How to determine the maximum speed(ensuring no collisions) of an omnidirectional robot with a ring of sequentially firing 'n'number of ultrasonic sonar sensors? Given the frequency 'f' of each sensor, the max acceleration 'a'.
The robot is i a world filled with sonar-detectable fixed (nonmoving) obstacles that can only be detected at 'x' meters and closer
Is this the maximum velocity that can be attained by the robot as in the 'ring cycle time'?
The way I approached it,
Consider: F=70kHz, a=0.5m/s2, x=5m, n=8.
Now considering the obstacle is at the furthest detectable distance, 5m
Considering the speed of sound as 300 m/s.
Time taken for the sensor to receive the signal = 2* ((1/70*103)/(300/70*103))*5 seconds. =1/30 seconds
Now as there are 8 sensors fired sequentially, total time taken= 8/30 seconds.
Therefore, ring cycle time=8/30 seconds.
Also overall update frequency of 1 sensor = 30/8 Hz.
Now is the maximum velocity , the velocity attained by the robot for t= 8/30 seconds?
Hi Marcus,
Well yes it doesn't but again the velocity is dependent on the obstacles in the robot's environment. And to detect those obstacles the sensor's frequency comes into play. I mean how quickly the sensors ll detect the obstacles in their path.
(2*5m)/300m/s = 33msec for a soundwave travelling 5m towards and backwards to the transducer (puls-echo mode)
A complete measurement over 8 transducer for max. distance of 5m needs 267msec.
[So the puls repetition rate for each transducer will be 3,75]
Your robot is able to detect changes in distances of obstacles every 8/30s. Within this time at the given acceleration of a=0,5m/s^2 a maximum velocity of 0,133m/s (0,48km/h) will be reached.
But, if your robot is able to travel for say 10s at constant acceleration without collision the speed at the end will be 10s*0,5m/s^2 = 5m/s = 18km/h. Now at each ring cycle time of 267msec a distance of 1,33m will be covered.
At least you have to give a better definition of your maximum speed.
If it would be okay to avoid collisions in the range of one detection for 5m range then v = 5m / 0,267s = 18,75m/s. That means a speed of 67,5 km/h. Perhaps too fast for collision avoidance.
By the way: bandwidth and risetime is only interesting if you want to calculate time-of-flight. In this example the speed of the robot seems to be in the focus of interest.
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