setup electronik

1

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Edited 3 time(s). Last edit at 10/26/2009 12:24AM by SebastienBailard.

What is this a class project?

It should be possible, this type of control more and likely should be implemented off of timer interrupt.

But as a exercise you could do it this way.

1. Save the desired number of steps for each axis
2. Calculate the number of steps per second requried for each axis.
3. Calculate the time delay between steps dividing steps per second into 1000
4. Take the current time from the millis(); function and add in the time delay and save as a next time to step value.
5. in the loop function check if there are steps remaining if so grab the current time.
6. compare it against the next time to step value for the desired axis.
6a. If greater or equal then issued the required step pulse.
6b. Decrement the step count.
6c. add time delay between steps to the next time to step value.

freds Wrote:
-------------------------------------------------------



Thanks for the answer!

Edited 2 time(s). Last edit at 08/26/2009 11:00AM by perpuna.

Thanks for the answer!

Edited 1 time(s). Last edit at 08/26/2009 11:00AM by perpuna.

#define s1_stepPin 53
#define s1_dirPin 51
#define s1_enablePin 49


#define s2_stepPin 47
#define s2_dirPin 45
#define s2_enablePin 43


void setup()
{
Serial.begin(9600);


pinMode(s1_stepPin, OUTPUT);
pinMode(s1_dirPin, OUTPUT);
pinMode(s1_enablePin, OUTPUT);
pinMode(s2_stepPin, OUTPUT);
pinMode(s2_dirPin, OUTPUT);
pinMode(s2_enablePin, OUTPUT);

digitalWrite(s1_dirPin, HIGH);
// digitalWrite(s1_stepPin, LOW);
digitalWrite(s1_enablePin, LOW);
digitalWrite(s2_dirPin, HIGH);
//digitalWrite(s2_stepPin, LOW);
digitalWrite(s2_enablePin, LOW);

}

void loop()
{

long t_end=1000000L;
int s1_steps=10000;
int s2_steps=20000;

int t1 = t_end/s1_steps; //Delay between steps of the first motor
int t2 = t_end/s2_steps; //Delay between steps of the second motor

int c1 = 0; //The counter of idle time of the first motor
int c2 = 0; //The counter of idle time of the second motor
long t_total = 0; //Time from the operation beginning

while (t_total++ < t_end) {
if (++c1 == t1) {
digitalWrite(s1_stepPin, HIGH);
c1 = 0;
}
if (++c2 == t2) {
digitalWrite(s2_stepPin, HIGH);
c2 = 0;
}
delayMicroseconds(1);
digitalWrite(s1_stepPin, LOW);
digitalWrite(s2_stepPin, LOW);
delayMicroseconds(100);
}
digitalWrite(s1_enablePin, HIGH);
digitalWrite(s2_enablePin, HIGH);
}

Edited 4 time(s). Last edit at 08/26/2009 10:24AM by perpuna.

It doesn't work because your loop only has delays when a motors steps.

This should work because it always delays 2us round the loop. It might be a bit inaccurate depending how long digitalWrite() takes. Making the second delay bigger would make it more regular but have less speed resolution.

void loop() 
{ 

	int t_end=1000000; 
	int s1_steps=10000; 
	int s2_steps=20000; 

	int t1 = t_end/s1_steps; //Delay between steps of the first motor 
	int t2 = t_end/s2_steps; //Delay between steps of the second motor 

	int c1 = 0; //The counter of idle time of the first motor 
	int c2 = 0; //The counter of idle time of the first motor 
	long t_total = 0; //Time from the operation beginning 

	while (t_total++ < t_end) { 
		if (++c1 == t1) { 
			digitalWrite(s1_stepPin, HIGH); 
			c1 = 0;
		}
		if (++c2 == t2) {
			digitalWrite(s2_stepPin, HIGH);
			c2 = 0;
		} 
		delayMicroseconds(1); 
		digitalWrite(s1_stepPin, LOW); 
		digitalWrite(s2_stepPin, LOW); 
		delayMicroseconds(1); 
	} 
}

---

[www.hydraraptor.blogspot.com]

#define s1_stepPin 53
#define s1_dirPin 51
#define s1_enablePin 49


#define s2_stepPin 47
#define s2_dirPin 45
#define s2_enablePin 43


void setup()
{
Serial.begin(9600);


pinMode(s1_stepPin, OUTPUT);
pinMode(s1_dirPin, OUTPUT);
pinMode(s1_enablePin, OUTPUT);
pinMode(s2_stepPin, OUTPUT);
pinMode(s2_dirPin, OUTPUT);
pinMode(s2_enablePin, OUTPUT);

digitalWrite(s1_dirPin, HIGH);
// digitalWrite(s1_stepPin, LOW);
digitalWrite(s1_enablePin, LOW);
digitalWrite(s2_dirPin, HIGH);
//digitalWrite(s2_stepPin, LOW);
digitalWrite(s2_enablePin, LOW);

}

void loop()
{

long t_end=1000000L;
int s1_steps=10000;
int s2_steps=20000;

int t1 = t_end/s1_steps;
int t2 = t_end/s2_steps;

int c1 = 0;
int c2 = 0; r
long t_total = 0;

while (t_total++ < t_end) {
if (++c1 == t1) {
digitalWrite(s1_stepPin, HIGH);
c1 = 0;
}
if (++c2 == t2) {
digitalWrite(s2_stepPin, HIGH);
c2 = 0;
}
delayMicroseconds(1);
digitalWrite(s1_stepPin, LOW);
digitalWrite(s2_stepPin, LOW);
delayMicroseconds(100);
}
digitalWrite(s1_enablePin, HIGH);
digitalWrite(s2_enablePin, HIGH);
}

Edited 1 time(s). Last edit at 08/26/2009 11:01AM by perpuna.