Bullet dropped lands the same time as a bullet fired..

Drop at range cannot be equated to velocity without some estimate of the ballistic coefficient of the bullet. This is pretty easy to see with any online ballistic calculator. Plug in any particular set of parameters you want and then vary only the BC the drop at various ranges changes even though the initial velocity remains constant.

Falling objects in our world accelerate at 32 FPS per second until they reach terminal velocity. Terminal velocity is the speed at which the air resistance equals the force of the gravitational pull. Terminal velocity is different for objects of different shape, density and orientation, but the acceleration up to that speed is constant. A feather and a chunk of tungsten both accelerate at the same rate, but the feather will have a far lower terminal velocity that is reached almost as soon as it's dropped, where the 1,200 pound per cubic foot metal will continue to accelerate for some time to a far higher terminal velocity.

Where a fired vs. falling bullet are concerned, within our atmosphere, terminal velocity could come into play if the elevation above ground were enough for the different orientations of the two projectiles to affect each one's terminal velocity (the dropped bullet would fall base down, the fired one more or less horizontal, so different cross sections and profiles). But dropped from just a few feet, they're still accelerating when they impact the ground, so will hit at the same time assuming a perfectly flat surface at least the distance of the trajectory.

mcb,

Imagine a series of sheets of paper, say 20' high by 20' across. Put these sheets of paper in a line, one every 100 yards (or meters, if you like metric system math better) for 1,000 yards (or meters). If we disregard any destablization or deceleration of the bullet caused by the paper,* you could use just the holes in those 10 sheets of paper to work out velocity and BC, no? Lower BC will cause more drop at the further distances relative to the closer one than a higher BC, for instance.

*Now, in order to make this practical, we could just shoot 10 rounds sequentially at those pieces of paper at those distances, but not in a line.

ATLDave said:

Imagine a series of sheets of paper, say 20' high by 20' across. Put these sheets of paper in a line, one every 100 yards (or meters, if you like metric system math better) for 1,000 yards (or meters). If we disregard any destablization or deceleration of the bullet caused by the paper, you could use just the holes in those 10 sheets of paper to work out velocity and BC, no? Lower BC will cause more drop at the further distances relative to the closer one than a higher BC, for instance.

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It would work but you would need to know the velocity at at least one know point somewhere along the trajectory. You would also need to know the exact angle of launch a very slight tip up or down on the barrel could add error to the calculation.

mcb said:

Drop at range cannot be equated to velocity without some estimate of the ballistic coefficient of the bullet. This is pretty easy to see with any online ballistic calculator. Plug in any particular set of parameters you want and then vary only the BC the drop at various ranges changes even though the initial velocity remains constant.

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Yes, you do need to know the BC of your bullet.

I think if you knew the launch angle was zero (imagine a raised firing platform), you could figure the velocity pretty well from just the first hole. Decay in velocity (as illustrated by drop) from there would give you the BC.

Brian Litz's work on this topic is interesting. He talks about what to do when your measured, real-world drop doesn't match your ballistics calculator predicted drop. If you adjust the velocity to get a match at one distance but still get discrepancies at other distances (and assuming you have eliminated the scope's elevation tracking as a factor, etc.), then you likely need to adjust your BC.

(Here's a non-Litz blog post discussing it: http://www.accuracy-tech.com/truing-ballistics-data/ )

ATLDave said:

I think if you knew the launch angle was zero (imagine a raised firing platform), you could figure the velocity pretty well from just the first hole. Decay in velocity (as illustrated by drop) from there would give you the BC.

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Nope in this case you would still need a good estimate of the BC or a velocity point.

IE:
A bullet with a .5 BC (a good 6.5mm - 30 cal bullet might have a BC like this) launched at 2000 fps from a perfectly level barrel would drop 4.91 inches (using a G1 ballistic model) at 100 yards
A bullet with a .095 BC (a light 30 cal pistol bullet might have a BC like this) launched at 2250 fps from a perfectly level barrel would drop 4.91 inches (using a G1 ballistic model) at 100 yards
A bullet with a .0572 BC (a 410 slug might have a BC like this) launched at 2500 fps from a perfectly level barrel would drop 4.91 inches (using a G1 ballistic model) at 100 yards

No doubt I can calculate any muzzle velocity I want (with in reason) simply by tweaking the BC to get me roughly a 4.91 inch drop.

