I have thougt a bit about this...but I don't know how to implement it, if it's possible. I'll use a ball to try explaning what I mean. Let's say this! In 2d a ball can move at 360 different angles. But on the screen it can only move in 8 different angles. ((x+1,y+0)(x+0,y+1)etc.) but then you can kind of represent all 360 angles. Then if you bounce this ball the reflection of the ball is depending on the angle of impact. Ok...this doesn't make any sense I think... But I assume my question is: How do you make a better representation of a bonce then just inverting x or y velocity? If anyone have done this I would be greatful for a formula or someting...but I am in my second year of High School and I have nor read much math and zero physics... And I know this post is kind of woolly and doesn't make so much sense...but please dont flame me about it...just ignore it =) //walle

Hi,

OK - here goes:

Quote:
In 2d a ball can move at 360 different angles.

Actually a ball could move in an infinite number of angles. There are 360 degrees in a circle, but the ball could bounce by 1/2 a degree, for example. As you get more exposed to maths, you'll find that angles can be expressed in many other ways - for example, on a calculator you'll typically find:

degrees (360 units to a circle)
radians (2 * pi units to a circle)
gradians (400 units to a circle)

So - for example, 180 degrees = pi radians = 200 gradians.

Quote:
But on the screen it can only move in 8 different angles.
((x+1,y+0)(x+0,y+1)etc.) but then you can kind of represent all 360 angles.

Why can't the ball move to (x+2, y+0)? You're probably thinking this doesn't make sense, as you're effectively working on a 2-d grid of pixels and it makes sense that the ball should only move one step at a time - but, quick quiz - is the distance travelled from (x, y) to (x+1, y) the same as (x, y) to (x+1, y+1) (hint : pythagorus' theorem).

This is quite a topic you're starting to break into. What I would suggest is starting to research Vectors (in particular, Vectors in 2-dimensions). I'm not quite sure how old you are (we don't use the term High School in the UK) and I'm guessing you're about 11-12 - in that case it might be worth chatting to your maths or physics teacher about some Vector references. Alternatively, try some of these websites:

MathWorld - everything you could even want to know about maths
GameDevs maths and physics articles - check the vectors and then the matrix sections

Some of this may seem a little daunting at times, but bear in mind you've got years to learn ahead of you - I've worked as a mathematician for many years and can thoroughly recommend it (even worked for a swedish company!)

Hope this helps,
Jim.

Quote:Original post by JimPrice

quick quiz - is the distance travelled from (x, y) to (x+1, y) the same as (x, y) to (x+1, y+1) (hint : pythagorus' theorem).

Yes. It has only travelled one pixel. But in the second example it have moved one pixel right and one down. But it's still just one pixel. Right??

And I'm 17. I used american school levels so as many as possible would understand. Or so I thougt...dont you go to high school when you are 16-19? What do you go to in Canada/UK?

//walle

Quote:But it's still just one pixel. Right??

The pythagorian theorem is especially useful for calculating distances in a cartesian coordinate system it states that for a right triangle, the length of the square of the hypotenuse is equal to the sum of the squares of the other two sides. Any diagonal line in cartesian coordinate space implies a right triangle because you can create one by adding two segments, aligned to the x and y axis respectively. in this case if you move the ball +1 in the x direction and +1 in the y direction, you imply a right triangle were the two shortest sides are of length 1 and 1. The hypotenuse of this triangle is the distance that the ball moves.

   ball's eventual position->    X                                /|                               / |                              /  | up one                             /   |                            /    | ball's original position-> X-----+                              over one

There for if the ball moves up one pixel and over one pixel, the actual distance traveled is the SQRT of 2 = 1.41 (approximately)
So the ball moves about 1.41 pixels in that case.

[Edited by - Kelly G on November 30, 2004 6:02:36 AM]

Ok. This is getting a bit over my head =( have to read the tutorials.

I don't understand this:

Quote: ball's eventual position-> X
/|
/ |
/ | up one
/ |
/ |
ball's original position-> X-----+

The math I have lerned is: one (kateter) one of the sides of the triangle (not the hypotenuse) ^2 + the other side(not the hypotenuse) ^2 = hypotenuse

In this case shouldn't the hypotenuse be 2?

