[Guest]

2 perfectly identical cannons aimed at each other.

First from the top of a castle and the second from the ground below.

They fire simultaneously at exactly the same speeds.

They are in range of each other, forget about wind resistance, will the cannon balls hit each?

Note that everything is similar, charge, ball etc no traps or other hidden things to look for.

simple question - simple answer y/n + theory

[Guest]

A few relevant points:

1. If we keep it simple and don't worry about air resistance, the cannonballs always trace a parabola where the horizontal speed is constant.

2. The horizontal distance covered by both cannonballs is equal.

3. Since the trajectory speed is the same for both cannons, the higher elevation cannon will always have an angle higher than the lower cannon. This is because the ball will spend more time in the air, and thus the horizontal vector of the trajectory will have to be lower.

Referring to the diagram: B will always be larger than A. Hence, no matter how you try and adjust the angles, the balls cannot meet in midair, but as long as the higher elevation cannon is within range of the lower elevation cannon, it is certainly possible for the cannons to both hit each other using the same muzzle speed, adjusting only the angle.

What would it take for the cannon balls to collide, variable angle or variable force (more/less charge).

Is it concievable that the balls can be aimed at the target and still collide mid air?

[Guest]

What would it take for the cannon balls to collide, variable angle or variable force (more/less charge).

Is it concievable that the balls can be aimed at the target and still collide mid air?

Interesting variation. Well, I'm pretty sure that two different parabolas can't intersect at more than two points, and since the trajectories have to intersect at the locations of the cannons, I would say that the only way they could collide is if they were following the exact same path. I don't see any reason that's not possible. You'd simply have to reduce the trajectory force of the higher cannon until its outgoing angle aligned with the incoming angle of the lower cannon ... and THUNK, no damage, unless some hapless bloke happened to be standing in the wrong place at the wrong time somewhere in between the cannons.

Edit: btw, my avatar is not a licorice scooby snack. It's duh puck!

Edited May 25, 2008 by Duh Puck

[Guest]

Interesting variation. Well, I'm pretty sure that two different parabolas can't intersect at more than two points, and since the trajectories have to intersect at the locations of the cannons, I would say that the only way they could collide is if they were following the exact same path. I don't see any reason that's not possible. You'd simply have to reduce the trajectory force of the higher cannon until its outgoing angle aligned with the incoming angle of the lower cannon ... and THUNK, no damage, unless some hapless bloke happened to be standing in the wrong place at the wrong time somewhere in between the cannons.

Edit: btw, my avatar is not a licorice scooby snack. It's duh puck!

It's not a variation DP- the cannons are aimed at each other, so it's simple math, do they meet midair with the same force? Perhaps a little math is required - ***** for drawing. Reason I persist is that I believe they do and have resisted calculations thus far myself so that others may input...

What flavour is the puck then?

My avatar is lemon and licorice with a somewhat spicy effect!

Edited May 25, 2008 by Lost in space

[Guest]

It's not a variation DP- the cannons are aimed at each other, so it's simple math, do they meet midair with the same force? Perhaps a little math is required - ***** for drawing. Reason I persist is that I believe they do and have resisted calculations thus far myself so that others may input...

What flavour is the puck then?

My avatar is lemon and licorice with a somewhat spicy effect!

It is a variation from your OP. Your OP said:

They fire simultaneously at exactly the same speeds.

As I explained in the post with the diagram, if they fire with the same speed, the higher elevation cannon will have to fire at a higher angle so that its trajectory has a smaller horizontal vector. In that case, it's impossible for the balls to collide midair.

You then said:

What would it take for the cannon balls to collide, variable angle or variable force (more/less charge).

Is it concievable that the balls can be aimed at the target and still collide mid air?

Force of the trajectory corresponds to the speed, so you have suggested a variation from the OP. Since parabolas can only intersect at two points, the only way the balls could collide in midair is if the force/speed of the higher cannon was reduced and its angle adjusted so that the trajectories of both balls are identical. Equal trajectories requires unequal speed, because one side has gravity helping out.

There is absolutely no need for math to prove this is the case. All that's needed to prove it is to know that trajectories always follow parabolas (discounting air resistance, that is, which is what your first point stated). Incidentally, it's probably not very difficult to use time-distance equations (Physics 101) to that this is the case, but I can't bring myself to waste that kind of time to prove something when it's not necessary.

In retrospect, I can't think of a flavor for a pure black puck other than licorice. Licorice it is!

[Guest]

In retrospect, I can't think of a flavor for a pure black puck other than licorice. Licorice it is!

Well I accept that without proof

My thought is that under normal circumstances cannon fire is for maximum effect (destruction), it's not like shooting a human cannon ball to reach a target - could you be missing something here? Two parabolas with the same start finnish Point, cannons fired to blast the opponent! - possibly the same paths????

[Guest]

Well I accept that without proof

My thought is that under normal circumstances cannon fire is for maximum effect (destruction), it's not like shooting a human cannon ball to reach a target - could you be missing something here? Two parabolas with the same start finnish Point, cannons fired to blast the opponent! - possibly the same paths????

How is it different than shooting a human cannonball? A human and a round piece of metal will follow the exact same rules of physics, especially if you're not considering air resistance.

