Help me settle this argument about bullet travel.

[DeputyMend]

Will a bullet shot straight up travel as far as a bullet shot horizontally? For the horizontal shot assume the bullet is shot at whatever angle maximizes the distance traveled.

I think that the bullet will not travel as far vertically as it will horizontally. I don’t know why I think that.

Don’t let me down guys!

[ForTehNguyen]

the vertical distance will be a lot less than the horizontal distance (assuming there isnt a ground to stop it prematurely). Force of gravity during vertical travel is far greater than the aerodynamic drag on the bullet when its traveling horizontally, thus the travel distances will be lower when fired vertically.

[chadbag]

the vertical distance will be a lot less than the horizontal distance (assuming there isnt a ground to stop it prematurely). Force of gravity during vertical travel is far greater than the aerodynamic drag on the bullet when its traveling horizontally, thus the travel distances will be lower when fired vertically.

I am too stupid to answer this (ie actually figure things out) but remember that gravity and air resistance affect both shots.

[NMBigfoot02]

I think the vertical shot will be shorter (only talking about the travel directly up, not when it runs out of steam and starts falling).

When the gun is fired horizontally, there is the drag force hindering its travel in the x direction while gravity pulls it toward the earth. In other words, the forces are perpendicular to each other, meaning that the drag force is really the limiting factor on the bullet's distance (assuming the bullet can fall forever.)

When you fire a bullet straight in the air, the drag force and gravity now act in the same direction, resulting in a much greater resistance force to the bullets travel.

This might be oversimplified, but there's no easy way to prove it or test it.

[Zhurdan]

Since when does the law of gravity change due to orientation of the mass's travel (at realistic distances from COE) or are you saying that the force of gravity is more of a factor in it's "flight time" than the aerodynamic drag? Either way you shoot it though, it's gotta cut thru the air regardless of if it is vertical or horizontal.

If you're only accounting for flight time to the apex of travel, I'd say that the vertical shot would cover more distance, but if accounting for maximum flight time including the back half of the trajectory (falling), I'd say a 45degree shot is going to cover more ground. But simply considering it from a "flight time" point of view, the vertically fired bullet will spend far more time in the air than one fired perpendicular to the ground on a perfectly flat surface. Dammit, now I'm gonna have to go digging for my books again!:D

[Complication]

It's been a while since high school physics (in which I got an A, if I recall ;))

Let's take a muzzle velocity of 425 meters/sec (handgun), but it should be the same for any velocity:
Note: these calculations don't take into account air resistance, earth curvature, wind, etc. etc.

Shot straight up, it'll take t seconds to reach a velocity of 0 m/s (where it stops going up and starts going down)
t = v/g
v is initial velocity (425m/s) and g is gravity (9.812865328m/s/s)
t = (425m/s) / (9.812865328m/(s^2))
t = 43.31048942s
And another t seconds to come back down. (This time v would be final velocity)
Total time = 2t = 86.62097885s
Now, distance traveled in that time is average velocity (1/2* 425m/s = 212.5m/s) multiplied by time (86.62097885s) = 18,406.958 meters.

Shot at an angle to optimize distance (45 degrees) you have two components: horizontal velocity (used to calculate distance traveled) and vertical velocity (used to calculate time to drop). They will be the same.
These form a triangle where the hypotenuse is 425m/s. Remember the Pythagorean theorem: A^2 + B^2 = C^2. And also that in a 90-degree triangle (where the angle we care about is 45 degrees), the base and height are equal.
C = 425 m/s
C^2 = 180,625
A^2 = 1/2 * 180,625 (A = B, A^2 = B^2)
A^2 = 90,312.5
A = Sqrt (90,312.5)
A = 300.520382 m/s
B = A
Now we know how fast it is going up (to cacluate time in air) and how fast it is going forward (to calculate how far it goes)
Same calculation as before, different numbers:
Time in air = 2t
t = v/g
t = (300.520382m/s) / (9.812865328m/(s^2))
t = 30.62514077s
2t = 61.25028154s
Now, total (horizontal) distance traveled = t * average velocity (300.520382m/s)
Total Distance traveled = 18,406.958m

The exact same distance.

If you want to compare VERTICAL distance shot up and HORIZONTAL distance shot at 45 degrees, they're the same.
If you want to compare VERTICAL distance shot up and VERTICAL distance shot at 45 degrees, the one shot upwards goes further.
If you want to compare TOTAL distance traveled, the one shot in an arc (45 degrees) travels further.

Now, even though the bullet is in the air for a shorter period of time (61 vs. 86 seconds) it is traveling (in the direction we care about) at a faster average speed (300 vs. 212 m/s). That is why the numbers come out the same.

Now, if you want an answer if you INCLUDE wind resistance, my guess is that the 45-degree shot will travel a shorter distance since it is in the air longer and therefore wind resistance will have a greater accumulated effect. Never learned that, though ;).

[Zhurdan]

As far as a cool experiment, check this out from Mythbusters (http://www.youtube.com/watch?v=D9wQVIEdKh8)

[NMBigfoot02]

Since when does the law of gravity change due to orientation of the mass's travel (at realistic distances from COE) or are you saying that the force of gravity is more of a factor in it's "flight time" than the aerodynamic drag? Either way you shoot it though, it's gotta cut thru the air regardless of if it is vertical or horizontal.

