How to calculate distance between 2 CFrame values

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#1

I need to get the distance of the Z value between 2 parts

I originally tried doing

Red.CFrame.Z - Blue.CFrame.Z

but this isn’t accurate when the parts Orientation changes

image

like in the image I want Black to be in front of Blue

but by doing
Black.CFrame = Red.CFrame*CFrame.new(0,0,distance)

2 Likes
#2

Subtract the positions and find the magnitude.

Code:

local distance = (Red.Position - Blue.Position).Magnitude
#3

This solution can work too

local distance = math.abs(Red.Position.Z - Blue.Position.Z)
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#4

this wouldn’t work if I rotate all the parts cause then black part would go in a different direction

#5

Then you would need trigonometry

I can help you with that, just tell me what kind of rotation you want to get

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#7

You need to use dotproduct.

local CFrameA = -- CFA
local CFrameB = -- CFB
local Dot = math.acos(CFrameA.LookVector:Dot(CFrameB.LookVector))
print(math.rad(Dot)) -- prints orientation related difference in angles.
-- For Positional:
local Distance = (CFrameA.Position-CFrameB.Position).Magnitude
print(Distance)
#8

I’m not sure on how to do that but basically what I need it is
image
the black part and blue part would be positioned correctly here

but if I move the blue part to the side I want the black part to stay in front of it

but by positioning it at the red parts position + the distance

image

to always be in front

by using CFrame also

I also want to know all the X Y Z distance between the parts

2 Likes
#9 
-- Defines the parts
local Red = workspace.Red
local Black = workspace.Black
local Blue = workspace.Blue

-- Loops
while true do
	-- Gets the positions X and Z from each part
	local X = Blue.Position.X
	local Z = Red.Position.Z
	
	-- Sets the position X and Z to the black position
	Black.Position = Vector3.new(X,0.5,Z)
	
	-- Waits a little bit for you pc not exploding
	wait()
end

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#10

the only thing is when rotating the parts all together then black position isn’t the same anymore

#11

¿Could you give me more explanations of how you want to it to work?

#12

Oh I see what you are trying to accomplish!

local red = -- red part
local black = -- black part
local blue = -- blue part

local z = Red.Position.Z

blue.Changed:Connect(function()
	local x,y = blue.Position.X,black.Position.Y
	black.Position = Vector3.new(x,y,z)
end)

For @JESUS5PACHECO Using while loops can cause a lot of ping issues, its better to use .Changed Events.

2 Likes
#13

Ah I see what your trying to say. Please wait a bit while I code it.

1 Like
#14

so what u showed me was close to what i wanted but
if you select all the parts and did ctrl+r
and have that script run
the black part will be positioned differently and not the same before you rotated it

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#15

¿Could you explain what is exactly what you want?, its hard to tell with only those images

#16

OP wants the part to stay attach no matter how they are oriented.

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#17

in your video the Black part was on the left of the Red part

I want it to always stay on the side its on even if you rotate the parts

#18

The positions are changed also, not only the rotations, so this problem will be a lot more complex of solving

#19

Could you explain more clearly how you want the parts to “stay on the side”

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#20


OP wants this ^^^

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#21

image
you see here the Black part is on the left of the Red part

image
then here I rotated it but it stays on the left side of the red part but always in front of the blue part

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