Author Topic: Chad pictures  (Read 42926 times)

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Offline elevenaugust

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Chad pictures
« on: May 24, 2008, 04:06:21 am »
Here, we (10538 and I) will post all the pictures and the infos needed for the calculations, regarding the Chad location.
« Last Edit: May 24, 2008, 08:03:22 am by 10538 »
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Offline Nemo492

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Re: Chad pictures
« Reply #1 on: May 24, 2008, 02:43:59 pm »
example :

http://ovnis-usa.com
The only motivation for the DRT is finding the truth.

Offline Nemo492

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Chad location 2
« Reply #2 on: May 24, 2008, 03:17:43 pm »

http://ovnis-usa.com
The only motivation for the DRT is finding the truth.

Offline Nemo492

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Chad location 4
« Reply #3 on: May 24, 2008, 03:22:01 pm »

http://ovnis-usa.com
The only motivation for the DRT is finding the truth.

Offline elevenaugust

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Re: Chad pictures
« Reply #4 on: May 24, 2008, 07:39:30 pm »
I will try today to take measurements for each photos with the laser range.
I will also upload all the photos in the ftp sever, maybe easier for you.

My first impression about the size of Chad's drone, looking to the pictures 4 and 5, is that this thing is really big!!
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Offline elevenaugust

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Re: Chad pictures
« Reply #5 on: May 25, 2008, 01:38:02 am »


All measurements are on meters
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Offline elevenaugust

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Re: Chad pictures
« Reply #6 on: May 25, 2008, 01:49:14 am »
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Offline elevenaugust

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Re: Chad pictures
« Reply #7 on: May 25, 2008, 01:52:48 am »
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Offline nekitamo

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Re: Chad pictures
« Reply #8 on: May 25, 2008, 10:51:31 pm »
My first impression about the size of Chad's drone, looking to the pictures 4 and 5, is that this thing is really big!!

This thing really seems enormous! I've noticed that two of your distance measurements in image #4 are almost exactly in-line with one of the drone's central axis and the drone seems to be very near the canopy, so I did some quick (and rough) calculations just to get overall impression of its size. I aligned Chad's image with one of yours (IMG_0943) and by using Field-of-View angle info from image's exif data I extrapolated the dimensions of two overlapping (due to the perspective) square image planes, actually separated by 14 meters vertex per your measurements. Then I calculated the size of the diagonal line from point A to B (22m in 3D space) and used it to judge the size of the central ring (perspective foreshortening ignored) - and here's the result:



Again, this is a very rough estimate - but 4 meters?! Are those trees really that big?
Must be something wrong with FoV angles... does this seem right to you?

Offline elevenaugust

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Re: Chad pictures
« Reply #9 on: May 26, 2008, 04:48:33 pm »
Hi Neki, just landing off the plane  :P and read all your great calculations.
Kudos for that!
I just want to add that, yes, that was my first impression when we made, Numbers and I, the distance estimation with the laser range, that this thing was:
1. VERY close to the canopy
2. VERY big

However, two things to consider:
1. When we went to the spot firstly, we wrongly interpreted Chad's position and placed it close to point n°5, and when we returned here, we both agree to place it further, to match the little branches on the left side and the general aspect of the vegetation on the right side, therefore it could be more useful and more accurate to use the photos
DSCF011
DSCF012 and
DSCF013




2. I wonder if Chad used the zoom.... And how it can affect the calculations?
« Last Edit: May 26, 2008, 06:21:45 pm by Nemo492 »
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Offline nekitamo

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Re: Chad pictures
« Reply #10 on: May 28, 2008, 10:43:52 am »
I've been checking and refining my calculations for the past few days and I'm going to repeat the previous example explaining every step of the way, but this time trying to be more accurate and using one of the new images suggested by elevenaugust. It would be great if someone tries to follow and verify my reasoning here, as I believe this method to be quite useful as an universal tool for "manual" 3D image analysis.

