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.
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.
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.
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.
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;) )
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.
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?
True, but the VELOCITY COMPONENTS are perfect triangles.
See my questions immediately above for when perfect triangles and force diagrams get sticky.
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.
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).
It’s what I do ;).
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.
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! 
I don’t know for sure, but I too am sure this is false.
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
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.
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.
You’re an evil man.
Yes I am:D
Look again. I edited to make it a little easier to calculate.