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Grand Challenge

background 
Kick Off Part 1 
Kick Off Part 2 
Preparations Part 1 
Preparations Part 2 
Preparations Part 3 
imaging 
aerial photo 1 
train photo 2 
aerial photo 3 
motion deblur 
Progress 1 
Progress 2 
 

Aerial photography: Part 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.

 

 

 

 

 


 

Aerial Imaging Technology

The maximum speed = VMAX in m/s that would cause a blurring of one pixel is given by:

h

altitude (height)

m

p

pixels in direction of travel

pixels

t

shutter time (often as a fraction of a second)

seconds

LCCD

dimension of the CCD in the direction of travel

mm

LF

focal length of the lens

mm

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 LCCD in your camera by working out the viewing angle a of your camera for a set lens focal length LF from:

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The mid-size Canon Powershot G9 camera

 

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.

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Camera calibration

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:

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Image optic flow analysis

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:

  • at the top of the image, the total displacement vector for  objects is 157 mm
  • at the middle  of the image, the total displacement vector for  objects is 169  mm
  • at the bottom  of the image, the total displacement vector for  objects is 175 mm

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 FV, where FV = 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:

  • in a plane at 37,000  feet (= 11,278 m)

  • having a CCD imager supporting 4,000  pixels in the direction of travel

  • with a shutter speed of 1/1,600 second

  • having a CCD imager that is calculated to be 38 mm wide in the direction of travel

  • with the  zoom telephoto lens set at a maximum focal length of 210 mm

we calculate an  aircraft speed that would cause the image to be blurred  by one pixel to be  816 m/s =  2,939 kph.

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The images

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

  • I was hoping for better images from the 12 MPixel, mid-sized, Canon G9 digital camera.
  • The results so far do not have sufficient clarity to enable us to detect a person at a window.
  • The importance of high shutter speeds is clear.
  • We might be losing resolution by using a lossy, compressed, JPEG image format, but that's all you can get from the G9.
  • You need to sit by the window in front of the wing to avoid image distortion caused by the hot exhaust from the turbine engines of the aircraft.

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Photographs from a moving train

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.

  • The Canon G9 is certainly struggling with taking photographs from moving platforms.
  • Reflections from the window are quite pronounced in some of the photographs.
  • No surprises that it tends to be the closer objects that are more blurred.

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