How about if we add the second impact data point? Only one of those would fit.

Also, you would know the average speed over the first 100 yards, regardless.

Theohazard said:

I'm going to call you out on this comment. Please reference the specific people in this thread who have displayed an unsatisfactory knowledge of physics. If you're not willing to name names and refer to specific posts, then your comment is entirely unproductive.

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I am not going to go back through the thread, but if you thought a bullet fired horizontally stays in the air longer than a bullet dropped from the same height, then you slept through high school physics and are part of the group to whom I was referring.

ATLDave said:

How about if we add the second impact data point? Only one of those would fit.

Also, you would know the average speed over the first 100 yards, regardless.

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Hmmm...

The average speed over that 100 yards has to be the same if the bullet drops the same amount. This goes back to the original premise of the thread. That said the velocity still varies wildly at any point in the trajectory when comparing them.

At the muzzle we have 2000 fps, 2250 fps, 2500 fps but at 100 yards its 1849 fps, 1473 fps, and 1245 fps respectively. The average is the same but the actually velocity at any particular distance is very different.

If we have the drop at two distances then we could solve for both a BC and an initial velocity. Our only assumption then would be which ballistic model we would be using G1, G7 etc.

*It important that we are measuring drop and not impact shift. This works if the bore is perfect level and we are measuring the drop from the line of the bore down to where the bullet passes at a particular range. Using an actual sighted in rifle at a given distance and measuring impact shift at other ranges will be a different animal I have not thought about really.

ETA: Thinking about it more I think I we are sighted in at one distance (say 100 yards) and then measure the drop at two additional distances we could so a similar calculation as above. We would need to account for sight offset also but that is easy.

Theohazard said:

No, it’s the same for any bullet at any speed.

Removing variables like air resistance, uneven terrain, and the curvature of the Earth, a bullet dropped from the hand will always hit the ground at exactly the same time as a bullet fired horizontally. It doesn’t matter what bullet and what velocity.

Think of it this way: What force would cause the bullet to “hang” in the air? Gravity works the same way on all objects, so the way to make your bullet work against the force of gravity and stay aloft longer is by having the bore of the firearm angled upwards. But if the bore is exactly perpendicular to the force of gravity (horizontal), it’s not working against or with gravity.

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However, if you shoot a bullet perfectly horizontally at 12,050 fps at the equator (assuming no mountains, land or air got in the way), the bullet would never hit the ground.

lysanderxiii said:

However, if you shoot a bullet perfectly horizontally at 12,050 fps at the equator (assuming no mountains, land or air got in the way), the bullet would never hit the ground.

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Only if the bullet would never slow down. This is one of the difficulties when measuring the velocity of any fired projectile....no matter where you do the measurement, it is only that fast at that exact spot and will be significantly faster closer to the muzzle and also greatly slower the farther away you get.

lysanderxiii said:

However, if you shoot a bullet perfectly horizontally at 12,050 fps at the equator (assuming no mountains, land or air got in the way), the bullet would never hit the ground.

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Correct but I believe the velocity you need is 25,950 fps to achieve orbit at see level, assuming no air drag. Subtract ~1500 fps if you fire east add 1500 fps to that if you fire west due to the earth's rotation. North or south add or subtract nothing.

RecoilRob said:

Only if the bullet would never slow down. This is one of the difficulties when measuring the velocity of any fired projectile....no matter where you do the measurement, it is only that fast at that exact spot and will be significantly faster closer to the muzzle and also greatly slower the farther away you get.

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Fire a rocket horizontally, about one foot off the ground and see if it doesn't hit the ground before the motor burns out. It would still be accelerating when it hits.

The reason a bullet fired at 12,050 fps from the top of My Everest, or similar, would never hit the ground is because it would be at orbital velocity. It's drop would match the curvature of the earth....

mcb said:

Correct but I believe the velocity you need is 25,950 fps to achieve orbit at see level, assuming no air drag. Subtract ~1500 fps if you fire east add 1500 fps to that if you fire west due to the earth's rotation. North or south add or subtract nothing.

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17,672 mph.

I think I messed up the conversion....

ATLDave said:

I don’t think that is correct. The REASON that bullets with a better BC drop less is BECAUSE their velocity decays less slowly. Bullets don’t generate lift.

