Horizon and Curvature Observations

[DaveC426913]

Thinking last night about what I should expect see of the Toronto skyline, which is 40 miles across the lake from me.

Starting with an ideal scenario (nape-of-shore to nape-of-shore), a chord calculator tells me that for r=3900 mi. d=40 mi. the chord height will be 270 ft.

I was surprised at that height, and doubted my calculations. A 270 foot mountain in the middle of the lake is a lot! But I soon realized that it's not really a lot over 40 miles. It's rise-over-run of only 1:780.


At first I assumed everything below that 270 foot mountain would be obscured, so I would only be able to see buildings from the 28th floor and up (assuming they are at the waterline). But upon reflection, I think I won't be able to see anything below double that, or 540 feet.

Confounding Factors:

• I don't think my height above the water will make a difference if it's small enough. I mean, if my sightline is 10 feet above the waterline, that'll mean I can see down to 530 feet instead of 540 feet, right?
  
  
  
  
• I also have to remember that the visible buildings are many metres above the waterline, so I can't just count building stories to estimate altitude.
• And I'll have to time the observations right, when atmospheric refraction is at a minimum.

Are my assumptions and calculations correct?


(Apologies for my wanton mixing of SI and metric)

 

[DaveC426913]

Researching how far above lake level the Toronto skyline is. Found this anomaly:

West L. Ontario is pretty consistently about 73m above sea level (turquoise) so that's good.

But this little blue-shaded area around the Harbourfront Towers reports as being 10m below lake level!

Guess I gotta take altitude heat maps with a grain of silt...

 

[tech99]

Light curves downwards when travelling across the earth under normal conditions, so you can see further than expected.. This requires a small correction to be applied when sighting the horizon for sextant observations. Similar effects occur with microwaves, and it is usual to assume a modified Earth radius of 4/3 for planning purposes.

 

[DaveC426913]

Light curves downwards when travelling across the earth under normal conditions,

Why? Do you mean because of refraction due to temp variations?

Would it factor in when the line of sight is basically nape-of Earth (i.e. no appreciale altitude) and over only 40 miles?

Ideally, I'd observe when water temp and air temp are similar, so that temp variations in the 540 foot air column are at a minimum.

 

[tech99]

There is normally a reduction in air temperature with height, and also a reduction in humidity. These create a vertical gradient whereby refractive index reduces with height.

 

[jbriggs444]

I don't think my height above the water will make a difference if it's small enough. I mean, if my sightline is 10 feet above the waterline, that'll mean I can see down to 530 feet instead of 540 feet, right?

A rule of thumb (without refraction correction) is 8 inches per mile squared.

10 feet is 120 inches. Divide by 8 and get 15. Take the square root and your line of sight brushes the water surface right about 4 miles out.

That leaves 36 miles to go. Square that and we get 1296. Times 8 inches is 864 feet.

 

[Ibix]

At first I assumed everything below that 270 foot mountain would be obscured, so I would only be able to see buildings from the 28th floor and up (assuming they are at the waterline). But upon reflection, I think I won't be able to see anything below double that, or 540 feet.

I think this calculation would be correct if you were on a Euclidean plane with a 270 foot wall 20 miles away. But you are on a sphere and, as the second and third diagrams in your OP show, your sightline touches the Earth considerably closer to you than half way.

Draw the Earth as a circle of radius $R$ on a piece of paper and you have $$\begin{eqnarray*}
y&=&\sqrt{R^2-x^2}\\
&=&R\sqrt{1-\frac{x^2}{R^2}}\\
&\approx&R-\frac{x^2}{2R}
\end{eqnarray*}$$where the last line is a first order Taylor expansion (the next term looks like $x^4/R^3$, and for a 40 mile distance this is $(40/4000)^2$ smaller than the first and can be neglected). So the surface is $x^2/2R$ below where it is at $x=0$. Plugging in the numbers, $1/2R$ works out to the $8\mathrm{\ inch/mile^2}$ rule cited by @jbriggs444.

Draw a horizontal line grazing the surface of the Earth at $x=0$. It is 10 feet above the parabolic approximation at $x=\pm\sqrt{10\mathrm{\ feet}\times2R}$ which works out to a shade under four miles. The buildings are therefore at $x=\mp 36\mathrm{\ miles}$, which makes their base about 850 feet below your sightline, in agreement with jbriggs444.

 

[Ibix]

The 4/3 rule @tech99 gave for optical effects would presumably lower that to a bit over 600 feet, though.

 

[A.T.]

The 4/3 rule @tech99 gave for optical effects would presumably lower that to a bit over 600 feet, though.

Some of the buildings in Toronto are taller than that:
https://en.wikipedia.org/wiki/List_of_tallest_buildings_in_Toronto#Ranking_by_standard_height

Thinking last night about what I should expect see of the Toronto skyline

What do you actually see?

