2D Side Scrolling Shooting Towards Mouse

[Silinter]

So, after 2-3 weeks of being on and around the site, I finally came up with a question that I can't seem to find an answer to.
I'm making my final project for my grade 10 Introduction to Programming (TIK20) class. I am, however, ahead of the class by far.
Now, I'm here to ask for somebody to write me a program that will gove me 100% and ace the course
No, My real purpose here is to take over the world! And, while I'm there, I'd wanted to create a character that can shoot. The only problem with that is that I can't for the life of me figure out how to make him shoot towards the mouse. I've tried to check out submitted work to see if any of them had the source for something similar, but only 3 or 4 of them showed up and they were compiled
While I'm on the topic of shooting, how would you figure out the angle of where the mouse is and display it similar to WireHang (http://games.flabber.nl/wire.hang/)?
Collision detection would be no problem, other then the collision detection of sprites, which I haven't really looked into. Althought if anyone can provide me with pointers to some good thread explaining this I would appreciate it.

Also, hello.

[ericfourfour]

Think of it this way: the character wants to shoot in a straigt line between his point (or the gun's) and the mouse's. In otherwords, you're dealing with a line segment (x1, y1, x2, y2). You can get your x and y components from that to make a vector. Then, the only thing the bullet has to do is follow the vector.

[Silinter]

And how would I go about doing that now?

[Sean]

If your shooting a bullet, you want it to track your mouse x and y variables.

Do to this, either increase or decrease until the bullet equals the mouse.

[A.J]

You use 'Mouse.Where', which returns the x value of the mouse and the y value of the mouse (and a 'button' which lets you know if it is being pressed).
here is an example I made for you (sry if it doesn't help. It took me 1 min to do it, so it might be bad.)

This program tells you the angle you form (with the mouse) and (0,0):

code:

View.Set ("offscreenonly") setscreen ("graphics") type coords :     record         x, y, x2, y2 : int     end record var x, y, button : int var points : flexible array 1 .. 0 of coords var count := 0 loop     loop         Mouse.Where (x, y, button)         var z := sqrt (x ** 2 + y ** 2)         Draw.FillOval (x, y, 5, 5, yellow)         Text.Locate (1, 1)         if button = 0 then             put x : 4, "  ", y : 4, "  ", arccosd ((y ** 2 - x ** 2 - z ** 2) / (-1 * (2 * z * x))), " degrees."             Draw.Line (0, 0, x, y, blue)             Draw.Line (x, 0, x, y, blue)         else             Draw.FillOval (x, y, 5, 5, blue)             put x : 4, "  ", y : 4, "  ", arccosd ((y ** 2 - x ** 2 - z ** 2) / (-1 * (2 * z * x))), " degrees."             Draw.Line (0, 0, x, y, blue)             Draw.Line (x, 0, x, y, blue)             new points, upper (points) + 2             points (upper (points) - 1).x := 0             points (upper (points) - 1).y := 0             points (upper (points) - 1).x2 := x             points (upper (points) - 1).y2 := y             points (upper (points)).x := x             points (upper (points)).y := 0             points (upper (points)).x2 := x             points (upper (points)).y2 := y             exit         end if         View.Update         delay (0)         Draw.Line (0, 0, x, y, white)         Draw.Line (x, 0, x, y, white)         Draw.FillOval (x, y, 5, 5, white)         for i : 1 .. upper (points)             Draw.Line (points (i).x, points (i).y, points (i).x2, points (i).y2, blue)         end for     end loop     delay (0) end loop

[CodeMonkey2000]

Your program crashes whe you hit 90 degrees, and I don't think it's very accurate.

[Bored]

Since your in grade 10 you propably don't have enough trig knowledge to develop your own forumula to calculate the angle between two points so I have provided you with one here:

Turing:

fcn GETANGLE (x1, y1, x2, y2 : real) : real     /*Code by Bored - Febuary 17, 2008     Returns the angle between two points (x1,y1) and (x2,y2)     0 Degrees is oriented to the right of the screen, 90 is upawrd*/     var opp := y2 - y1     var adj := x2 - x1     if adj = 0 then         result 0     elsif opp >= 0 and adj > 0 then         result arctand (opp/adj)     elsif adj < 0 then         result 180 + arctand (opp/adj)     elsif opp < 0 and adj > 0 then         result 360 + arctand (opp/adj)     end if end GETANGLE

This will return the angle with 0 degrees being to the right and rotating counter clockwise making 90 degrees up. This is the general way in which angles are expressed in two dimension because it is based of of standard position.

[A.J]

CodeMonkey2000 @ Sun Feb 17, 2008 9:26 pm wrote:

Your program crashes whe you hit 90 degrees, and I don't think it's very accurate.