Here are the basic ingredients:
- unknown image for analysis, with unknown parameters, but shot at known location
- calibrated image from the same location, as similar as possible to the previous one
- physical measurements from the location (laser measured distances)

The calibrated image is simply an image with known parameters from which we can deduce the precise angle of view. For this we can either use EXIF metadata (if present) or calculate it by using an object of known size and at known distance in the image. As final precision of this calculations depends mainly on precision of this image, it should be as good as possible (high resolution, undistorted, high contrast, low noise...), though it doesn't have to be perfect if its flaws are known and corrigible - i.e. if it was shot with the angle-of-view-calibrated camera with known barrel distortion (this calibration can also be done later). Here's one of the suggested calibration images with basic info and illustration of the angle of view concept:



It was resized to 50% of its original size which reduces precision (btw, can I get hold of the original image?), though unlike clipping it doesn't affect the angle of view. As EXIF data survived, I extracted it with Phil Harvey's EXIFtool which automatically calculates horizontal FoV (48.9°) if the required input data fields (focal length and 35mm scaling factor) are present. The other way to do it would be to ask elevenaugust how far from the camera he was standing and what is the format of images he's holding. Assuming it was plain A4 paper (0.3m longer size), by measuring it in PS or similar program with a ruler tool (75px) we can deduce the pixel size for the image distance plane (another important concept here) containing the paper (75/0.3=250px/m), and also the width of the whole image (Xres=800px) at the specified distance plane, so 800/250=3.2m. This seems about right, if you can imagine elevenaugust's full height and rotate him horizontally to measure the width of the image. If we enter this value along with the previously determined HFoV (48.9°) into the on-line angular size calculator, it returns the distance of 3.5m from the camera.

The next step is to overlay the image with an unknown object (Chad's image #4 per elevenaugust's numbering notation) onto the known setup of our calibration image and align it by using reference points present in both images in order to translate objects dimensions into known and measurable environment - this is the basic postulate on which my calculations depend, and if it's not correct I'm sorry for wasting your time. But as I believe that was also the reason why you tried to match the original images, we are probably safe in using this assumption :)

As you can see in the following image, I used rotation, perspective correction, resizing and whatever else was necessary to achieve as good alignment as possible:



I believe this also answers elevenaugust's question about zooming - as long as unknown image can be precisely aligned with the calibrated one, it doesn't matter what parameters were used for its recording because by translation into our known setup we are effectively replacing its parameters with known values. Naturally, it would be better if calibration image already matches the unknown without adjustments, but that is not the case with either of images from the location. Even if they were perfectly aligned from the exactly same spot, there's wind, vegetation growth, etc...

But here's another thing, specific for this case: the alignment of images is not even necessary! It will serve as an illustration, but for actual measurement I will use the original image with a line drawn between two selected reference points (marked A and B) identified in both images. I've entered  laser distance measurements for these reference points and marked the square that I'm going to use in the next part of my calculations, especially selected because, as you can see, its imaginary diagonal line is passing very close to craft's horizontal axis along the smaller side fins, and it looks as if the craft is very close to the tree canopy - so, hopefully, by calculating the size of this diagonal line we could use it to measure the actual dimensions of the craft.

To be continued...

Offline elevenaugust

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Re: Chad pictures
« Reply #11 on: May 28, 2008, 11:24:04 am »
Hi Neki,

Amazing work! :)
To answer your questions:
1- These photos were taken from 10538's camera, I'll ask him to post the original here without resizing
2- Yes, the paper I was holding is an A4 format
3- You're absolutly right about my distance from the camera, 3.5 meters is a very good estimation.
« Last Edit: June 05, 2008, 08:04:07 pm by 10538 »
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Offline spf33

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Re: Chad pictures
« Reply #12 on: May 28, 2008, 01:37:53 pm »
really nice work, nekitamo.
while i haven't done 3d studies with the chad image\location yet, right now in 3ds max i can corroborate the estimated FOV of ~48º(max calculates 48.217º), so you are definitely on the right track.