If you're only accounting for flight time to the apex of travel, I'd say that the vertical shot would cover more distance, but if accounting for maximum flight time including the back half of the trajectory (falling), I'd say a 45degree shot is going to cover more ground. But simply considering it from a "flight time" point of view, the vertically fired bullet will spend far more time in the air than one fired perpendicular to the ground on a perfectly flat surface. Dammit, now I'm gonna have to go digging for my books again!:D

I was referring to a horizontal shot with the barrel oriented at 0 deg compared to a vertical shot with the barrel at 90 deg. I missed the part where the OP stated that is would be at an angle to maximize the distance.

When the barrel is at 0 deg, the force of gravity will be acting directly downward, or 270 deg and the drag will be acting in the opposing direction of the bullet's travel, or 180 deg. When the barrel is at 90 deg, both the gravity and drag force will be acting downward at 270 degrees. Since the acceleration of gravity is constant and the drag force is related to the velocity of the bullet, they should not change with the barrel's orientation. Basically, the forces will now "stack" on each other, slowing the bullet faster.

[skyugo]

It's been a while since high school physics (in which I got an A, if I recall ;))

Let's take a muzzle velocity of 425 meters/sec (handgun), but it should be the same for any velocity:
Note: these calculations don't take into account air resistance, earth curvature, wind, etc. etc.

Shot straight up, it'll take t seconds to reach a velocity of 0 m/s (where it stops going up and starts going down)
t = v/g
v is initial velocity (425m/s) and g is gravity (9.812865328m/s/s)
t = (425m/s) / (9.812865328m/(s^2))
t = 43.31048942s
And another t seconds to come back down. (This time v would be final velocity)
Total time = 2t = 86.62097885s
Now, distance traveled in that time is average velocity (1/2* 425m/s = 212.5m/s) multiplied by time (86.62097885s) = 18,406.958 meters.

Shot at an angle to optimize distance (45 degrees) you have two components: horizontal velocity (used to calculate distance traveled) and vertical velocity (used to calculate time to drop). They will be the same.
These form a triangle where the hypotenuse is 425m/s. Remember the Pythagorean theorem: A^2 + B^2 = C^2. And also that in a 90-degree triangle (where the angle we care about is 45 degrees), the base and height are equal.
C = 425 m/s
C^2 = 180,625
A^2 = 1/2 * 180,625 (A = B, A^2 = B^2)
A^2 = 90,312.5
A = Sqrt (90,312.5)
A = 300.520382 m/s
B = A
Now we know how fast it is going up (to cacluate time in air) and how fast it is going forward (to calculate how far it goes)
Same calculation as before, different numbers:
Time in air = 2t
t = v/g
t = (300.520382m/s) / (9.812865328m/(s^2))
t = 30.62514077s
2t = 61.25028154s
Now, total (horizontal) distance traveled = t * average velocity (300.520382m/s)
Total Distance traveled = 18,406.958m

The exact same distance.

If you want to compare VERTICAL distance shot up and HORIZONTAL distance shot at 45 degrees, they're the same.
If you want to compare VERTICAL distance shot up and VERTICAL distance shot at 45 degrees, the one shot upwards goes further.
If you want to compare TOTAL distance traveled, the one shot in an arc (45 degrees) travels further.

Now, even though the bullet is in the air for a shorter period of time (61 vs. 86 seconds) it is traveling (in the direction we care about) at a faster average speed (300 vs. 212 m/s). That is why the numbers come out the same.

we need some clever person to add air resistance to this equation. looks like you're on to something so far.

[Complication]

we need some clever person to add air resistance to this equation. looks like you're on to something so far.

First thoughts on air resistance:
It decelerates the bullet so the longer the bullet is in the air, the more it should slow it down. The problem is complicated, however, by trying to figure out drag coefficients and all that jazz.

But on second thought:
You've got terminal velocity issues. I calculated the vertical component of the 45-degree shot at 300 m/s. I'm fairly certain that the terminal velocity (the point at which the force of drag = the force of gravity and the bullet no longer accelerates but reaches its maximum--terminal--velocity) of a bullet would be much less than 300 m/s. That means that it takes longer to fall to earth than it does to get to it's zenith (peak).

Now, terminal velocity doesn't effect vertical distance for a 90-degree shot. It just takes longer to fall. But on a 45-degree shot, once it reaches it's zenith it doesn't keep accelerating down until it hits the ground, terminal velocity would basically be like it gliding down after a time.

Think of a model rocket: it shoots up really really fast and then pops its parachute. Think of the parachute as air resistance. Once it parachute deploys, the rocket still accelerates downwards until the drag from the 'chute steadies it at a constant velocity and it glides to earth.

So considering only terminal velocity (affecting only vertical distance traveled), it will stay in the air longer and go further.

Factoring in air resistance's horizontal component, even though it stays in the air longer, it's always slowing down so it travels less far.

The question now becomes which magnitude is greater? I have no clue.

[ForTehNguyen]

the deciding factor would be calculating how long gravity takes to slow a bullet to zero velocity vs how long it takes aerodynamic drag to slow a bullet to zero velocity. Zero velocity being the point where max distance has been travelled. The one with the longer time is the one that went the furthest distance. Its more complicated because drag is proportional to the speed, so as the speed slows the drag force gets lower, so it isnt a constant force. Gravity tho is always constant. I have the feeling you'll have to use an integral to find the average drag force as it is not a constant force at all speeds.

