[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]

Let's see...

Gravity is constant for both balls. Assuming that they would have hit each other in a zero-g environment, they should hit each other in an environment with gravity. Here's the breakdown in terms of ball coordinates:

-No matter the angle of the cannons, the balls will strike or pass each other at the mean of the cannons' x-coordinates.

-were there no gravity, they would strike each other at the mean of their y-coordinates, as well

-Gravity adds an equal force downward on each ball, so each ball is an equal distance lower than the zero-g y value when their x-value is equal.

Does that make sense, w/o any math or diagrams?

Edited April 30, 2008 by Budouka

[Guest]

They will not. Although they are moving the same speed in the x -direction they are not moving at the same speed in the y - direction except for at the instant they are fired. Gravity does add an equal downward force on each ball, however this means the accelerations are the same, not the velocities. The cannonball fired from the ground will be moving slower in the y-direction than the ball fired from the top throughout the entire flight of each ball.

This means that they will not reach the mean of their y-coordinates at the same time, the cannonball fired from the top of the cliff will reach it first. The cannonball fired from the ground will pass over the top of the cannonball from the top of the cliff in most situations.

Of course, there exist situations in which the cannonballs will not be fired at high enough speeds to even reach the midway point of there x-coordinates before hitting the ground, so there will be no chance in them hitting each other. This would be the easy way to prove that the cannonballs will not always hit, although it doesn't prove that they never do.

Of course, we are talking about theoretical models here, so we are considering only whether the center of masses of the objects will pass the same point at the same time. In real life if the cannonballs are big enough, the distance between the cannons is small enough, and the balls are fired fast enough, it is possible that they may hit, but it will not be a direct head on collision.

I think.

Edited April 30, 2008 by steve10

[onetruth]

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

Wouldn't it be fun if we could have a brainden get together every once in a while and actually test some of these ideas out to see whose answer held out the best? Of course, we'd need unlimited resources and an emergency/evacuation crew at hand.

[akaslickster]

yes unless you can provide a good reason why not. I found no reason why not.

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]

so the cannonballs are guilty of collision until proven innocent, akaslickster? I don't think that works...

in response to Steve10--

we can ignore the "x" coordinate of the meeting point--it will be the same no matter what. So will the amount of time that each ball has been in the air, because they both have the same exit speed. Gravity is acceleration. Acceleration=distance*time^2. No matter what, the two cannon balls will be translated an equal distance downward from the no-gravity meeting point at the time of collision/passing, because they have been in the air for the same amount of time. Since we accept that the cannonballs will meet in a zero-g environment, it follows that they will collide in an environment with gravity, as well. QED.

To be totally thorough, consider the two extreme cases (for logical consideration only--they don't quite fit the OP) of two cannons facing each other on the same level, and of one directly above the other. In the latter case, the cannonballs will collide, regardless of gravity; any fool could tell you that. In the former, you can imagine the symmetry of the ball paths, and the fact that they are predestined to collide regardless of any constant force on the y-axis. Were what you are positing true, the balls would not hit each other in the latter case.

[Guest]

I guess I am right

They will hit each other as they are under the same acceleration g.

They will be following eq

S= Vi*t + (1/2)a*t^2

If a=0 they will meet(as directed towards each other). So the change in vertical distance when a=g is same for both

[akaslickster]

so the cannonballs are guilty of collision until proven innocent, akaslickster? I don't think that works...

in response to Steve10--

we can ignore the "x" coordinate of the meeting point--it will be the same no matter what. So will the amount of time that each ball has been in the air, because they both have the same exit speed. Gravity is acceleration. Acceleration=distance*time^2. No matter what, the two cannon balls will be translated an equal distance downward from the no-gravity meeting point at the time of collision/passing, because they have been in the air for the same amount of time. Since we accept that the cannonballs will meet in a zero-g environment, it follows that they will collide in an environment with gravity, as well. QED.

To be totally thorough, consider the two extreme cases (for logical consideration only--they don't quite fit the OP) of two cannons facing each other on the same level, and of one directly above the other. In the latter case, the cannonballs will collide, regardless of gravity; any fool could tell you that. In the former, you can imagine the symmetry of the ball paths, and the fact that they are predestined to collide regardless of any constant force on the y-axis. Were what you are positing true, the balls would not hit each other in the latter case.

I presented the statement without looking at other spoilers beforehand, and solely regarding the OP. Looks like my guess has not been denied as of yet. Of course, I'm new with this and assuming cannons are able to be fired straight upward. Edited May 1, 2008 by akaslickster

[Guest]

"Wouldn't it be fun if we could have a brainden get together every once in a while and actually test some of these ideas out to see whose answer held out the best? Of course, we'd need unlimited resources and an emergency/evacuation crew at hand. "

Yes, I agree. Bonanova's Beer Bottle Balance Mobile is the one I'd love to build. (It's an oldie, do a search).

[Guest]

Somehow I read this as one cannon aimed STRAIGHT down and the other STRAIGHT up. That was confusing...

[Guest]

I'm pretty sure the answer will vary depending on how far away [the ground cannon is from the cannon atop the castle] in proportion to how far the cannons can shoot. For instance:

1=

||

||

||

||

||............................................................................=2

This scenario could yield a different answer than:

1=

||

||

||

||

||.......=2

I'm no physics wiz, and I may be completely wrong, but that seems logical to me.

[Guest]

Somehow I read this as one cannon aimed STRAIGHT down and the other STRAIGHT up. That was confusing...

