Calculate vector trajectory based on start point and end point, withsimple gravity

This question is in regards to a Scorched Earth style artillery game.

My environment consists only of gravity (constant) and a Y plane/floor
for projectiles. There is no air drag, wind, or obstacles.

I have a known origin point, and destination point.

I want to calculate the needed artillery vector from the origin point
to land on the destination point. I don't know how to solve this
because of the gravity.

How would I solve this, assuming origin and destination points are on
the same Y value?

How would I solve this, if origin and destination may not be on the
same Y value?

Thanks for any help.
- Aaron

On Tuesday 26 August 2008 17:50, Aaron Beall wrote:

> This question is in regards to a Scorched Earth style artillery game.
>
> My environment consists only of gravity (constant) and a Y plane/floor
> for projectiles. There is no air drag, wind, or obstacles.
>
> I have a known origin point, and destination point.
>
> I want to calculate the needed artillery vector from the origin point
> to land on the destination point. I don't know how to solve this
> because of the gravity.

Aaron,

Hint: write an equation for a projectile given a beginning angle and
"power" (velocity) (it will be a second degree equation having trig
functions of the angle in the coefficients). Now solve the equation
for angle. You should be able to plug in the origin, destination, and
power to get an angle.

-paul-

On Aug 28, 12:13*pm, "Paul E. Black" <p.bl...@acm.org> wrote:
>
> Hint: write an equation for a projectile given a beginning angle and
> "power" (velocity) (it will be a second degree equation having trig
> functions of the angle in the coefficients). *Now solve the equation
> for angle. *You should be able to plug in the origin, destination, and
> power to get an angle.
>
> -paul-

That makes sense, unfortunately I don't know how to write such an
equation. Any tips on how? Where I get completely lost is how to
formulate gravity, since it's just a constant applied over time. But
time is an unknown, so it can't be part of the formula, can it?

A second degree equation... you mean like this?
y = ax² + bx + c

I'm not really sure what to do what that... I did just find this
though, my understanding is very limited but it seems to be solving
exactly what I'm trying to solve:
http://hyperphysics.phy-astr.gsu.edu/Hbase/traj.html

On Aug 28, 12:13*pm, "Paul E. Black" <p.bl...@acm.org> wrote:
>
> Hint: write an equation for a projectile given a beginning angle and
> "power" (velocity) (it will be a second degree equation having trig
> functions of the angle in the coefficients). *Now solve the equation
> for angle. *You should be able to plug in the origin, destination, and
> power to get an angle.
>
> -paul-

That makes sense, unfortunately I don't know how to write such an
equation. Any tips on how? Where I get completely lost is how to
formulate gravity, since it's just a constant applied over time. But
time is an unknown, so it can't be part of the formula, can it?

A second degree equation... you mean like this?
y = ax² + bx + c

I'm not really sure what to do what that... I did just find this
though, my understanding is very limited but it seems to be solving
exactly what I'm trying to solve:
http://hyperphysics.phy-astr.gsu.edu/Hbase/traj.html

On Aug 28, 9:46*pm, Miss Elaine Eos <M...@your-pants.PlayNaked.com>
wrote:
>
> To move a bullet:
>
> startx = 0, starty = 0 * * * *// starting location
> ang = degressToRadians(25) * *// 25 degress up on rifle
> speed = 1200 * * * * * * * * *// 1200 fps bullet
> time = x * * * * * * * * * * *// number of seconds after firing.
>
> // Figure out new position
> endx = startx * * * * * * * * // start location...
> * *+ cos(ang) * speed * time *// ...plus ballistics
>
> endy = starty * * * * * * * * // start location
> * *+ sin(ang) * speed * time *// ...ballistics
> * *- (time * time) * 32.15 * *// ...minus gravity
>
> Now, to solve YOUR problem...
>
> First work the above equation backwards to find out how many seconds
> from now your bullet will be at the correct X location. *(There are
> often 2 answers for this, btw.)
>
> Then work the other half of the equation backwards to figure out what
> angle you need to point at to get the Y location to be what you want.
>
> arcsine & arccosine will be involved.
>
> Luck!
>

Thanks Miss Elaine! Unfortunately my brain is still not wrapped around
this.

> First work the above equation backwards to find out how many seconds
> from now your bullet will be at the correct X location.
What exactly do you mean by "work backwards"? If I solve for time, I
still don't know angle so I can't solve the formula? And won't the
time also be affected by the angle?

