Technical explanation required

[shotgun sam]

I had a recent conversation with my father yesterday where He asked me a question about shooting “ what hits the ground first a bullet fired or a bullet dropped” My reply based on My school days was they both hit the ground at the same time My old physics teacher was next to us and said that was the right answer.

 

However I said the real answer is the dropped bullet as the fired one actually rises slightly before it begins to drop.

 

Can any of you technically minded people explain how this happens in a format I can pass to the physics teacher.

 

Thanks

 

Sam

[Billy.]

It was tested not long ago. They still hit the ground at the same time, as gravity acts equally on both objects.

 

Your statement about it rising (I'm guessing to have a zero) would invalidate the test.

 

Edited September 29, 2011 by Billy.

[Beardo]

However I said the real answer is the dropped bullet as the fired one actually rises slightly before it begins to drop.

 

don't be fooled into thinking the bullet rises before it drops by looking at a ballistic table - that's only in relation to the scope's bore - the scope is in fact angled down in relation to the barrel's bore. It's just easier to look at a table drawn that way.

[shotgun sam]

Someone said to me it is to do with the rifling in the bore that spins it in an upward direction also there is a fraction of a second difference while the bullet is in the barrel the one dropped is dropping due to gravity.

[Billy.]

Someone said to me it is to do with the rifling in the bore that spins it in an upward direction also there is a fraction of a second difference while the bullet is in the barrel the one dropped is dropping due to gravity.

 

You would time the bullet dropping from when it leaves the barrel, not when it is inside it.

 

It's the same with ballistics programs, which time from the muzzle, not from when the trigger is pulled.

Edited September 29, 2011 by Billy.

[timps]

The geeky answer is.

 

In a vacuum with a floor designed to take into account the curvature of the earth both bullets would hit at the same time, but the factor to take into consideration is air resistance and its effect on the moving bullet.

 

The greater velocity of the fired bullet would be effected more by air resistance (drag) than the dropped bullet.

 

At the start both bullets have the same force in the y-direction (vertical).

However,the fired bullet is moving horizontally so the air resistance is initially only in the x-direction (Horizontal), but the air resistance on the y-direction for the fired bullet still depends on the fired speed of the bullet since it is proportional to its velocity.

 

Thus the drag experienced in the y-direction by the fired bullet is much higher than the drag experienced by the dropped bullet affecting how quickly it falls to earth.

 

This greater drag would mean the fired bullet would hit the ground later.

 

The physical size of the bullet (calibre) and the initial velocity (Super-sonic and subsonic) of the bullet would alter how much this drag would affect the bullet.

 

What this equates to in time difference I don’t know as the formulas get complicated, but it is mathematically provable that velocity will affect the drag in both the X and Y axis simultaneously.

[-Mongrel-]

The geeky answer is.

 

In a vacuum with a floor designed to take into account the curvature of the earth ...

This greater drag would mean the fired bullet would hit the ground later...

 

... but it is mathematically provable that velocity will affect the drag in both the X and Y axis simultaneously.

 

So, care to explain the Mythbusters clips results?

 

As I understood it, the velocity of the fired bullet and resultant horizontal resistance will only slow its horizontal velocity. Gravity will still pull them down at equal speeds, resulting in them landing together.

[timps]

So, care to explain the Mythbusters clips results?

 

As I understood it, the velocity of the fired bullet and resultant horizontal resistance will only slow its horizontal velocity. Gravity will still pull them down at equal speeds, resulting in them landing together.

 

For a start there was a difference in time between the two bullets landing in the mythbusters clip no matter how small (I never said it was large).

If the bullet was a faster velocity fired from a greater height over a longer distance there would be a bigger difference.

Also I don’t know efficacy of mythbusters test but I have read the formulas.

 

Fluid dynamics is very complicated, gravity exerts the same force on both bullets, but the simple fact is velocity affects drag on x & Y planes as its proportional according to the formula. The fact the fired bullet has a higher velocity means it has higher drag even though gravity is pulling them down with the same force, the higher the drag the more force need to pull it down at the same speed or the same force but lower speed.

