paper: joystick sensitivity (gain) adjustment

[AustinSchuh]

Originally Posted by Joe Ross

By default, RobotDrive squares the inputs before it outputs the PWM, so you should consider the affects of that on any input adjustments.

If you are referring to the data that I provide, that was generated using the victor.Set function, rather than passing it through the RobotDrive.

[Roger]

Originally Posted by Ether

What kind of drivetrain are you using, and what type of driver interface?

It's our 2009 robot with skid steer/tank drive; see this photo. We're using LabVIEW and the standard KOP joysticks and driver station. A single joystick does the driving. The robot drive has the drill-type gear shifter (1st and 2nd gear only; 3rd gear was too fast for Lunacy's floor.)

Adding the second axis to the formula won't be a problem. It was more of "let's get one working first, then add the second." But now that you wrote it out, it makes sense that full power is always possible. We just ran out of time to do the changes.

This was a perfect project as a warm-up, with just enough learning joystick interfacing, sub-vi creation, and math programming, to get done in an hour then debug and change and test after. Adding the different modes and deciding which is "best" will be fun.

There are also two different drivers that we have to program for, and not for the same time. The first (and obvious) is the high school competition driver, who has practice driving with the robot. The second, more unpredictable, driver is the off-season "carnival" driver, a wanna-drive-the robot seven-year-old that throws the joystick all over the place. The former is easy, the latter we can only slow down the action so the robot doesn't crash too much.

[jhebbel]

Sorry to revive a more than dead thread but its really the only relevant thread I could find. Im intrigued by the b + (1-b)*[a*x^3 + (1-a)*x] equation, but also confused, doesnt this need to be formated to solve for x and not y? Has anybody done this yet?

[jhebbel]

For anybody else that lands here from Google like I did wondering the similar, The reason I though the equation needed to be formatted to solve for X is I didn't realize this was for inverse deadzone, to calculate for traditional deadzone it is y=G*((x-D)/(1-D))^3+(1-G)*(x-D)/(1-D); Where X is in value, D is Deadzone and G is Gain.