question about autonomous timing
 

im bit confused about 26.2 ms loop area part

does that mean 1 = 26.2 ms when i set up a counter or something? :confused:

Re: question about autonomous timing
 

It means that the code within that area is executed approx. every 26.2ms, so if you increment a variable it should indicate time in approx. 26.2ms steps.

Re: question about autonomous timing
 

a better way to set up a timer, if you are comfortable with it, is to use the processor interrupts. www.kevin.org/frc has some sample code that can be cut and pasted into your code. I knew nothing about interrupts and timers and such and after a few hours staring blindly at the screen everything clicked. its really sweet how everything works. search chiefdelphi for other threads related to interrupt driven timers

Re: question about autonomous timing
 

Interrupts can be very dangerous and difficult to debug. I wouldn't recommend their use for rookie teams, espically at this late date.

Re: question about autonomous timing
 

i want to know what 1 second equals to in this part
1000?
400?

what is it!?!?!?

Re: question about autonomous timing
 

The processor runs at 10 MHz, 10,000,000 clock cycles per second. The joystick info (etc) is only updated every 10 MHz / (2^18). That's 10,000,000 divided by 2 raised to the 18th power. 2^18 is 262144. So 10,000,000 divided by 262144 is 38.1469... cycles per second. One cycle has a duration of 1 / 38.14... = 0.0262144 second, or 26.2 milliseconds (ms).

So in one second, there are 38 or 39 cycles of the "main loop".

You can pre-calculate certain fixed periods of time (using pocket calculator or the Windows calculator accessory, not the robot controller), converting from seconds to cycles of 38 Hz:

To know when a certain pre-determined amount of time has passed, say, 11.7 seconds, figure out how many counts that would be:

11.7 / .0262144 = 446.3195801

and round to the nearest integer, because the controller works fastest with integers: 446.

A quicker way? Mutliply seconds by 38:

11.7 * 38 = 444.60000

which is 445 when rounded. A little different than 446. An even faster way that you could do in your head:

11.7 * 38 is about: 12 * 40 = 480

Close enough? Think like an engineer and make that decision yourself!

Remember, to count these cycles, declare a "static" variable. "static" means that the variable isn't lost every 26.2 ms cycle, which could happen otherwise.

You might have some code that looks like this:

-Norm

Re: question about autonomous timing
 

oooo ok thx for clearing that up ;)