Isn't it going to make 0.2m side?

Course Support ROS Basics In 5 Days (Python)
1

Respected Prof, I solved the quiz and got full marks(10) already in part 6 of ROS-Basics. But, as I was going through solution, I got a doubt.

radius=request.side
for i in range(request.repetitions):
rospy.loginfo(“Moving forward…”)
move_circle.linear.x = 0.2
move_circle.angular.z = 0
my_pub.publish(move_circle)
time.sleep(radius)

As it is demanded, the square should be at least 1sqm. But, here the robot velocity is 0.2m/s and it continues only for 1 sec(As the side is 1m). Is’nt it going to move only 0.2m before it rotates?..and hence, the square made will be 0.2sqm

2

Prof @albertoezquerro can give more insight into this.

3

Yeah Professor Alberto @albertoezquerro
Would you please tell me if this is only for understanding purpose. Whether the code is provided just to get the concept Or Is this code actually making a 1sqm square ?
If it is actually making a 1sqm square, then would you please break down the code I mentioned

radius=request.side
for i in range(request.repetitions):
rospy.loginfo(“Moving forward…”)
move_circle.linear.x = 0.2
move_circle.angular.z = 0
my_pub.publish(move_circle)
time.sleep(radius