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Modeling a projectile's motion

As a somewhat active member of the Scripting Helpers discord one of the most common questions I see is how to have a projectile travel an arc. Most people want to know how to do this for things like basketballs or cannon balls and so forth. Since this is such a popular question I thought it would be worth writing a blog post on it and talking about a few other things we can extend from our findings.


Deriving the equation

Using basic knowledge of physics and simple calculus we can derive the basic equation of motion that a projectile will take.

We know that the derivative of position with respect to time is velocity and that the derivative of velocity with respect to time is acceleration. Thus, the reverse is also true, the integral of acceleration with respect to time is velocity and the integral of velocity with respect to time is position.

In our case we know that the only acceleration affecting our objects is gravity. Therefore, all we must do is integrate gravity with respect to time twice to give us an equation for position.

eq1

Due to the rules of indefinite integration we are left with an unknown constant, C, but using some intuition we can figure this out easily enough. We know that at t = 0 our velocity should be equal to our initial velocity which we will define as v0.

eq2

Next, we’ll integrate velocity with respect to time to find position.

eq3

Similar to before we’re left with some constant, C, and again we know that at t = 0 we should be at our initial position which we define as x0.

eq4

Awesome! We have a time-based equation for position!

Basketball example

One of the neat things we can do with this equation is find the initial velocity needed to reach a certain target given time. To figure this out we define our target as x1 and rearrange the position function from above for v0.

eq5

gif1

local t = 1;
local mouse = game.Players.LocalPlayer:GetMouse();
local hrp = game.Players.LocalPlayer.CharacterAdded:Wait():WaitForChild("HumanoidRootPart");
local bball = script.Parent:WaitForChild("Handle");

mouse.Button1Down:Connect(function()
    local g = Vector3.new(0, -game.Workspace.Gravity, 0);
    local x0 = hrp.CFrame * Vector3.new(0, 2, -2)

    -- calculate the v0 needed to reach mouse.Hit.p
    local v0 = (mouse.Hit.p - x0 - 0.5*g*t*t)/t;

    -- have the ball travel that path
    local c = bball:Clone();
    c.Velocity = v0;
    c.CFrame = CFrame.new(x0);
    c.CanCollide = true;
    c.Parent = game.Workspace;
end)

Alternatively we could do this directly with CFrames if we didn't care as much about the collisions.

local t = 1;
local mouse = game.Players.LocalPlayer:GetMouse();
local hrp = game.Players.LocalPlayer.CharacterAdded:Wait():WaitForChild("HumanoidRootPart");
local bball = script.Parent:WaitForChild("Handle");
local rs = game:GetService("RunService").RenderStepped;

mouse.Button1Down:Connect(function()
    local g = Vector3.new(0, -game.Workspace.Gravity, 0);
    local x0 = hrp.CFrame * Vector3.new(0, 2, -2)

    -- calculate the v0 needed to reach mouse.Hit.p
    local v0 = (mouse.Hit.p - x0 - 0.5*g*t*t)/t;

    local c = bball:Clone();
    c.CanCollide = true;
    c.Anchored = true;
    c.Parent = game.Workspace;

    -- have the ball travel that path
    local nt = 0;
    while (nt < t*2) do
        c.CFrame = CFrame.new(0.5*g*nt*nt + v0*nt + x0);
        nt = nt + rs:Wait();
    end
end)

Jump power and height

For those of you who are unaware the property Humanoid.JumpPower can be thought of as the v0 used to model a jump. We can use this information with the position equation we derived to find the height of our jump or even the jump power we need to reach a certain height.

To find the jump height we need again to draw on some basic calculus intuition. We know that the derivative of a function allows us to calculate the slope at any given point. Looking at an example position function we can see that since it’s a parabola, its maximum height coincides with a slope of zero.

gif2

We already know the derivative of the position function is v(t) so if we set it equal to zero and solve for t then we know the time value when the position function has reached its maximum height.

eq6

We can take this time value and plug it into the position function to have a formula for jump height.

eq7

Noting that the initial height (position) is zero we’re left with the following formula to find jump height which we’ll define as h.

eq8

We can rearrange this formula to find v0 (jump power) given a non-negative height.

eq9

We can also look at the original position equation and solve for x(t) = 0 with the quadratic equation to find the time it takes for the jump land.

eq10

Unsurprisingly, this gives us the max height time value doubled since it takes the same amount of time to go up as it does down.

local humanoid = game.Players.LocalPlayer.CharacterAdded:Wait():WaitForChild("Humanoid");

-- note that typically gravity should be a negative value, but in this case I leave it positive hence the lack of -2 throughout the code
-- set the jump height to 10
humanoid.JumpPower = math.sqrt(2*game.Workspace.Gravity*10);

humanoid.StateChanged:Connect(function(old, new)
    if (new == Enum.HumanoidStateType.Jumping) then
        local h = humanoid.JumpPower^2/(2*game.Workspace.Gravity);
        local t = 2*humanoid.JumpPower/game.Workspace.Gravity;
        wait(t);
        print(string.format("Jump took %s seconds and reached a max height of %s studs", t, h));
    end
end)

Mapping the projectile’s path with a beam

One of the common things people like to do with this equation is use it to help players aim. The simplest way to do this would be to draw parts between intervals on the curve. However, another way to do this is to use beams!

