# Any flickr members here?

by fluidistic
Tags: flickr, members
 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). $\tan \alpha(t) = \frac{d_{\parallel}+vt}{d_{\perp}}\\ \frac{\mathrm{d}\alpha(t)}{\mathrm{d}t} = \frac{1}{1+(\frac{d_\parallel+vt}{d_{\perp}})^2} \cdot \frac{v}{d_\perp}$ For the front, $d_\parallel=2\mathrm{m}$. For the rear $d_\parallel=6\mathrm{m}$, 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.