I will guarantee you gravity will slow it faster than aerodynamic drag. You can reasonably assume gravity is already much stronger (in this particular situation) than aerodynamic drag since we are no where near terminal velocity of a bullet. Since gravity is much stronger, drag has little bearing on the vertical scenario.

If you want the angle to maximize the horizontal distance (assuming there is a ground and is infinitely flat) it is always 45 degrees. The 45 degree angle maximizes both flight time and horizontal speed to give you maximum horizontal distance travelled.

http://en.wikipedia.org/wiki/Drag_coefficient

[pilotguyo540]

Okay, these answers are getting way too complicated. It is really simple, so don't overthink it.

The constants are weight, muzzle velocity, gravity, and drag coefficient. Therefore we do not need to include them in this discussion.

A bullet traveling perpindicular to level will travel farther than one shot at an angle to maximize horizontal distance.I believe that angle is around 30 degrees.

A projectile moving only up only has to overcome gravity and decrerasing drag in its travel. Also all of the energy used to propel the bullet is consumed as velocity drops to zero. Then gravity overtakes inertia and the projectile falls all the way back to earth. This travel would be more than double in length to the optimum distance travel.

The projectile aimed for maximum distance would never come close to the travel distance of the vertical shot on even just the upwards leg. The bullet will impact the ground with forward momentum provided by initial thrust only.

The exact numbers don't matter a damn. The theory is as true for rocket science and aviation in general as it is in firearms.

[Complication]

The constants are weight, muzzle velocity, gravity, and drag coefficient. Therefore we do not need to include them in this discussion.
Drag coefficient and terminal velocity need to be taken into consideration (and behave differently) in the 90- and the 45-degree shots.



A bullet traveling perpindicular to level will travel farther than one shot at an angle to maximize horizontal distance.I believe that angle is around 30 degrees.
The question was vertical-shot (90 degrees) vs. optimal shot (which is 45 degrees). Not sure where 30 is coming from.


The exact numbers don't matter a damn.
False. Which numbers you use for constants don't matter a damn. But, as you can see, using numbers gets you an answer based somewhat in reality while pulling stuff outta thin air gets you an uneducated guess.

You're right in that I don't need the exact numbers that I used above. You're absolutely wrong, however, when it comes to disregarding equations and basic physics and just making stuff up. I don't mean to be rude, but what you said isn't just wrong... it's the opposite of true.

[Zhurdan]

Drag coefficient and terminal velocity need to be taken into consideration (and behave differently) in the 90- and the 45-degree shots.



What differences are there in drag coefficients in regards to a 90-, 45- and 0- elevation shot? Same air right? Given perfect/same conditions (mainly no wind) for both shots, what can change the dynamics of drag or were you referring to terminal velocity? Terminal velocity won't become a factor in the 0- shot as it will fall to the ground far earlier than it slowing to the terminal velocities at an angled shot.

Serious question, not hacking on you. You seem pretty damn smart, and I like to learn.

[ForTehNguyen]

i guarantee you the vertical distance will be much less than the horizontal distance due to the fact that gravity will be much stronger than air resistance at these speeds. We are nowhere near terminal velocity of an freefalling object the size of a bullet, dont even bother think about TV.

[pilotguyo540]

Drag coefficient and terminal velocity need to be taken into consideration (and behave differently) in the 90- and the 45-degree shots.


The question was vertical-shot (90 degrees) vs. optimal shot (which is 45 degrees). Not sure where 30 is coming from.


False. Which numbers you use for constants don't matter a damn. But, as you can see, using numbers gets you an answer based somewhat in reality while pulling stuff outta thin air gets you an uneducated guess.

You're right in that I don't need the exact numbers that I used above. You're absolutely wrong, however, when it comes to disregarding equations and basic physics and just making stuff up. I don't mean to be rude, but what you said isn't just wrong... it's the opposite of true.

No offense taken. Let me explain further.

Terminal velocity would only affect the vertical shot on its return to earth, not the horizontal shot. A propelled bullet would travel faster than terminal velocity. This only comes into play for time, not distance. The distance is determined by the altitude gained from the propellant.

The 30 degrees comes from the supposed optimal distance trajectory. not sure what the exact number is. I thought it was thirty. It has been many years since I heard it and I was surprised it was not 45 degrees myself. for argument sake we can say 45 degrees.

Atmospheric lapse rate is an advantage for the vertical bullet. for argument sake we can disregard the drag benefits for the vertical shot. Gravity works on both bullets equally. both bullets will use all of the energy not used to overcome drag to overcome gravity. The vertical shot will use its energy to overcome gravity until all forward velocity is lost. The bullet will then reverse course and fall to the earth. The forward projectile will continue to travel with forward velocity until it impacts the earth.

Equations are important. I am not going to dispute that. My facts are solid though. I did make one assumption that I can see as misleading. I ASSumed that we were only taking into account the hypotenuse, or the distance from shot fired to impact.

[Zhurdan]

i guarantee you the vertical distance will be much less than the horizontal distance due to the fact that gravity will be much stronger than air resistance at these speeds. We are nowhere near terminal velocity of an freefalling object the size of a bullet, dont even bother think about TV.

