy math quiz right now, and arguing with your textbook.

Go to the website he used (I posted it earlier). See the instructions / diagram. You will quickly see why he's making a mistake.
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Except for the fact that the terrain in from of him, as he demonstrates on a pic of the view, the terrain would certainly obscure the PEAK. Again, he's not getting down and viewing the mountain from sea level. If he is sitting at 5400 feet and will view the mountain from there, that is why he's putting in "0" as the observers height. He is only interested in calculating the amount of drop and then viewing it from exactly where he is at.

All you guys need to realize is that the drop is the drop. If the peak drops by 770 feet he shows you where the peak should be. It would not drop below the horizon at sea level, but it would certainly drop below the terrain where he shouldn't be able to see it from where he was sitting.
 
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Shawn Armstrong said:
He'll actually be able to see a peak height of 5385 ft above sea level for a distance of about 178 miles, if he's standing on a lookout point of 5400 ft in height above sea level.

The power of math.....
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That's only if he's viewing the entire mountain with no obstructions in front of him.

Like I said, go view the vid on Lake Michigan from sea level. Exact same calculations, except the buildings also don't disappear below the horizon like they should. But my guess is, none of you are actually interested in reality, but only being able to try to point to some technically that tells yourselves that you're right....and the curvature is really hiding things below it at exactly 8 inches per mile squared.
 
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Get_A_Grip said:
Go view the other vid if you prefer a viewing from sea level. You obviously can't understand that the PEAK of the mountain is what should drop 770 feet and should be obscured by the terrain in front of him. The peak won't drop 5400 feet and be invisible.
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I've watched both vids. I'm not sure you actually understand trig. Maybe there's something I can explain to you? Would you like to know why a guy standing in a boat 34 miles from Chicago can see buildings that are over a certain height?

I can even give you the math, if you'd like. Bob provided the link to "horizon".
 
Get_A_Grip said:
Except for the fact that the terrain in from of him, as he demonstrates on a pic of the view, the terrain would certainly obscure the PEAK. Again, he's not getting down and viewing the mountain from sea level. If he is sitting at 5400 feet and will view the mountain from there, that is why he's putting in "0" as the observers height. He is only interested in calculating the amount of drop and then viewing it from exactly where he is at.

All you guys need to realize is that the drop is the drop. If the peak drops by 770 feet is shows you where the peak should be. It would not drop below the horizon at sea level, but it would certainly drop below the terrain where he shouldn't be able to see it from where he was sitting.
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Is the Earth flat, according to you?
 
Get_A_Grip said:
That's only if he's viewing the entire mountain with no obstructions in front of him.

Like I said, go view the vid on Lake Michigan from sea level. Exact same calculations, except the buildings also don't disappear below the horizon like they should. But my guess is, none of you are actually interested in reality, but only being able to try to point to some technically that tells yourselves that you're right....and the curvature is really hiding things below it at exactly 8 inches per mile squared.
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Maybe I was watching a different video than you. Could you see the BASE of Greybeard Mountain....? I could only see the peak, with about 80%+ blocked by other terrain....
 
Get_A_Grip said:
Except for the fact that the terrain in from of him, as he demonstrates on a pic of the view, the terrain would certainly obscure the PEAK. Again, he's not getting down and viewing the mountain from sea level. If he is sitting at 5400 feet and will view the mountain from there, that is why he's putting in "0" as the observers height. He is only interested in calculating the amount of drop and then viewing it from exactly where he is at.

All you guys need to realize is that the drop is the drop. If the peak drops by 770 feet he shows you where the peak should be. It would not drop below the horizon at sea level, but it would certainly drop below the terrain where he shouldn't be able to see it from where he was sitting.
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Again, go to the website yourself. Read the diagram. You are arguing with basic math my friend. What you're saying is that the instructions on the website are wrong.

Yes, "drop is drop". Nobody is arguing otherwise. What you are not understanding is that your horizon changes as your altitude increases. You can see further at 10 feet altitude than at 0. And further at 5400 feet than at 10.

And WAIT A MINUTE, what is the height of the other mountain peak??? Surely it's greater than 770 feet? So no matter where he's standing, with a declination of only 770 feet, he'd be able to see the mountain peak no matter his own altitude at 34 miles! all that would be obscured, even if standing at ground level, is the bottom 770 feet of the mountain. The peak would still be totally visible!
 
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