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by fluidistic
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Maui
#19
Feb15-13, 06:42 PM
P: 724
Quote Quote by fluidistic View Post
I'd understand if the back of the car was out of focus while the front was right on the focus, but it doesn't look like it. It looks like the back is not only blurry but also "moving" (same for the head of the passenger). How is that possible?


I would guess that it's due mainly to focus being locked only on the front of the car. Look how blurry the car by the left headlamp is. My camera has a defocus mode and can take similar pictures on purpose:

fluidistic
#20
Feb15-13, 07:41 PM
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Hmm Maui, I don't see the similarity between your photo and M quack's.
I've been thinking about http://en.wikipedia.org/wiki/Rolling_shutter, rolling shutter. Namely the left side of the photo is scanned before the right side. If we add the focus on the front part of the car, I think it makes sense.
M Quack
#21
Feb16-13, 07:46 AM
P: 662
I don't think AF tracking is the culprit.

Assuming that the car moved at 10 km/h, it moved by ~30cm during the 1/10sec exposure time. A DOF calculator (http://www.dofmaster.com/dofjs.html) gives sharpness from 2.15m to 28.5m for 28mm at f/5.6 with focus at 4m. This easily covers the entire car, moving or not.

I don't think that this is a rolling shutter effect either. For that the exposure time is much too long. Also, remember that this is a DSLR with a real, mechanical shutter.

That leave the panning motion.

Draw the car as a box, say 2x4m, and the photographer as a point, say 2m ahead and 2m sideways. Draw a line from the point closest to the photographer, and a line from the rear of the car to the photographer.

Now displace the car forward a bit and redraw the lines. The angle between the two front lines is much larger than that between the two rear lines.

So by tracking the front of the car (rotating by the angle between the two front lines), the rear gets smeared out (by the difference in angle).

[itex]
\tan \alpha(t) = \frac{d_{\parallel}+vt}{d_{\perp}}\\
\frac{\mathrm{d}\alpha(t)}{\mathrm{d}t} = \frac{1}{1+(\frac{d_\parallel+vt}{d_{\perp}})^2} \cdot \frac{v}{d_\perp}
[/itex]

For the front, [itex]d_\parallel=2\mathrm{m}[/itex]. For the rear [itex]d_\parallel=6\mathrm{m}[/itex], so the sweep rate is much larger for the front (v/4m) than for the rear (v/20m). Parallel and perpendicular refer to the motion of the car.

The effect is so noticable because the length of the car is quite large compared to the distance to the photographer.
fluidistic
#22
Feb16-13, 10:58 AM
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I think you got it here M Quack, that makes sense.
Also putting panning motion in google picture shows similar pictures. There's even a photo where the moving object's closest to the camera point is in its middle. So both the rear and front looks like "they are moving". (http://www.dailytravelphotos.com/archive/2009/12/01/)
M Quack
#23
Feb17-13, 04:33 AM
P: 662
I don't think it depends on which point of the object is closest to the camera.

What is important is that different points have different distances. I could have tried to track the rear of the car, then the front would have been smeared (in the opposite direction). Have to try this one day...


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