OK armchair physicists, explain this....

J

Jack 99

New member
This was a little bit of trivia I picked up somewhere:

If you were on a perfectly flat plane, and fired a bullet perfectly horizontal to the plane, and simultaneously simply dropped an identical bullet from the same height as the fired bullet, which would hit the ground first, the dropped bullet or the fired bullet? The answer is they hit the ground at the same time, gravity affects each bullet identically.

Just about the time I think I understand the physics involved, I think about an actual scenario with a real gun and real bullets.

For argument's sake, let's set the parameters like this:

A)
You're on a perfectly flat planet, no curvature of the earth to deal with.
B)
The bullets are both 180 grain FMJ and the fired bullet is moving at a muzzle velocity of 3000 ft/second (just a tad over 30-06 velocity).
C)
All objects fall at 32ft/second per second so the platform is exactly 32 ft high (one second for the bullet to impact the ground).
D)
There is no air or other resistance factors.

Now, you're 32 feet above the plane, fire the bullet at 3000 ft per second and simultaneously drop an identical bullet from the exact same height. Both bullets will impact the ground at the same time one second later.

The dropped bullet I'm fine with, but will the fired bullet really drop 32 feet over 3000 ft? I find this a little hard to swallow. That's a little over 1/2 mile. I know that the 30-06 drops pretty fast after 600 yds or so, but that fast?

Anybody have any insights before I go insane?

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"Put a rifle in the hands of a Subject, and he immediately becomes a Citizen." -- Jeff Cooper
 
It is hard to fathom, isn't it? I often drive myself nearly insane trying to mentally work through and visualize such things. Because of this, and the fact that it's been a while since I've done such work, I will let someone else field this.

I do want to point out one thing though. You made a faulty assumption that from 32ft the bullet will take one second to fall. This would be true only if the bullet was already travelling downward at a constant velocity of 32ft/s at the time of the firing of the second bullet. In reality, it starts off with a velocity of zero, and after being dropped, it accelerates at 32ft/s/s. It will, therefore, take more than a second to hit the ground, and the fired bullet will travel farther than 3000ft before it hits the ground. I don't have a calculator with me or I'd work the time out for you. I did, however, use the following kinematic equation:

D=V(int)*T + 0.5*A*(T*T), where D is distance in meters, V(int) is initial velocity in m/s (which is zero in the case of the dropped bullet), T is time in seconds, and A is the acceleration due to gravity, in m/s/s

with a time of 1 second, and figured in that timespan the dropped bullet would travel 4.9m, which is something like 16 feet or so. So, think of your fired bullet moving at 3000ft/s for one second dropping only 16ft.

Make any sense? :)

Please, somebody correct me if I'm wrong.

Update: okay, here it is another way...I solved for time using the 32ft high platform, and, under many assumptions and roundings, came up with a time of 1.415s for the fall of the dropped bullet from 32ft, meaning that the fired bullet travels (3000ft/s*1.415s)=4245ft=0.8 miles before hitting the ground, if I did it right, that is (and that's a big IF!)


[This message has been edited by BAB (edited November 23, 1999).]
 
That sounds a little better, a 16 ft. drop over a one second period.

I guess my High School physics is a little rusty, though. I thought objects fell at a rate of 32ft per second per second. So the first second, an object drops 32 ft, the second second, it drops 64 ft for a total of 96 ft over 2 seconds, third second, 96 ft for a total drop of 192 ft and so on. Am I totally wrong?
 
Jack, the problem is that you're confusing speed with acceleration. Things don't *fall* at 32ft/sec/sec, they *accelerate* to terminal velocity (the point at which they no longer gain speed) at 32ft/sec/sec.

HTH.

------------------
"The right of no person to keep and bear arms in defense of his home, person and property,
or in aid of the civil power when thereto legally summoned, shall be called into question.."
Article 11, Section 13, CO state constitution.
 