 

Drag is proportional to the square of the speed and opposite to the velocity vector, then adding horizontal speed increases vertical drag.

 

 

 

If you wish to read the formula then read the links and text below as they are a lot more knowledgeable than I am at explaining the maths and no point in me repeating their work.

 

Drag HAS to considered a force with a vector opposite direction of travel (in still air).

 

To keep the math easy, here is an order-of-magnitude calculation.

 

Gravity is 10.0 m/s.

 

Bullet dropped is going 10.0 m/s at t = one second.

Drag (fdis roughly square of velocity, or 100. Vertical component of that (dvert) is 100 (all of it).

 

Bullet fired at 500 m/s. It has a horizontal velocity (vh) of 300 m/s at t = one second. We'll say it also has a vertical velocity (vv) of 10m/s at t = one second.

 

Velocity = square root of (vh2 + vv2) = 300.167 m/s.

 

fd, Total Drag, is the square of velocity, or about 90,100.

 

The vertical component of that drag has the same ratio to total drag as the vertical component of velocity to total velocity, or

 

dvert/90100 = 10/300

 

dvert = 3003.

 

The squaring makes all the difference.

 

http://scienceblogs.com/dotphysics/2009/10/mythbusters-bringing-on-the-physics-bullet-drop.php

 

 

http://www.nennstiel-ruprecht.de/bullfly/faq.htm#Q9

 

 

http://hyperphysics.phy-astr.gsu.edu/hbase/grav.html#bul

 

In the above link he quantifies it by saying "If air friction is neglected, then the drop of a bullet fired horizontally can be treated as an ordinary horizontal trajectory. The air friction is significant, so this is an idealization."

Air friction is drag so affects the drop time.

Edited September 29, 2011 by timps

[casts_by_fly]

The geeky answer is.

 

In a vacuum with a floor designed to take into account the curvature of the earth both bullets would hit at the same time, but the factor to take into consideration is air resistance and its effect on the moving bullet.

 

The greater velocity of the fired bullet would be effected more by air resistance (drag) than the dropped bullet.

 

At the start both bullets have the same force in the y-direction (vertical).

However,the fired bullet is moving horizontally so the air resistance is initially only in the x-direction (Horizontal), but the air resistance on the y-direction for the fired bullet still depends on the fired speed of the bullet since it is proportional to its velocity.

 

Thus the drag experienced in the y-direction by the fired bullet is much higher than the drag experienced by the dropped bullet affecting how quickly it falls to earth.

 

This greater drag would mean the fired bullet would hit the ground later.

 

The physical size of the bullet (calibre) and the initial velocity (Super-sonic and subsonic) of the bullet would alter how much this drag would affect the bullet.

 

What this equates to in time difference I don’t know as the formulas get complicated, but it is mathematically provable that velocity will affect the drag in both the X and Y axis simultaneously.

 

 

timps,

 

This is absolutely incorrect. If you want to break it down into vector functions, you have an x-component (direction of travel perpendicular to the pull of gravity) and a y-component of travel (direction of travel parallel to gravity). In the X direction, you have velocity pushing the bullet forward and drag from the bullets velocity slowing it down. In the y-direction you have the pull of gravity pulling the bullet down and the air resistance of the bullet in that direction (caused by the bullets y-velocity) resisting gravity.

 

As soon as the bullet is fired (or dropped) gravity starts to cause an increase in velocity in the y direction (i.e. move the bullet toward the earth). The velocity the bullet is moving in that direction determine the air resistance in that direction. Both bullets will start from zero velocity in that direction, so the air resistance for both bullets at the start is 0 and grows with increasing velocity. The air resistance against gravity will solely depend on the velocity imparted by gravity and for both bullets will be identical. The x-component velocity does not impart a drag function in the y-direction and thus has no bearing on the speed at which the bullet drops to the earth.

