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A cannon ball is fired from a cannon horizontally
A cannon ball is fired from a cannon horizontally
04-16-2018, 11:48 AM
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#121
Originally Posted By untranslatedZA⏩
This is correct, and if this occurs, then the balls certainly will take different times to hit the ground.If the canonball is shot with enough velocity it could theoretically escape earth's gravity even if fired horizontally due to the earth being round
04-16-2018, 11:58 AM
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#122
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Originally Posted By Jacobcapra⏩
The effect of air resistance will typically be small that experiments won't pick it up. But there is a difference caused by the vertical component of air resistance being different in the two cases. If you follow the argument above, it is clear why this is the case. It is best seen just drawing a diagram of the forces acting on the ball (we are considering 2 here- gravity and air resistance). The air resistance force vector will be directly opposite the trajectory of the ball.yes I meant Vy, also realized you were taking the drag coefficient and density in the drag force equation and calling it a constant K, not calculating a resultant vector from the vertical and horizantal velocity components which would introduce the square root.
That being said, it is not true that horizontal velocity affects vertical drag.... otherwise the hundreds of experiments exactly like the OP would not have found the exact opposite conclusion.
Contemplating actually crunching the numbers to prove it
That being said, it is not true that horizontal velocity affects vertical drag.... otherwise the hundreds of experiments exactly like the OP would not have found the exact opposite conclusion.
Contemplating actually crunching the numbers to prove it
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04-16-2018, 12:00 PM
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#123
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Originally Posted By untranslatedZA⏩
If the canonball is shot with enough velocity it could theoretically escape earth's gravity even if fired horizontally due to the earth being round
Some of the answers ITT are so lulzy.
04-16-2018, 12:02 PM
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#124
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Originally Posted By wincel⏩
I don't think many in here are disagreeing with this. It's pretty clear that a ball shot vs dropped with no air resistance, flat earth, no wind, even ground etc will hit the ground at the same time. It's when air resistance is added, that some are saying itstillwill hit the ground at the same time, which is not quite correct.Alright. This is ridiculous. Let me show you *******s how you do this so you can all stfu. Let's assume there is no air resistance. Let's assume the cannonball is at a height h above the ground at the initial time t=0. We will set the origin at the initial position of the ball. Thus, the initial position of the ball will be (0,0), and the final position of the ball will have y coordinate -h. The ball will have initial horizontal velocity v0x, and initial vertical velocity v0y. The acceleration due to gravity of the ball will be assumed to be constant with a magnitude of g and a sign of -1 due to our coordinate system, the acceleration horizontally will be 0 since there is no rocket booster or any other such chit on the ball, and we will treat the ball as a point particle. Air resistance will be neglected. We will assume the Earth is approximately flat and the velocity of the cannonball is much lower than the escape velocity of the Earth. Then, the equations of motion are simply given as follows:
x_f=v0xt+0
-h=v0yt-(1/2)gt^2
Now note that t is the same for the system of equations since it is the travel time to get from the initial to the final position for the ball. Now, we note that if the ball is fired horizontally, it has no y component for its initial velocity so v0y=0. Then, if we write the y equations for each situation, we have 2h=gt^2 for the cannonball if fired horizontally and 2h=gt^2 for the cannonball if dropped from rest. Notice that the mass (and therefore, the weight) of the balls is not relevant at all under these approximations. (It would only become relevant if it were large enough relative to the mass of the Earth.) The ball takes approximately sqrt(2h/g) units of time to hit the ground.
x_f=v0xt+0
-h=v0yt-(1/2)gt^2
Now note that t is the same for the system of equations since it is the travel time to get from the initial to the final position for the ball. Now, we note that if the ball is fired horizontally, it has no y component for its initial velocity so v0y=0. Then, if we write the y equations for each situation, we have 2h=gt^2 for the cannonball if fired horizontally and 2h=gt^2 for the cannonball if dropped from rest. Notice that the mass (and therefore, the weight) of the balls is not relevant at all under these approximations. (It would only become relevant if it were large enough relative to the mass of the Earth.) The ball takes approximately sqrt(2h/g) units of time to hit the ground.
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04-16-2018, 12:03 PM
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#125
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Originally Posted By AquilaConfido⏩
your mom knows about projectile ballistics since I fired my cok into her pussy like a ballistic missile2:1 Mathematics degree here, and you're wrong. Speed matters when cannonball is fired horizontally, because F=ma.