If you want maximum effect, you need to increase speed at the point of impact. The only way to do this is to increase speed at the point of release. That's muzzle velocity. If your velocity is high enough, you can aim almost directly at the target (like you would with a gun). Even so, the path will still be a parabola, and I'm pretty sure my previous conclusions still hold. I already stated they can follow the same path and collide in midair, but not if they fire with equal force. Consider this diagram:

Both follow the same parabolic arc, and the balls can collide. However, this is not possible if they are fired with the same muzzle velocity. In this case, since there is no upwards vector of the higher cannon's trajectory, its ball would actually spend less time in the air (gravity is speeding it up), and its muzzle velocity would have to be higher than the lower cannon's ... uh oh. Guess that means I was missing something, doesn't it? If the higher cannon requires a higher velocity to match paths in this instance, but required a lower velocity in the previous case, then it would follow that there's some point where the two solutions intersect and you could indeed have matching trajectories with identical muzzle velocities. Interesting. Unfortunately don't have time to try it tonight. Maybe tomorrow, if somebody else doesn't get there first.

[Guest]

Both follow the same parabolic arc, and the balls can collide. However, this is not possible if they are fired with the same muzzle velocity. In this case, since there is no upwards vector of the higher cannon's trajectory, its ball would actually spend less time in the air (gravity is speeding it up), and its muzzle velocity would have to be higher than the lower cannon's ... uh oh. Guess that means I was missing something, doesn't it? If the higher cannon requires a higher velocity to match paths in this instance, but required a lower velocity in the previous case, then it would follow that there's some point where the two solutions intersect and you could indeed have matching trajectories with identical muzzle velocities. Interesting.

Looks like there is more to consider.

What would make them meet mid air. while still aimed destined to reach target? Is that possible?

What would be equal force considerring gravity will pull on both balls?

[Guest]

I think the simplest way to look at it is to look at the motion of the first cannon relative to the second. Asume the second ball is stationary, effectively making the velocity of the first one double. Its still aimed at the second one so if you neglect gravity it will hit it. If you add gravity in, both balls are on the same accelerating reference frame, so its like you shot a puck on an air hockey table and then lifted the table up. The table acts on both the same so it cancels each other out.

So yes it will hit.

I like this question! It's properly confusing. For a while I was sure Dredvard had hit the nail on the head. As Einstein observed, the effects of gravity are exactly the same as the effects of acceleration. Suppose the cannons were positioned on opposite sides of a space rocket which was accelerating at 1g. The paths of the cannonballs would be the same, relative to the rocket, as they would be if there were normal gravity. HOWEVER, the cannonballs (if they didn't collide) would not necessarily hit their targets at the same time. The cannons have to aim to hit where the other cannon will be when the cannonball hits it (allowing for acceleration), and since that will not be at the same time, it is not the same path. That way of looking at it really confuses the issue! My head hurts.

Rewind... start again... putting it simply there would be 2 different parabolas (the lower cannon will launch the cannonball at a higher velocity than the velocity it ends at, vice versa for the other cannon, and since they launch at the same speed they are not the same parabola) and I'm pretty sure that on its own guarantees they will not cross.

But we need maths to prove it! Maths coming up...

[Guest]

I like this question! It's properly confusing. For a while I was sure Dredvard had hit the nail on the head. As Einstein observed, the effects of gravity are exactly the same as the effects of acceleration. Suppose the cannons were positioned on opposite sides of a space rocket which was accelerating at 1g. The paths of the cannonballs would be the same, relative to the rocket, as they would be if there were normal gravity. HOWEVER, the cannonballs (if they didn't collide) would not necessarily hit their targets at the same time. The cannons have to aim to hit where the other cannon will be when the cannonball hits it (allowing for acceleration), and since that will not be at the same time, it is not the same path. That way of looking at it really confuses the issue! My head hurts.

Rewind... start again... putting it simply there would be 2 different parabolas (the lower cannon will launch the cannonball at a higher velocity than the velocity it ends at, vice versa for the other cannon, and since they launch at the same speed they are not the same parabola) and I'm pretty sure that on its own guarantees they will not cross.

But we need maths to prove it! Maths coming up...

So the race is on between you and DP!

Take the zero gravity moving rockets with same accelertaion but one is a 1000 meters or yards ahead as Big shots 2 if you wish!

[Guest]

OK, here's proof I think.

The paths of the cannonballs are parabolic (proof of this is a bit trivial so I'll leave it out). So they be expressed as:

y₁=a₁x+b₁x²

y₂=a₂x+b₂x²

where y is the height relative to the 1st cannon and x is the distance from the 1st cannon, so they must both pass through (0,0).

The difference in height at any point x is

y₁-y₂=(a₁-a₂)x+(b₁-b₂)x²

which is zero only at the origin and where x = (a₁-a₂)/(b₂-b₁), unless a₁=a₂ and b₁=b₂ in which case it's zero everywhere (same parabola).

So either they only meet at 2 points, which is where the 2 cannons are (cannot meet in the middle), or a₁=a₂ and b₁=b₂ and the 2 paths are the same.

Are they the same path? On the parabola y=ax+bx², the rate of change of gradient d²y/dx²=2b, a constant. So another cannonball following the same parabola, from a different position, must have the same value of b. A little calculation reveals the value of b to be -1/2(g/h²) where h is the horizontal velocity of the cannonball. Since g is constant, to get the same value of b, you need the same value of h. For that to happen the 2 cannonballs must have been shot at the same angle. But the only other place where the gradient of the parabola is the same is at y=0, so the other cannon must have been at the same height. Which it wasn't. So the parabola could not be the same! Ta da! QED

Edited May 26, 2008 by octopuppy