# Physics Formula

#### wraithseeker

##### Tired.
If I were to give time and displacement as user inputted variables, which formula should I use to get vi and vf?

I tried to derive formulas from here but to no avail #### kingkingyyk3

##### Visitor (Welcome to the Jungle, Baby!)
Velocity = Displacement/time

Vi = Initial velocity
Vf = Final velocity

Vi is counted at the starting and Vf is counted at the end.

This is what I had learn in my physic class.

• 1 person

#### uberfoop

If you're using the variables of time, displacement, vi, and vf, you would use the formula relating those:

d = ((vi + vf)/2)*t

Solve for vi and vf and you get:

vi = 2d/t - vf
vf = 2d/t - vi

You cannot solve vf or vi from simply inputting time and displacement, because 2 leftover variables means not enough information has been plugged in to get a definite answer.

It of course gets much more messy if acceleration isn't constant though, since then vf is equal to vi plus the integral of a(t)dt evaluated from ti to tf.

• 1 person

#### wraithseeker

##### Tired.
accleration is constant and is the value of gravity.

So I need something like time, displacement and initial velocity?

EDIT: Wait no, final velocity is always 0 becaus I am making a jump.

#### uberfoop

Or time, displacement, and final velocity, to get initial velocity, yeah.

#### wraithseeker

##### Tired.
But final velocity is always 0 for a jump or a knockback.

EDIT: How about taking only time? I will only move the unit by the Z axis and not touch x & y.

EDIT: I did some hand written stuffs shouldn't equation be

vi = ((2*d)-vf)/(t*t)

#### uberfoop

Well, sort of. It's 0 after impact with the ground, but that means you just ditched constant acceleration. Assuming no air drag occurs and the total change in displacement is 0, final velocity should be equal to the velocity at the begining of the jump.

I'm not entirely sure what to get at though, what exactly do you need the numbers for?

#### wraithseeker

##### Tired.
JASS:
``````private function Periodic takes nothing returns nothing
local integer i = 0
local Jump d
loop
exitwhen i &gt;= Count
set d = FlyZ<i> // gravity = -981. TIMEOUT = 0.03
set d.vz= d.vz+ 0.5 * d.gravity * TIMEOUT
set d.vi= d.vi+ d.vz *TIMEOUT
call SetUnitFlyHeight(d.target,d.vi,0.0)
if d.vi &lt; 0. or GetWidgetLife(d.target) &lt;= 0.0451 then
call d.destroy()
call d.Code.execute(d.target)
set Count = Count - 1
if Count &gt; 0 then
set FlyZ<i> = FlyZ[Count]
set i = i - 1
else
call PauseTimer(Timer)
endif
endif
set i = i + 1
endloop
endfunction

function Heh takes unit u,real displacement, real time returns nothing
local Jump d
local real velocity = 0.
if IsUnitInGroup(u,HitGroup) then
set d = Jump<u>
set velocity = 2.00*displacement/time - 0.00 // yr formula
set d.vz = d.vz+velocity
else
if Count == 0 then
call TimerStart(Timer,TIMEOUT,true,function Periodic)
endif
set velocity = ((2.00*displacement)- 0.00)/(time*time)  // err.. the formula im trying to test.
call BJDebugMsg(R2S(velocity))
set d = Jump.create(u,velocity)
set Jump<u> = d
set FlyZ[Count] = d
set Count = Count + 1
endif
endfunction</u></u></i></i>``````

I took out parts of my jump library.

How did I ditched constant accleration? No air drag, a free fall and displacement is not 0 as the unit is moving to a direction and not coming back.

#### uberfoop

I'm assuming you're talking about z velocity, and, in that case, assuming that a unit is stationary when it lands means you are ditching constant accelleration. When an object is at rest on the ground, it has no z accelleration. When it is propelled into the air, extreme nigh-instant accelleration gives it a high velocity in the positive z direction. In the air, the object suffers from a strong gravitational accelleration in the negative z direction. When it's about to hit the ground, its z velocity will be roughly equal to its initial z velocity but in the negative z direction. Upon striking the ground, it suffers VERY HIGH accelleration to stop its descent.
Thus, non-constant accelleration.
This is, of course, unless the jump ends at the maximum z height, at which point vf is indeed equal to 0.

Also, I don't need to see the code. Rather, what is the 'velocity' variable representing? Initial upward velocity? That's the question.
Also, usually in jumps, vi and vf aren't really used, rather it is simulated bit by bit with something derived from:
(delta)s = Vi*t + .5at^2

#### wraithseeker

##### Tired.
That velocity variable represents initial velocity so the unit will fly up and down automatically.

EDIT: Yes I am talking about the Z axis.
EDIT2: How would I do it if I have no constant acceleration? My knockback reduces speed by decrement every interval tick period of 0.03.

#### uberfoop

Ok, in that case, if you're going to use that formula, use the formula assuming values at halfway through the jump (when the unit is at its highest)(this is so that (delta)d isn't 0, since all that does is create a condition where the only requirement is that Vi = -Vf and thus the equation is useless); d will be the max jump height, Vf will be 0 at that height, and t will be half of the total time for the entire jump:
Thus, considering the lack of Vf:

Vi = 2d/t

#### Rainther

##### I guess I should write something of value here...
I'm not quite sure how a jump would be simulated however, my method would be:

JASS:
``````function timer takes nothing returns nothing
//Get data
local MaxHeight = 150
set data.integer = data.integer + 1
call SetUnitFlyHeight(data.unit, ( MaxHeight * 4 * data.integer / TurnOff ) - ( MaxHeight * 4 data.integer * data.integer / TurnOff / TurnOff ), 0)