When cannonball is dropped, equation becomes F=mg because gravity.
When fired from horizontal, acceleration must be taken into account because that will have an effect on how long the projectile will travel before it hits the ground.
Basic projectile ballistics.
When cannonball is dropped, equation becomes F=mg because gravity.
When fired from horizontal, acceleration must be taken into account because that will have an effect on how long the projectile will travel before it hits the ground.
Basic projectile ballistics.
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04-16-2018, 12:08 PM
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#126
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Originally Posted By smashedurgfx10⏩
When did you stop being a red?your mom knows about projectile ballistics since I fired my cok into her pussy like a ballistic missile
04-16-2018, 12:09 PM
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#127
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Originally Posted By TugOfPeace⏩
Well that post is wrong..I can't believe people actually think anything different from what is stated in this post would actually happen
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04-16-2018, 12:11 PM
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#128
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Originally Posted By TugOfPeace⏩
except he already admitted he was wrong dumb@ss. Tempted to start negging ITT jesus lolI can't believe people actually think anything different from what is stated in this post would actually happen
The only scenario where the horizontal would be in the air longer is if it was shot with enough speed the curvature of the earth was significant enough to dramatically alter the distance between the cannon ball and the ground. The speed necessary for that would have to be rather close to escape velocity though.
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04-16-2018, 12:12 PM
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#129
04-16-2018, 12:16 PM
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#130
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It would depend on how the cannonball leaves the barrel. If it has topspin then it'll be pushed into the ground due to Magnus force. If it has backspin the it'll "float" and stay in the air a bit longer than the dropped ball.
04-16-2018, 12:21 PM
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#131
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Lol miscers are still fighting over this huh.
To Aquila, to make you feel better, I got negged too and it's OK since I'm wrong (I guess).
Lol at miscers who wont let the time excuse pass. Oh well, I graduated college 21 yrs ago, a lot of those I learnt are gone lol. These young *******s will understand the situation in years to come.
But the good thing is that, we can always neg back
To Aquila, to make you feel better, I got negged too and it's OK since I'm wrong (I guess).
Lol at miscers who wont let the time excuse pass. Oh well, I graduated college 21 yrs ago, a lot of those I learnt are gone lol. These young *******s will understand the situation in years to come.
But the good thing is that, we can always neg back
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04-16-2018, 12:25 PM
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#132
Originally Posted By numberguy12⏩
I'll need to work through the equations for this assuming linear or even quadratic drag. I am not sure whether this is the case or not. I want to say that it should still be the same travel time, but I am going to say that is purely a guess since I haven't worked this out. I remember studying projectile motion with air resistance in classical mechanics, but it has been a while.I don't think many in here are disagreeing with this. It's pretty clear that a ball shot vs dropped with no air resistance, flat earth, no wind, even ground etc will hit the ground at the same time. It's when air resistance is added, that some are saying itstillwill hit the ground at the same time, which is not quite correct.
Edit: I think I agree with your argument after thinking about it a little.
Originally Posted By gotskillsgivjob⏩
This is another interesting point. if there is rotational motion involved, we can get pressure gradients forming in the air due to the boundary layer motion of the fluid around the ball.It would depend on how the cannonball leaves the barrel. If it has topspin then it'll be pushed into the ground due to Magnus force. If it has backspin the it'll "float" and stay in the air a bit longer than the dropped ball.
04-16-2018, 12:28 PM
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#133
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Originally Posted By jinda628⏩
Reps on recharge, those guys are idiots.Lol miscers are still fighting over this huh.
To Aquila, to make you feel better, I got negged too and it's OK since I'm wrong (I guess).
Lol at miscers who wont let the time excuse pass. Oh well, I graduated college 21 yrs ago, a lot of those I learnt are gone lol. These young *******s will understand the situation in years to come.
But the good thing is that, we can always neg back
To Aquila, to make you feel better, I got negged too and it's OK since I'm wrong (I guess).
Lol at miscers who wont let the time excuse pass. Oh well, I graduated college 21 yrs ago, a lot of those I learnt are gone lol. These young *******s will understand the situation in years to come.