 

Thanks,

Rick

[timps]

timps,

 

This is absolutely incorrect. If you want to break it down into vector functions, you have an x-component (direction of travel perpendicular to the pull of gravity) and a y-component of travel (direction of travel parallel to gravity). In the X direction, you have velocity pushing the bullet forward and drag from the bullets velocity slowing it down. In the y-direction you have the pull of gravity pulling the bullet down and the air resistance of the bullet in that direction (caused by the bullets y-velocity) resisting gravity.

 

As soon as the bullet is fired (or dropped) gravity starts to cause an increase in velocity in the y direction (i.e. move the bullet toward the earth). The velocity the bullet is moving in that direction determine the air resistance in that direction. Both bullets will start from zero velocity in that direction, so the air resistance for both bullets at the start is 0 and grows with increasing velocity. The air resistance against gravity will solely depend on the velocity imparted by gravity and for both bullets will be identical. The x-component velocity does not impart a drag function in the y-direction and thus has no bearing on the speed at which the bullet drops to the earth.

 

Thanks,

Rick

 

 

Because the flow of air around the fired bullet is turbulent the air resistance is proportional to v^2. Your statement would be correct if the air resistance were proportional to v.

In simple terms as the fired bullet cuts through the air at speed it affects the air around it, this turbulence affects the drag on its downward motion as it’s not the same still air as in the case of the dropped bullet.

 

If the above were not the true and both X & Y were totally separate in air, (they are separate in a vacuum) then the principle of lift on aeroplane wing would not exist, no matter how fast you went down the runway.

There is no lift on a bullet but the fact it is displacing air out of the way at speed means it’s affecting the air all around its length (similar to a wing). The affect this has on the bullets vertical drag is highlighted in the equation below.

 

Fluid dynamics is very complicated, the formulas I listed above and also in my post above are from professors, academics and are listed on many academic web sites & physics forums, if you accept they are correct (I do) then you have to accept horizontal speed affects vertical drag.

 

If you think the formulas are incorrect then highlight were they are wrong, if you don’t understand them enough to point out where they are wrong you have to accept the professors and academics who wrote them know more than you or I on this subject.

 

I come from a science background, I understand the theory and accept the formulas make sense but I do accept them on face value (not seen anyone prove they are incorrect) but then again I accept F=MA on face value without doing my own tests.

 

Cheers

Craig

[poontang]

:o

 

Ctaig,

 

You are officially the Geekmeister General.

 

I salute you

[Catweazle]

Also, the fired bullet will be spinning very fast. Would this spin mean that the path towards the ground would not be perfectly vertical ? I think it would drop at an angle, meaning the fired bullet would have further to travel.

[Dekers]

I had a recent conversation with my father yesterday where He asked me a question about shooting “ what hits the ground first a bullet fired or a bullet dropped” My reply based on My school days was they both hit the ground at the same time My old physics teacher was next to us and said that was the right answer.

 

However I said the real answer is the dropped bullet as the fired one actually rises slightly before it begins to drop.

 

Can any of you technically minded people explain how this happens in a format I can pass to the physics teacher.

 

Thanks

 

Sam

 

If you could replicate this with all the variations (twist, drag, angle, atmospherics, etc) removed the answer is YES!

 

But Guys, we are talking the detail here, and the answer is actually NO, because you can't do that!

 

Forget the angle of shot, twist rate, air resistance, BC, etc, etc

 

The difference is tiny and will change by minute fractions with circumstances.

 

If you have the kit to measure it, or the brain to be concerned by it ATB!

 

[timps]

:o

 

Ctaig,

 

You are officially the Geekmeister General.

 

I salute you

Thanks I love the fact everyone thinks because I am big, raced bikes, worked the doors & into martial arts with a dumb northern accent I was not a geek.

 

Also, the fired bullet will be spinning very fast. Would this spin mean that the path towards the ground would not be perfectly vertical ? I think it would drop at an angle, meaning the fired bullet would have further to travel.

 

Don’t get me started on centrifugal now known as not a real force or centripetal which does apparently still exist. I know a spinning object falls to earth quicker than a non spinning object due to my motocross days. Tabletop jump hit the back brake stop the rear wheel spinning it rises front wheel drops for the landing. Endo rev the engine rear wheel spins more it drops and front end rises.