Jack,

No, you're right about the 32ft/s/s. But, like Coinneach said, the key is the extra "per second"...this is an acceleration. So, in this case, the dropped bullet has an initial velocity of zero as it sits in your hand. Once you release it, it starts to accelerate at 32ft/s/s, meaning that at the *end* of the first second, it is travelling at 32ft/s, and at the *end* of the second second, it is travelling at 64ft/s, etc. The velocity is not constant...it's constantly increasing (at the rate of 32ft/s every second). So, for the time between the drop and one second later, the downward velocity of the bullet is somewhere between zero (Vel. at time=0), and 32ft/s (Vel. at time=1 sec.). Only at the *end* of 1 second, and only for that instantaneous moment, will the bullet be travelling at 32ft/s. And so and and so forth.

Fun stuff, huh? :)


[This message has been edited by BAB (edited November 23, 1999).]
 
If you were on a perfectly flat plane, and fired a bullet perfectly horizontal to the plane, and simultaneously simply dropped an identical bullet from the same height as the fired bullet, which would hit the ground first, the dropped bullet or the fired bullet?

(I'll ignore air resistance.)

They'll both hit the ground at the same time. Horizontal velocity has no effect on vertical velocity.

Try this: take a couple of similar objects (bullets, rocks, phone books, whatever - anything that will not suffer non-negligable air resistance). At the same time and from the same distance, drop one and toss the other horizontally. They'll hit the ground simultaniously.
 
One fired one dropped, which would hit the ground 1st?

In the absence of air resistance both would hit the ground at the same time. In air the surprising thing is that the one that was dropped would hit first.

Why? Because the one that was fired is travelling initially much faster experiences the transonic drag rise (close to Mach 1). As drag is higher it takes longer to hit the ground, at least in theory.

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"Quemadmoeum gladius neminem occidit, occidentis telum est."
("A sword is never a killer, it's a tool in the killer's hands.") -
Lucius Annaeus Seneca "the Younger" (ca. 4 BC-65 AD).
 
Something to remember is that guns don't fire bullets horizontally. They are fired at an upward angle (it's slight, but that extra three inches of rise makes a difference). For example, if I remember my charts correctly, a .223 bullet tends to climb to about 250 yds, at which point it's something like 3-4 in above the boreline. Since the bullet is doing about 1000yds/s that's a quarter second before the bullet even starts to fall.
 
Oh, acceleration and speed are not identical. Duh. Proving once again that a little knowledge is a dangerous thing.

Zach, bullets don't really "climb," that's an old wives tale. If you shoot perfectly flat, the bullet will travel downward almost as soon as it exits the barrel (in a pure Newtonian universe, it does). The AR 15 has a sight plane that's elevated a couple of inches so the illusion of climb is pretty profound. If you zero the AR at 25 yds, its also zeroed at 250 yds, as memory serves. That's just because the bullets arc intersects the sight plane at both those distances.
 
Coin and Zach nailed it. To put it a little different than Zach -- In your scenario, the gun barrel is perfectly parallel to the plane of the ground; therefore the bullet begins to drop immediately upon exiting the muzzle. In reality, when shooting a rifle, even if your line of sight is perfectly parallel to the ground, the barrel is aimed slightly upward, so that the bullet rises for 60-150 yards or so (to an apex of 2 to 4+ inches above the line of sight), depending on how the rifle is zeroed and the velocity, then begins to fall, then crosses the plane of the line of sight at 100-300 yards (at the point of zero), then continues to fall below the plane of the line of sight, faster and faster toward the ground - but remember, it didn't "drop" ANY (below the plane of sight, that is) until this point of "zero" which could be 100-300 yards, give or take.

[This message has been edited by Futo Inu (edited November 23, 1999).]
 
If an '06 is fired exactly horizontally, with the bore pointed exactly at the center of a target at 100 yards, it will fall below the center of the target. Per my Sierra book, 1.88 inches. Remember, when you are zeroing at any range, the line of the bore approaches or even crosses the line of the sight(s). The bullet rises above the line of sight and then drops back onto target.

Also per my Sierra book, a 180gr Spitzer BT bullet with a muzzle velocity of 3,000 ft/sec will drop 309.34 inches at 1,000 yards. Let's call that 25.8 feet. Wind resistance is part of the deal.

In your example, with you being 32 feet above ground, and one G = 32.16 ft/sec/sec, the bullet will fall only 16.08 feet the first second. It begins its fall at zero feet per second, remember, and while its VELOCITY is 32.16 feet per second at the end of one second, the vertical DISTANCE it has travelled is only 16.08 feet.

