Ask Your Question

Using laser_scan_matcher output

asked 2012-02-29 02:04:59 -0500

alfa_80 gravatar image

updated 2012-02-29 21:51:47 -0500

I would like to have a confirmation on my understanding in order to use the output of laser_scan_matcher(lsm) in a correct way. Kindly refer to the provided figure(sorry for providing a mere sketch). For convention, please assume any subscript of 1 refers to the previous data(means x1= previous x value from lsm and so on). Likewise, subscript of 2 refers to the current data. P1 and P2 is the position of the x and y values returned by pose2D of the previous and current values respectively that have been mapped into a cartesian coordinate.

My questions are:

  1. Is that correct the representation of variables in the context of lsm?

  2. Calculation of Q:

    i) Either:

    Q = sqrt[(x2-x1)² + (y2-y1)²] x sin(theta1)  OR 
    Q = sqrt[(x2-x1)² + (y2-y1)²] x sin(theta2) OR
    Q = sqrt[(x2-x1)² + (y2-y1)²] x sin(theta2-theta1)

    ii) OR Q = x2-x1

This sounds like a fundamental math problem, but with respect to the values returned by lsm makes me confused. Perhaps someone who is experienced using the package or preferably the author himself can give some comments regarding this.

In my opinion, the correct one should be Q = sqrt[(x2-x1)² + (y2-y1)²] x sin(theta1), but I don't know if this one is correct or not in this regard.

Thanks in advance.

image description


image description

edit retag flag offensive close merge delete

1 Answer

Sort by » oldest newest most voted

answered 2012-02-29 05:39:29 -0500

  1. Yes, that is a correct interpretation of the output.

  2. The way you have drawn the figure, Q is equal to (y2 - y1), or the change of the y component of the robot frame with respect to the fixed frame

What is Q, in the context of your application?

Also, this is not really related to laser_scan_matcher, but looks more like a general trigonometry question, or at best a question on robot motion models. Maybe this reference would be useful.

edit flag offensive delete link more


@Ivan Dryanovski: I doubt about what I've represented regarding the theta from "lsm" in the drawn image is right. Could it be the case, like I edited in the figure? I mean, the theta is the angle how much it rotates after it translates. The black part of the robot, say, its front.

alfa_80 gravatar image alfa_80  ( 2012-02-29 21:55:51 -0500 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools


Asked: 2012-02-29 02:04:59 -0500

Seen: 303 times

Last updated: Feb 29 '12