I did not mean for it to read as directly below, they can of course be placed anywhere that is in range of both - incidentally why would any one fire a cannon directly downward instead of dropping it?

Also, I believe a cannon has to point at least a little up to project the ball or it falls/ rolls out. Further cannons are usually fired from a distance - you can use 100 or 1000 yards if you like, it matters not - they are exactly the same (twins in every way and so are the charges and balls - totally identical)

[akaslickster]

I did not mean for it to read as directly below, they can of course be placed anywhere that is in range of both - incidentally why would any one fire a cannon directly downward instead of dropping it?

Also, I believe a cannon has to point at least a little up to project the ball or it falls/ rolls out. Further cannons are usually fired from a distance - you can use 100 or 1000 yards if you like, it matters not - they are exactly the same (twins in every way and so are the charges and balls - totally identical)

Then what are the exact measurements between the 2 cannons using degrees, north ,south, east ,and west.?

[Guest]

Then what are the exact measurements between the 2 cannons using degrees, north ,south, east ,and west.?

choose your own to theorize with, say 2000 yards 10lb ball, 300 ft per sec, they will travel in arc, degrees are up to you they are bth aimed at each other and are accurate in their targeting.

Pretend you are Napoleone di Buonoparte training at the academy if you wish.

[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.

[akaslickster]

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.

Yes I will agree with Dredvard. If its wrong then who is correct. I missed my cannon classes. Wahhhh!

[Guest]

Being a bow hunter, I know that at the exact moment the arrow leaves the bow gravity takes over, thus arrow starts to "fall" immediately,(as with any firearm) and also shooting at something that is downhill from you may actually seem farther than what it is, and uphill seems closer than actuality, so my thinkin is that if the two cannons are aimed directly at each other, one of the balls will be high the other low, so they will pass each other and not collide.

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]

What if you wanted the cannon balls to collide at the halfway point - that's another way to look at the problem - Gravity does come into play!

[Guest]

Yes, They will hit. If we knew speed and distance apart, it would also be possible to find where they hit.

[Guest]

assuming the both cannons shoot the balls with the same amount of energy, it is pretty obvious that they don't meet, since it takes more energy to go against the gravity, therefore the lower cannon's ball results in falling short. If they have the same angles, they will not meet once again, for obvious reasons.

[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

The wording is ambiguous: are the cannons aimed in the sense that their rounds are hitting each other, or aimed in the sense that they are in each others' cross-hairs. In the former case, the answer is "possibly", and in the latter case, the answer is "definitely".

[Guest]

I don't consider myself smart enough to explain my answer scientifically but

No, the cannon balls would not collide due to trajectory, gravity and the earth being round and such.

I believe the cannon ball shot downward from the castle will pass below the canon ball being shot upward from the fround.

[Guest]

The wording is ambiguous: are the cannons aimed in the sense that their rounds are hitting each other, or aimed in the sense that they are in each others' cross-hairs. In the former case, the answer is "possibly", and in the latter case, the answer is "definitely".

They are aimed in the sense that they are to land the cannon ball on the others cannon - lets assume it is to land directly in the mouth of the other cannon and that they are the most precise experts in ballistics on both sides.

You can't use the cross hair comparison because they were aimed to curve and fall and allow for gravity and air resistance (external balistics is where you may need to look if you are curious).

[Guest]

They are aimed in the sense that they are to land the cannon ball on the others cannon - lets assume it is to land directly in the mouth of the other cannon and that they are the most precise experts in ballistics on both sides.

You can't use the cross hair comparison because they were aimed to curve and fall and allow for gravity and air resistance (external balistics is where you may need to look if you are curious).

Er, right. My question exactly. Wasn't clear whether you meant they were pointed at each other, or they were firing for effect. Also wasn't clear from several answers.

Anyway, after pondering this a bit more, I'm changing my previous answer (which was "possibly") to an emphatic "NO," apart from the trivial solution where one cannon is directly above the other, or where both are at the same elevation (which would be a kind of useless fortification...). Otherwise, the trajectories of the rounds are not only different, but they also do not intersect (except on impact), so under no circumstances can there be a collision at some intermediate point. You can do it with some math, but it's easier to explain as follows:

If the gunners at the bottom attempt to fire back in the exact direction from which they are receiving fire, their rounds will fall short. This is because the round fired at higher elevation trades potential energy for kinetic. Thus, on impact, the higher ground rounds (boo!) are traveling faster than when they were fired. Since the guns are identical, the crew below is unable to give their rounds enough kinetic energy to make it to the target along that trajectory. Therefore, those on the bottom have to adjust their trajectory to find the proper range. I think this is essentially what steve10 was trying to explain in his original post.

Another way to look at it is that the gun on top of the wall (let's call him Bert) obviously has a greater range than the one at ground level (Ernie). So if Bert can drop rounds on Ernie, while staying out of Ernie's range, there is clearly no trajectory for which Ernie can hit Bert, which obviously includes the one he is getting peppered from.

As a final note, unless the target is at the maximum range of your weapon, there are two inclination angles with which you can reach your target. In practice, the higher (closer to vertical) trajectory is almost always chosen, to minimize the range error.

[Guest]

It is still interesting that people are theorising over this but do not do math. If any body does like a little bit of mathh made easy, then here is a place to acquire assistance which takes out the guess work and 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.

Edited May 25, 2008 by Duh Puck