 

[DaveC426913]

What do you actually see?

Well, most of the skyline, but I haven't viewed it under controlled conditions to see how much is eclipsed by the curvature of the Earth. I'm about 12m above the waterline here.

 

[jbriggs444]

Well, most of the skyline, but I haven't viewed it under controlled conditions to see how much is eclipsed by the curvature of the Earth. I'm about 12m above the waterline here.

Ahh. So your eyes are 12 m above the water line, not 10 feet.

That is about 480 inches. At 8 inches per mile squared, that is 60 miles squared. Taking the square root we have almost 8 miles. Multiplying by 4/3 for refraction, we get a bit over 10 miles to the point where your sight line touches the water.

That leaves 30 miles to go. Dividing by 4/3 for refraction that is effectively about 22.5 miles. Squaring and multiplying by 8 inches per mile squared we get about 4000 inches (340 feet or 100 meters).

[I am not absolutely sure that is how the 4/3 works]

 

[DaveC426913]

Ahh. So your eyes are 12 m above the water line, not 10 feet.

No. A.T. asked what I "actually see" (i.e. now - as opposed to what I said - which is what can I "expect to see" when I go down to nape-of-shore - as outlined in my OP).

That's why this is anticipatory. I have not done the observations-under-controlled-circumstances yet. When I make the observations I will go down to the shoreline.

 

[DaveC426913]

I am disappointed at this need to account for refraction that apparently cannot be narrowed-down let alone eliminated. It means this is not an empirical observation - I have account for something based - not on what I see with my eyes - but what I cannot see and have to take on-faith.

It kind of throws everything out the window.

Granted, real-world observations always have confounding factors in them, but this one seems intractible. We're just sort of guessing (to one sig dig) how much to compensate for refraction. Which means the margin of error will be much larger than the level of accuracy the experiment needs.

Any results I get - from zero feet to a thousand feet - will simply be recorded as "oh well ... I guess the refraction is much larger/smaller than we thought - and there's no way to confirm."

 

[A.T.]

No. A.T. asked what I "actually see" (i.e. now - as opposed to what I said - which is what can I "expect to see" when I go down to nape-of-shore - as outlined in my OP).

I obviously meant what you see in the setting you are asking about, not what you see as you read my post (which would be a screen presumably).

Anyways, there are plenty of images of Toronto from across the lake:
https://www.google.com/search?sca_e...&biw=2560&bih=1283&dpr=1#imgrc=Kg8POzr5HdgDFM

From Niagara (seems to be from the beach):

From Olcott:

 

[DaveC426913]

I obviously meant what you see in the setting you are asking about, not what you see as you read my post

Sure, but I hadn't done the observation part of this yet, so best I could offer was what I know so far - which is currently from 12m altitude.

 

[DaveC426913]

I see the flaw in my plan now

I went out today and took some pix.

Here is the view from the waterline:

Pity that's not Toronto (33mi**). That's Mississauga in the centre-right (25mi) (whose skyline is miles from the shore) and Bronte (11mi) on the left.

** I was wrong about the distance estimation; I had looked it up instead of measuring it myself.

To be sure I could see Toronto at all, I went to the top of the mountain (+100m altitude from waterline and an addtional 1.5mi from shore). This is the view:

Bronte is top-centre and Toronto is far, far right.

In other words, I'm going to have to wait for a much clearer day to see anything at all, nevermind attempting to estimate altitudes at that distance.

 

[A.T.]

In other words, I'm going to have to wait for a much clearer day to see anything at all, nevermind attempting to estimate altitudes at that distance.

You might also need a good camera with a tele-lens.

 

[russ_watters]

I am disappointed at this need to account for refraction that apparently cannot be narrowed-down let alone eliminated.

It would surprise me if this was a thing. You're only looking though 50 miles horizontally and less than a thousand feet of elevation. Not a lot for lensing. Refraction corrections were made for celestial navigation in the Navy (looking through the entire atmosphere) but I don't think we did that for height of eye vs horizon distance. It was just a simple table:

https://thenauticalalmanac.com/Bowditch-%20American%20Practical%20Navigator/TABLE%2012-%20DISTANCE%20OF%20THE%20HORIZON.pdf&ved=2ahUKEwjqjfGG_vGEAxVeLFkFHYbhC1MQFnoECBIQAQ&usg=AOvVaw2okllekPmHnaaDrasb1oao

 

[DaveC426913]

It would surprise me if this was a thing. You're only looking though 50 miles horizontally and less than a thousand feet of elevation. Refraction correction were made for celestial navigation in the Navy (looking through the entire atmosphere) but I don't think we did that for height of eye vs horizon distance.