What?

If you shoot a bullet straight up in the air, it will stop, duhh. And, it will reach a maximum terminal velocity as it falls. That much is a given. It will not accelerate (during the fall) any more than it's terminal velocity. Everything has a terminal velocity. Not everything will reach it because it takes a longer fall for some things to attain it, but a bullet fired from a gun, straight up, will most definitely get high enough to attain a terminal velocity on its return trip to Terra Firma. A bullet would need to fall about 8-10 seconds (depending on the density and shape) to attain its terminal velocity. That would be fairly easy considering how high a bullet would go in the first place giving it ample flight time on the return trip to attain terminal velocity according to several sources. (Just google terminal velocity of a bullet/marble, you could spend all day reading)

[danpass]

It's been a while since high school physics (in which I got an A, if I recall ;))

Let's take a muzzle velocity of 425 meters/sec (handgun), but it should be the same for any velocity:
Note: these calculations don't take into account air resistance, earth curvature, wind, etc. etc.

Shot straight up, it'll take t seconds to reach a velocity of 0 m/s (where it stops going up and starts going down)
t = v/g
v is initial velocity (425m/s) and g is gravity (9.812865328m/s/s)
t = (425m/s) / (9.812865328m/(s^2))
t = 43.31048942s
And another t seconds to come back down. (This time v would be final velocity)
Total time = 2t = 86.62097885s
Now, distance traveled in that time is average velocity (1/2* 425m/s = 212.5m/s) multiplied by time (86.62097885s) = 18,406.958 meters.

Shot at an angle to optimize distance (45 degrees) you have two components: horizontal velocity (used to calculate distance traveled) and vertical velocity (used to calculate time to drop). They will be the same.
These form a triangle where the hypotenuse is 425m/s. Remember the Pythagorean theorem: A^2 + B^2 = C^2. And also that in a 90-degree triangle (where the angle we care about is 45 degrees), the base and height are equal.
C = 425 m/s
C^2 = 180,625
A^2 = 1/2 * 180,625 (A = B, A^2 = B^2)
A^2 = 90,312.5
A = Sqrt (90,312.5)
A = 300.520382 m/s
B = A
Now we know how fast it is going up (to cacluate time in air) and how fast it is going forward (to calculate how far it goes)
Same calculation as before, different numbers:
Time in air = 2t
t = v/g
t = (300.520382m/s) / (9.812865328m/(s^2))
t = 30.62514077s
2t = 61.25028154s
Now, total (horizontal) distance traveled = t * average velocity (300.520382m/s)
Total Distance traveled = 18,406.958m

The exact same distance.

If you want to compare VERTICAL distance shot up and HORIZONTAL distance shot at 45 degrees, they're the same.
If you want to compare VERTICAL distance shot up and VERTICAL distance shot at 45 degrees, the one shot upwards goes further.
If you want to compare TOTAL distance traveled, the one shot in an arc (45 degrees) travels further.

Now, even though the bullet is in the air for a shorter period of time (61 vs. 86 seconds) it is traveling (in the direction we care about) at a faster average speed (300 vs. 212 m/s). That is why the numbers come out the same.

Now, if you want an answer if you INCLUDE wind resistance, my guess is that the 45-degree shot will travel a shorter distance since it is in the air longer and therefore wind resistance will have a greater accumulated effect. Never learned that, though ;).

I read the OP as sending a bullet to a new destination (e.g.: downrange).

Hence I did not take into account the return of the bullet (being shot vertically) as the horizontal bullet would not be returning.

Therefore the horizontal bullet travels "farther".

[Complication]

What differences are there in drag coefficients in regards to a 90-, 45- and 0- elevation shot? Same air right? Given perfect/same conditions (mainly no wind) for both shots, what can change the dynamics of drag or were you referring to terminal velocity? Terminal velocity won't become a factor in the 0- shot as it will fall to the ground far earlier than it slowing to the terminal velocities at an angled shot.

Serious question, not hacking on you. You seem pretty damn smart, and I like to learn.

The drag coefficients should be the same. The difference is that you've got two totally different math problems DESPITE the fact that the air is the same, the bullet is the same, and gravity and muzzle velocity are the same.

(Again, I had high school physics and then a year of physics in college--and that was a few years ago, so this is just my simple understanding.)

With the vertical bullet, you have a 1-dimensional problem.
Muzzle velocity: UP
Gravity: DOWN
Drag: DOWN on the way up, UP on the way down.

With the arc'd bullet, you have a 2-dimensional problem.
Muzzle velocity: UP component and HORIZ component (the forces look like a triangle)
Gravity: DOWN
Drag: UP/DOWN component and HORIZ component (triangular forces, again)

Now, here is why it matters:
Calculations for how long it stays in the air are 1-dimensional. So for the arc problem you need to strip out the horizontal component of drag and muzzle velocity.
Calculations for how far it travels are 1-dimensional. So for the arc problem you need to strip out the vertical component of drag and muzzle velocity.

The same was true of the "frictionless vacuum" problem but it turned out that it came out the same anyways. So why is drag different? My GUESS (and it's just a guess) is that terminal velocity throws a wrench in the works and makes it different. If that's true, then you can't ignore drag. If it's not true, then someone smarter than me should probably step in with an explanation or a link describing where everything cancels everything else out.

