How to Show the Horizon is Below Eye Level, Using Actual Eyes

Bravo, sir. One question:

Did you draw lines along all the mortar lines and see where they meet up, or did you find a point where a few of the lines meet and than draw lines from that to the points at the right?
I first adjusted the vertical pitch of the 'grid' to fit the bricks on the right side where the dots are. Then I moved the X and Y coordinates of the vanishing point to get the best fit, as the grid 'morphed'. It is more convincing when you are moving it around. The vertical position is the most important/sensitive, but there you have a good long row of bricks.
 
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Rory

Senior Member
I first adjusted the vertical pitch of the 'grid' to fit the bricks on the right side where the dots are. Then I moved the X and Y coordinates of the vanishing point to get the best fit, as the grid 'morphed'. It is more convincing when you are moving it around. The vertical position is the most important/sensitive, but there you have a good long row of bricks.
I guess that's an okay way to do it, if we assume that the bricks are all perfectly evenly spaced, and that the thickness of the mortar between each row is also consistent.

I think the more honest way to do it, though, is to do it visually, drawing lines across the tops of the rows, or through the centre of the mortar, and seeing where they meet. It's unlikely that they'll all meet at the same exact point - the ones in the WTC pictures above don't - but they should give a good idea of where eye level is, and it safeguards against ideas of manipulation, and fiddling the results. After all, it only takes a tiny adjustment to have the angles change substantially.

I think this is why you really need: a) a high res image, so you can ensure the line you draw is as accurately aligned as possible; and b) a clear shot of the bricks (or whatever it happens to be) so you can see where it is you're drawing.

Though I do think, going by that last photo you posted, if the image was high res enough, the fence would be less of an issue, and you'd be able to make out the courses clearly enough.

Good for you for persevering with this. :)
 
I guess that's an okay way to do it, if we assume that the bricks are all perfectly evenly spaced, and that the thickness of the mortar between each row is also consistent.

I think the more honest way to do it, though, is to do it visually, drawing lines across the tops of the rows, or through the centre of the mortar, and seeing where they meet. It's unlikely that they'll all meet at the same exact point - the ones in the WTC pictures above don't - but they should give a good idea of where eye level is, and it safeguards against ideas of manipulation, and fiddling the results. After all, it only takes a tiny adjustment to have the angles change substantially.

I think this is why you really need: a) a high res image, so you can ensure the line you draw is as accurately aligned as possible; and b) a clear shot of the bricks (or whatever it happens to be) so you can see where it is you're drawing.

Though I do think, going by that last photo you posted, if the image was high res enough, the fence would be less of an issue, and you'd be able to make out the courses clearly enough.

Good for you for persevering with this. :)
I might have another go. It makes is more exciting doing it in dribs and drabs, when you don't know what the final answer will be, and one might make a fool of oneself. :)

I make it 0.5 degrees in my head, but I will check properly. No I don't! 0.306 degrees.
 
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Mick West

Administrator
Staff member
I make it 0.5 degrees in my head, but I will check properly. No I don't! 0.306 degrees.
My calculator says:
https://www.metabunk.org/curve/?d=6.437377991570253&h=114&r=6371&u=m&a=a&fd=60&fp=3264
 
My calculator says:
https://www.metabunk.org/curve/?d=6.437377991570253&h=114&r=6371&u=m&a=a&fd=60&fp=3264
Thanks, I have it bookmarked, but was hoping to get the same answer twice before looking. I thought it was roughly in the right ballpark.
My paper OS map says alt. = 122 m but it is not so critical when you are up a bit. I'll get the computer to spit out the answer, after finding some antique A4. and a pencil.
 
Though I do think, going by that last photo you posted, if the image was high res enough, the fence would be less of an issue, and you'd be able to make out the courses clearly enough.
Go for it!
And here is the view NW from the spot as well: DSCF4761.JPGDSCF4762.JPG DSCF4762- x20 vert zoom.jpg

ps. I took the car, not the dog, but it is a rough track, and I risked getting locked behind a gate..
 
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What's your equation for working out the dip to the horizon?
dip ( radians) = dip in brick units* brick height pitch/ distance to brick.*
You can work out the camera to brick distance from knowing the length of the wall in bricks*brick horizontal pitch and using similar triangles.

In this case I did it iteratively by tweaking the distance to the near brick corner until the near and far bricks gave the same answer. I can be sure that the angular dip is constant for bricks at both corners. I did not allow for the angle of the wall, as the cosine would be nearly one for this shallow angle.

From the size and shape of the triangles you can see that the far bricks appear to be about half the height of the near bricks, so in this case they must be twice as far away, which means the camera is another wall length away from the near corner.
 
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Rory

Senior Member
dip ( radians) = dip in brick units* brick height pitch/ distance to brick.*
You can work out the camera to brick distance from knowing the length of the wall in bricks*brick horizontal pitch and using similar triangles.
I'm finding that a little difficult to follow. Not sure what 'brick height/horizontal pitch' means in this context. I think if I knew the distance to the wall, though, I could work out the dip to the horizon. Can you explain how to 'use similar triangles'?