Aaron Beall wrote:
> This question is in regards to a Scorched Earth style artillery game.
>
> My environment consists only of gravity (constant) and a Y plane/floor
> for projectiles. There is no air drag, wind, or obstacles.
>
> I have a known origin point, and destination point.
>
> I want to calculate the needed artillery vector from the origin point
> to land on the destination point. I don't know how to solve this
> because of the gravity.

Read this tutorial:

http://en.wikibooks.org/wiki/High_sc...jectile_motion

It has all the equations you need.

John Nagle
Animats

On Aug 29, 2:02*am, John Nagle <na...@animats.com> wrote:
>
> * * Read this tutorial:
>
> * * * *http://en.wikibooks.org/wiki/High_sc...jectile_motion
>
> It has all the equations you need.
>
> * * * * * * * * * * * * * * * * John Nagle
> * * * * * * * * * * * * * * * * Animats

Thanks so much. This looks like my answer:
http://upload.wikimedia.org/math/7/7...4665b01f35.png

"This may be solved for angle è to give the "angle" equation to hit a
target at range dh"

I think at this point I'm just having some issue implementing the math
into programming syntax. I've tried a few combinations and they all
gave close results under some situations but didn't work in others.
var angle = 0.5 * Math.sin(-1) * ((GRAVITY*range)/
(velocity*velocity));
var angle = Math.pow(Math.sin((GRAVITY*range)/
(velocity*velocity)),-1) / 2;

On Aug 29, 11:07*am, Miss Elaine Eos <M...@your-pants.PlayNaked.com>
wrote:
> In article
>
> You have confused sin^-1 (aka "arcsine") with sin(-1), which is a
> completely different thing. *I'm surprised you get it to work under ANY
> circumstances!
>
> Also, I'm a big fan of intermediate variables for equations you don't
> understand. *This facilitates stepping through with a debugger and
> understanding what's going on.
>
> Even though arcsine is written as sine to the -1 power, that's not
> really how it's calculated. *Use Math.asin(value).

Thank-you, I was definitely confused about arcsine. Somewhat
embarrasing how rusty my highschool physics are!

I seem to still be off somewhere:

GRAVITY = 3
range = target.x - origin.x
velocity = 35
angle = 0.5 * Math.asin((GRAVITY*range)/(velocity*velocity))

http://abeall.com/files/flash/tests/Game/trajectory.swf

Gray & Blue = iterative approximation to find closest angle
Red = trajectory from origin -> mouse
Green = motion formula from above, the one I'm trying to solve
Drag the origin arrow and target polygon, but note that range is
determined as x difference between origin and target at the moment. If
I can solve this for same Y position first I'll worry about different
Y elevations later.

On Aug 29, 9:49*pm, Miss Elaine Eos <M...@your-pants.PlayNaked.com>
wrote:
> In article
> <b715ad9e-11d5-4764-9a91-2f1729109...@v39g2000pro.googlegroups.com>,
> *Aaron Beall <cont...@abeall.com> wrote:
>
>
>
> > On Aug 29, 11:07*am, Miss Elaine Eos <M...@your-pants.PlayNaked.com>
> > wrote:
> > > In article
>
> > > You have confused sin^-1 (aka "arcsine") with sin(-1), which is a
> > > completely different thing. *I'm surprised you get it to work underANY
> > > circumstances!
>
> > > Also, I'm a big fan of intermediate variables for equations you don't
> > > understand. *This facilitates stepping through with a debugger and
> > > understanding what's going on.
>
> > > Even though arcsine is written as sine to the -1 power, that's not
> > > really how it's calculated. *Use Math.asin(value).
>
> > Thank-you, I was definitely confused about arcsine. Somewhat
> > embarrasing how rusty my highschool physics are!
>
> Some notes:
>
> X & Y in alphabetical order.
> cos & sin in alphabetical order.
> On the unit circle, cos = x value, sin = y value.
>
> arc[cos]sin = the angle that, on the unit circle, has that [x]y value.
>
> arc[cos] is typically limited to 0-pi (0-180°). *You have to do a tiny
> am amount of juggling if you're dealing with the other two quadrants. *
> It's simple mirror-images, though.

Thanks again for your help! I'll try to explain the reasoning behind
my madness below:

>
> > I seem to still be off somewhere:
>
> > GRAVITY = 3
>
> 3 *what*?! *gravity = ~9M/s^s or ~32ft/s^s. *I don't know any units
> where it's ~3. *I suppose you can make gravity whatever you want for
> your game...