1^2+1^2=2

//walle

It's actually:

"one of the sides of the triangle (not the hypotenuse) ^2 + the other side(not the hypotenuse) ^2 = hypotenuse ^2 "

so its SQRT (1^2+1^2);

Morning,

Aargh - just posted a long reply and then managed to delete it - oh well, here goes again.

OK - so KellyG has explained why a move from (x, y) to (x+1, y+1) is a physically longer step than the move from (x, y) to (x+1, y). If you can accept this, then the next thing to accept is that we could step from any pixel to any other, so long as we understand what the step length is. If you think about it in relation to speed this makes sense also - for example, how would you model a ball travelling twice as fast? You'd have it go twice the distance (so from (x, y) to (x+2, y), for example). If you can do this, why can't you take any arbitrary step of length 2?

OK - time for some ASCII art. Here is the first example:
o : empty pixel
b : ball
w : wall



This is the example you originally gave. The ball is approaching the wall at an angle of 45 degrees, the bounce angle is correct, and the ball is travelling at 1.4 units / frame.

Now let's double the ball speed:



Exactly the same example, only now the ball is traveling twice as fast. So now we can accept that the ball could travel at any speed we wish - and can therefore step to any pixel if it's next step that is consistent with the angle it's travelling at, so long as that step has distance equal to it's speed.

So let's now assume that the ball is travelling like a Knight in chess - so from (x, y) to (x+2, y+1), or some rotation thereof. We now have:



Can you see how this travels in a shallower angle, but still accurately represents bouncing?

In practice, we tend to record co-ordinates as real numbers, because in order to calculate angles we have to use trigonometric functions like sin and cos. But, when the image is rasterised to the screen, those co-ordinates are turned into integers so they can be physically represented - but that's a whole 'nother topic.

This is the basis of vectors that I spoke about in my first reply. A vector (as used here) is just a representation of direction of travel. For example, if we know that in one step the ball travels from (x, y) to (x+2, y+1), then the direction of travel is the vector (2, 1). Once you understand vector nomenclature and then start to learn more about them, you'll find they're immensely powerful - and vectors and their bigger brothers, matrices, underpin loads of things that go on in computer programming.

Quote:
And I'm 17. I used american school levels so as many as possible would understand. Or so I thougt...dont you go to high school when you are 16-19? What do you go to in Canada/UK?

No offence meant, I was just trying to get a bead on what your background was likely to be and what environment you were in - I emmigrated to Canada from the UK recently, so don't really have any knowledge of the North American school system, and we have different names for schools in the UK. Same caveat applies - talk to your maths and /or physics teachers about vectors, see what they come through with.

Any more questions, let us know.
Jim.

Edited to add source tags to preserve spacing in diagrams!

[Edited by - JimPrice on November 30, 2004 11:55:25 AM]

JimPrice: Thanks for the long reply!

Kelly G: Thanks for pointing that out...forgot about the last ^2...

Ok. This is really educational, I've seen all the exampels by testing. But it's nice to see the math behind.

I have used two variables to change the balls position: xvel and yvel.
And then changed the speed by increasing them...I'm starting to feel that this is a very easy way to do this. And collisiondetection doesn't work well with higher speeds.

Quote:No offence meant, I was just trying to get a bead on what your background was likely to be and what environment you were in

Non taken. =)


//walle

Ah, collision detection is a whole new topic!

The problem I expect you're hitting is this - using the same bouncing ball example, let's say the ball is travelling at a 45 degree angle to the wall, but is travelling very quickly - so that now, in one frame, it's travelling from (x, y) to (x+4, y+4). Time for some ascii art again:



The problem is the ball is never immediately next to the wall, but is moving so quickly it seems like it passes through the wall.

This is again where vectors come in useful. Both the wall and the ball's direction of travel during the last frame can be represented by 2 different lines (the wall by the 'w's in the picture above, and the ball by a line travelling from top-left to bottom-right corners). Each of these lines can be represented by a vector. We can immediately see that the 2 lines cross, and therefore that there must have been a collision during the balls travel. Using a neat bit of maths, you can also compare the 2 vectors and find out not only if the ball collided with the wall, but also at exactly what co-ordinates.

So - the morale of the story is still the same. Time to learn about Vectors!

All the best,
Jim.