Adjusting Coordinates Given Angle / Window Size in JFrame

Question

So this may be a fairly simple question, however I've been writing this code for over 8 hours so my brain is slightly pooped. Basically, I have a frame that looks like this: (https://i.stack.imgur.com/YkbD1.jpg)

If you'll excuse my crude drawing, I was wondering about the maths. Given the angle theta, what I want to do is, given a specific Point object, that will be the bullet I shoot, how do I adjust the x/y coordinates in parallel to move along that given angle? There is a window size of 700x500 inside the JFrame. I'll post the code in a pastebin if it is quite necessary but I felt that this question was purely arithmetic (sin/cos/tan) related.

Basically I have this:

        if(pinballAngle > 90){
            pinballCoordinate.x+=(pinballAngle/3);
            pinballCoordinate.y-=20;
        }
        else if(pinballAngle < 90){
            pinballCoordinate.x+=(pinballAngle/3);
            pinballCoordinate.y-=20;
        }
        else{
            pinballCoordinate.y-=20;
        }

So pinballCoordinate is the Point object that I am trying to adjust given the angle of theta, that is the 'gun' that shoots the pinball. The 'pinballAngle' variable is the angle of the gun.

So, the question could boil down to a linear path following problem. That is, you have a start point, you have an angle and you want to follow a path away from the start point. Here I'm falling back to calculating a point on a circle to generate my target point and then I just need to follow a straight line between those points :P - It's probably a little convoluted, but I fall back on things I know how to solve when presented with issues like this ;) — MadProgrammer, Mar 21 '17 at 23:35
See the example cited [here](http://stackoverflow.com/a/4550667/230513). — trashgod, Mar 22 '17 at 08:12

score 0 · Answer 1 · answered Mar 21 '17 at 23:42

0

first, you will need your angle in radian

int speed = 20; // the speed of the bullet

double rad = pinballAngle * 2 * Math.PI / 360;
pinballCoordinate.x += (int) ( Math.cos(rad)*speed );
pinballCoordinate.y += (int) ( Math.sin(rad)*speed );

You will probably need to use the opposite in x or y, replace with some -= where it is needed.

the Math documentation: https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html

answered Mar 21 '17 at 23:42

User9123

396
3
8

`-=` is not needed, as sine and cosine already produce the correct positive/negative values. – GameGibu Mar 21 '17 at 23:51
You'll probably need `-=` for `y` but `+=` for `x`. – Dawood ibn Kareem Mar 21 '17 at 23:58
@DavidWallace that's true in the case that they are correcting for the inverted y-axis used by most layout/drawing utilities. The geometries are, in these cases, merely flipped vertically, such that rotations are represented in the opposite direction. – GameGibu Mar 22 '17 at 00:06
@GameGibu Yes, and it also depends on what pinballAngle means in the OP source code, aka if 0° is the left or right direction (in the picture, it's the left direction, so `-=` for x) – User9123 Mar 22 '17 at 00:23
How do you determine the vertical (y) or horizontal (x) distance when all the parameters (sides) of triangle are unknown? – hashmap Mar 22 '17 at 02:10
@NimitPatel what do you mean ? obviously you can't determine a unique triangle when no side lenght is fixed. Here, the hypotenuse is arbitrarily set to 20 – User9123 Mar 22 '17 at 02:16

Adjusting Coordinates Given Angle / Window Size in JFrame

1 Answers1