Distance between players
13 replies
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30.03.11 05:29:54 pm
Hello there. I have an idea, but I can't realise that - I don't know how can I get distance between two players (distance in pixels). 
Maybe someone of you can explain how can I get distance? Thanks a lot for your answers. 
Maybe someone of you can explain how can I get distance? Thanks a lot for your answers. 
30.03.11 05:49:12 pm
The Pythagoream Theorem states that the hypotenuse is equal to the adjacent + the opposite squared, so therefore if you find the vector of two points ( x1 - x2, y1 -y2 ) you can find the length of the hypotenuse.
So basically you can do it like this:
Just replace id1 and id2 with the variables you are using.
So basically you can do it like this:
Code:
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distance = math.sqrt(player(id1,"x") - player(id2,"x") + player(id1,"y") - player(id2,"y"))
Just replace id1 and id2 with the variables you are using.
30.03.11 06:40:35 pm
@DannyDeth
You didn't "square" the variables:
You didn't "square" the variables:
Code:
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distance = math.sqrt((player(id1,"x") - player(id2,"x"))^2 + (player(id1,"y") - player(id2,"y"))^2)
30.03.11 06:44:32 pm
That is wrong, to calculate the hypotenuse, you need the length of the other 2 sides of the triangle, the code is something like this
You cant use math.sqrt on a negative number
Code:
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side1 = math.max(player(id1,"x"),player(id2,"x"))-math.min(player(id1,"x"),player(id2,"x"))
side2 = math.max(player(id1,"y"),player(id2,"y"))-math.min(player(id1,"y"),player(id2,"y"))
distance = math.sqrt(side1^2+side2^2)
side2 = math.max(player(id1,"y"),player(id2,"y"))-math.min(player(id1,"y"),player(id2,"y"))
distance = math.sqrt(side1^2+side2^2)
You cant use math.sqrt on a negative number
Should I come back and make one last map for CS2D?
30.03.11 07:00:19 pm
Let's try your code with an example
player 1 X100 Y100
player 2 X200 Y300
sqrt(100 - 200 ^ 2 + 100 - 300 ^ 2)
sqrt(100 - 40000 + 100 - 90000)
sqrt(-39900 - 89900)
sqrt(-129800)
no number multiplyed by himself can be a negative number
example
2^2 = 2*2 = 4
-2^2 = -2*-2 = 4
player 1 X100 Y100
player 2 X200 Y300
sqrt(100 - 200 ^ 2 + 100 - 300 ^ 2)
sqrt(100 - 40000 + 100 - 90000)
sqrt(-39900 - 89900)
sqrt(-129800)
no number multiplyed by himself can be a negative number
example
2^2 = 2*2 = 4
-2^2 = -2*-2 = 4
Should I come back and make one last map for CS2D?
30.03.11 07:18:23 pm
@Vectar666 No, I didn't see them. Sry 
, I have problems with brackets when I make luas. Anyway, both codes should work.
, I have problems with brackets when I make luas. Anyway, both codes should work. Should I come back and make one last map for CS2D?
30.03.11 08:13:36 pm
Unknown_Soldier has written:
@Vectar666 No, I didn't see them. Sry , I have problems with brackets when I make luas. Anyway, both codes should work.
, I have problems with brackets when I make luas. Anyway, both codes should work.and, I'd prefer math.abs instead of math.max with math.min.
Just a little less space eaten by it.
31.03.11 01:40:25 am
Just use this
Just call it with two player IDs.
Ex.:
Code:
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function math.dist(p1, p2)
local x, y = player(p1, "x"), player(p2, "y");
local x2, y2 = player(p2, "x"), player(p2, "y");
return math.sqrt((y2-y)^2 + (x2-x)^2);
end
local x, y = player(p1, "x"), player(p2, "y");
local x2, y2 = player(p2, "x"), player(p2, "y");
return math.sqrt((y2-y)^2 + (x2-x)^2);
end
Just call it with two player IDs.
Ex.:
Code:
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dist = math.dist(1, 32);
We must secure the existence of our people and a future for white children. 14/88
31.03.11 12:59:42 pm
Vectar666 has written:
You didn't "square" the variables
Fudge cakes. It was late and I was tired, what can I say?
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