[Tom Bishop]

I used the Pythagorean Theorem (a^2 + b^2 = c^2) to calculate the drop over 4.4 miles.



Suppose that the earth is a sphere of radius 3963.1676 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangel as in the diagram.

3963.1676^2 + 4.4^2 = 15706716.7857

When we square root that figure we get 3963.17004249

Thus your position is 3963.17004249 - 3963.1676 = 0.00244248999 miles above the surface of the earth

Converting to feet, 0.00244248999 miles = 5280 * 0.00244259 = 12.8963471472 feet

Hence after 4.4 miles the earth drops approximately 12.9 feet, or over 1 story of a building.

[Tom Bishop]

Judging from the pictures presented, it does not appear that the beach drops over one story below the water line.

[Tom Bishop]

Tintagel is correct, there does seem to be some sort of optical effect going on.

Why is the first story of the house compressed into the beach?

[Hoppy]

Tintagel is correct, there does seem to be some sort of optical effect going on.

Why is the first story of the house compressed into the beach?

That is what I am asking, why is all of the beach and over 1/2 of the first floor covered with what seems to be a wall of water. Especially considering the "wall of water" is the approximate size predicted by RET. Judging by the white railing, I did mislabel the 1st floor as 2nd floor in the original image. When I first took the picture I thought it verified FE, and labeled the photo.



This image is from the same rock as the original image, camera 32" above the water.


This image is from a pier , camera is 8' above the water.


Here are two pictures from today,  Today the pictures look entirely different. It looks flat today.     WTF
I also want to add, the water level was within 6" of the first image. Low wind conditions, but the haze in the area seemed thicker today. There was heavy chemtrail spraying today.

[Tintagel]

Interesting documentation of an optical anomaly, and not an uncommon one if the RET folks are any indication.  The side-by-side at the end is pretty compelling, though, I'd say. 

Perhaps the tides could be contributing somehow to the anomaly in the first photo?

[markjo]

Here are two pictures from today,  Today the pictures look entirely different. It looks flat today.     WTF

Different atmospheric conditions produce different atmospheric refractive results.  That and yes, the tide can be a factor as well.

[rottingroom]

I used the Pythagorean Theorem (a^2 + b^2 = c^2) to calculate the drop over 4.4 miles.



Suppose that the earth is a sphere of radius 3963.1676 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangel as in the diagram.

3963.1676^2 + 4.4^2 = 15706716.7857

When we square root that figure we get 3963.17004249

Thus your position is 3963.17004249 - 3963.1676 = 0.00244248999 miles above the surface of the earth

Converting to feet, 0.00244248999 miles = 5280 * 0.00244259 = 12.8963471472 feet

Hence after 4.4 miles the earth drops approximately 12.9 feet, or over 1 story of a building.

Nobody as been in dispute over what the drop would be from 0 elevation. Hoppy was at 2 ft making the drop significantly less at 4.75 ft.

[rottingroom]

Tintagel is correct, there does seem to be some sort of optical effect going on.

Why is the first story of the house compressed into the beach?

That tan sliver is not the beach and this can be verified from Hoppy's close ups and from Google Earth. The beach does not extend across the entire waterfront.

[rottingroom]

Hoppy the new drop at 32 inches in elevation is 3.47 ft while the 8 ft elevation shot is .589 ft.

You figure this out by using this calculator.

You use the first calculator by putting in your elevation. It gives you distance which you subtract from your 4.4 miles.

Then you take that distance and put it in the second calculator which gives you your drop.

[Hoppy]

Hoppy the new drop at 32 inches in elevation is 3.47 ft while the 8 ft elevation shot is .589 ft.

You figure this out by using this calculator.

You use the first calculator by putting in your elevation. It gives you distance which you subtract from your 4.4 miles.

Then you take that distance and put it in the second calculator which gives you your drop.

I already told you what the drop should be on RE, 12.9 feet across 4.4 miles. Then you can subtract the nearly 3' elevation of the camera = Total drop of 9.9'.

[rottingroom]

Hoppy the new drop at 32 inches in elevation is 3.47 ft while the 8 ft elevation shot is .589 ft.

You figure this out by using this calculator.

You use the first calculator by putting in your elevation. It gives you distance which you subtract from your 4.4 miles.

Then you take that distance and put it in the second calculator which gives you your drop.

I already told you what the drop should be on RE, 12.9 feet across 4.4 miles. Then you can subtract the nearly 3' elevation of the camera = Total drop of 9.9'.

Problem is that that is completely incorrect. Did you miss math class or something?

[rottingroom]

We even drew you pictures which clearly explain why. If you think you just subtract then show a good diagram about why. I can tell you now it is impossible.

[Hoppy]

We even drew you pictures which clearly explain why. If you think you just subtract then show a good diagram about why. I can tell you now it is impossible.