 

If you could replicate this with all the variations (twist, drag, angle, atmospherics, etc) removed the answer is YES!

 

But Guys, we are talking the detail here, and the answer is actually NO, because you can't do that!

 

Forget the angle of shot, twist rate, air resistance, BC, etc, etc

 

The difference is tiny and will change by minute fractions with circumstances.

 

If you have the kit to measure it, or the brain to be concerned by it ATB!

 

In the real world we would not see the difference but I am a geek so ner :lol:

[timps]

ps thanks for using my drinking name poontang

Edited October 1, 2011 by timps

[Kes]

Craig, you signed yor post above with yor drinking name but as others have said what about the 'Coreolis effect' ?

Surely this would increase the path travelledand add minutely to the time of travel of a fired bullet?

Also what about the micro texture of the bullets surface - this should have a greater drag effect at high velocity and less at lower 'dropped' velocities.

I'm sure the dropped bullet hits the ground first but the time difference is, other than to the scientist, a negligeable proportion of the total travel time.

[Catweazle]

Don’t get me started on centrifugal now known as not a real force or centripetal which does apparently still exist. I know a spinning object falls to earth quicker than a non spinning object due to my motocross days. Tabletop jump hit the back brake stop the rear wheel spinning it rises front wheel drops for the landing. Endo rev the engine rear wheel spins more it drops and front end rises.

 

That's a different effect. When you put your back brake on you link the spinning wheel to the bike, so it rotates with the wheel, hence nose goes down. When you blip the throttle the bike tries to rotate in the opposite direction to the wheel, opposite reaction and all that, hence nose goes up.

 

Road bikes are interesting too, engines are often designed to rotate in the opposite direction to the wheels, to cancel out some of the gyroscope effect and aid quick steering.

[mad1]

Because the flow of air around the fired bullet is turbulent the air resistance is proportional to v^2. Your statement would be correct if the air resistance were proportional to v.

In simple terms as the fired bullet cuts through the air at speed it affects the air around it, this turbulence affects the drag on its downward motion as it’s not the same still air as in the case of the dropped bullet.

 

If the above were not the true and both X & Y were totally separate in air, (they are separate in a vacuum) then the principle of lift on aeroplane wing would not exist, no matter how fast you went down the runway.

There is no lift on a bullet but the fact it is displacing air out of the way at speed means it’s affecting the air all around its length (similar to a wing). The affect this has on the bullets vertical drag is highlighted in the equation below.

 

Fluid dynamics is very complicated, the formulas I listed above and also in my post above are from professors, academics and are listed on many academic web sites & physics forums, if you accept they are correct (I do) then you have to accept horizontal speed affects vertical drag.

 

If you think the formulas are incorrect then highlight were they are wrong, if you don’t understand them enough to point out where they are wrong you have to accept the professors and academics who wrote them know more than you or I on this subject.

My head hurts

I come from a science background, I understand the theory and accept the formulas make sense but I do accept them on face value (not seen anyone prove they are incorrect) but then again I accept F=MA on face value without doing my own tests.

 

Cheers

Craig

 

My head hurts

[timps]

Craig, you signed yor post above with yor drinking name but as others have said what about the 'Coreolis effect' ?

Surely this would increase the path travelledand add minutely to the time of travel of a fired bullet?

Also what about the micro texture of the bullets surface - this should have a greater drag effect at high velocity and less at lower 'dropped' velocities.

I'm sure the dropped bullet hits the ground first but the time difference is, other than to the scientist, a negligeable proportion of the total travel time.

Poontang is a friend of mine on facebook, the local pub sent me a load of drinks vouchers with my name spelt Ctaig.

 

Getting back on topic, the original poster asked for a technical explanation to give to a scientist (his old physics teacher) as to why the fired and dropped bullets did not hit the ground at the same time.

 

To a scientist the equation I gave makes sense, vertical drag is proportional to velocity therefore the drag on the bullets is not the same. Because of this (from a scientific point of view) they can never hit at the ground at same time whilst in air or any fluid for that matter.