Offline 10538

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Re: Chad pictures
« Reply #13 on: May 28, 2008, 03:46:47 pm »
Hi Neki,

Amazing work! :)
To answer your questions:
1- These photos were taken from 10538's camera, I'll ask him to post the original here without resizing
2- Yes, the paper I was holding is an A4 format
3- You're absolutly right about my distance from the camera, 3.5 meters is a very good estimation.

Here it is:   http://home.comcast.net/~dl1027/files/object/08052313.JPG
« Last Edit: June 05, 2008, 08:04:33 pm by 10538 »

Offline nekitamo

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Re: Chad pictures
« Reply #14 on: May 29, 2008, 03:36:46 pm »
Here it is:   http://home.comcast.net/~dl1027/files/object/08052313.JPG

Thanks, I will use this one for my further calculations. I assume that laser measurements were also done from that exact same spot? This is also an important detail I've neglected to emphasize in my previous post, but I can compensate for offset if you can provide the details.


while i haven't done 3d studies with the chad image\location yet, right now in 3ds max i can corroborate the estimated FOV of ~48º(max calculates 48.217º), so you are definitely on the right track.


Well, I've just spent hours trying to figure out the right way to precisely calculate FoV from EXIF data. Various tools give different results, but which one is right? Here's what I learned, maybe you'll find it useful, too.

Wikipedia states that for standard rectilinear lenses the FOV is calculated by using this formula:

FOV = 2 arctan ( image area size / (2*focal length))

So if you want horizontal, vertical or diagonal FoV, you simply use horizontal, vertical or diagonal image area size, respectively. With digital cameras the image area is actually CCD sensor size, and Fuji's specs for 10538's camera (FinePix2800ZOOM) list its CCD sensor as 1/2.7-inch with 2.1 million pixels. By 1/2.7" they are referring to an old (since 50's) TV camera tube standard size, with the following dimensions:
Code: [Select]
Type Aspect Dia. (mm) Diagonal Width Height
1/2.7" 4:3 9.407 6.721 5.371 4.035

By using 5.37mm sensor width and 6mm focal length from the above image, I get the same value of HFoV like you did in 3DS: 48.217°. However, it seems that CCD manufacturers don't really use those standard sizes, and I've learned that the actual size of CCD sensor used in 10538's camera is 5.27 x 3.96 mm. Calculation for this value with 6mm focal length returns the following HFoV: 47.419°.

But I've also learned that neither of this values is actually correct, as not all of CCD's 2.1 million pixels are used in the resulting image - 10538's camera uses only 1.9 million effective pixels, so the effective sensor size used for calculation should be even smaller. But then where did EXIFtool get its, as I'm aware now - completely wrong value of 48.9 degrees? I've analyzed its source code and I believe I've identified the problem - wrong value is calculated only for non-35mm cameras with aspect ratio of 4:3 - and I'm going to write to the author to correct this. However, after studying the source code now I know how to do it manually. Here's what you need for manual FoV calculation (actual data from 10538's image):

Focal Plane X Resolution        : 3053
Focal Plane Y Resolution        : 3053
Focal Plane Resolution Unit     : cm
Image Width                     : 1600 px
Image Height                    : 1200 px
Focal Length                    : 6.0 mm

The first three values hide the key to the true effective size of the imaging sensor. The same value for both X and Y plane focal resolution is normal for square pixels, so we can use either value to calculate the size of a single pixel. With indicated centimeter as resolution unit, this is done by dividing 10 with 3053 = 0,0032 millimeters/pixel. If it were inches, we would divide 25.4 with the focal resolution value for the same result.

The rest is easy:
0.0032 x 1600 = 5.24mm (effective horizontal image area size)
0.0032 x 1200 = 3.93mm (effective vertical image area size)
Horizontal FoV = 47.179°
Diagonal FoV = 57,254°
Vertical FoV = 36,267°

This should finally be the correct values, but I'll still double check them with my camera (also Fuji) to see how much precision can actually be achieved this way.
« Last Edit: June 05, 2008, 08:06:17 pm by 10538 »