(Though bearing in mind this may never be that accurate - but does do enough to show what's needed: that the horizon is below eye level.)
 
Not sure what 'brick height/horizontal pitch' means in this context
(dip in brick units)* (brick height pitch) means the height (ft.) between the horizon and the vanishing point horizontal lines, at the particular corner. Where brick vertical pitch is 0.25 (ft.) and horizontal pitch is 0.75 (ft.)

See also the edited version above.
 
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I think it reduces to:
Actual distance from 3 to camera =( Actual length of wall from 2 to 3) x A/B
Someone should check. It is a bit counterintuitive that you can use a point that represents infinity and the far end of a wall to tell you how far from the near end you are!
I may have got my A, B and A+B combobulated 50% of the time when doing it my head..
crazy clouds' brick2 pic.jpg
 
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Rory

Senior Member
The way I thought I could do it was:
  • draw a vertical line on a brick on the near corner at eye level
  • on the same course, find the point where the brick height is half the apparent size
  • using your brick dimensions figure out how far that point is from the near corner
  • that is therefore how far the near corner is from the camera
  • using that, create a right-angled triangle formed by the camera; the bottom of the near wall; and the point on the near wall at eye level
  • using your brick dimensions, work out the height of the wall at eye level
  • calculate the angle to the bottom of the wall
  • using your brick dimensions, calculate the number of degrees per pixel (or vice versa)
  • using that, work out the viewing angle down to the horizon
I gave it a quick try, and quickly gave up. I think it became pretty much immediately apparent that measuring the bricks just isn't accurate enough.

I think the best thing is just to be happy that it very clearly shows that the horizon is below eye level. :)
 
The way I thought I could do it was:
  • draw a vertical line on a brick on the near corner at eye level
  • on the same course, find the point where the brick height is half the apparent size
  • using your brick dimensions figure out how far that point is from the near corner
  • that is therefore how far the near corner is from the camera
  • using that, create a right-angled triangle formed by the camera; the bottom of the near wall; and the point on the near wall at eye level
  • using your brick dimensions, work out the height of the wall at eye level
  • calculate the angle to the bottom of the wall
  • using your brick dimensions, calculate the number of degrees per pixel (or vice versa)
  • using that, work out the viewing angle down to the horizon
I gave it a quick try, and quickly gave up. I think it became pretty much immediately apparent that measuring the bricks just isn't accurate enough.

I think the best thing is just to be happy that it very clearly shows that the horizon is below eye level. :)
On average buildings are level, but you cannot trust any particular one to be accurately level enough, considering the small dip to be measured.

Fitting the grid means I am using floting point virtual bricks for the measurements. :)
 

Rory

Senior Member
What y'all think of these photos, with regard to using parallel lines to establish the vanishing point, and therefore eye level?

eye level infinity pool.jpg

upload_2019-1-19_9-19-53.png

(Hint: the sides of the pool and the markings on the runway all converge at a point some way above the horizon.)
 
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Mendel

Active Member
What y'all think of these photos, with regard to using parallel lines to establish the vanishing point, and therefore eye level?
They're appealing, but don't rise to scientific standards. There is no proof that the sides of the pool are parallel.
We don't know the airport, if we did, we could get the data for the elevation of either end of the runway to determine the slope of the runway [see below], and regulations (plus experience) would probably make a strong case for the runway markings to be parallel. Not everything in aviation is parallel (the approach lighting's "vanishing point" is the touchdown area), but runway markings are.

If you have pictures showing the airport from both directions [see next post], and the vanishing point of the runway is above the horizon in both cases, that would constitute fairly good proof (similar to looking through a fixed tube with a crosshair at both ends and finding the horizon lower than that both ways) -- you'd have to argue camera distortion to raise doubt, for an image with no visible distortion.

Edit 1: the airport is MLE / VRMM on the Maledives, the original picture (I think) is at http://www.airliners.net/photo/-/-/451959
Edit 4: The image shows the airport from a southerly direction. It was taken on September 17, 2003.

Edit 2: the aeraodrome chart for VRMM is at http://www.aviainfo.gov.mv/downloads/aip/aerodrome/ad_2_vrmm.pdf . "VRMM AD 2.12 Runway physical characteristics" shows the southern end (THReshold) to be at 1.62m and the northern end at 1.73m. Runway midpoint is at 1.62m (AD 2.8).
Edit 3: The runway is 45m wide and 3200m long including the concrete aprons at the end, the asphalt is 2960m long.
 
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Mendel

Active Member
View of Maledives airport (VRMM) from the North, the vanishing point is exactly on the wing edge. The seawall visible on the left is not distorted, it is curved in reality.

Malé_Airport_(8684876813).jpg
Source: https://commons.wikimedia.org/wiki/File:Malé_Airport_(8684876813).jpg
 
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