Beams are represented by cubic Bezier curves which is great because it means they can perfectly match any polynomial that is at most of third degree (meaning cubic). Our function is second degree (quadratic) so we’re all set.

Our first step is going to be to head over to the Wikipedia page on Bezier curves. Luckily for us they already have the cubic Bezier curve equation and the first derivative which we’ll also be needing.

eq11

A few things we should take note of here if we’re trying to find a Bezier curve equivalent to our projectile motion function. We know that P0 = x0 and that P3 = x1, thus we have two unknowns we want to solve for, P1 and P2. We can also recognize that the Bezier curve function is normalized such that s = [0, 1]. As such we’ll need to similarly normalize our position and velocity functions as we’ll be using them soon.

We’ll define t1 as the time value that can be used to find x1, that is x(t1) = x1.

eq12

We will now set these equal to the two Bezier functions and solve for the unknowns. We’ll make sure to pick s values that make this process as easy as possible.

eq13

eq14

Great, we now have all four points that make up the equivalent Bezier curve. Our next step is to look at the beam wiki page and see how these points can be used to find the correct values and properties we need to adjust to find our beam object.

Looking at the page it seems that P1 and P2 are defined by the following:

P1 = attachment0.WorldCFrame.rightVector * beam.CurveSize0

P2 = -attachment1.WorldCFrame.rightVector * beam.CurveSize1

As such the last step is going to be to calculate these world CFrame values such that the right vectors are parallel with P1 – P0 and P3 – P2. This is easy enough with knowledge of the cross product and how the CFrame rotation matrix is formed.

As such our final code is:

gif3

local attach0 = Instance.new("Attachment", game.Workspace.Terrain);
local attach1 = Instance.new("Attachment", game.Workspace.Terrain);

local beam = Instance.new("Beam", game.Workspace.Terrain);
beam.Attachment0 = attach0;
beam.Attachment1 = attach1;

local function beamProjectile(g, v0, x0, t1)
    -- calculate the bezier points
    local c = 0.5*0.5*0.5;
    local p3 = 0.5*g*t1*t1 + v0*t1 + x0;
    local p2 = p3 - (g*t1*t1 + v0*t1)/3;
    local p1 = (c*g*t1*t1 + 0.5*v0*t1 + x0 - c*(x0+p3))/(3*c) - p2;

    -- the curve sizes
    local curve0 = (p1 - x0).magnitude;
    local curve1 = (p2 - p3).magnitude;

    -- build the world CFrames for the attachments
    local b = (x0 - p3).unit;
    local r1 = (p1 - x0).unit;
    local u1 = r1:Cross(b).unit;
    local r2 = (p2 - p3).unit;
    local u2 = r2:Cross(b).unit;
    b = u1:Cross(r1).unit;

    local cf1 = CFrame.new(
        x0.x, x0.y, x0.z,
        r1.x, u1.x, b.x,
        r1.y, u1.y, b.y,
        r1.z, u1.z, b.z
    )

    local cf2 = CFrame.new(
        p3.x, p3.y, p3.z,
        r2.x, u2.x, b.x,
        r2.y, u2.y, b.y,
        r2.z, u2.z, b.z
    )

    return curve0, -curve1, cf1, cf2;
end

game:GetService("RunService").RenderStepped:Connect(function(dt)
    local g = Vector3.new(0, -game.Workspace.Gravity, 0);
    local x0 = hrp.CFrame * Vector3.new(0, 2, -2)
    local v0 = (mouse.Hit.p - x0 - 0.5*g*t*t)/t;

    local curve0, curve1, cf1, cf2 = beamProjectile(g, v0, x0, t);
    beam.CurveSize0 = curve0;
    beam.CurveSize1 = curve1;
    -- convert world space CFrames to be relative to the attachment parent
    attach0.CFrame = attach0.Parent.CFrame:inverse() * cf1;
    attach1.CFrame = attach1.Parent.CFrame:inverse() * cf2;
end)
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Commentary

Leave a Comment

incapaz says: September 5, 2018
Good article but just a note: Instance.new parent argument is deprecated
Wafflecow321 says: September 8, 2018
Lol, I spent like 3 days teaching myself physics a few days ago to solve this. Wish I had bother reading this instead.
magicguy78942 says: September 8, 2018
Woah! I just made a question about this not long ago. I was really disappointed then that there were no answers. But this is EXACTLY WHAT I REALLY NEEDED. THANK YOU!
INOOBE_YT says: September 13, 2018
Ego is a great guy