Need an algorithm for positionning an object and shoot

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Need an algorithm for positionning an object and shoot

(9 posts) (3 voices)

Started 2 years ago by Lapa
Latest reply from dbolli

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Lapa
Member

Hi everybody,

I'm new in iPhone's developments but not in developments.
Actually I have a two problems :
- Positionning an object on the screen
- Shoot in one direction.

The player can move himself on the screen with his finger, so basically he will drag his avatar on the screen with his finger. That is pretty easy to do with cocos2d.
I need to position a cross next to the avatar by following this basic rules :
- when the player go left, the cross will be at his right
- when the player go up, the cross will be at his bottom
- when the player go right, the cross will be at his left
- when the player go down, the cross will be at his top
- the cross need to be at a constant distance from the player

I can create a code for this basci code, but I want to be able to manage multiple movement like :
- go up and left in the same time, the cross will be at the bottom right of the player and at an unique distance.
That is he hard part of my problem, I try many algorithm an nothing want to works.

The only thing who works "properly" is the easing technics with the player position, but if you move to fast the distance between the player and the cross will not be constant and some time you can have the player on the top left of the screen and the cross on the bottom right of the screen...

My second problem is related with the first one.
I want my player to shoot a ball in the direction giving by his position and the cross position !
So basically the player will shoot in the exact opposite direction than his movement.
I was able to get the vector between the player and the cross, after that I can say that I want a movement between the player position and X times my vector (x in a constant) in Y seconds. Thats works.

But I would like a variation of that, I want to be able to shoot 3 balls one in the opposite direction, one other with -5° and the other one +5°.

I have no idea how I can do that !

Could you help me for that ?

PS : I'm sorry for my english, it's not my main language ;)

Posted 2 years ago #
onedayitwillmake
Member

Hi there, this might be helpful to you.
I wrote a little tutorial, on how to throw an object - the principle is really very similar except you'll be ignoring the previous touch and using your characters location for the bullet.

http://www.learningiphone.com/?p=71

All are welcome to correct me on my code, as I'm also just learning myself.

http://onedayitwillmake.com
http://learningiphone.com

Posted 2 years ago #
Lapa
Member

Thanks for this answer I should be able to throw my ball with a variation of my first angle.

Now I still need to figure out how to resolve my first problem...

Posted 2 years ago #
dbolli
Member
For the issue with synching the cross sprite position with the avatar sprite, I suggest you do this in your step: function.

Add the sprites to your layer with a tag (e.g. kMyAvatarSpriteTag defined in an enum ), make sure you save your last move direction in a global struct to allow multiple move directions and then reference your sprites in step: and keep synched e.g.
```
Sprite *myAvatarSprite = (Sprite*)getChildByTag:kMyAvatarSpriteTag;
Sprite *myCrossSprite = (Sprite*)getChildByTag:kMyCrossSpriteTag;

if ( gLastMove.up && gLastMove.right ) {
  myCrossSprite.position.x = myAvatarSprite.position.x - kOffsetXDist;
  myCrossSprite.position.y = myAvatarSprite.position.y - kOffsetYDist;
}
else ...
```
For the extra balls at +5 and -5 degrees there will be some trigonometry needed
Let's say player position is x0,y0 and the end point of the first ball vector
is down and to the left of this at x1,y1
Imagine a right angled triangle between the start position of the first ball vector x0,y0 the end position x1,y1 and the third point of the triangle at x0,y1.
From this we can calculate the radius of a circle whose centre is x0,y0 and where x1,y1 is a point on the circumference using Pythagoras' Law h^2 = a^2 + o^2 so
h = sqrt( ( abs(x0 - x1) ^ 2 ) + ( abs(y0 - y1) ^ 2 ) )
and using the cosine rule to find the angle A (i.e. theta) at x1,y1
cos A = (b² + c² - a²)/2bc
A = acos( ( h^2 + abs( x1 - x0 ) - abs( y1 - y0 ) ) / ( 2 * h * abs( y1 - y0 ) )
then x2,y2 i.e. point on circle 5 deg more than x1,y1 is h * cos(A+5),h* sin(A+5)
and x3,y3 i.e. point on circle 5 deg less than x1,y1 is h * cos(A-5),h * sin(A-5)
there may be some conversion between rads and degs involved :)

hth
Derek.
Posted 2 years ago #
Lapa
Member

Thx dbolli, but for the first problem I was looking for something more general, to be abale to manage very situation. Indeed between the move left and the move up, the move left and up is only one of the infinite possibility.
So I have :
- the last player's position
- the new player's position

We can draw a circle around the new location and draw a line from the new position to the last position. The cross's position will be where the line get on the circle...

But I don't know how to do that... I can't remember of my math classes (Too far).

Your solution for the second problem may work, I need to try it.

Posted 2 years ago #
dbolli
Member

I see :)

As you suggest you can use the solution for the second problem to solve the first
If r is the radius of your circle around new player location x1,y1
then you can find the theta angle and the x and y values between x0,y0 old player location and x1,y1 new player location as per above
then location of cross is at r * cos(A),r * sin(A) where A is the theta angle as above

Posted 2 years ago #
Lapa
Member

Thanks I will try that :)

Posted 2 years ago #
Lapa
Member

I don't know why it doesn't work... For the first problem the cross stay en bottom left corner.
For the second probleme, the player shoot on the bottom left corner :/

Do you have an idea why ?

Posted 2 years ago #
dbolli
Member

Here's some working code for a sprite orbitingSprite that will rotate around a given sprite originalSprite (update the position via schedule or step: etc)

#define kOrbitRadius 60.0f

short orbitangle = 0;

[orbitingSprite setPosition:ccp( originalSprite.position.x + ( kOrbitRadius * cosf( CC_DEGREES_TO_RADIANS( orbitangle ) ) ), originalSprite.position.y + ( kOrbitRadius * sinf( CC_DEGREES_TO_RADIANS( orbitangle ) ) ) )]; // :dbolli:20110703 13:08:04 As per parametric form equation at http://en.wikipedia.org/wiki/Circle

orbitangle = ( orbitangle + 1 ) % 360; // :dbolli:20110703 13:08:04 Increase angle with wraparound

similarly the cross on the line between current position and last position is at r * cos(A),r * sin(A) where A is the theta angle as above

Posted 7 months ago #

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