Re: time-discrete odometry
Yes, you are correct, I messed up the sign there => $$\theta' = -50°$$
Omg, the more complicated the task gets, the more simple stuff I seem to not think of :D
edit:
Ok, I figured out my mistake, was another sign error... But now I'm good. Thank you again!
Hello, I have another question regarding my cognitive robotics class. Here is my current task:
And this is the slide from the lecture about this topic:
Now just plugging in everything I know into the formulas, I get this:
$$\boxed{3 = \sqrt{x'^{2} + y'^{2}}\\
-20 = atan2(y',x')\\
-30 = \theta'...
I asked about this, and this was the reply:
The noise has a shift of 1 (the mean), so if the real reading is 10, then the noisy reading (which will be the most probable) will be 11. Therefore, the difference between the expected reading at a certain x_i location (taking into account the noise...
Since you're asking like that, probably yeah :D.
But I'm not 100% sure how yet...
Can I just use this formula now?
prob(z1|x2) = this
prob(z2|x2) = this
Hey guys, I need some help. Here's my task:
Here's what I came up with: I need to calculate prob(x1|z1), prob(x1|z2) and so on and then just write down which has the highest probability. To do that, I would use Bayes rule, but for that, I need p(z1|x1) and p(z1). How do I get these?
I have...