## What impact does a rolling shutter camera have on passby measurements?

When passing-by objects are measured, the camera's RGB sensor reads color information for each single pixel. Cameras with rolling shutters read line by line while global-shutter cameras read this information for all lines at once.

Fig. 1 clearly shows the offset of the ICE train's door caused by the rolling shutter. In fig. 2, however, the door is shown without any offset as expected.

An overview of the geometric offset values for differently sized objects and varying moving speeds is provided in table 1. The values describe the offset of an object's lower edge to its upper edge and are rounded to full cm.

If the moving object covers about 50 % of the total image height, the offset becomes significant at a speed of 50 km/h. At frequencies <500 Hz, the impact of the offset on the acoustic map can be less significant. Therefore, we recommend that you use global-shutter cameras when measuring fast-moving objects.

Table 1: Geometrical offset caused by rolling shutter
Object height in pixels (% of image) /
Speed
108 (10 %)216 (20 %)324 (30 %)432 (40 %)540 (50 %)648 (60 %)756 (70 %)864 (80 %)972 (90 %)1080 (100 %)
50 km/h5 cm9 cm14 cm19 cm23 cm28 cm32 cm37 cm42 cm46 cm
100 km/h9 cm19 cm28 cm37 cm46 cm56 cm65 cm74 cm83 cm93 cm
150 km/h14 cm28 cm42 cm56 cm69 cm83 cm97 cm111 cm125 cm139 cm
200 km/h19 cm37 cm56 cm74 cm93 cm111 cm130 cm148 cm167 cm185 cm
250 km/h23 cm46 cm69 cm93 cm116 cm139 cm162 cm185 cm208 cm231 cm
300 km/h28 cm56 cm83 cm111 cm139 cm167 cm194 cm222 cm250 cm278 cm

#### Example calculation

We use the following formula to determine the offset.

Let's s be the horizontal distance between upper and lower edge of the object, v the object's velocity, tline the sensor's reading speed per line and h the object's height in lines. Then

$$s = v \cdot t_{line} \cdot h.$$

A speed of 100 km/h, an object height of 540 lines and a reading speed of 1/32400 s per line, result in an offset of

$$s= 100 \text{ } \frac{\text{km}}{\text{h}} \cdot \frac{\text{1}}{\text{32400}} \text{ } \frac{\text{s}}{\text{line}} \cdot 540 \text{ } \text{lines} = 27,\overline{7} \text{ } \frac{\text{m}}{\text{s}} \cdot \frac{1}{32400} \text{ } \text{s} \cdot 540 \equiv 42 \text{ } \text{cm}.$$

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