How can I use trigonometry to find this angle?

xDeltaXen · November 29, 2020, 2:03pm

I want to use trigonometry to find this angle I have looked at this thread:

But the code provided did not work:

local Adjacent = game.Players.LocalPlayer.Head.CFrame
local Hypotenuse = workspace.CurrentCamera.CFrame
--The both are CFrames, not Vectors, but it should help you to understand

local FinalVector = Adjacent:ToObjectSpace(Hypotenuse).LookVector --We need to make the 2 vectors relative, ldk as without this it dont work. Still Remember we are using 2 CFrames and the function only takes 1 Parameter

local Theta = math.acos(FinalVector)

Here is my angle

how can I find this angle please let me know and provide resources!

xDeltaXen · November 29, 2020, 2:06pm

Here is my code right now

local Lighting = game:GetService("Lighting")
local Players = game:GetService("Players")
local LocalPlayer = Players.LocalPlayer
local LocalPlayerCharacter = LocalPlayer.Character or LocalPlayer.CharacterAdded:Wait()
local LocalPlayerHead = LocalPlayerCharacter:WaitForChild("Head")
local CurrentCamera = workspace.CurrentCamera

function VisRay(RayNew)
	local my_ray = RayNew

	local part = Instance.new("Part")
	part.Anchored = true
	part.CFrame = CFrame.new(my_ray.Origin:Lerp(my_ray.Direction, 0.5), my_ray.Direction)
	part.Size = Vector3.new(1, 1, (my_ray.Origin - my_ray.Direction).Magnitude)
	part.Parent = workspace
end

local Magnitude = (Lighting:GetSunDirection() - CurrentCamera.CFrame.LookVector).Magnitude
print(math.deg(math.acos(Magnitude)))

dthecoolest · November 29, 2020, 2:43pm

I don’t believe that triangle is a right-angle triangle so you can’t directly use the code made by @Eternalove_fan32 to measure the angle between those two parts of using math.acos.

I suggest instead using vector trigonometry instead in the form of using the dot product between vectors as derived from the law of cosines mathematically as explained by this very helpful post by @ThanksRoBama:

Thanks RoBama

math.acos( a:Dot(b)/(a.Magnitude * b.Magnitude) )

Or if you use a CFrame look vector since CFrame vector components are unit vectors we can just do this which is very helpful to shorten code:

math.acos( a:Dot(b) )

xDeltaXen · November 29, 2020, 2:48pm

Like this? But i dont know why its creating so many blureffects?

local Lighting = game:GetService("Lighting")
local Players = game:GetService("Players")
local LocalPlayer = Players.LocalPlayer
local LocalPlayerCharacter = LocalPlayer.Character or LocalPlayer.CharacterAdded:Wait()
local CurrentCamera = workspace.CurrentCamera

local ViewSunDirection = false

while wait() do
	if ViewSunDirection and not Lighting:FindFirstChild("BlurEffect") then
		local BlurEffect = Instance.new("BlurEffect")
		BlurEffect.Parent = Lighting
	elseif Lighting:FindFirstChild("BlurEffect") then
		Lighting.Blur:Destroy()
	end
	
	local Difference = CurrentCamera.CFrame.Position - Lighting:GetSunDirection()
	local Degree = math.deg(math.atan2(Difference.X, Difference.Z))
	
	if Degree > 120 and Degree < 125 then
		ViewSunDirection = true
	else
		ViewSunDirection = false
	end
end

xDeltaXen · November 29, 2020, 2:55pm

also that just gives this

Eternalove_fan32 · November 29, 2020, 8:36pm

but you can split a regular triangle into 2 right-angle triangles, like egomoose do this!

github.com

EgoMoose/Articles/blob/master/3d triangles/3D triangles.md

# 3D Triangles

On occasion you may find it necessary to draw triangles given three points in 3D space. This article aims to go over one method that can be used to create those triangles using knowledge of trigonometry and coordinate frames.

## Table of contents

* [Triangle decomposition](#triangle-decomposition)
* [Manipulating the rotation matrix](#manipulating-the-rotation-matrix)
* [Optimization](#optimization)

## Triangle decomposition

The first thing we need to be aware of is that Roblox does not have a singular default shape that can be used to create any 3D triangle. However, a shape that Roblox does have is right-angle triangles, otherwise known as wedges. Thus, we can decompose any triangle into a maximum of two right-angle triangles and draw from there. We can easily draw this process by placing a perpendicular line on the longest edge and having it intersect with the opposite vertex.

![img1](imgs/img1.png)

Now that we understand the basics of our goal we have to ask ourselves how we can achieve this mathematically. We'll start this process off by finding some information about the triangle depending on what side is the longest.

![img2](imgs/img2.png)

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Eternalove_fan32 · November 29, 2020, 8:38pm

what do you exactly want to do?

xDeltaXen · November 29, 2020, 8:39pm

when the player is looking at the sun it adds a blur effect

Eternalove_fan32 · November 30, 2020, 6:15am

Then you may find this interesting:
https://github.com/boatbomber/Distortion-Screen-Rain/blob/master/ScreenRain.lua#L265
It adds dropplets if you look in the sky, but you can change this that it adds blur effect instead of waterdropplets!

(Credit to @boatbomber)

xDeltaXen · November 30, 2020, 4:16pm

I get it now but can you explain why you are using acos? And how is :Dot being used?

Eternalove_fan32 · November 30, 2020, 5:28pm

Hope this helps:

github.com

EgoMoose/Articles/blob/master/Vectors/Dot product.md

# Dot product

The dot product is an operation performed on two vectors in any dimension that returns a single number. It is also sometimes called the scalar product, the inner product, or rarely the projection product. The dot product is considered a way to multiply two vectors. Performing the dot product operation is simple enough, but in order to use it in code you will need a good understanding of the mathematics behind the operation. This article aims give a better understanding of the dot product so that you can use it in your games.

## Table of contents

* [Geometric definition](#geometric-definition)
* [Algebraic definition](#algebraic-definition)
* [Case study: Reflecting lasers](#case-study-reflecting-lasers)

## Geometric definition

To start off let’s have a definition for the dot product given vectors U and V.

*“The scalar projection of U onto V multiplied by the magnitude of V”*

*“The scalar projection of V onto U multiplied by the magnitude of U”*

This definition may of course leave you wondering what a scalar projection is, and more importantly how to calculate it. A scalar projection is the amount that one vector travels in another vector’s direction. So if we say that we want the projection of U onto V we want to know how much of vector U is going in the same direction as vector V and vice versa for the projection of V onto U.

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