 

On a slow moving heavy bullet the drag will have a very small influence but on a lighter faster bullet the influence would be greater.

 

There are many other factors to take into consideration as you rightly point out, but unless they exactly cancel out the difference in drag between the two bullets then my technical answer for a technical person (physics teacher) the op asked for still stands. The bullets cannot hit at the same time.

 

It’s like the old joke.... engineer :- I work to a thousand of an inch ......Bricklayer :- that’s nothing I have to be spot on.

 

To the average Joe they hit at the same time but to a scientist there is a provable difference. If a scientist ignores this equation on a bullet because it’s a small difference, he would then have to ignore the same equation on something a lot lager travelling a lot faster where the difference would be visible to the naked eye.

[timps]

That's a different effect. When you put your back brake on you link the spinning wheel to the bike, so it rotates with the wheel, hence nose goes down. When you blip the throttle the bike tries to rotate in the opposite direction to the wheel, opposite reaction and all that, hence nose goes up.

 

Road bikes are interesting too, engines are often designed to rotate in the opposite direction to the wheels, to cancel out some of the gyroscope effect and aid quick steering.

 

yep you are right, but so am I there is more than one force at work when you stop the spinning wheel.

 

Rotating objects falling in a gravitational field are accelerated at a rate greater than "G", the commonly accepted rate for non-rotating objects falling in a vacuum.

 

Throw a push bike wheel in the air with spin it will fall a lot quicker than the same wheel without any spin.

 

My head hurts

 

so does mine but that is due to the fact I was out drinking last night

[timps]

Just to answer your questions as I didn't before.

 

as others have said what about the 'Coreolis effect' ?

Surely this would increase the path travelledand add minutely to the time of travel of a fired bullet?

It does not matter how far the bullet travels horizontally as long as the curvature of the earth is taken into consideration the vertical drop is the same for both bullets. The rotation of the bullet will make it fall to earth quicker as rotating objects falling in a gravitational field are accelerated at a rate greater than "G".

 

Also what about the micro texture of the bullets surface - this should have a greater drag effect at high velocity and less at lower 'dropped' velocities.

That's my point, that's what the equation is all about.

 

Point is that in a vacuum with a none rotating bullet they hit the deck at the exact same time no matter how fast the bullet is travelling.

[HW682]

 

Its like the old joke.... engineer :- I work to a thousand of an inch ......Bricklayer :- thats nothing I have to be spot on.

 

 

That reminds me of the old joke about the Science student and the Engineering student placed at one side of a room and an attractive young member-of-the-opposite-sex (this is the modern PC version ) at the other side.

 

They are told they can cross half way across the room, then half the remaining distance, then half the remaining distance etc etc.

After a few seconds thought the Science student looks depressed and announces that there is no point as even with infinite time he will never actually get there.

The Engineering student replies "maybe not, but I will pretty soon get close enough!"

 

I agree with the last few comments. It is a case of a little knowledge being dangerous.

To someone with no or very little knowledge of the situation it seems "common sense" or intuition tells you it is obvious that one gets there before the other.

As you get a bit more knowledge you can do the schoolboy calculations (which usually start by excluding air resistance etc ) and treat the vertical and horizontal resloved components of velocity as independent and show that they both land at the same time. In the real world, there will be lots of effects which will influence the result. As already said, there will be a small difference - but too small for most people to measure.

 

Lets face it - if you dropped two identical bullets at the same time, if you could measure with enough precision, you would find that they wouldn't land at EXACTLY the same time. If they were truely identical then over a series of tests you would expect bullet 1 to hit first 50% of the time and bullet 2 to hit first 50% of the time.

If you repeated the experiment with one dropped and one fired I am confident that the split would shift from 50/50. If dropped from a bigger height and fired with a higher speed (to maximise the effects and give them longer to take effect) then the split would move to 100% or at least much closer to it in favour of one of them.

Edited October 1, 2011 by HW682

[keg]

Timps, nowt wrong with a northern accent, it's just that you're from the wrong side of the hills, we in the broad acres are far superior

[Muddy Funker]

Jesus, my head nearly exploded reading this thread!