That explains it. Technically any single point or pixel in the photo is exposed for the specified shutter speed time. However, the time that it takes to expose the overall photo (or even just the length of the bullet streak) is greater than the specified shutter speed time. Since the shutter “roll” apparently moves horizontally when the camera is in landscape orientation, I imagine that if Mills had had the camera turned 90-degrees (portrait orientation) that the shutter roll would have been vertical and a horizontal bullet streak would have been closer to the correct length for that 1/8000 of a second shutter speed setting. Thanks Jerry!
The camera used was a Sony ILCE-1 camera which used a
vertical travel mechanical shutter:
But that is not a problem since the camera was turned so that the vertical travel of the shutter curtain was in the same direction as the bullet travel.
While the
exposure time is effectively 1/8000th of a second, the time it takes for the shutter opening to travel from top to bottom of the image sensor (shutter speed), appears to be 1/400th or 1/500th of a second (depending on whether it was 35 mm full frame or APS-C). This means that the space between shutter curtains as it passes over the sensor is 1/20th or 1/16th of the vertical image captured. So each photo is essentially a series of exposures from top to bottom of 1/8000th second snippets. The time between the beginning of exposure of the top snippet and the end of exposure of the bottom one would be 1/400th of a second or 1/500th of a second.
If the camera used a shutter speed of 1/400th of a second (2.5 ms) and an exposure time of 1/8000th of a second, the time between one end of the streak and the other would be the width of the streak divided by the frame width x 2.5 ms. So in this case, I measure the streak to be 21.4% of the total width of the image. So the total exposure time is:
Exposure time for streak = .214 x 2.5 = .535 milliseconds.
If the camera used a shutter speed of 1/500th of a second (2 ms), the exposure time for the streak would be .214 x 2 = .428 ms.
Bullet speedIf the muzzle speed of the bullet was 3300 feet per second, by the time it travelled 450 feet it would be down to around 2700 fps. So in 1 ms it would travel 2.7 feet. In .428 ms it would travel 1.16 feet or about 14 inches. In .535 seconds it would travel about 17 inches.
So it looks to me like that is the real bullet streak is consistent with being captured by a shutter speed of 1/400th or 1/500th of a second with an exposure time of 1/8000th of a second (shutter curtain window of 1/16th or1/20th of the sensor width travelling across the image sensor in the direction of the bullet travel (ie direction of bullet travel in the image).