Physic 101 of bullet and gravity.

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doctorhumbert

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Or Physic 102 if you didn't flunk 101 class.

This is something I've learned in High school Physic class (yes was actually awake during this one). provided that they have same weight and air resistance upon them, two balls dropped straight down, and thrown horizontally to the ground level will reach the ground at the same time.

Will this apply to bullets in real life?
IF: say .45 bullet was fired from 1911 horizontally from the ground.
and drop .45 bullet (not cartridge). At the same time.

Will they reach the ground same time? (fired bullet about couple thousand miles away from the spot of course)

My guess is that droped bullet would reach ground first. Rotating bullet would have lift effect from the air, thus stay in flight longer.

If this is true my estimate is that bullet with 1400 fps velocity would stay air born longer before dropping to ground 1400 feet away. (provided that bullet was fired from height it took One second for bullet to drop naturally. If so how much longer does it stay in flight?

You get 10 extra point added onto this weeks exam if you get the answer correctly (You will need it!).
 
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Very simply...In real life...no. Like you probably figured out, in real life you have pesky factors like drag to worry about.*

For more information than you could ever want on the subject, might I suggest Rinkers book on understanding ballistics.

Good read.

http://www.amazon.com/exec/obidos/t...103-4980645-7099067?v=glance&s=books&n=507846

*But it is an easy reference for teachers to use to teach about gravity.

SC
 
There are two ways that an object traversing thru a fluid can get "lift".

One is by the shape of the object relative to it's path thru the fluid (air). The best example is a wing on an airplane, which is shaped and oriented intentionally to get lift.

The other is by rotating the object such that the top surface of the object is moving TOWARDS the direction that the overall object is travelling. This is called the "Magnus effect" (if I remember by sophmore college aerodynamics class correctly), and is the trick used by baseball pitchers to create a curve ball.

The key to using the magnus effect is that the rotation as to be in an axis PERPENDICULAR to the vector of travel for the object. Bullets spinning as they leave the barrel due to rifling are spinning PARALLEL to the vector of travel, so the magnus effect does NOT help "lift" the bullet.

So, to answer your question: the bullet that is traveling forward, having been fired, has a lower coefficient of drag than the bullet dropped. (This assumes you don't drop the bullet "pointy end first".)

If the CD is lower for the bullet fired, it will reach the ground soonder than the dropped bullet, due to less resistance, at least in theory.

In reality, there are so many other small variables we are ignoring here, it's hard to say what would really happen without some real detailed, careful experiments.
 
The bullet that's fired has more than just gravity to work with.

In other words, you've forgotten about momentum.

"An object in motion wants to remain in motion, in a straight line".

The powder charge kick to the fired round gives it a hell of a lot of sideways momentum. Gravity will have to fight that momentum. Gravity will win, but momentum will slow it down some.

Gravity will have air resistence on it's side, however, so the downwards curvature of the round will be far greater in atmosphere than a vacuum.
 
Hmmm. Just thought of something:

There's a complication because the rate of fired bullet fall towards the ground is a "curve" rather than a straight line. The longer the round has been flying, the slower it's going (wind resistance) and the closer to "straight down" it's falling.

So there's a potentially huge difference between doing the "drop and fire" test with both dropped round and fired bullet 5 inches above the ground versus 5 feet, yards, miles.

I'd want a computer program to calculate the variables :). But just working in my brain, I think that in ALL cases the fired bullet stays airborn longer.
 
If I am correct, Momentum still would not affect the gravity if it is horizontal to the ground. Yes, bullet wants to stay in forward motion, but not away from the gravity. Also, the gravity is a constant factor.


some contrasting answers to this question... I guess you guys are ready for this weeks Exam without extra 10 points....:cool:
 
Jim,

If you do a "free body diagram" from physics on the two bullets, you will find that the "momentum" you refer to plays no real part in the downward motion of the bullet, and, at the same time, plays NO part in helping keep the bullet "up" either.

