0
$\begingroup$

I have a 'Baron' robot frame with 4 static wheels, all driven by a motor. At the moment I'm thinking of handling it like a 2 wheel differential drive. Left and right wheels would receive the same signal. Actually you can interpret it as a tank on caterpillars, exept there is no link between the two tires. Does anyone have a different idea about this?

Ps: The purpose of the robot will be to know it's exact location. I will use a kalman filter (EKF) to do sensor fusion of the odometry and an IMU with accelero, gyro and magnetometer. So in the kalman filter I add the odometry model of a differential drive robot.

$\endgroup$
  • $\begingroup$ What is your question exactly? $\endgroup$ – Chuck Jun 13 '16 at 18:58
0
$\begingroup$

Yes, you can treat a skid-steer or tracked vehicle as a 2 wheel differential drive vehicle. The only downside is that you will not know the exact point of rotation. On a 2 wheel differential drive robot it is right between the wheels, but on your robot it could be anywhere on the centerline of the robot between the axles. The location will depend on the center of mass of the robot and the friction between the wheels and the ground.

I think there is no reason why you can't feed the odometry into a kalman filter. You should give it a fairly high covariance though. Actually, when driving straight the covariance can be low. It is only when turning that it should be large.

$\endgroup$
0
$\begingroup$

I think you should read some of ROS stuff. However, how to generate odometry in this case? I mean in a way that is accurate.

If you switch to skid steering I think it will be better... however, I don't really understand it well. All I know is that it makes different speed values for the 4 motors... so you have 4 motors instead of 2. I don't know how to make it in ROS too.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.