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English Distance between players

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old Distance between players

Echo-Charlie
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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.

old Re: Distance between players

DannyDeth
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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:
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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.

old Re: Distance between players

Vectarrio
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@DannyDeth
You didn't "square" the variables:
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distance = math.sqrt((player(id1,"x") - player(id2,"x"))^2 + (player(id1,"y") - player(id2,"y"))^2)

old Re: Distance between players

Unknown_Soldier
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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

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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)

You cant use math.sqrt on a negative number

old Maths

Unknown_Soldier
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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

old Re: Distance between players

Vectarrio
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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.

and, I'd prefer math.abs instead of math.max with math.min.
Just a little less space eaten by it.

old Re: Distance between players

archmage
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Just use this
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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

Just call it with two player IDs.
Ex.:
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dist = math.dist(1, 32);
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