The bullet being dropped is being acted upon by the following forces in the vertical axis
1. gravity pulling it down
2. aerodynamic drag resisting downward motion

The bullet being dropped is being acted upon by the following forces in the horizontal axis:
**NONE**

Now for the bullet that was fired:
The bullet fired is being acted upon by the following forces in the vertical axis
1. gravity pulling it down
2. aerodynamic drag resisting downward motion

The bullet fired is being acted upon by the following forces in the horizontal axis:
1. aerodynamic drag resisting forward motion

So, as you can see, for up and down motion, the forward action of the bullet plays NO part in that behavior.

That's it, unless my mental physics skipped a force somehow. Maybe someone who has done free body diagrams more recently than 25+ years ago can help here.
 
Even considering drag and the incredibly slight curve of the earth, they'll hit at the same time.

Objects drop due to gravity without respect to any other movement. In the example, there is air drag, but the downward vector of drag do to gravity is the same for the two objects. The forward drag is huge, but that determines horizonal deceleration only, not vertical.

If you want to figure how much a bullet will drop, look down the bore at what it's pointing at. Figure the time the bullet will be in the air until it gets there. That time, multiplied by 9.8 meter per second squared will give you the horizontal drop from the line of sight at the target, regardless of barrel inclination.


The slippery application of all this is when shooting a rifle up or down slope. In either case, the holdover is less, for a given range, than a horizontal shot. Is the bullet dropping less? NO! It's just that the shooter is viewing that vertical drop from an angle, which makes it appear shorter (distended), since the reticle is no longer parallel to the vertical drop.
 
"The powder charge kick to the fired round gives it a hell of a lot of sideways momentum. Gravity will have to fight that momentum. Gravity will win, but momentum will slow it down some. "

Nope, momentum ain't got a thing to do with it. The bullets will hit the ground at exactly the same time regardless of muzzle velocity of the fired bullet. The "dropped" bullet has zero horizontal velocity, the fired bullet may have 1000f/s horizontal velocity, but it's downward motion in space is completely independent of whatever horizonatl velocity it has. I guarantee it. They will hit the ground at the same time.

As for bullet "lift" due to air? The bullet is symmetric so the air can not impart lift. The wing of a plane gets lift becaus the top surface is longer in distance forcing the air to flow a longer path from front to rear of the wing surface. The Bernoulli principle states that it will therefore create a lower pressure than the bottom where the air has a shorter path: Ergo, lift. The bullet just gets slowed down in it's horizontal velocity some. Unless it starts tumbling or changes angle in the air, there will be no lift effect.
 
Quater bore gunner, here is simple physic lesson for you.

"momentum": Think of Det. Frank (the old white haired guy) from the Naked gun Poster.


Notice how he can ride on the fired cartridge (instead of bullet). He can even jump up and down on the .357 (I think) round with ease because of this thing called momentum. Otherwise, if he jumped, bullet, I mean cartridge, would leave him behind!:eek:

"Magnus Effect" uhhhh, isn't that like force that allows you to ride bicycle without falling off?
 
No AZ, you got it right. Jim, look at it this way, a slight cross breeze can push the bullet because it only has to deal with momentum in the horizontal direction (zero horizontal velocity means it only has to push the mass at rest).

I'm sure there's some weird effects for a bullet spinning at 30K RPM but I doubt there's a net lift, probably more of a random directional benefit. Same deal with headwind versus tailwind. Whatever lift exists has to be hard to measure.

Curvature of the Earth is probably is bigger effect that those air flows (the ground is 0.0x inches lower a few hundred yards away).
 
zak. Let's assume that bullet is in ball shape (to eliminate different air resistant factor of butt first dropped bullet), but fired with rifling, not smooth bore to apply spinning factor on the fired bullet.

I think everyone agrees that gravity affects both same, like I suggested in the beginning.

Only question is how much lift (possibly from air resistance and spinning effect), if any, does fired bullet has, and whether it would stay in flight considerably longer.
 
