Why can't I see mountains?

I'm 110km from Scafell Pike, the tallest mountain in England. It has a height of slightly more than 900m. This means it's angular diameter from where I am is 0.45°, which is easily visible (it's abut half the size of a full moon).

I can't see Scafell Pike, even when I climb to the highest point around. It can't be a perspective effect (0.45° is easily big enough to see), and it can't be atmospheric degradation (according to FE proponents, when the moon is near the horizon you are looking through 1000's km of atmosphere and you can still see it fine), and it can't be refraction (Snell's law predicts it would be more visible, not less).

So, if the Earth is flat, why can I not see Scafell Pike?

I can't even see mountains that are less than 30 miles from me.  Damn that atmosphere!

according to FE proponents, when the moon is near the horizon you are looking through 1000's km of atmosphere and you can still see it fine

I laugh at you... atmosphere gets much much thinner just a couple of hundred feet up

I laugh at you... atmosphere gets much much thinner just a couple of hundred feet up

Yeah, and Scafell Pike isn't more than a couple of hundred feet high, is it?

You sir, have no cerebral cortex. All that is within your cognitive power is breathing, heartbeat and some simple speech. Logic, however, seems to be upon you, and Blackadder II is turning in his grave.

you cant see moutains because you have no eyes

I say this to clarify if you have eyes if you dont this is a pointless topic.

They're right of course, the atmolayer obscures your vision at a certain distance, but it's good to keep in mind that it's working in sychronity with natural perspective. Simply put, stuff also gets smaller in your field of vision the further away it is, until it is no longer visible. Even if there were a completely transparent atmolayer (there isn't), you'd still only be able to see so far with the naked eye.

He's back!

He's back!

thanks to me

*bows*

*Throws rose*

They're right of course, the atmolayer obscures your vision at a certain distance, but it's good to keep in mind that it's working in sychronity with natural perspective.

When the moon is on the horizon, according to FET you are looking at it through 1000s km of "atmolayer" but you can still see it fine. So why can't I see a mountain just 100km away?

Simply put, stuff also gets smaller in your field of vision the further away it is, until it is no longer visible. Even if there were a completely transparent atmolayer (there isn't), you'd still only be able to see so far with the naked eye.

Yes, which is why I made sure I could see it at this distance (it would appear about half the soze of the full moon, easily visible). It would still be close enough to see.

Just to confirm with everyone, I've done a quick check and can definately say yes, i can see. I did a control test and I can indeed see my wardrobe (which isn't so far away that it would be hidden behind the curvature of the Earth).

When the moon is on the horizon, according to FET you are looking at it through 1000s km of "atmolayer" but you can still see it fine. So why can't I see a mountain just 100km away?

No. The moon is on the horizon because of refraction.

No. The moon is on the horizon because of refraction.

How many times must you be told, Tom! The moon would appear higher, not lower, due to refraction.

Even so, that doesn't answer my question.

How many times must you be told, Tom! The moon would appear higher, not lower, due to refraction.

Even so, that doesn't answer my question.

I've come to agree that as light curves into the earth celestial bodies would appear higher than they actually are. However, they would still descend in altitude as the bodies recede into the distance. Here's an example:

Noon:



1pm:



2pm:



As we can see, the sun would descend at a constant pace as it passes by overhead. If the earth had no atmosphere we would see the sun's apparent speed change constantly like a jet passing by overhead. It would be whooshing by at its fastest pace directly overhead, and slowest as it is near the horizon. We do not see this. Since the virtual sun is at a cosine to the curved path of light, the sun moves at a constant rate across the sky.

When the sun recedes its virtual image descends into the horizon as the path of light becomes increasingly shallow. As the edges of the sun's spotlight passes overhead of the observer the sun will appear to refract into the surface of the earth.

Even so, that doesn't answer my question.

The answer to the question is that the sun and moon are bright enough to shine through the density of the atmosphere. The effect is similar to fog lights through fog. If you look at the sun when it is near the horizon line you will see that its intensity becomes an order of magnitude dimmer. It's possible to look directly at the sun with the naked eye without squinting. This is because the sun must shine through a great length of the earth's atmosphere, becoming much dimmer in the process.

There is a similar effect with the moon. If you've ever see the moon when it is near the horizon, you will see that the moon is much dimmer than it is overhead.

I've come to agree that as light curves into the earth celestial bodies would appear higher than they actually are.

I'm glad you saw your mistake Tom.

However, they would still descend in altitude as the bodies recede into the distance.

Yep.



Now that we agree that the sun appears higher than it actually is due to refraction let's see what this implies for the setting of the sun.  In my diagram, I have drawn the sun and its image as it should be at sunset.  According to estimates you have made before, the sun needs to be about 10000 km away when it appears to set.  The height of the sun is about 4800 km.  These distances are to scale on my drawing.  However, the size of the sun and the thickness of the atmosphere are not.

Since the light of the sun always curves downwards, we see that the angle of the image of the sun at this particular location must be greater than tan^-1(4800/10000) = 25.6^o.  The sun would not appear even close to setting.

As we can see, the sun would descend at a constant pace as it passes by overhead. If the earth had no atmosphere we would see the sun's apparent speed change constantly like a jet passing by overhead. It would be whooshing by at its fastest pace directly overhead, and slowest as it is near the horizon. We do not see this. Since the virtual sun is at a cosine to the curved path of light, the sun moves at a constant rate across the sky.

When the sun recedes its virtual image descends into the horizon as the path of light becomes increasingly shallow. As the edges of the sun's spotlight passes overhead of the observer the sun will appear to refract into the surface of the earth.

I think the opposite effect occurs.  On a FE (and indead RE), the virtual sun would appear to slow down even more (than a "jet plane") as it approached the horizon, because the further the real sun goes away, the more the distance between the real and the virtual increases.

Remember that it is already well understood that refection increases the hours of daylight by a small amount, thereby implying a virtual "slowing down" in apparent speed across the sky.

Quote

How many times must you be told, Tom! The moon would appear higher, not lower, due to refraction.

Even so, that doesn't answer my question.

I've come to agree that as light curves into the earth celestial bodies would appear higher than they actually are.

Finally, you have seen the light, so to speak. Now, that view out of your window?

When the sun recedes its virtual image descends into the horizon as the path of light becomes increasingly shallow. As the edges of the sun's spotlight passes overhead of the observer the sun will appear to refract into the surface of the earth.

Go on then. Shows us what sunset would look like. Ideally to scale.

As we can see, the sun would descend at a constant pace as it passes by overhead. If the earth had no atmosphere we would see the sun's apparent speed change constantly like a jet passing by overhead. It would be whooshing by at its fastest pace directly overhead, and slowest as it is near the horizon. We do not see this. Since the virtual sun is at a cosine to the curved path of light, the sun moves at a constant rate across the sky.

Ho, ho, your ineptitude in all its glory.
I rather think you mean the virtual sun is 'at a tangent' not 'at a cosine'.

No, I think cosine works. Not quite sure though. If he is talking about horizon and angle from the sun's position to you, than it is cosine.

No, I think cosine works. Not quite sure though. If he is talking about horizon and angle from the sun's position to you, than it is cosine.

'At a tangent' is a recognised phrase, 'at a cosine' is not.

The line to the virtual sun is 'at a tangent' to the curved light path in the figure.

The angle at midday in his graphic is 90º. Cosine 90º is zero.
As the sun moves further away the angle decreases and the cosine increases.
What has that go to do with the sun touching the horizon at sunset?

Everything.

I like the smiley faces on the suns

What are you talking about?

Tom's diagrams

Oh, ok.