Thread: My new project

Hi, my new project is a turret that locks on the target...
The target is in a 2D space and has 3 coordinates, x and y...
How to calculate the angle of which turret has to move?

Is it something like this:

Code:

...
angle = 90;
scanf("%d", &x);
scanf("%d", &y);
for(i=0;i<x;i++){
         angle = angle - (angle/2);
         }
for(i=0;i<y;i++){
         angle = angle + (angle/2);
         }
...

I know this isn't the whole thing because what would i do if angle was 0 at the start?

This page will help sort it out:
Mathwords: Equation of a Line

slope = (y1-y2)/(x1-x2)

When you know the location of two points along the (imaginary) line (like you and the targets y and x points).

A line can have 0 slope, that's no problem. Vertical lines are no good for slope of a line though, since you can't divide by zero. Their slope isn't zero, it's just "no slope".

Once you find the slope of the line, from you to your target, I believe you're set. If the target is moving, and you have to "lead" it a bit, in order to hit the target, then the math increases.

so...ma position is 0, 0, and targets 2, 1...so

slope = (0-0)/(2-1)... the result is 0!?

This is my code...it doesn't work, i think i need a few "ifs" to see if x<y, x>y and stuff...

Code:

#include <stdio.h>

struct pos_s{
    int x;
    int y;
    };

float angle;
struct pos_s pos;
int i;
int main(){
    angle = 45;
    scanf("%d", &pos.x);
    scanf("%d", &pos.y);
    printf("processes:\n");
    for(i=0;i<pos.x;i++){
        angle = angle - (angle/2);
        printf("%f\n", angle);
        }
    for(i=0;i<pos.y;i++){
        angle = angle + (angle/2);
        printf("%f\n", angle);
        }

    printf("final angle:");
    printf("%f", angle);
    return 0;
    }

Thank you, sounds very good, but what is "pi"?

I did this in my code, it calculates degrees OK but only if starting position is 40 degrees and if number is negative( - ).

Code:

#include <stdio.h>
struct pos_s{
    int x;
    int y;
    };

float angle;
struct pos_s pos;
int i;
int main(){
    angle = 40;
    scanf("%d", &pos.x);
    scanf("%d", &pos.y);
    printf("processes:\n");
    if(pos.x > pos.y){
        angle = angle + ((pos.x - pos.y) * 5);
       }
    else{
        angle = angle - ((pos.x - pos.y) * 5);
       }
    printf("final angle:");
    printf("%f", angle);
    return 0;
    }

Is pi = 3,14159265?

Pi is a mathematical constant. 3.14159.....

Just follow what Epy said, and you shouldn't have any trouble getting the angle to your target.

EDIT: Check this page out for more on PI:

Pi - Wikipedia, the free encyclopedia

Oh well, my dad is an phi and pi freak, always talks about it...i know what pi is, just didn't knew it was what Epy said...but thanks!

And what if (y1-y2)/(x1-x2) = (0 - 1)/(0 + 2) = -0.5, that will be negative degrees...or?

OK... I did it, it works well, but only if both coordinates are positive... How can i fix this?

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define pi 3.14159265

struct pos_s{
    int x;
    int y;
    };

float angle;
struct pos_s pos;
int i;
int main(){
    angle = 40;
    scanf("%d", &pos.x);
    scanf("%d", &pos.y);
    printf("processes:\n");
    angle = ((atan((0 - pos.y) / (0 - pos.x)))*(180/pi));
    printf("final angle:");
    printf("%f", angle);
    return 0;
    }

The piece of code that is red is the angle calculation...

OK, i fixed these things...

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define pi 3.14159265

struct pos_s{
    int x;
    int y;
    };
int main(){
    float angle;
    struct pos_s pos;
    int i;
    scanf("%d", &pos.x);
    scanf("%d", &pos.y);
    printf("processes:\n");
    angle = ((atan((0 - pos.y) / (0 - pos.x)))*(180/pi));
    printf("final angle:%f", angle);
    return 0;
    }

Whoops... It still can't do it if x is smaller than y, and if x or y is negative ...
How to fix this?

assume x1, y1 are your coordinates, and x2,y2 are the coordinates of the target

atan((y2-y1)/(x2/x1)) gives you the angle, the signs of x2-x1 and y2-y1 give you the quadrant.

as nonoob noted, atan2 may be better, unless you check for a zero in x2-x1, in which case the angle is 90 or 270