# Applying parabollic shooting

xev
Hi!

As a part of a subject project I'm developing a game. It is a kind of real-time strategy game where two teams of tanks have to destroy each other (not very original, but interesting enough).

I have come across a problem I cannot solve: applying parabollic shooting to simulate their bullets.

I use the typical formulas:
x=x0+vx*t
vx=v0*cos(a)
y=y0+vy*t-0.5*g*(t)^2
vy=v0*sin(a)

where a=alpha, g=gravity (9.8 default) and v0=initial velocity modulus

What happens is that I had always used these formulas to calculate some typical things like: maximum height reached, ending x, etc.

But now, given an initial point, a final point and an initial velocity, I need to calculate the alpha that would take the bullet the minimum time.

I need to have a function f(t), to minimize it to know the minimum time and then get the alpha. But I always get struck in the depths of the calculus.

Can anybody help me, please? Thank you.

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Tom Mattson
Staff Emeritus
Gold Member

Originally posted by xev
Hi!
Hi, and welcome to PF. Same Xev from sciforums and philosophyforums I presume?

I use the typical formulas:
x=x0+vx*t
vx=v0*cos(a)
y=y0+vy*t-0.5*g*(t)^2
vy=v0*sin(a)

But now, given an initial point, a final point and an initial velocity, I need to calculate the alpha that would take the bullet the minimum time.

I need to have a function f(t),
Actually, what you need is t as a function of &alpha;. You will want to differentiate t(&alpha;) with respect to &alpha; and set derivative equal to 0, yadda yadda yadda.

to minimize it to know the minimum time and then get the alpha. But I always get struck in the depths of the calculus.
]

It sounds like you've got the calculus and are getting lost in the depths of the algebra. I've highlighted the function you need above. You could alternatively use the equation for y, but it is more complicated so I'm sticking with the equation for x. What you do is solve it for t.

x-x0=v0cos(&alpha;)t
(x-x0)/(v0cos(&alpha;))=t

Now, can you fill in the yaddas?

xev
ballistics(2)

Hi there

>>Same Xev from sciforums and philosophyforums I presume?

I am afraid I don't even know of the existence these sci/philosofyforums.

Since English is not my native language, I don't know what you mean by yadda yadda (maybe it means etc or something more specific?)

-> my progress: i've taken care of your funcion, trying to minimize it. But I reach the following statement:
cos(a) + x * sin(a)=0

I should get the alpha as a function of x, but I cannot find any way to be able to isolate the alpha and to continue working on it. Do you know how these equations are solved? Thank you!

>>Welcome to PF

Thanks!!

Tom Mattson
Staff Emeritus
Gold Member

Originally posted by xev

Since English is not my native language, I don't know what you mean by yadda yadda (maybe it means etc or something more specific?)
Yes, it means "et cetera". It's from Seinfeld. :D

-> my progress: i've taken care of your funcion, trying to minimize it. But I reach the following statement:
cos(a) + x * sin(a)=0
t as a function of &alpha; is:

t(&alpha;)=((x-x0)/v0)(cos&alpha;)-1

You need to take the derivative of that with respect to &alpha;. Use the chain rule:

t'(&alpha;)=((x-x0)/v0)(-1)(cos&alpha;)-2(-sin&alpha;)
t'(alpha;)=((x-x0)/v0)sin&alpha;/cos2&alpha;

That is what you should set equal to zero. Solve for &alpha;, and you have your angle that minimizes the time.

I should get the alpha as a function of x, but I cannot find any way to be able to isolate the alpha and to continue working on it. Do you know how these equations are solved? Thank you!
Why are you looking for &alpha; as a function of x? x is fixed, and anyway you are trying to minimize the time. If I am understanding you correctly, you definitely need t as a function of &alpha;, as I outlined above.

xev
Ballistics(3)

t'(alpha;)=((x-x0)/v0)sin¥á/cos2¥á

then I do:

((x-x0)sin¥á)/(v0*cos2¥á)) = 0

then:

(x-x0)sin(a)=0

we get: a=k*PI

referring to the original function, I substitute the reached alpha:

t(k*PI)=((x-x0)/v0)/(cos(K*PI))

Because we need positive times, I restrict alpha to being 2*k*PI

then t(2*k*PI)=t(¥á)=((x-x0)/v0)/(cos(2*k*PI))= (x-x0)/v0

So the conclusion is: the best alpha is 0¨¬
This isn't useful for me, am I doing something wrong or maybe I amb approaching the problem with the incorrect method?

Thank you !

Tom Mattson
Staff Emeritus
Gold Member
OK, I think I see the problem. What you have here is an overdetermined system, and this will become apparent when I eliminate the t from the parametric equations for x and y.

x-x0=v0cos(&alpha;)t
y-y0=v0sin(&alpha;)t-(1/2)gt2

Solve the first equation for t.

t=(x-x0)/(v0cos(&alpha;))

Plug this into t in the second equation.

y-y0=v0sin(&alpha;)((x-x0)/v0cos(&alpha;))-(1/2)g((x-x0)/v0cos(&alpha;))2

I could clean this up a bit, but I don't need to to make my point. y, y0, x, x0, v0 and g are all fixed. &alpha; is the only thing left, so it is fixed too. You don't have freedome to let it vary to optimize the time.