Why are distant skylines at the same angle upwards if the Earth is curved ?

Why not? If the Earth were transparent, we would be seeing the bottom of everything on the opposite side, wouldn't we? And looking at a structure about an eighth of the way around the globe away from us, we should see it incredibly far away, very low below the horizon, and at an apparent slant away from us, with the bottom also visible. That should be testable with a transparent orb in a 3D modeling program.

 

A pole at 60ft?

 

There's a semantical issue here, I suppose. I think of an apparent lean being to the left or the right. If a 100' conifer is leaning straight away from me at a 10 degree angle I'll never see it without moving left or right, but with a skyscraper at the same angle you could figure it out if you knew its height/width ratio.

 

Pardon me, if I have lost track of this conversation. I was away for a while.

I assume we are still talking about the apparent lean due to the curvature of the earth. If that is the case, there will never be an apparent lean of ten degrees for a sky scraper.

You will always be looking head on at the skyscrapers unless you were looking right or left of the skyscraper and seeing them out of the corner of your eye.

In order to see a difference in angle of ten degrees between two skyscrapers due to the curvature of the earth, they would have to be nearly seven hundred miles apart.

 

Actually when you are on shore line or out at sea, you get a better idea of the curvature of Earth.

We were taught this crap in elementary school.

When standing on the beach close to the waterline, how far is the horizon ? About 3 miles. But you can see upper structure or mast of a ship that is further out to sea but not the hull of the ship.

How to Calculate the Distance to the Horizon
Have you ever been out on a leisurely cruise and suddenly wondered, "How far it is to the horizon?" Or maybe your destination is a port that has a lighthouse and you wonder "How far away will I be when I see the lighthouse?" (Well, you're in luck, even if you are a sick unit that thinks of these sorts of things - so are we.) We have the answer!

Of course you can find tables that do the calculation for you in numerous navigation books, almost every book that talks about passagemaking, the Coast Pilot, almanacs, etc. But what if you didn't have any of these references onboard? How could you calculate the distance to the horizon or the "distance off" if you know the height of an object?

It's simple, really. If you want to know the distance to the horizon you simply have to know your height of eye. That is the distance that your eyes are off the surface of the water.

If you're in a jon boat, that would probably be about three feet (if you are sitting like you should be in a jon boat). Of course, if you were in a jon boat you probably wouldn't care how far the horizon was.

Anyway, I digress. If you are on the tuna tower of a sport fishing boat you may be 15, 20, 25 feet above the surface of the water.

Once you know your height of eye you simply plug that into the following formula:

1.17 times the square root of your height of eye = Distance to the horizon in nautical miles

For example, if your height of eye is 9 feet above the surface of the water, the formula would be:

1.17 times the square root of 9 = Distance to the horizon in nautical miles.

1.17 * 3 = 3.51 nautical miles

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If you want to calculate the distance at which an object becomes visible, you must know your height of eye and the height of the object. You then do the same calculation for your distance to the horizon and the object's distance to the horizon and add the distances together. For example:

You have the same height of eye of 9 feet so your distance to the horizon is still 3.51 nautical miles. You're approaching a port that has a lighthouse that is shown on your chart to have a height of 81 feet. Using the same formula you would find that 1.17 times the square root of 81 (1.17 * 9) = 10.53 nautical miles (the light house can be seen 10.53 nautical miles over the horizon)

By adding the two together: 3.51 + 10.53 = 14.04 nautical miles, you should be able to see the lighthouse when you are 14.04 nautical miles away.
http://boatsafe.com/nauticalknowhow/distance.htm

If one were to use a laser, not a level or water level that are affected by gravity but a laser it would show that a building 25 or 50 miles away aren't standing at a true 90 degree right angle while on the surface but actually an acute or obtuse angle. But so small undetectable with the naked eye.

But using a plumb and your typical lever it would show the building was at a 90 degrees right angle.

Does that make sense ?

 

That formula is good only for short distances. The further away, the bigger the error in the formula.

 

That's because you don't live on the transparent Earth I dreamed up.

Durandal and I were talking about a single skyscraper, which, given a transparent Earth, would never appear to lean left or right, irrespective of the observer's position on Earth.

 

And CNN, NBC News, CBS News, ABC News, MSNBC, the WaPo and the NYT all sent reporters to Wasilla, Alaska climbing on top of buildings looking for Russia.

Except Fox News who sent a reporter and cameraman to Little Diomede Island in Alaska and Sarah Palin was 100% right, you can see Russia from Alaska.

But the Zogby exit polling on the day a community organizer got elected revealed those who voted for McCain were well educated, well informed and not low information voters while those who voted for Obama were low information or misinformed and 86.9 % thought that Palin said that she could see Russia from her "house," even though that was Tina Fey who said that.
http://mediamalpracticemovie.com/research-zogby.asp

 

Back during my sailing days I wasn't so much concerned with distance but used a pair of binoculars, sexton, chronograph watch, sexton, navigation charts and "Reed's Nautical Almanac and Coast Pilot."

 

Back during my sailing days, I wasn't concerned with that at all. Except for a small amount on a Florida inland waterway, I did most of my sailing on a lake. I spent most of my time keeping the sails trimmed in the shifting wind.