All values are measured in pixels. So a value of 3 in this case simply
means the objects y velocities are accelerate downward at a rate of
3px per iteration. An iteration is basically a single frame, so at
30fps this value represents 90px/sec. This might not be physically
accurate, but it gives the effect I'm happy with.

>
> > range = target.x - origin.x
> > velocity = 35
>
> It's important that all your units be the same. *You can't have gravity
> in "meters per second" and velocity in "miles per hour" or any of that. *
> Pick a units & stick to it. *I typically use meters & seconds when doing
> Open-GL work. *Other architectures get whatever is convenient.

Velocity is again in pixels per iteration/frame. So at 30fps this
becomes 1050px/sec. The velocity may vary quite a bit, but it will be
a known value at the time I need to solve the projectile trajectory.

>
> > angle = 0.5 * Math.asin((GRAVITY*range)/(velocity*velocity))
>
> Where did you come up with the 0.5 factor?

I came up with 0.5 based on the part in formula where it says: ½sin-¹
from here:
http://upload.wikimedia.org/math/7/7...4665b01f35.png

Am I misreading the notation?


>
> Don't forget my suggestion to use intermediate variables. *They really
> help you understand things a lot.

When you say "intermediate variables" you mean mean something like
this?

velocity2 = velocity*velocity;
gravityRange = GRAVITY*range;
angle = Math.asin(gravityRange/velocity2)/2;

>
> >http://abeall.com/files/flash/tests/Game/trajectory.swf
>
> Ok, this looks about right. *Where's the problem?
>
> > Gray & Blue = iterative approximation to find closest angle
> > Red = trajectory from origin -> mouse
> > Green = motion formula from above, the one I'm trying to solve
>
> Oh -- Green = your calculated answer? *Well, it's wrong

Yeah, I'm smart enough to figure that part out, at least!

>
> Break it down into smaller parts. *Use intermediate variables. *Se my
> previous message (the part just before the stuff you quoted, above) for
> how you have to work-backwards to calculate 2 different parts. *Do them
> separately, in separate, intermediate variables. *(Don't optimize while
> you're still trying to get the algorithm right!)
>
> Remember that Math.foo() trig functions typically deal in radians. *I
> don't know what language you're using, but asin() & acos() might want
> you to normalize (fit onto the unit-circle) your factors.

At the moment I'm using ActionScript (Flash, hence the .swf preview I
posted), and you are right, it deals in radians.

>
> I notice in your calculation, you completely omitted time. *Time
> matters. *That's why I said you have to figure the X-part first. *(See
> previous message.) *I'm guessing that your green line is the angle you
> WOULD shoot if the bullet arrive in 1 unit of time (1s), but it doesn't,
> so it's wrong.

This is where I get the most confused, I think. While I understand
time must matter, I don't understand how it plays in to my problem. I
can't know the time it will take to reach the target until I know the
trajectory, can I? And where does time play into this function:
http://upload.wikimedia.org/math/7/7...4665b01f35.png

Thanks again for your time!

>
> > Drag the origin arrow and target polygon, but note that range is
> > determined as x difference between origin and target at the moment. If
> > I can solve this for same Y position first I'll worry about different
> > Y elevations later.
>
> Sure.
>
> --
> Please take off your pants or I won't read your e-mail.
> I will not, no matter how "good" the deal, patronise any business which sends
> unsolicited commercial e-mail or that advertises in discussion newsgroups..

>
> > Remember that Math.foo() trig functions typically deal in radians. *I
> > don't know what language you're using, but asin() & acos() might want
> > you to normalize (fit onto the unit-circle) your factors.
>
> At the moment I'm using ActionScript (Flash, hence the .swf preview I
> posted), and you are right, it deals in radians.

Ok, now I feel stupid. Lost in my confusion about the formula I forgot
to convert radians to degrees. After doing so, the result is very
close:
http://abeall.com/files/flash/tests/Game/trajectory.swf

But for some reason it is still not perfect. I would expect the
trajectory to intersect the target point exactly, instead it seems to
shift to the right as it approaches the maximum range of the specified
velocity. Any idea? Currently I have:

range = target.x - origin.x
angle = 0.5 * Math.asin((GRAVITY*range)/(velocity*velocity))
angle = angle/(Math.PI/180)