Finding distance between 2 objects

[RottenMilk]

UGGGHHHHHHH

I knew this before.

But now i forgot.

how would you find the diffrence between the player's _x coordinate and the enemy's _x coordinate?

i'm using as2.

I know it has something to do with atan....????

help pls.

[Montycarlo]

Math.abs(obj1.x - obj2.x)

[Glaiel-Gamer]

Math.tan(Math.atan(enemyx-playerx))

[Montycarlo]

At 12/18/09 12:26 AM, Glaiel-Gamer wrote: Math.tan(Math.atan(enemyx-playerx))

hahahahahahahahahahaha, thumbs up.

[RottenMilk]

I trust glaeil. thanks.

[Nisas]

they say the hypotenuse is equal to the square root of the sum of both sides squared.

A^2 + B^2 = C^2

IE:

var xdif:Number = obj1.x - obj2.x;
var ydif:Number = obj1.y - obj2.y;
dist = Math.sqrt(xdif*xdif + ydif*ydif)

[HonterGames]

At 12/18/09 12:34 AM, RottenMilk wrote: I trust glaeil. thanks.

Actually all of the replies u got would be acceptable depending on what you really wanted (eg the distance between objects, the difference between x pos, etc)

[Cojones893]

At 12/18/09 12:34 AM, RottenMilk wrote: I trust glaeil. thanks.

well you shouldn't...

Do you know what atan and tan do?

[Cojones893]

At 12/18/09 12:26 AM, Glaiel-Gamer wrote: Math.tan(Math.atan(enemyx-playerx))

You missed a part

Math.PI*(Math.pow(Math.pow(Math.tan(Math.atan(obj1.x-obj2.x)),2),.5))/Math.PI

More accurate.

[Yambanshee]

just so you dont get totaly confused, everyone except for Nisas is screwing with you

Basicly, its like a right angled triangle. think of the y as the upright line, x as the horozontal line and the line connecting the two is a, or the distance. now to work out a, the formula is this:

a^2 = X^2+Y^2
so
a = square root(x+y)
so (in AS)

xdist = mc2.x - mc1.x
ydist = mx2.y - mc1.y
distance = Math.sqrt((xdist*xdist)+(ydist*ydist))

[4urentertainment]

At 12/18/09 02:16 PM, Cojones893 wrote:

At 12/18/09 12:26 AM, Glaiel-Gamer wrote: Math.tan(Math.atan(enemyx-playerx))

You missed a part

Math.PI*(Math.pow(Math.pow(Math.tan(Math .atan(obj1.x-obj2.x)),2),.5))/Math.PI

More accurate.

Can someone explain all that?

[Yambanshee]

Math.PI*(Math.pow(Math.pow(Math.tan(Math .atan(obj1.x-obj2.x)),2),.5))/Math.PI

Can someone explain all that?

if you read it carefully, you'll see its just a long complex formula which cancels out, giving you a end result of obj1.x-obj2.x i believe

[4urentertainment]

At 12/18/09 02:58 PM, Yambanshee wrote:

Math.PI*(Math.pow(Math.pow(Math.tan(Math .atan(obj1.x-obj2.x)),2),.5))/Math.PI

Can someone explain all that?

if you read it carefully, you'll see its just a long complex formula which cancels out, giving you a end result of obj1.x-obj2.x i believe

Funny. So they're just messing with the dude. I knew you couldn't possibly get the distance without taking the y distance into account.

[Montycarlo]

At 12/18/09 03:01 PM, 4urentertainment wrote: Funny. So they're just messing with the dude. I knew you couldn't possibly get the distance without taking the y distance into account.

He never asked for the vector difference, he merely asked for the x axis distance, not displacement.

[4urentertainment]

At 12/18/09 03:10 PM, Montycarlo wrote:

At 12/18/09 03:01 PM, 4urentertainment wrote: Funny. So they're just messing with the dude. I knew you couldn't possibly get the distance without taking the y distance into account.

He never asked for the vector difference, he merely asked for the x axis distance, not displacement.

Really...Now that I read the post again, that does seem as what he wanted. I just figured he couldn't have wanted something that simple.

[RottenMilk]

I just want to know the difference between 2_x's. Very simple.

Not the difference between two objects. thanks.

[zrb]

At 12/19/09 12:22 PM, RottenMilk wrote: I just want to know the difference between 2_x's. Very simple.
Not the difference between two objects. thanks.

Well its been stated above (unless you don't need help anymore cause I can't tell):

xDifference = object2._x-object1._x;

Replace the objects by enemy or player or whatever you want.