"There's a complication because the rate of fired bullet fall towards the ground is a "curve" rather than a straight line. The longer the round has been flying, the slower it's going (wind resistance) and the closer to "straight down" it's falling."

As I recall, the shape of the curve is a section of a parabola. It's because the horizontal velocity is pretty constant, the downward velocity is accelerating at a fixed rate.


"So there's a potentially huge difference between doing the "drop and fire" test with both dropped round and fired bullet 5 inches above the ground versus 5 feet, yards, miles.

I'd want a computer program to calculate the variables . But just working in my brain, I think that in ALL cases the fired bullet stays airborn longer."

NOPE. Vertical motion is independent of horizontal velocity.

Want proof? Drop a ball and watch it fall straight down. Did you know while it was falling "straight down" it was actually going about 35,000 mph horizontally relative to space because the Earth's surface is spinning that fast? Since all our points of reference are going the same speed, it falls "straight down" as far as we know. The bullet question is similar. If we and the bullet were in a lab that was shot at the same velocity and instant as the bullet, we would see it as falling straight down and would measure the exact same time to hit the ground as a person measuring the time of the one he just let drop from his hand toward th ground..
 
Trust me, if other variables are constant, ball dropping in straight vertical line, and ball dropping in horizontal motion drop at the same time.
Just remember that horizontal resistance has nothing to do with gravity.

This is a fact. You may try doing experiment, but it is difficult to keep variables constant.

Only variable I am now concerned with is possible effect of the bullet spinning, and what it might cause.
 
Just remember that little experiment that a couple of astronauts performed on the moon to prove Newton's theories, and that is a good basis to start from.

As long as the fired bullet and the dropped bullet are not given any initial arc, they would both hit hte ground at the same time regardless of how fast or slow the fired bullet went. I do not think that gyroscopic spin would have anything to do with it, just so long as the bullet fired is not shaped to change it's direction or is a flechette-type projectile. If they were, then you would be using the forward momentum, redirected into another area, potentially causing the fired bullet to move from it's original path.

Essentially, IN A VACUUM, they would both hit the ground at the same time. In any other situation, there are a lot of variables that can/will change throughout the length of the time the fired bullet leaves the barrel and finally touches the ground. The effects would be minimal, but might be measurable. Even so, any variation, if all contingencies are accounted for, you will find that the bullet should hit the ground same time as dropped.
 
Doctorhumbert,

Read the article on the Magnus Effect that I posted the link to. You will see that bullet spin has NO effect on the bullet's drop rate.

The spinning has to be in the same direction as the flight for it to impact the drop rate, and last time I checked, all modern projectiles spin about an axis that is parallel to the flight vector.
 
I think I came to a fairly satisfying conclusion on the previous question.

Those who answered correctly received 10 points... Rests, good luck.

Here is another... Will 9mm bullet dropped by Soy milk Latte drinking snotty yuppy New Yorker on the top of the Empire State building seriously hurt the backward redneck tourists who aimlessly wonders while unnecessarily congesting the Manhattan?

in other word, will it gain enough velocity from such height?

5 extra point on this one.
 
My high school physics teacher taught that it only applied in a vacuum. He had worked with von Braun, so I'll believe him.

It must have been 1967 when I saw his favorite classroom demonstration for the first time. He used an air pump to fire a LARGE brass bullet through a pipe. It was aimed at a coffee can hanging from an electromagnet on the ceiling at the far end of the classroom.

When he threw the switch the gun fired and the power was cut to the magnet dropping the can. The bullet hit the falling can - since the bullet and the can were falling at the same rate.

The bullet was large enough to crush a one-pound coffee can if it hit the side.

He never missed. There would have been a hole or two in the cinderblock wall behind the magnet if he'd missed. He was using a thumb-sized bullet.

I know it doesn't prove much, but it was a rare treat to have a teacher who was interesting. (Another good one - I had a history teacher at the same year who had been living in Peking when it fell to Mao.)

John
 
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