Drop over distance is driven by velocity; drop over time is constant. BC predicts how much velocity decays. It only flattens trajectory by keeping velocity higher.

In fact, rather than needing BC to calculate velocity from drop, knowing drop at multiple distances would allow you calculate the BC of the bullet!

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Hmmm....don't believe I ever said or even implied that a flying bullet generates lift. We are in agreement about how BC affects velocity....but again without knowing either velocity or BC how are you going to predict one or the other just from a drop at a known distance? Slower will drop more as will the less aerodynamic BC, and let's not forget that many bullets will display multiple BC's as their velocity decays...as well as varying BC depending on their initial velocity.
I suppose if you had enough data points you could match up velocity and BC with a good ballistic calculator and make a decent assumption of what both had to have been to create that trajectory, but if you only have a couple targets it would be impossible to say for certain just what happened.

I suspect that either 2 or 3 data points is the required theoretical minimum, but additional data points would probably be needed in light of unavoidable randomness/indeterminacy in the data.

lysanderxiii said:

Fire a rocket horizontally, about one foot off the ground and see if it doesn't hit the ground before the motor burns out. It would still be accelerating when it hits.

The reason a bullet fired at 12,050 fps from the top of My Everest, or similar, would never hit the ground is because it would be at orbital velocity. It's drop would match the curvature of the earth....

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Even at the top of Mt. Everest there is enough atmosphere to rapidly slow any projectile you could fire. Launching a rocket is a totally different discussion being as it can maintain or increase speed as it flies so in that case...yes, it would be theoretically possible to orbit a rocket at 29,000 feet if it could sustain the needed velocity and withstand the heat. But neither is possible. Even maintaining Mach 3 at 80,000 ft. heats metal to bright orange or thereabouts and orbit velocity that low would melt anything I'm aware of. The Shuttle re-entering from orbit velocity only survived by gradually encountering the edges of the atmosphere and slowing down as it descended....had it plunged too steeply it would have incinerated itself.

Yeah in the real world orbital velocities are not possible inside the atmosphere due to frictional heating and stress. Skin friction would probably tear most objects apart while melting them at the same time. Look what happens to most orbital objects who's orbit decays and they re-enter. Very few objects can reenter unscathed.

Mn Fats said:

I must of slept through that class and became part of the "group" to whom you were referring to.

I started this thread to ask a simple question to a (not so simple to me) answer. I thought there was a way it could be beat. I was wrong. Very educational posts and I really do appreciate the time and knowledge that was extended my way.

Once again, thanks THR members.

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A very High Road response from MnFats as always.

This reminds me of the question about whether you can float a battleship with a quart of water. The answer is “yes, but...”. Similarly, the question of whether a bullet hits the ground later fired out of a gun vs. dropped is the same. You probably can’t set up a test rig that’s accurate enough to really tell, so the everyday approximation of “no” is more correct than not.

As the old gag goes: “in theory, theory is the same as practice; In practice, they’re different.” It’s funny because it’s true!

Glad to see everyone taking the high road in the discussions, mostly.

Balrog said:

This thread is proof that public schools have ruined the brains of many Americans.

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I knew that was going on when I went to high school. I decided not to participate. My mind was a blank sheet of paper when I graduated.

I filled it up later in life.

Variation in gravity and apparent gravity[edit]
A perfect sphere of uniform mass density, or whose density varies solely with distance from the centre (spherical symmetry), would produce a gravitational field of uniform magnitude at all points on its surface, always pointing directly towards the sphere's centre. The Earth is not spherically symmetric, but is slightly flatter at the poles while bulging at the Equator: an oblate spheroid. There are consequently slight deviations in both the magnitude and direction of gravity across its surface. The net force (or corresponding net acceleration) as measured by a scale and plumb bob is called "effective gravity" or "apparent gravity". Effective gravity includes other factors that affect the net force. These factors vary and include things such as centrifugal force[3] at the surface from the Earth's rotation and the gravitational pull of the Moon and Sun.

Effective gravity on the Earth's surface varies by around 0.7%, from 9.7639 m/s2 on the Nevado Huascarán mountain in Peru to 9.8337 m/s2 at the surface of the Arctic Ocean.[4] In large cities, it ranges from 9.7760[5] in Kuala Lumpur, Mexico City, and Singapore to 9.825 in Oslo and Helsinki.