But the good thing is that, we can always neg back
04-16-2018, 12:32 PM
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#134
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Originally Posted By numberguy12⏩
The vertical component of air resistance is not different in the two cases, and you can simply choose an object with a large drag coefficient and small mass so that the air resistance is not negligible compared to the force of gravity, and run the experiment. Which has been done, by the way.The effect of air resistance will typically be small that experiments won't pick it up. But there is a difference caused by the vertical component of air resistance being different in the two cases. If you follow the argument above, it is clear why this is the case. It is best seen just drawing a diagram of the forces acting on the ball (we are considering 2 here- gravity and air resistance). The air resistance force vector will be directly opposite the trajectory of the ball.
I even ran the numbers myself and found that the vertical drag on a ball with horizontal velocity is actually a littlelessthan one simply dropped, but only because equations of motion are not perfect. In reality, the vertical drags are the same.
can't believe I'm having to discuss this... I mean have you studied fluid mechanics? particle motion through a fluid is one of the most interdependent and complex models you could imagine.
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04-16-2018, 12:36 PM
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#135
04-16-2018, 12:40 PM
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#136
for anyone curious about the spin causing lift, it just wouldn't happen. With a smooth bore it's pure propulsion with next to no movement/spin. The amount of force needed to spin something with the mass of a cannonball, to the point of altering it's path, would have to beincredible and deliberate.
Skip to about 1:40 and take a look at how the ball barely moves even a fraction of an inch.
https://www.youtube.com/watch?v=zsN1GvqSG2I
Skip to about 1:40 and take a look at how the ball barely moves even a fraction of an inch.
https://www.youtube.com/watch?v=zsN1GvqSG2I
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04-16-2018, 12:43 PM
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#137
04-16-2018, 12:45 PM
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#138
Originally Posted By Jacobcapra⏩
I think there's a lot more going on in a real world application that may have affected the results, but the equations for a point particle with air resistance are pretty clear on this, and numberguy's argument seems fine to me.The vertical component of air resistance is not different in the two cases, and you can simply choose an object with a large drag coefficient and small mass so that the air resistance is not negligible compared to the force of gravity, and run the experiment. Which has been done, by the way.
I even ran the numbers myself and found that the vertical drag on a ball with horizontal velocity is actually a littlelessthan one simply dropped, but only because equations of motion are not perfect. In reality, the vertical drags are the same.
can't believe I'm having to discuss this... I mean have you studied fluid mechanics? particle motion through a fluid is one of the most interdependent and complex models you could imagine.
I even ran the numbers myself and found that the vertical drag on a ball with horizontal velocity is actually a littlelessthan one simply dropped, but only because equations of motion are not perfect. In reality, the vertical drags are the same.
can't believe I'm having to discuss this... I mean have you studied fluid mechanics? particle motion through a fluid is one of the most interdependent and complex models you could imagine.
04-16-2018, 12:46 PM
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#139
Originally Posted By chino3⏩
Good to know!for anyone curious about the spin causing lift, it just wouldn't happen. With a smooth bore it's pure propulsion with next to no movement/spin. The amount of force needed to spin something with the mass of a cannonball, to the point of altering it's path, would have to beincredible and deliberate.
Skip to about 1:40 and take a look at how the ball barely moves even a fraction of an inch.
https://www.youtube.com/watch?v=zsN1GvqSG2I
Skip to about 1:40 and take a look at how the ball barely moves even a fraction of an inch.
https://www.youtube.com/watch?v=zsN1GvqSG2I
04-16-2018, 12:49 PM
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#140
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Originally Posted By wincel⏩
Agreed. Biggest factor is probably that air is not totally homogeneous, so the particles in the air the collide with a projectile to impede it's motion, which causes drag, are not uniform and two projectiles dropping through air in different locations will actually collide with a different number of particles.I think there's a lot more going on in a real world application that may have affected the results, but the equations for a point particle with air resistance are pretty clear on this.
Also a projectile with a high velocity will cause cavitation in the surrounding air, which reduces drag. But those even being the biggest factors are negligible.
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04-16-2018, 12:50 PM
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#141
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Originally Posted By Jacobcapra⏩
lol you can talk about experiments, crunching numbers all you want. This means nothing and is unsubstantiated.The vertical component of air resistance is not different in the two cases, and you can simply choose an object with a large drag coefficient and small mass so that the air resistance is not negligible compared to the force of gravity, and run the experiment. Which has been done, by the way.