My point is that no one has yet explained why terminal velocity is irrelevant (going to read pilotguy's latest post when I get home from work to see if it explains it) and I'm not ready to toss it out without understanding why first.

[Zhurdan]

Complication:

Got it. I thought you were assigning equal value (when you said difference) to drag and terminal velocity.

Terminal velocity should only play a roll in the vertical shot as the bullet would actually slow down past (and eventually stop) the terminal velocity because it's going straight up. At 45 degrees, I seriously doubt that it'd lose enough energy to drop below the terminal velocity. I'm still searching for the link where I read that (with hefty explanation) but I'm at work so it'll probably have to wait till I get home.

[ForTehNguyen]

Yall are confusing what terminal velocity is. TV is not when the bullet stops, its a specific velocity value where the force of weight = force of aerodynamic drag in a falling object, and resulting in no gain in speed, zero acceleration. Think skydiving and top speeds for cars. The car cant gain anymore speed more because of aerodynamic/drivetrain/tire drag = engine power.

Dont even think about it here because the true TV of a bullet is going to be much greater than muzzle velocity.

[orionz06]

Real world conditions say the bullet fired horizontally will travel longer as the bullet that has drag and gravity acting in the same direction. This ignores the fact that the bullet will really skip off of the ground before stopping.

[Zhurdan]

Yall are confusing what terminal velocity is. TV is not when the bullet stops, its a specific velocity value where the force of weight = force of aerodynamic drag in a falling object, and resulting in no gain in speed, zero acceleration. Think skydiving and top speeds for cars. The car cant gain anymore speed more because of aerodynamic/drivetrain/tire drag = engine power.

Dont even think about it here because the true TV of a bullet is going to be much greater than muzzle velocity.

I completely understand what Terminal velocity is... 3 years of physics kinda lends itself to that. (albeit 17 years ago, but I don't think it's changed since then ;) )

Although, I think your understanding in this case is a bit off. Just because an object can exceed it's terminal velocity under the direct power of propellant, that doesn't mean it's going to simply free fall anywhere near those speeds. In fact, that's why a bullet is affected (slowed down) by drag, because it is currently past it's free fall terminal velocity. The problem with a horizontal example of a bullet fired, is that it will be overcome by gravity (aka fall to the ground) long before it expends it's velocity to the point of reaching terminal velocity.

[ForTehNguyen]

not really, gravity is downwards, drag is going to be against the bullet path. The fact that it travels further horizontally is because drag is a much lower force in magnitude to gravity in the vertical scenario.

[pilotguyo540]

Another thought on Pythagorean's theorem. We are not dealing with perfect triangles. We are dealing with a degrading curvature. The velocity is reducing as we go farther down range. On the back half of the horizontal shot we are losing velocity and altitude at ever increasing rates. The bullet is not traveling as far for the given gravitational pull. This is why 45 degrees is not the ideal angle for distance. Again, this does not play into my oversimplified theory. The trig figures are commonly used in these scenarios to give a rough idea, but are not accurate. They are used because they are exponentially easier to figure than the alternative.

Complication did a lot of work in this and I in no way want to discredit that. I think we understood the question differently.

Either way, I had fun! (even if I lose at the end of the day;) )

[pilotguyo540]

drag is a much lower force in magnitude to gravity in the vertical scenario.

This is not necessarily true. Drag doubles at about the rate of 10 knots increase.
A plane traveling at 100 knots will have roughly twice the drag as a plane traveling at 90 knots. This exponential increase can easily outweigh the magnitude of gravity at bullet speeds.

[Complication]

Another thought on the drive home (7-minute commutes rock):
Bullet stabilization. It's no surprise that bullets are stabilized when they come out of the barrel and destabilize after a while. This means the drag coefficients change over the flight of the bullet.
Essentially, this means that the time it takes to reach the zenith is shorter than the time to fall back to earth (at 90 or 45 degrees) because drag increases after the bullet becomes destabilized.

This is how I imagine it:
Take the frictionless vacuum version.
The trajectory looks like a perfect arc, right? (shooting left to right)

Now add the vertical component of drag (terminal velocity, increased drag from destabilization, whatever).
Now the trajectory looks like a sideways half-teardrop. More circular on the left, stretched out to the right.

Go back to the perfect arc. No vertical drag.
Add in the horizontal drag. Faster on the left, slower on the right.
Now it looks like a sideways half-teardrop again, shorter and stubbier on the right side this time.

Now add both components of drag (from destabilization/air/terminal velocity).
Does it look like a perfect arc again? Or is it oblong to the left? Or to the right? I have no clue. This is just the thinking that makes me believe drag needs to be factored in somehow.

If I give you money, you've got more money.
If I take money from you, you've got less money.
If I give you money and then take money from you, do you have more or less? (You'd have to know how much I gave and how much I took)
[Either that or you'd "give" me one of your bullets and take all your money back]

I'm just wondering how much drag "gives" to horizontal trajectory (the idea of floating down via increased vertical drag) and how much it "takes" from horizontal trajectory (horizontal drag).


No offense taken. Let me explain further.

Terminal velocity would only affect the vertical shot on its return to earth, not the horizontal shot. A propelled bullet would travel faster than terminal velocity. This only comes into play for time, not distance. The distance is determined by the altitude gained from the propellant.

