How to make a size of y change depending on how high a number value is?

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#1

So I want the size of this part to go the smallest bit bigger, depending on the ammount of value there is in this numbervalue, but I do not know where to start.

How to keep only 1 sized size change?
#2

Do you want to tween the part or just size while loop it? The basic answer to your problem if your looping a size is to just add the numbervalue to the size. if your tweening it then just set the end size to the part size plus a numbervalue.

#3

No tween, I would just like the size to increase, lets say per 1 value to the int value, add 0.01 size to the y of xyz.

#4

You can just multiply the integer by 0.01, put that into the Y input of a Vector3 constructor, and then add the Vector3 to the size value of the part.

#5

So I have this code so far, what am I doing wrong?

local moneyValue = script.Parent.MoneyValue

script.Parent.Equipped:Connect(function()

local dog = moneyValue.Value * 0.01

script.Parent.Handle.Size.Y = script.Parent.Handle.Size.Y + dog

end)
#6

I can’t notice anything wrong right now, what is the difference between what you are expecting to happen and what actually happens?

#7

It says Y cannot be assigned that.

#8

Oh, I didn’t notice that on the first time around, but you cant set Y. You have to set the size property as a whole, but you can add a y value by setting the size to Vector3.new(0,script.Parent.Handle.Size.Y+dog,0)

#9

Any way to do this without setting the value to 0, for x z/.

#10

Not that I know of, but using this method really isn’t that complicated

#11

Not so much :confused: Char 30 limited

#12 
local moneyValue = script.Parent.MoneyValue

script.Parent.Equipped:Connect(function()

local dog = moneyValue.Value * 0.01

script.Parent.Handle.Size = script.Parent.Handle.Size + Vector3.new(script.Parent.Handle.Size.X,dog,script.Parent.Handle.Size.Z)

end)

Hope this works for you :slight_smile:

#13

Sorry that I didn’t make this post earlier, I was able to get this fixed a while ago.