I even ran the numbers myself and found that the vertical drag on a ball with horizontal velocity is actually a littlelessthan one simply dropped, but only because equations of motion are not perfect. In reality, the vertical drags are the same.
can't believe I'm having to discuss this... I mean have you studied fluid mechanics? particle motion through a fluid is one of the most interdependent and complex models you could imagine.
I even ran the numbers myself and found that the vertical drag on a ball with horizontal velocity is actually a littlelessthan one simply dropped, but only because equations of motion are not perfect. In reality, the vertical drags are the same.
can't believe I'm having to discuss this... I mean have you studied fluid mechanics? particle motion through a fluid is one of the most interdependent and complex models you could imagine.
If you reread post #100.....you will realize why the vertical component of drag force (hence dVy/dt, hence Vy) will vary with Vx in the presence of drag force. Please specify exactly where in that post there is an error (you already tried once, and your objection was shown to be false).
To wincel- a canon ball shot in the medium of typical earth atmosphere would be best modeled with quadratic drag. Air being not so viscous of a medium, and the cannon ball being relatively large, is why this is the case (high Reynolds number).
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04-16-2018, 12:52 PM
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#142
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Here's a question...
I see that the general consensus is that they land at the same time.
HOWEVER. Wouldn't the shot bullet always land a little bit later even if it is like 0.000001 of a second?
For example, lets say a bullet flies 30000x fast than it currently does. At that speed the earths curvature would come into play, no? Because it would have a higher distance to drop... so if you shorten that distance, wouldn't the shot bullet land just a very small amount of time after the dropped bullet?
I see that the general consensus is that they land at the same time.
HOWEVER. Wouldn't the shot bullet always land a little bit later even if it is like 0.000001 of a second?
For example, lets say a bullet flies 30000x fast than it currently does. At that speed the earths curvature would come into play, no? Because it would have a higher distance to drop... so if you shorten that distance, wouldn't the shot bullet land just a very small amount of time after the dropped bullet?
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04-16-2018, 12:55 PM
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#143
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Originally Posted By numberguy12⏩
my response remains the same, you didn't show it to be false.lol you can talk about experiments, crunching numbers all you want. This means nothing and is unsubstantiated.
If you reread post #100.....you will realize why the vertical component of drag force (hence dVy/dt, hence Vy) will vary with Vx in the presence of drag force. Please specify exactly where in that post there is an error (you already tried once, and your objection was shown to be false).
To wincel- a canon ball shot in the medium of typical earth atmosphere would be best modeled with quadratic drag. Air being not so viscous of a medium, and the cannon ball being relatively large, is why this is the case (high Reynolds number).
If you reread post #100.....you will realize why the vertical component of drag force (hence dVy/dt, hence Vy) will vary with Vx in the presence of drag force. Please specify exactly where in that post there is an error (you already tried once, and your objection was shown to be false).
To wincel- a canon ball shot in the medium of typical earth atmosphere would be best modeled with quadratic drag. Air being not so viscous of a medium, and the cannon ball being relatively large, is why this is the case (high Reynolds number).
If you calculate the drag resultant vector, and break it down into it's x and y components, you will find that the y component of the drag force for a projectile in motion is always equal to the drag force on the same projectile that was dropped with no horizontal velocity.
That is exactly what I calculated, too btw. Took a 1 Kg ball with a 1 m^2 drag area, used equations of motion to determine its velocity and angle after 1 second of flight, calculated a drag force at that instant acting in the opposite direction with the same angle, broke it down into it's x and y components, and compared the y component of the drag force to the drag force on the same projectile after 1 second of free fall.
They were essentially the same, and only different because newtonian equations of motion are linear when in fact velocity changes exponentially with drag force.
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04-16-2018, 12:57 PM
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#144
Originally Posted By areyoucereal⏩
gravity gonna gravity bruhHere's a question...
I see that the general consensus is that they land at the same time.
HOWEVER. Wouldn't the shot bullet always land a little bit later even if it is like 0.000001 of a second?
For example, lets say a bullet flies 30000x fast than it currently does. At that speed the earths curvature would come into play, no?Because it would have a higher distance to drop...so if you shorten that distance, wouldn't the shot bullet land just a very small amount of time after the dropped bullet?
I see that the general consensus is that they land at the same time.
HOWEVER. Wouldn't the shot bullet always land a little bit later even if it is like 0.000001 of a second?