It obviously should crash when it hits 90 degrees, since it is impossible for two 90 degree angles to coexist in one triangle.
Maybe you didn't know what angle it measures.

it measures the angle formed at the bottom left side of the screen.
(So the x value of the mouse if one side, the y value is the other, and the hypotenuse is sqrt(x**2-y**2))

[CodeMonkey2000]

I know what it measures. Your program only works when you are n the 1st quadrant though. Anything else, its wrong.

[Sean]

The only reason why it works in the 1st quadrant, is because Turing is only in the first quadrant.

[Mackie]

Vilament @ Mon Feb 18, 2008 8:38 am wrote:

The only reason why it works in the 1st quadrant, is because Turing is only in the first quadrant.

Turing has all the quadrants, it only shows the first.

[CodeMonkey2000]

No, where the quadrants are vary with your reference point. There are always four, even if they are off screen. I can use the middle of the screen as my reference point and all four quadrants will be on screen.

[McKenzie]

You do not need any knowledge of Trigonometric relationships (sin, cos, tan) to solve this problem. Go back to what eric44, and Vilament said. You do want to see the velocity of your shot as a vector. Suppose your shot starts at 100,100 and it is going to spot 500,400. Then this would make sense:

code:

shotX := 100 shotY := 100 shotVx := 4 shot Vy := 3

Where Vx, Vy is the velocity of the shot. If you add Vx, Vy to your X, Y each time around a loop then your X,Y will progress from 100,100 to 500,400.

The key question is how do we compute Vx, Vy, given x1,y1 x2,y2? Use similar triangles. Put your points on a graph and draw a triangle where (x1,y1) -> (x2,y2) is the hypotenuse. Now draw a smaller triangle starting at (x1,y1) with the hypotenuse equal to how far you want the shot to travel every update. In my example I used 5. All I need to do now is use ratios to find how long the other two sides of your triangle is, these will be the x,y components of your velocity vector. In my example I said smallX is to bigX as 5 is to dist(x1,y1,x2,y2), or ... x/400 == 5/500.

[Silinter]

McKenzie @ Mon Feb 18, 2008 1:04 pm wrote:

Suppose your shot starts at 100,100 and it is going to spot 500,400. Then this would make sense:

code:

shotX := 100 shotY := 100 shotVx := 4 shot Vy := 3

Where Vx, Vy is the velocity of the shot. If you add Vx, Vy to your X, Y each time around a loop then your X,Y will progress from 100,100 to 500,400.

Wow. You weren't lying. It actually does get there Which is pretty damn good considering that's what I'd want to do and that's a pretty big development right there.

code:

setscreen ("graphics") View.Set ("offscreenonly") var shotX, shotY, shotVx, shotVy : int shotX := 100 shotY := 100 shotVx := 4 shotVy := 3 loop     drawline (100, 100, shotX, shotY, black)     shotX += shotVx     shotY += shotVy     put shotX, "x ", shotY, "y"     exit when shotX = 500 and shotY = 400     View.Update     delay (50)     cls end loop

Quote:

The key question is how do we compute Vx, Vy, given x1,y1 x2,y2? Use similar triangles. Put your points on a graph and draw a triangle where (x1,y1) -> (x2,y2) is the hypotenuse. Now draw a smaller triangle starting at (x1,y1) with the hypotenuse equal to how far you want the shot to travel every update. In my example I used 5.

Now, when you say draw a smaller triangle starting at (x1,y1) with the hypotenuse equal to how far you want the shot to travel, do you mean the length of the hypotenuse being 5 or the hypotenuse end points being (x1,y1) and (4,3). Or would they be the same thing...

Quote:

All I need to do now is use ratios to find how long the other two sides of your triangle is, these will be the x,y components of your velocity vector. In my example I said smallX is to bigX as 5 is to dist(x1,y1,x2,y2), or ... x/400 == 5/500.

I'd already finished gr.10 math, in which we do learn briefly about sine, cosine and tangents. However, we didn't start on anything like 'vectors'. What are they? They sound pretty damn important but I'm not very familiar with them... at all. Anybody know a good source that they can link to or explain to me briefly on what they are?
Thx for all the help!

[McKenzie]

Vectors are measurements in a direction. 10m is a distance, 10m at a 30 degree angle is a vector. You can always break down a vector into it's horizontal and vertical components. So if I have vector 10m from the origin at an angle of elevation of 30 degrees, I could draw a triangle with 30 as the angle and find the x and y values with trig, like:
sin(30) = y/10
y = 10 * sin(30)
y = 5

The thing is your mouse position is already the x,y components of a vector (and getting the vector from some point to the mouse is straight forward as well). All you need is smaller vector that is proportionate. Consider this simple code:

code:

var mx,my,mb:int loop     mousewhere(mx,my,mb)     cls     for i: 1..10         var x := round(mx/10 * i)         var y := round(my/10 * i)         drawfilloval(x,y,3,3,red)     end for     delay(20) end loop