Bullets falling back to earth

Why wouldnt it be spin stablized? Granted there are some rotational friction forces (skin layer) slowing it down, including air pressure against the meplat and base going up and down..but give rotational speeds of the common bullet are in excess of 100,000 rpm...... it would take a lot to slow it down.

Gunner

" ..The world has gone crazy. Guess I'm showing my age... I think it dates from when we started looking at virtues as funny. It's embarrassing to speak of honor, integrity, bravery, patriotism, 'doing the right thing', charity, fairness. You have Seinfeld making cowardice an acceptable choice; our politicians changing positions of honor with every poll; we laugh at servicemen and patriotic fervor; we accept corruption in our police and bias in our judges; we kill our children, and wonder why they have no respect for Life. We deny children their childhood and innocence- and then we denigrate being a Man, as opposed to a 'person'. We *assume* that anyone with a weapon will use it against his fellowman- if only he has the chance. Nah; in our agitation to keep the State out of the church business, we've destroyed our value system and replaced it with *nothing*. Turns my stomach- " Chas , rec.knives

No kinetic energy remaining, they have potential energy due to their increased height, relative to some lower height. The potential energy is proportional to the height.

I can go out in my fromt yard

If you are considering air resistance, any bullet fired up is going to have sufficient kinetic energy and attain a sufficient height to return at terminal velocity. We have to either simplify the problem by neglecting air resistance or consider all factors, which would get pretty complicated. In any case, energy is conserved and theoretically can all be accounted for, in terms of heat, work done on the greater environment, etc.. As far as physics class, the important thing in any lab class is to snag a decent looking girl as a lab partner early on...

The article didn't give a lot of details, it was an initiation, they were shooting a guy with paintball guns and someone fired the real gun to provide sound effects. Daniel Carver didn't approve.

Regardless of weight, a falling body will accelerate at 33 ft. per second, per second.

I think gunner has already stated that the terminal velocity for common commercial shell sizes is way, way far below the muzzle velocity. This says you are right in my book.

Jim

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Need to put the engines on O2 as well - I think none of the ww1 vintage aircraft had turbochargers - that was probably the limiting factor for useable altitude.

Jim

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As an aside I should add that I don't and wouldn't fire into the air, as I've allways told my kids, you are personally responsible for every round you fire. Period.

regards, JTMcC.

Yes......but you have a brain.

A month or so ago, the MythBusters on the Discovery Channel tested the penny legend and determined that the terminal velocity of a penny tossed off the Empire state Bldg was about 60 MPH, as I recall. There is not enough kinetic energy in a penny at that speed to do much damage.

Randy

No, I'm not mistaken about a "little slower". This is a relative term with wide variation in value; in other words, I have not done the calculations but the velocity will be slower. If you real want the velocity of the bullet apply the following:

(1) v = v(0)+at (2) x = x(0)+1/2(v(0)+v)t (3) x = x(0)+v(0)t+ 1/2at^^2 (4) v^^2 = v(0)^^2 + 2a(x-x(0))

Also note that these calcuations (from Physics 101; Halliday and Resnick/Wiley and Sons Inc) do not include atmospheric losses.

have fun. S. Evan

JTMcC wrote:

No. As Hatcher reported, the army tests showed the bullets were still spin stablized when they came down (base first). You have to realize that a .30 bullet leaves the muzzle spinning nearly 200,000 RPM. Even after gravity has slowed its upward velocity to zero, it is still spinning at an incredible rate. So gyroscopic forces easily overcome any aerodynamic tendency for it to nose over and come down point first.

Gary

Tumbling which way? Bullets tumble when they lose a percentage of their rpm. Bullets shot straight up always tumble. There go the old aerodynamics...

FWIW, before WWII the US Army tested .30 cal military bullets shot straight up. They returned to Earth at a velocity of 200 - 220 mph. It didn't matter how high they were shot, what their initial muzzle velocity was, etc.

Specific sectional density doesn't matter much. Aerodymanics go out the window once they start to tumble. What matters is their mass relative to

*effective* projected cross-sectional area, times some factor also related to mass. This is a backhanded way of getting at the Reynolds number for a given-size bullet. The "effective" area is a factor of shape and the way it tumbles.

I understand that the terminal velocity of a spread-eagled human is something like 120 mph. It doesn't matter if you fall from 2,000 feet or

50,000 feet, you hit the ground at around 120 per. Does anyone have better data on that?

Ed Huntress

I don't think so, although it's been many years since I read the research. I thought they were tumbling.

Ed Huntress

Nope.

This is frequently the case and examples in high school Physics have carefully chosen conditions so this will be ok. In order to do so, we apply the principle of superposition which requires that the total effect of a set of forces is the sum of the effects of the individual forces. This requires linearity. Drag is proportional to velocity squared.

Set up the differential equations of motion for a projectile fired at some angle to the horizontal in a vertical gravitational field and experiencing a retarding force alligned with but in opposite direction to the motion vector that is proportional to the magnitude of the velocity squared. It's a neat problem and fun to work. (I first did it studying arrow flight.) You will discover two things: 1) The resulting set of differential equations has no closed form solution (elliptic integrals) and must therefore be solved numerically. (Runge-Kutta is a good method if you wish to try it.) 2) Attempting to separate the forces into vertical and horrizontal components gives an approximation of variable quality but is, in fact, wrong. Speaking vaguely, since the aerodynamic forces depend on the square of velocity, a change in horizontal velocity affects vertical as well as horizontal force and vice versa.

Ted

32 ft/sec^2 "near" Earth's surface in vacuum.

Ted

BTW, if you find this entertaining, here's some newer data, based on Doppler radar research:

Ed Huntress

FWIW Frictional losses are very significant at high velocities(i.g. shuttle reintry). When I was in infantry school at Ft. Jackson during the last 'action' in S.E. Asia I had an opportunity to get one of those info cards that came in each case of 50 cal. machine gun ammo. On that card, range, angle, velocity data were given WITH atmospheric effects. If I just used the equations listed above and optimum 45 deg angle I calculated the range of the 50 cal. to be IIRC miles but the tables showed about 20% what I calculated w/o allowances for air resistance. If one is really interested in this stuff you might look at some of the work Galileo(1564-1642) did with cannon trajectories years before Newton(1642-1727) Just Google _Galileo projectile motion_ or some such. Larry