For example, lets say a bullet flies 30000x fast than it currently does. At that speed the earths curvature would come into play, no?Because it would have a higher distance to drop...so if you shorten that distance, wouldn't the shot bullet land just a very small amount of time after the dropped bullet?
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04-16-2018, 12:58 PM
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#145
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Originally Posted By areyoucereal⏩
Yes. The earth's curvature will be a factor (actually local variations of height of ground- hills/valleys are even more relevant), and other factors exist such as air resistance and wind.Here's a question...
I see that the general consensus is that they land at the same time.
HOWEVER. Wouldn't the shot bullet always land a little bit later even if it is like 0.000001 of a second?
For example, lets say a bullet flies 30000x fast than it currently does. At that speed the earths curvature would come into play, no? Because it would have a higher distance to drop... so if you shorten that distance, wouldn't the shot bullet land just a very small amount of time after the dropped bullet?
I see that the general consensus is that they land at the same time.
HOWEVER. Wouldn't the shot bullet always land a little bit later even if it is like 0.000001 of a second?
For example, lets say a bullet flies 30000x fast than it currently does. At that speed the earths curvature would come into play, no? Because it would have a higher distance to drop... so if you shorten that distance, wouldn't the shot bullet land just a very small amount of time after the dropped bullet?
The point is in an idealized setting, with lots of variables neglected, the balls should hit at the same time.
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04-16-2018, 12:58 PM
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#146
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Originally Posted By areyoucereal⏩
yeah, pretty much. the projectile with horizontal velocity will actually land later, all else the same because of the curvature. But if you do this experiment with Styrofoam balls inside a classroom, which will have essentially a perfectly flat floor, you can remove that variable.Here's a question...
I see that the general consensus is that they land at the same time.
HOWEVER. Wouldn't the shot bullet always land a little bit later even if it is like 0.000001 of a second?
For example, lets say a bullet flies 30000x fast than it currently does. At that speed the earths curvature would come into play, no? Because it would have a higher distance to drop... so if you shorten that distance, wouldn't the shot bullet land just a very small amount of time after the dropped bullet?
I see that the general consensus is that they land at the same time.
HOWEVER. Wouldn't the shot bullet always land a little bit later even if it is like 0.000001 of a second?
For example, lets say a bullet flies 30000x fast than it currently does. At that speed the earths curvature would come into play, no? Because it would have a higher distance to drop... so if you shorten that distance, wouldn't the shot bullet land just a very small amount of time after the dropped bullet?
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04-16-2018, 12:59 PM
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#147
Originally Posted By TugOfPeace⏩
Good god....the misc is full of retards.I can't believe people actually think anything different from what is stated in this post would actually happen
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04-16-2018, 01:02 PM
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#148
Originally Posted By numberguy12⏩
I see, and yes, after thinking about it more, I agree with your reasoning. It is interesting to think about how many different things can go into actually realistically modeling a simple ball being thrown.lol you can talk about experiments, crunching numbers all you want. This means nothing and is unsubstantiated.
If you reread post #100.....you will realize why the vertical component of drag force (hence dVy/dt, hence Vy) will vary with Vx in the presence of drag force. Please specify exactly where in that post there is an error (you already tried once, and your objection was shown to be false).
To wincel- a canon ball shot in the medium of typical earth atmosphere would be best modeled with quadratic drag. Air being not so viscous of a medium, and the cannon ball being relatively large, is why this is the case (high Reynolds number).
If you reread post #100.....you will realize why the vertical component of drag force (hence dVy/dt, hence Vy) will vary with Vx in the presence of drag force. Please specify exactly where in that post there is an error (you already tried once, and your objection was shown to be false).
To wincel- a canon ball shot in the medium of typical earth atmosphere would be best modeled with quadratic drag. Air being not so viscous of a medium, and the cannon ball being relatively large, is why this is the case (high Reynolds number).