Your experiment, performed on one of those laser-levelled fields I've seen in California, could be set up. Bore parallel to the ground. To "measure", I guess a video camera would be set up behind the apparatus, to record the two puffs of dust. After you find where the fired bullet hits, you could put a large steel plate there, as well as under your platform, and electronically record the times of the impacts.

Lights, camera, action! You push a button, a pair of solenoids actuate the gun and the drop-release. Wind resistance might make the fired bullet hit short of where we might expect, but I would expect them to hit the ground at the same time.

Hope this BS helps.

:), Art
 
Ya know, the next time someone tries to paint all gun owners with the same illiterate-redneck brush, I'm gonna point them to this thread. :D

------------------
"The right of no person to keep and bear arms in defense of his home, person and property,
or in aid of the civil power when thereto legally summoned, shall be called into question.."
Article 11, Section 13, CO state constitution.
 
Ever think about the RPM the bullet is spining at?

How does it effect bullet flight?

We know that the spin stabilizes the bullet by creating a gyroscopic effect, but wouldn't the spining bullet create a low or high pressure spots from the wind drag trying to slow it's spin?

------------------
The new guy.

"I'm totin, this pistol because my dang SKS won't fit in my holster"
 
Jimmie:

That's the difference between theoretical physics and actual. Other factors will influence the calculation. Updrafts, downdrafts, air pressure a whole lot of little variables can influence the results. The example given is theoretical, given a perfect world. But we all know that we don't live in one. Besides we all know that if the bullet we drop is attached to a cat with a piece of buttered toast tied to it's back it will never hit the ground. For all of you who don't know about the catera-toast perpetual motion machine, I suggest that you do a search on TFL for buttered toast. :)



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Richard

The debate is not about guns,
but rather who has the ultimate power to rule,
the People or Government.
RKBA!
 
This really is fundamental h.s. physics. Nothing mysterious about it. I don't see all the excitement over the fact that, in a vacuum, all objects fall at the same rate.

An easy test for the falling bullet theory is to have a friend hold an apple at 100 yds.Boresight (with a laser) on the apple, making sure that your rifle is at the same height as the apple. Have your friend drop the apple at exactly the same time as you fire your rifle. After your friend cleans up all the brown stuff from the surrounding area, he'll find that you hit the apple, every time.

On the other hand, if you fire a bullet in space, without the benefit of gravity, the bullet will eventually return to its point of origin. Or at least a ray will, so I'd assume that a bullet would also.
 
Ok, I think I can answer this. Being an engineering student I've taken more physics classes than I care to remember. If we are in agreement that both bullets will hit the ground at the same time, this is how I solved the problem. Using the same equation as Bab the bullet dropped from rest, with no initial vel. will hit the ground in 1.41 sec. Now using the same equation D=Vi(t)+ 1/2 a t^2, and assuming the same time and that the acceleration due to gravity is negative, I calculated the bullet leaving with an initial vel. of 3000 fps will travel a distance of 4198 ft.
I'll have to check a couple of textbooks to make sure but that's the best I can do off the top of my head.
However I'm not sure what answer you are looking for.
If you are wanting the drop at 3000', to make a long story short, I used the same equation, solved for the time it took to reach 3000'. From there I calculated the drop. I found it to be 20'.
This could be wrong, but after checking the Hornady Hnadbook of cartridge reloading, they say the drop of a 180gr bullet for 600 yards will be 76.4", with a 100 yard zero and 3000 fps initial vel. this is factoring in resistance however.
Now just for funnies if you just want to know how far the fired bullet will travel, using the same time it took the dropped bullet to hit the ground, you use the equation D=RT(distance = rate x time), this comes out to a traveled distance of 4230'.
I hope this helps. I'll check my calculations and formulas and see if I can come up with something better.

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As to marksmanship, it is not what you once did, rather it is what you can do on demand.
 
If you shoot one bullet and drop the other, who'd pester you first, the local politician or greedy scumbag lawyer?

Sorry couldn't resist.
 
I've been biting my tongue trying to stay out of this, especially since BAB answered it correctly with the very first response. But, Mute you have brought up the one truism that can't be ignored. Now the question is, "What if this experiment was done as a science fair project - how long would the student be suspended?"
 
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