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how to calculate the odometry for irobot create robot

asked 2013-02-26 14:08:36 -0500

vincent gravatar image

updated 2013-02-27 03:16:52 -0500


I'm trying to understand how the odometry is calculated in irobot_create_2_1 driver. I spent lots of time trying to figure out where the minus sign comes from in y=-sin(th)*d, on the 53th line in I've included the code snippet. I also included the odometry calculation part in from turtlebot stack, which does the same thing but easier for me to understand.

Can anyone give me a hint or let me know where I can find good resource on mobile robot odometry/kinematics in order to understand the whole scenario. Thanks


00046                 d = self.create.d_distance / 1000.
00047                 th = self.create.d_angle*pi/180
00048                 dx = d / elapsed
00049                 dth = th / elapsed
00051                 if (d != 0):
00052                         x = cos(th)*d
00053                         y = -sin(th)*d
00054                         self.x = self.x + (cos(*x - sin(*y)
00055                         self.y = self.y + (sin(*x + cos(*y)
00057                 if (th != 0):
00058                = + th


    # this is really delta_distance, delta_angle
    d  = sensor_state.distance * self.odom_linear_scale_correction #correction factor from calibration
    angle = sensor_state.angle * self.odom_angular_scale_correction #correction factor from calibration

    x = cos(angle) * d
    y = -sin(angle) * d

    last_angle = self._pos2d.theta
    self._pos2d.x += cos(last_angle)*x - sin(last_angle)*y
    self._pos2d.y += sin(last_angle)*x + cos(last_angle)*y
    self._pos2d.theta += angle

I've attached my calculation as below, can someone point out where went wrong? image description

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answered 2013-02-26 14:59:39 -0500

updated 2013-03-02 03:09:03 -0500

I don't know if I can answer your specific question, but this is a good source for learning about how to calculate the odometry:


(EDIT): It appears that the global coordinate frame (self.x, self.y and in has +y to the left. But the coordinate frame for the change in position (x,y, and th in has +y to the right.

It seems unlikely to me that that is what the original author intended. It's probably either a bug that doesn't matter in the end, or maybe an artifact of the odometry that supplies data to

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thank you Jon, i've read through this tutorial but still didn't get the answer for my question.

vincent gravatar image vincent  ( 2013-02-26 15:09:15 -0500 )edit

I think it just comes down to the definition of the axes and the angle. If I've got it right, I think +x is forward, +y is to the right, and +angle is counter-clockwise.

Jon Stephan gravatar image Jon Stephan  ( 2013-02-26 15:48:02 -0500 )edit

ok i see. Is it left-handed coordinate system?

vincent gravatar image vincent  ( 2013-02-27 02:00:30 -0500 )edit

based on my calculation, it came down to this: self._pos2d.x += cos(last_angle)x + sin(last_angle)y; self._pos2d.y += - sin(last_angle)x + cos(last_angle)y. (the rest is the same with the examples). still different signs... not sure if you got the same result...

vincent gravatar image vincent  ( 2013-02-27 02:19:57 -0500 )edit

i've edited my question with my calculation attached. could you please have a look and let me know if anything went wrong? thx.

vincent gravatar image vincent  ( 2013-02-27 02:59:16 -0500 )edit

Wow, I thought I understood this, but I guess not. It looks to me like the coordinate frame of x and y (the change of x and y) has +y to the right, while self.x and self.y has +y to the left. I also confirmed on my robot that +y (the global +y) is to the left.

Jon Stephan gravatar image Jon Stephan  ( 2013-02-27 15:39:36 -0500 )edit

I also confirmed that rotating to the right gives a positive y, and a negative self.y.

Jon Stephan gravatar image Jon Stephan  ( 2013-02-27 15:54:56 -0500 )edit

Great! that explains. Is it like a convention to define these two coordinate frame differently or is it just specific to the create robot?

vincent gravatar image vincent  ( 2013-02-28 04:34:42 -0500 )edit

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Asked: 2013-02-26 14:08:36 -0500

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Last updated: Mar 02 '13