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04-16-2018, 01:13 PM
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#149
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Assuming cannonC1is dropped from heightHand anotherC2is fired with a constant velocityu2, both from same height H, we have:
Initial Conditions:
(Apostrophe denotes vector notation)
C1 :
initial velocity vectoru1' = 0i + 0j
acceleration vectora' = 0i -gj
g : acceleration due to gravity
i, j : x and y axis / two directions of movements
C2 :
u2' = u2i + 0j
For C1 its straight forward, applying equations of motion:
s = vt + (1/2)at^2; v = u + at;
we get,
H = (gt) - (1/2)gt1^2
t1 = (2h/g)^(1/2)
Applying same equations to C2, we have:
v2' = (u2)i + (-gj)t2;
||v2|| = v2 = (u2^2 + (gt2)^2)^(1/2)
H = v2t2 - (1/2)gt2^2
this gives,
H = t2(u2^2 + (gt2)^2)^(1/2) - (1/2)g(t2)^2
we have ignored all other forces other than gravity (like air friction but that doesnt affect the final conclusion because its a simple force to model in and these equations will remain the same only the values of certain params will change),
We can clearly see that t2 (time of Cannon C2 to hit ground) is dependent on u2 (initial velocity of C2)
Initial Conditions:
(Apostrophe denotes vector notation)
C1 :
initial velocity vectoru1' = 0i + 0j
acceleration vectora' = 0i -gj
g : acceleration due to gravity
i, j : x and y axis / two directions of movements
C2 :
u2' = u2i + 0j
For C1 its straight forward, applying equations of motion:
s = vt + (1/2)at^2; v = u + at;
we get,
H = (gt) - (1/2)gt1^2
t1 = (2h/g)^(1/2)
Applying same equations to C2, we have:
v2' = (u2)i + (-gj)t2;
||v2|| = v2 = (u2^2 + (gt2)^2)^(1/2)
H = v2t2 - (1/2)gt2^2
this gives,
H = t2(u2^2 + (gt2)^2)^(1/2) - (1/2)g(t2)^2
we have ignored all other forces other than gravity (like air friction but that doesnt affect the final conclusion because its a simple force to model in and these equations will remain the same only the values of certain params will change),
We can clearly see that t2 (time of Cannon C2 to hit ground) is dependent on u2 (initial velocity of C2)
*Canadian Crew*
04-16-2018, 01:19 PM
-
#150
Originally Posted By olympianspirit⏩
wrongAssuming cannonC1is dropped from heightHand anotherC2is fired with a constant velocityu2, both from same height H, we have:
Initial Conditions:
(Apostrophe denotes vector notation)
C1 :
initial velocity vectoru1' = 0i + 0j
acceleration vectora' = 0i -gj
g : acceleration due to gravity
i, j : x and y axis / two directions of movements
C2 :
u2' = u2i + 0j
For C1 its straight forward, applying equations of motion:
s = vt + (1/2)at^2; v = u + at;
we get,
H = (gt) - (1/2)gt1^2
t1 = (2h/g)^(1/2)
Applying same equations to C2, we have:
v2' = (u2)i + (-gj)t2;
||v2|| = v2 = (u2^2 + (gt2)^2)^(1/2)
H = v2t2 - (1/2)gt2^2
this gives,
H = t2(u2^2 + (gt2)^2)^(1/2) - (1/2)g(t2)^2
we have ignored all other forces other than gravity (like air friction but that doesnt affect the final conclusion because its a simple force to model in and these equations will remain the same only the values of certain params will change),
We can clearly see that t2 (time of Cannon C2 to hit ground) is dependent on u2 (initial velocity of C2)
Initial Conditions:
(Apostrophe denotes vector notation)
C1 :
initial velocity vectoru1' = 0i + 0j
acceleration vectora' = 0i -gj
g : acceleration due to gravity
i, j : x and y axis / two directions of movements
C2 :
u2' = u2i + 0j
For C1 its straight forward, applying equations of motion:
s = vt + (1/2)at^2; v = u + at;
we get,
H = (gt) - (1/2)gt1^2
t1 = (2h/g)^(1/2)
Applying same equations to C2, we have:
v2' = (u2)i + (-gj)t2;
||v2|| = v2 = (u2^2 + (gt2)^2)^(1/2)
H = v2t2 - (1/2)gt2^2
this gives,
H = t2(u2^2 + (gt2)^2)^(1/2) - (1/2)g(t2)^2
we have ignored all other forces other than gravity (like air friction but that doesnt affect the final conclusion because its a simple force to model in and these equations will remain the same only the values of certain params will change),
We can clearly see that t2 (time of Cannon C2 to hit ground) is dependent on u2 (initial velocity of C2)
"It won't get better, just different."
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