|Aerial photography: Part 1||- Aerial Photo 1|
We have taken the opportunity on recent flights, such as on this flight on Alitalia AZ201 from London to Rome, to test the capabilities of the Canon G9 digital camera in the aerial photography area.
The maximum speed = V MAX in m/s that would cause a blurring of one pixel is given by:
The nominal size of a CCD imager in a digital SLR camera is 36 mm wide by 24 mm high. You can calculate the size of the imager L CCD in your camera by working out the viewing angle a of your camera for a set lens focal length L F from:
The 12 MPixel Canon Powershot G9, OFF in upper left, and at max optical zoom on upper right. See Payload I for more details on this camera. Total weight is 320 grams.
This camera was used on a recent flight from London to Rome, with the Alitalia Airbus A320-100/200 clocked by a handheld GPS unit travelling at 848 kph. In each pair of photographs shown next, the first is the JPEG image with the shutter speed set at 1/1,600 second, and the telephoto zoom lens set at the maximum zoom magnification. The second image is the magnification of a portion shown inside a white outline box of the first, after which Adobe Photoshop was used to apply Auto Level and Auto Contrast to improve on the magnified images.
We used a Leica GeoSystems Disto A5 laser rangefinder, with a specified error of less than 1.5 mm, to locate the camera at a position 2,109 mm from the tape measure shown above. With the lens of the Canon G9 camera set at maximum zoom, corresponding to a focal length of 210 mm, we have taken the above photograph, which as been cropped vertically but not horizontally. The field of view at 2,109 mm is seen to be 191 mm, which corresponds to an angle of 10.350 degrees. We calculate the width of the imaging CCD to be 38 mm from the next expression:
We use several overlapping photographs and a knowledge of the aircraft speed to calculate the height of the aircraft above ground level.
The above photograph was time stamped by the digital camera at 10:40:59 am.
The above photograph was time stamped four seconds later at 10:41:03 am, during which time the aircraft had travelled 953.3 m. The aircraft is travelling east from London to Rome at 858 kph, as measured on a hand held Garmin GPSmap 60CSx GPS unit. In fact, we have used four sequential photographs taken over a total of 3 + 4 + 4 = 11 seconds in our calculations. In 11 seconds the aircraft covers 2,621.7 m.
Because the camera is further away from the houses and buildings at the top of the photograph, and closer to those at the bottom of the photograph, we find that the distance covered, namely 2,621.7 m, corresponds to a shorter length at the top of the photograph relative to the corresponding length at the bottom of the photograph.
Because the camera is not pointing vertically downwards, we measure a difference in the relative location of objects at the bottom of the sequential images relative to those in the middle and to those at the top of the sequential images. Over the 11 seconds covered by the four images, the relative locations of identifiable objects is displaced as follows:
The situation we have is as shown in plan view (upper diagram) and side view (lower diagram) above. It is as if we were projecting a rectangular box from the observer onto a screen that had a slope to it. From this diagram, you can see that we are fitting more length into the top of the photograph relative to the length being fitted along the bottom of the photograph. Consequently, as we physically move on, we find that the shapes at the top of the photograph move less from frame to frame relative to those objects at the bottom of the photographs. One can perform some simple trigonometry to work out the following relationships.
The vertical field of view in degrees, measured from camera calibration, is F V , where F V = A2 - A1
where in this case:
FV = 7.895 degrees (measured for the Canon G9 camera at maximum zoom)
W1 = 157 mm
W2 = 175 mm
from which we calculate that A1 = 34.21 degrees
At the bottom of the photograph, the measured displacement vector is 157 mm and this corresponds to a distance covered in 11 seconds of L1 = 2,621.7 m. The width of the photograph on A4 size paper was 159 mm, so the width of the photograph corresponded to L1 = 2,621.7 * 159 / 157 = 2,655 m. Using some more trigonometry, we derive the relationship:
where in this case:
A1 = 34.21 degrees
the horizontal field of view = FH = 10.35 degrees for the Canon G9 camera at maximum zoom
L1 = 2,655 m
from which we calculate the aircraft height above ground to be = h = 11,374 m = 37,317 feet.
The take home message is that from optic flow analysis , knowing the aircraft speed and the field of view of the camera lens one can calculate both the height above ground and the pointing angle of the camera from two overlapping photographs as long as the time difference between the two photographs is also known.
For the mid-size Canon G9 camera:
we calculate an aircraft speed that would cause the image to be blurred by one pixel to be 816 m/s = 2,939 kph.
Image 1 at a nominal 37,000 feet Above Ground Level (AGL)
Image 2 at a nominal 37,000 feet AGL
Image 3 at a nominal 30,000 feet AGL
Image 4 at a nominal 30,000 feet AGL
Image 5: coming in to land: still several thousand feet AGL
Image 6: coming in to land: a few thousand feet AGL
Image 7: coming in to land: a few thousand feet AGL
Spot the four people we believe can be seen in the above magnified image.
Conclusion so far
These photographs were taken from the 08:35 train from Waterloo Station in London to Weymouth on Thursday 1st November, 2007, using the Canon Powershot G9 digital camera.