The 30 degrees comes from the supposed optimal distance trajectory. not sure what the exact number is. I thought it was thirty. It has been many years since I heard it and I was surprised it was not 45 degrees myself. for argument sake we can say 45 degrees.

Atmospheric lapse rate is an advantage for the vertical bullet. for argument sake we can disregard the drag benefits for the vertical shot. Gravity works on both bullets equally. both bullets will use all of the energy not used to overcome drag to overcome gravity. The vertical shot will use its energy to overcome gravity until all forward velocity is lost. The bullet will then reverse course and fall to the earth. The forward projectile will continue to travel with forward velocity until it impacts the earth.

Equations are important. I am not going to dispute that. My facts are solid though. I did make one assumption that I can see as misleading. I ASSumed that we were only taking into account the hypotenuse, or the distance from shot fired to impact.

I'm not sure I follow. (Another reason I use numbers is because it helps me visualize and that helps me understand). Can you explain this like I am a toddler?

Boy, this has turned into a real hot topic, eh?

[Complication]

Another thought on Pythagorean's theorem. We are not dealing with perfect triangles. We are dealing with a degrading curvature.
True, but the VELOCITY COMPONENTS are perfect triangles.
See my questions immediately above for when perfect triangles and force diagrams get sticky.


the true TV of a bullet is going to be much greater than muzzle velocity.
I don't know about that. That situation is, essentially, taking a bullet and dropping it off a tall building or an airplane. Even with MILES and MILES, will the drag be so low that it accelerates to 1400 feet per SECOND? Maybe. I'm skeptical. Not my area of expertise, though.

[Zhurdan]

not really, gravity is downwards, drag is going to be against the bullet path. The fact that it travels further horizontally is because drag is a much lower force in magnitude to gravity in the vertical scenario.

magnitude has nothing to do with it. If you drop a bullet from 6 feet, and fire a bullet from 6 feet, it doesn't matter how fast the bullet is leaving the muzzle, it'll still hit the ground at the same time as the dropped bullet (measurably anyways). It's still always falling at 32ft/s regardless of it's horizontal velocity, just like the vertical shot is slowing at 32ft/s until it stops.

Oh, and forces don't stack like things in a game of WoW or Pokemon cards. ;) The forces are what they are, and they act in a manner appropriate to the object. In a vertical shot, gravity and drag are the same as they are in a horizontal shot, they're just pointing in different directions (the drag that is).

[Complication]

these answers are getting way too complicated.
It's what I do ;).

[pilotguyo540]

Another thought on the drive home (7-minute commutes rock):
Bullet stabilization. It's no surprise that bullets are stabilized when they come out of the barrel and destabilize after a while. This means the drag coefficients change over the flight of the bullet.
Essentially, this means that the time it takes to reach the zenith is shorter than the time to fall back to earth (at 90 or 45 degrees) because drag increases after the bullet becomes destabilized.

This is how I imagine it:
Take the frictionless vacuum version.
The trajectory looks like a perfect arc, right? (shooting left to right)

Now add the vertical component of drag (terminal velocity, increased drag from destabilization, whatever).
Now the trajectory looks like a sideways half-teardrop. More circular on the left, stretched out to the right.

Go back to the perfect arc. No vertical drag.
Add in the horizontal drag. Faster on the left, slower on the right.
Now it looks like a sideways half-teardrop again, shorter and stubbier on the right side this time.

Now add both components of drag (from destabilization/air/terminal velocity).
Does it look like a perfect arc again? Or is it oblong to the left? Or to the right? I have no clue. This is just the thinking that makes me believe drag needs to be factored in somehow.

If I give you money, you've got more money.
If I take money from you, you've got less money.
If I give you money and then take money from you, do you have more or less? (You'd have to know how much I gave and how much I took)
[Either that or you'd "give" me one of your bullets and take all your money back]

I'm just wondering how much drag "gives" to horizontal trajectory (the idea of floating down via increased vertical drag) and how much it "takes" from horizontal trajectory (horizontal drag).



I'm not sure I follow. (Another reason I use numbers is because it helps me visualize and that helps me understand). Can you explain this like I am a toddler?

Boy, this has turned into a real hot topic, eh?

Touche! It sure is fun though.

The arc curvature would be oblong right. The trajectory would stay pretty straight(ish) until we start the down hill run.

Damn good question though! I need to think about this a little more.

Velocity will decrease on the horizontal bullet until TV is attained. Then it will stay the same. It could actually drop lower than TV if we lose too much momentum going up. Then we will will see velocity steadily drop, level off, than increase, then plateau. Kind of like a ski jump with a platform on the end. This would make a tremendous difference in what the curvature would look like, but either way we would still have to be oblong left for max distance.

As far as TV's affect on either bullet. The TV will have no effect on the vertical shot. Returning speed in no way equates to distance traveled. On The horizontal shot TV will come to play only as an eventual known entity. We know that the bullet is dropping at X fps, while it is traveling forward at Y fps. Y will always be slowing, X will eventually equalize at TV, thus oblong left.

just food for thought.

Maybe we should forward this to mythbusters.

[Zhurdan]

Well, TV does have a place IF we were including the return trip as part of the overall traveled distance, but it seems most of the conversation is not including that in the mix. I was, perhaps that's why I got stuck on TV being a factor in flight time of the vertical shot.

