At what resolution ( in terms of cm per pixel) Phantom 3 Professional or Advanced flying at 300' AGL can capture aerial photography ? The speed should be as fast as possible to get best resolution for large area real estate photography. I would like 15 miles per hour. Is this speed possible for real estate photography?
Resolution of phantom 3 at 300' AGL?
- Thread starter zameer
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Simple trigonometry shows that for a camera with a field of view θ at an altitude, h, the width, w, of the image on the ground is going to be given by:
w = 2h.tan(θ/2) .
If the camera sensor has n pixels in that dimension and we make the assumption of an undistorted image then the average resulting spatial resolution across the object plane, r, will be given simply by:
r = 2h.tan(θ/2)/n .
The field of view of the P3 camera is 94˚ and at its highest resolution, the P3 images are 4096 pixels in width, and so the resolution of the image at 300 ft (91 m) will be:
2 x 91 x tan(47) / 4096 = 0.048 m per pixel = 4.8 cm per pixel.
w = 2h.tan(θ/2) .
If the camera sensor has n pixels in that dimension and we make the assumption of an undistorted image then the average resulting spatial resolution across the object plane, r, will be given simply by:
r = 2h.tan(θ/2)/n .
The field of view of the P3 camera is 94˚ and at its highest resolution, the P3 images are 4096 pixels in width, and so the resolution of the image at 300 ft (91 m) will be:
2 x 91 x tan(47) / 4096 = 0.048 m per pixel = 4.8 cm per pixel.
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Nice work sar104, now make a radio out of a coconut and get us off this island.
Apologies, then, but it really is as simple as it gets. High School mathematics applied to everyday problems - a much under-appreciated tool.
Nice work sar104. My chart below indicates:
Distance 300ft = 91.44m
Aspect ratio 16:9
->Object's max X 170.93m
->Object's max Y 96.15m
UHD: 4096 × 2160
-> X 17093cm / 4096pixels = 4.17cm/pixel
-> Y 9615cm / 2160pixels = 4.45cm/pixel
I guess the slight difference in our results is explained by rounding errors and by diagonal vs horizontal/vertical dimensions.
Distance 300ft = 91.44m
Aspect ratio 16:9
->Object's max X 170.93m
->Object's max Y 96.15m
UHD: 4096 × 2160
-> X 17093cm / 4096pixels = 4.17cm/pixel
-> Y 9615cm / 2160pixels = 4.45cm/pixel
I guess the slight difference in our results is explained by rounding errors and by diagonal vs horizontal/vertical dimensions.
Thanks. But I'm not sure where you get the x dimension of 170.93 m. It should be 196 m. And the pixels are square, so the resolution is the same on the x and y axes.
At 91.44m the diagonal dimension is 196.12m (AFAIK the 94° FOV is diagonal), so x 170.93m, and y 96.15m.
...yes it disturbed me that the resolution with square pixels should be the same for x and y...
...yes it disturbed me that the resolution with square pixels should be the same for x and y...
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Good point - the FOV obviously has to be diagonal. So my calculation was incorrect - in the x-direction the width of the field of view is less as you noted. But, checking the specifications, the 94˚ FOV on the camera actually refers to the full effective sensor size, which is 4000 x 3000, so you need to use those values to translate diagonal to x or y.
That gives an image width of 156.9 m and an image height of 117.7 m. Thus the resolution is 156.9/4000, or 117.7/3000, which are both equal to 3.92 cm per pixel.
Revising the equation accordingly, for this particular 4:3 sensor (thus a 3:4:5 triangle for the diagonal), gives, for x and y:
r = 8h.tan(θ/2)/5n
AFAIK:
Camera sensor Sony EXMOR 1/2.3” (diagonal 7.81mm, 6.16 x 4.62mm)
Effective pixels: 4072 × 3046 = 12.40 M (total pixels: 4168 × 3062 = 12.76 M, active pixels: 4024 × 3036 = 12.22 M)
Lens FOV 94°, focal length 3.61mm (or 20mm @ 35mm format equivalent) f/2.8
ISO Range 100-3200 (video) 100-1600 (photo)
Electronic Shutter Speed 8s - 1/8000s
Image Max. Size 4000 × 3000
Video Recording Modes:
UHD: 4096 × 2160p 24/25, 3840 × 2160p24/25/30
FHD: 1920 × 1080p 24/25/30/48/50/60
HD: 1280 × 720p 24/25/30/48/50/60
Max. Bitrate Of Video Storage 60 Mbps
...so the image sensor resolution is somewhat vague. The OP wasn't quite clear if s/he was doing images or video and in which settings ... OP wanted 15 mph which is about 50% of P3P max speed but didn't mention lighting, ISO, shutter speed which contribute to image blur...
Those are the quoted Sony specs - I was using the DJI numbers which, presumably, are what they base their optical specs on: 4000 x 3000 as the effective area used for the 12 MP images. Everything else is a crop, and the 4096 pixel dimension on the UHD may be slightly interpolated, or perhaps DJI just round 4096 x 3046 down to 4000 x 3000. If the latter is true then I am slightly underestimating the resolution.
So the simple calculation above indicates the maximum linear resolution based on the optics and sensor size. Other recording modes are effectively either simple crops such as the 16:9 UHD (same resolution) or downsampled, such as FHD and HD (lower resolution).
Nice! Thanks!
I like this thread. Always trying to figure out fov at my given height when I plan a flight on Litchi hub.
Not to highjack the thread, I want to do a true super wide image by planning in Litchi to go y height- stay there. Then move x number of feet take an image, move x number of feet on the same line take an image Etc. Then stitch in Photoshop. I'll try to figure out the x distance between shots w your formula...
Sent from my iPad using PhantomPilots mobile app
I like this thread. Always trying to figure out fov at my given height when I plan a flight on Litchi hub.
Not to highjack the thread, I want to do a true super wide image by planning in Litchi to go y height- stay there. Then move x number of feet take an image, move x number of feet on the same line take an image Etc. Then stitch in Photoshop. I'll try to figure out the x distance between shots w your formula...
Sent from my iPad using PhantomPilots mobile app
Thanks, that 4000x3000 full effective sensor size explains it.
So, for example, at 100 m distance the 4000x3000 resolution is 4.29 cm/pixel.
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Yes - that's what I get. Or, alternatively, if it is really 4096 x 3046, then it would be 4.19. Close enough not to matter.
I originally ran these calculations for a similar problem - gridline separation for grid searching with the flight AGL determined by required resolution to detect appropriate clues or subjects.
Once you figure out the altitude required, you can simply use the x-direction FOV reduced by whatever amount of overlap that you want. I use at least 10% overlap, so that gives you a separation of 0.9 x 8h.tan(θ/2)/5 where θ = 94˚, which evaluates to approximately 1.5h.
Thanks. ...I usually prefer terse to-the-point-figure-it-out comments but you really beat me with the formula r = 8h.tan(θ/2)/5n. I gave it to my high-school son to digest ;-)
That was just the original r = 2h.tan(θ/2)/n modified to get the width from the diagonal length of a 3:4:5 triangle - hence just multiplied by 4/5.
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