if data.integer &gt;= TurnOff then //Turn off can be Distance of the 2 points divided by the movement occured of the unit each intervall
//Destroy Timer
endif
endfunction``````

I know I've posted this before, but haven't received any responds.

This is a rather easy way to make a unit fly. It's possible to make the unit fly faster at start, or slower are start.

• 1 person

#### wraithseeker

##### Tired.
@Rainther I don't really understand what you are doing though.

Vi = 2d/t

Did not work too well, the unit flew too high and the knockback long ended

#### Rainther

##### I guess I should write something of value here...
Lemme try to explain. data.integer will eventually reach the same value as TurnOff.

Meaning that data.integer / TurnOff will eventually become the value 1.

So ( MaxHeight * data.integer / TurnOff ) will at the end become only MaxHeight. ( MaxHeight * (data.integer / TurnOff = 1) )

Same thing goes for ( MaxHeight * data.integer * data.integer / TurnOff / TurnOff ) which also will become only MaxHeight, however is does this at a slower speed since the values multiplied with themselves get's lower. x < 1 gives x * x < x

When we are half way through the flying it'll be (MaxHeight * 4 * (data.integer / TurnOff) = 0.5)) - (MaxHeight * 4 * (data.integer / TurnOff) = 0.5) * (data.integer / TurnOff) = 0.5)) which is the heighest we'll reach.

That's why I do MaxHeight times 4. Because MaxHeight * 0.5 - MaxHeight * 0.25 = MaxHeight / 4

And at the end they'll become (MaxHeight - MaxHeight = 0) The orginial height.

EDIT: Seperated it so it'd be easier for the eye • 1 person

#### Strilanc

##### Veteran Scripter
If the jump starts and stops at the same height (ie. flat ground), then you also know that the initial z velocity is opposite the z final velocity:

Vz(final) = -Vz(initial)

So our total change in velocity is 2*Vz(initial), caused by acceleration from gravity over time:

g*t = 2*Vz(initial)
t = 2*Vz(initial)/g

Half way through the jump the unit will be at its highest point, which is:

dz = t/2*Vz(initial) - g*(t/2)^2/2
= 2*Vz(initial)/g/2*Vz(initial) - g*(2*Vz(initial)/g/2)^2/2
= Vz(initial)^2/g - Vz(initial)^2/g/2
= Vz(initial)^2/g/2

Put it all together:

Vz(initial) = sqrt(2*dz*g)
Vz(initial) = t*g/2

You can't specify both the time and the height; only one or the other. If you mean the actual distance d the unit will travel, then just set the unit's speed along the ground to be d/t.

#### Light Alkmst

##### New Member
If it's user inputted, then vi should be given. Otherwise, you could essentially start with whatever vi you want... EDIT:
If I were to give time and displacement as user inputted variables, which formula should I use to get vi and vf?

vi * vi = vf * vf - 2 * a * d
Hmm...that's still 2 vars we don't have...

#### wraithseeker

##### Tired.
Vz(initial) = t*g/2

Didn't work. It displayed negative values for me.

#### saw792

##### Is known to say things. That is all.
Because gravity is negative.

#### wraithseeker

##### Tired.
Well so how to fix? Making it positive didn't solve the problem.

#### Strilanc

##### Veteran Scripter
What do you mean it displayed negative values? If you don't even understand the effects of changing the sign of gravity then I'm not sure I can help.

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