Thanks. That was my intuition too.

 

[jbriggs444]

It would surprise me if this was a thing. You're only looking though 50 miles horizontally and less than a thousand feet of elevation. Not a lot for lensing. Refraction corrections were made for celestial navigation in the Navy (looking through the entire atmosphere) but I don't think we did that for height of eye vs horizon distance. It was just a simple table:

https://thenauticalalmanac.com/Bowditch-%20American%20Practical%20Navigator/TABLE%2012-%20DISTANCE%20OF%20THE%20HORIZON.pdf&ved=2ahUKEwjqjfGG_vGEAxVeLFkFHYbhC1MQFnoECBIQAQ&usg=AOvVaw2okllekPmHnaaDrasb1oao

Yowza. This link (the one below, not the one above) will put you to sleep with pages of formulae. It agrees with the sentiment that rules of thumb for celestial navigation are not directly applicable.

I. INTRODUTTION

Existing astronomic refraction equations cannot be satisfactorily applied to objects at distances up to a few thousands of miles from the earth because they were developed for a special use in which the object is at an infinite distance. In this paper, equations are derived that are applicable to objects both inside and outside the atmosphere.

Rather than trying to extract meaning from that document, I went to a simpler reference.

the coefficient k, measuring the ratio of the radius of the Earth to the radius of the line of sight, is given by

where temperature T is given in kelvins, pressure P in millibars, and height h in meters. The angle of refraction increases with the coefficient of refraction and with the length of the line of sight.
[...]
This yields

where L is the length of the line of sight in meters and Ω is the refraction at the observer measured in arc seconds.

Typical temperature gradient ($\frac{dT}{dh}$) is about 6.5 kelvins per 1000 meters. We are talking about a distance ($L$) of 40 miles or 65000 meters. Let us assume a temperature ($T$) of about 300 kelvin and an atmospheric pressure of about 1000 millibars.

Edit: Did not evaluate $k$ in the first draft

Let me first evaluate $k$, the ratio between the radius of the Earth and the radius of the line of sight:$$k = 503 \frac{P}{T^2}(0.0343 + \frac{dT}{dh}) = 503 \frac{(1000)}{(300)^2}(0.0343 - \frac{6.5}{1000}) = 0.16$$That hits in the ball park of the $\frac{4}{3}$ heuristic that was posted previously by @tech99.

Evaluating $$\Omega = 8.15 \frac{LP}{T^2}(0.0342 + \frac{DT}{dh}) = 8.15 \frac{(65000)(1000)}{(300)^2}(0.0342 - \frac{6.5}{1000}) = 160 \text{ arc seconds}$$This is the difference between the angle you think you are looking and the actual angle to the target. Over 40 miles (65 kilometers), the actual angle would be about 0.0055 radians or about 1100 arc seconds below the horizontal. As explained in the Wiki reference, the angle of the line of sight is half of the total curvature. The apparent angle would be only 940 arc seconds below the horizontal.

Edit: Typical temperature gradient is negative, not positive. -0.0065 kelvin per meter rather than +0.0065 kelvin per meter. That leads to a factor of 1.46 correction. Let me edit that into the computations above.

 

[russ_watters]

Yowza. This link will put you to sleep with pages of formulae. It agrees with the sentiment that rules of thumb for celestial navigation are not directly applicable.

Sorry, it didn't seem to parse right when posting from my cell. It went straight to the table when I opened it from Google. I'll try to fix it when I get home.

 

[DaveC426913]

Typical temperature gradient (dTdh) is about 6.5 kelvins per 1000 meters. We are talking about a distance (L) of 40 miles or 65000 meters.

Woah. Have you confused vertical with horizontal distances here?

Surely the relevant distance for measuring temperature gradient is 80 netres (270 feet), not 65000 metres.

 

[DaveC426913]

The apparent angle would be only 860 arc seconds below the hohorizontal.

Or 14.3 arc minutes. Half the width of the moon.

 

[jbriggs444]

Woah. Have you confused vertical with horizontal distances here?

Surely the relevant distance for measuring temperature gradient is 80 netres (270 feet), not 65000 metres.

No. I have not confused anything.

The $L$ figure is the length of the horizontal line of sight. That is 65 kilometers or 65000 meters.

The $\frac{dT}{dh}$ figure is for the vertical temperature gradient. A little Googling shows 6.5 kelvins per 1000 vertical meters.

However, I did manage to ignore the fact that the temperature gradient is negative. It usually gets colder as you go up. As we know from desert mirages, light travels more rapidly in high temperature low density air. It curves away from high temperatures and toward low temperatures. A positive gradient (high temperatures at high altitudes) would increase the downward refraction. A negative gradient (low temperatures at high altitudes) tends to reduce it.