Great conversation though, thanks for the mental workout! :D

[pilotguyo540]

Y
Dont even think about it here because the true TV of a bullet is going to be much greater than muzzle velocity.

I don't know for sure, but I too am sure this is false.

[ForTehNguyen]

my assumption of TV >>>> MV was false

http://www.loadammo.com/Topics/March01.htm


Bullets in the Sky

We frequently get questions about firing bullets vertically into the air. The most frequent question is, "Will bullets fired into the air return to the earth at the same speed they left the gun?" Other questions asked are; "How far does the bullet travel when fired vertically and how long does it take to come down, or does the falling bullet have enough energy to be lethal should it strike someone on the ground?"

Some have tried vertical shooting, but very few have had any luck hearing the bullet come back and strike the ground. When a bullet is fired vertically it immediately begins to slow down because of the effects of gravity and air drag on the bullet. The bullet deceleration continues until at some point the bullet momentarily stops and then it begins to fall back toward earth. A well-balanced bullet will fall base first. Depending on bullet design, some bullets may tumble on their way down and others may turn over and come down point first.

The bullet speed will increase until it reaches its terminal velocity. The bullet reaches terminal velocity when the air drag equals the pull of gravity or stating it another way, the bullet weight and drag are balanced. Once this velocity is achieved the bullet will fall no faster.

In 1920 the U.S. Army Ordnance conducted a series of experiments to try and determine the velocity of falling bullets. The tests were performed from a platform in the middle of a lake near Miami, Florida. The platform was ten feet square and a thin sheet of armor plate was placed over the men firing the gun. The gun was held in a fixture that would allow the gun to be adjusted to bring the shots close to the platform. It was surmised that the sound of the falling bullets could be heard when they hit the water or the platform. They fired .30 caliber, 150 gr., Spitzer point bullets, at a velocity of 2,700 f.p.s. Using the bullet ballistic coefficient and elapsed time from firing until the bullet struck the water, they calculated that the bullet traveled 9,000 feet in 18 seconds and fell to earth in 31 seconds for a total time of 49 seconds.

As a comparison, the .30 caliber bullet fired in a vacuum at 2,700 f.p.s. would rise nearly 21.5 miles and require 84 seconds to make the ascent and another 84 seconds to make its descent. It would return with the same velocity that it left the gun. This gives you some idea of what air resistance or drag does to a bullet in flight.

Wind can have a dramatic effect on where a vertically fired bullet lands. A 5 mile per hour wind will displace the 150 gr. bullet about 365 ft based on the time it takes the bullet to make the round trip to earth. In addition the wind at ground level may be blowing in an entirely different direction than it is at 9,000 feet. It is no wonder that it is so difficult to determine where a falling bullet will land.

Out of the more than 500 shots fired from the test platform only 4 falling bullets struck the platform and one fell in the boat near the platform. One of the bullets striking the platform left a 1/16 inch deep mark in the soft pine board. The bullet struck base first.

Based on the results of these tests it was concluded that the bullet return velocity was about 300 f.p.s. For the 150 gr. bullet this corresponds to an energy of 30 foot pounds. Earlier the Army had determined that, on the average, it required 60 foot pounds of energy to produce a disabling wound. Based on this information, a falling 150 gr. service bullet would not be lethal, although it could produce a serious wound.

Many other experiments have been made to find the amount of air drag on a .30 caliber bullet at various velocities and it was found that the drag at 320 f.p.s. balances the weight of the .021 lb. (150 gr.) bullet and terminal velocity is achieved. For larger calibers the bullet terminal velocity is higher since the bullet weight is greater in relation to the diameter. Major Julian Hatcher in his book Hatcher’s Notebook estimates that a 12 inch shell weighing 1000 pounds and fired straight up would return with a speed of 1,300 to 1,400 feet per second and over 28 million foot pounds of striking energy.

Watch our web site for the next topic of interest "How Far Will My Gun Shoot." Until then, shoot safely and know where you bullets are going.

Sincerely,

The Ballistician

[mnoe82]

my assumption of TV >>>> MV was false

http://www.loadammo.com/Topics/March01.htm

As a comparison, the .30 caliber bullet fired in a vacuum at 2,700 f.p.s. would rise nearly 21.5 miles and require 84 seconds to make the ascent and another 84 seconds to make its descent. It would return with the same velocity that it left the gun. This gives you some idea of what air resistance or drag does to a bullet in flight.


Is that right?

[pilotguyo540]

Is that right?

Pure theory, but yes. Drag is such a huge deal, that scientists used to say that man will never travel faster than the speed of sound. We blew that theory out of the water a long time ago. See my above post about velocity versus speed and you get the idea.

[Complication]

Is that right?

One-way:
t = v/g
v = 2,700 ft/s
g = ~32 f/s
t = 84.375 s

x2 for the round trip

Distance traveled is simply average velocity (2,700 ft/s at the muzzle, 0 ft/s at zenith = 1,350 ft/s) times time.
One way:
d = v * t
113,400 ft = 1,350 ft/s * 84 s
113,400 ft / 5,280 ft (in a mile) = 21.47 miles.

x2 for the round trip


NOW, with drag:
d = v * t
t = 18 s (as calculated in the article)
v = 1,350 ft/s (still leaves the barrel at 2,700 ft/s, still comes to a rest at 0 ft/s)
d = 1,350 ft/s * 18 s
d = 24,300 ft
24,300 ft / 5280 ft/mi = 4.6 miles.