Look man. The most important variable is the drop across the distance. Even your calculator gave the same answer as Rowbotham's chart. So I think we can agree that the drop over 4.4 miles is 12.9'. Just because I move the camera up 2' feet doesn't change the drop over 4.4 miles. So the end of the drop is going to 10.9' instead.

Think of it like this, if the drop over a distance was 10', and blocked the entire view of a 10' bldg sitting right at the water. If you climbed a 10' ladder you would be able to see the entire building.

[Rama Set]

I think Rottingroom is saying that:

1. You can see the entire building at the horizon

2. The building would only be obscured by the drop between the horizon and the object.

So at a distance if 4.4 miles and an altitude of 2ft the object is unobscured for 1.73 miles and then is obscured by the curvature of the remaining 2.67 miles.

Hoppy, let me know what part of this you disagree with.

[Hoppy]

I think Rottingroom is saying that:

1. You can see the entire building at the horizon

2. The building would only be obscured by the drop between the horizon and the object.

So at a distance if 4.4 miles and an altitude of 2ft the object is unobscured for 1.73 miles and then is obscured by the curvature of the remaining 2.67 miles.

Hoppy, let me know what part of this you disagree with.

I think what he is saying on the points you bring up are correct.

What I am saying is correct also:

"Think of it like this, if the drop over a distance was 10', and blocked the entire view of a 10' bldg sitting right at the water. If you climbed a 10' ladder you would be able to see the entire building."

You have to think of the total distance, don't divide it up to 2 different distances. If you do divide it, you will get the wrong answer.

[Rama Set]

If what I am saying is correct it cannot also be wrong.  Your analogy does not exclude the truth of what I said either. The wall analogy is also not entirely apt since you are not dealing with a curve.

Basically if everything I have asserted is true, then my conclusion must be true. The drop that you claimed is not incorrect, but I think should be classified as an upper limit of drop, not the correct drop.

[markjo]

You have to think of the total distance, don't divide it up to 2 different distances. If you do divide it, you will get the wrong answer.

Incorrect.  You must divide the total distance into two different distances to get the correct answer.  First there is the distance from the observer to the horizon, then there is the distance from the horizon to the house.

[rottingroom]

Even your calculator gave the same answer as Rowbotham's chart.

Yes, because Rowbotham's chart is completely correct for an elevation of 0 FT.

Just because I move the camera up 2' feet doesn't change the drop over 4.4 miles. So the end of the drop is going to 10.9' instead.

It absolutely does change the drop. It isn't a matter of simple subtraction.

Think of it like this, if the drop over a distance was 10', and blocked the entire view of a 10' bldg sitting right at the water. If you climbed a 10' ladder you would be able to see the entire building.

Yes because climbing to 10 ft in elevation would account for the entire height of the building but the problem is that between the elevation of 0 and 10 ft there is an exponential curve.

I have made 2 charts representing the relationship in Rowbotham's chart between height and distance. \

HERE THEY ARE

One of them shows the relationship for elevations 0-10 and the other shows relationship for elevations 0-20. In both examples distance grows exponentially. For what you are saying to hold true... that we simply subtract a number then the growth in distance should not be exponential but linear. This should tell you that there is more involved that just subtracting the height.

I hope this clears things up.

[Hoppy]

You have to think of the total distance, don't divide it up to 2 different distances. If you do divide it, you will get the wrong answer.

Incorrect.  You must divide the total distance into two different distances to get the correct answer.  First there is the distance from the observer to the horizon, then there is the distance from the horizon to the house.

Marrrrkjo. That how to get the wrong answer.
That is why the drop over 3 miles is 6' = 72".              Correct
 Not a drop of 1 mile(8") +(8") + (8") = 24"                Incorrect

[rottingroom]

You have to think of the total distance, don't divide it up to 2 different distances. If you do divide it, you will get the wrong answer.

Incorrect.  You must divide the total distance into two different distances to get the correct answer.  First there is the distance from the observer to the horizon, then there is the distance from the horizon to the house.

Marrrrkjo. That how to get the wrong answer.
That is why the drop over 3 miles is 6' = 72".              Correct
 Not a drop of 1 mile(8") +(8") + (8") = 24"                Incorrect

That isn't what Markjo is saying. When you raise the elevation of the observer. You get a new distance to the horizon. Then you measure the drop from that horizon to the target. You don't just divide things to your liking in the same way as your crude example. You have to try and think about this.

Have a good look at 29silhouette's picture again:



For a distance of 4 miles there should be a drop of 10.7 ft
The top of the image shows the observers horizon (when viewing from an elevation of 6 ft) to be 3 miles.

From this point we can use Rowbotham again because NOW we are at 0 elevation for the remaining mile giving us the 8 inch drop.