With drag, it traveled 1/5 of the distance, vertically.

Looks like we got ourselves some real-world examples, now.

[Complication]

If you want to get real specific, Air loses 3% density per 1000 ft gain in altitude. Sea level density is 14.7 (or 14.5 depending on who you ask) psi.
You're an evil man.

[pilotguyo540]

One-way:
t = v/g
v = 2,700 ft/s
g = ~32 f/s
t = 84.375 s

x2 for the round trip

Distance traveled is simply average velocity (2,700 ft/s at the muzzle, 0 ft/s at zenith = 1,350 ft/s) times time.
One way:
d = v * t
113,400 ft = 1,350 ft/s * 84 s
113,400 ft / 5,280 ft (in a mile) = 21.47 miles.

x2 for the round trip


NOW, with drag:
d = v * t
t = 18 s (as calculated in the article)
v = 1,350 ft/s (still leaves the barrel at 2,700 ft/s, still comes to a rest at 0 ft/s)
d = 1,350 ft/s * 18 s
d = 24,300 ft
24,300 ft / 5280 ft/mi = 4.6 miles.

With drag, it traveled 1/5 of the distance, vertically.

Looks like we got ourselves some real-world examples, now.
If you want to get real specific, Air loses 3% density per 1000 ft gain in altitude. Sea level density is 14.7 (or 14.5 depending on who you ask) psi.
Millibars would be easier. Sea level is 1,000 millibars, 18,000 ft is 500 millibars, and 100,000, and 100,000' is about 10 millibars.

[pilotguyo540]

You're an evil man.

Yes I am:D

Look again. I edited to make it a little easier to calculate.

[pilotguyo540]

Will a bullet shot straight up travel as far as a bullet shot horizontally? For the horizontal shot assume the bullet is shot at whatever angle maximizes the distance traveled.

I think that the bullet will not travel as far vertically as it will horizontally. I don’t know why I think that.

Don’t let me down guys!

Are you disappointed yet?:D

and pardon the digital stutter on my last post when I typed 100,000 twice.

[Complication]

Yes I am:D

Look again. I edited to make it a little easier to calculate.

I was right in the middle of plugging numbers into excel, calculating instantaneous height, and velocity and then was going to factor in terminal velocity and all of that good stuff and you have to drop the density of air on me! ARGH!

That's about when I realized I'd much rather be enjoying a good run. So that's what I'm going to do.

[pilotguyo540]

I was right in the middle of plugging numbers into excel, calculating instantaneous height, and velocity and then was going to factor in terminal velocity and all of that good stuff and you have to drop the density of air on me! ARGH!

That's about when I realized I'd much rather be enjoying a good run. So that's what I'm going to do.

Good call. to be continued.

[Buckaroo]

Help me settle this argument about bullet travel.

You engineers! No wonder he is in an argument LOL!

43 posts and no conclusion yet, Rum is better than calculus I agree!

Buckaroo (who has two amazing kids who "get" math)

[DeputyMend]

I am amazed at how much thought has gone into this. Needless to say, this thread has not helped settle the debate. I thought for sure there would be some definitive answer out there that someone could point too, but I guess this thread will be what people point to in the future to answer this question.

Keep it up guys, we need an answer!

[xfyrfiter]

Brain hurts, gonna quit readin' now, this is way too complicated for us normal dummies.

[pilotguyo540]

Touche! It sure is fun though.

The arc curvature would be oblong right. The trajectory would stay pretty straight(ish) until we start the down hill run.

Damn good question though! I need to think about this a little more.

Velocity will decrease on the horizontal bullet until TV is attained. Then it will stay the same. It could actually drop lower than TV if we lose too much momentum going up. Then we will will see velocity steadily drop, level off, than increase, then plateau. Kind of like a ski jump with a platform on the end. This would make a tremendous difference in what the curvature would look like, but either way we would still have to be oblong left for max distance.

As far as TV's affect on either bullet. The TV will have no effect on the vertical shot. Returning speed in no way equates to distance traveled. On The horizontal shot TV will come to play only as an eventual known entity. We know that the bullet is dropping at X fps, while it is traveling forward at Y fps. Y will always be slowing, X will eventually equalize at TV, thus oblong left.

just food for thought.

Maybe we should forward this to mythbusters.

Please note that my brain was really tired by this time ant all ov the "oblong left" remarks should be oblong right.

Complication, did you ever put these in to excel and see what the final outcome is? I can't believe I actually lost sleep over this question!

I have been on the road ever since this day, and frankly, my calculus is a little rusty.

[Complication]

Please note that my brain was really tired by this time ant all ov the "oblong left" remarks should be oblong right.

Complication, did you ever put these in to excel and see what the final outcome is? I can't believe I actually lost sleep over this question!

I have been on the road ever since this day, and frankly, my calculus is a little rusty.

No, I sorta gave up. I was in the middle of calculating drag when you dropped air density on me. The obsessive-compulsive in me doesn't want to add in drag without adding in air density too. And that's totally not happening. Especially because of all the work I've got to do (full-time masters + 2 part time jobs = tons of free time)