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The bullet would have to create lift to overcome the curvature of the earth.
It's all moot anyway because the earth is flat, right?
The Shop
I think terminal velocity would be somewhere between 150-170 MPH maybe?
.45s are sub-sonic.
The speed of sound is 767 miles per hour, and that the distance to the Sun is 92,960,000 miles. So, 5,049.98 Days.
Which is 13 yrs and 10 months.
About 8 minutes we would still see it after the disappearance.
That is very likely to happen too.
Hoseclamp, I believe the bullet will fall to the ground in the same amount of time, regardless of which direction the wind is blowing. The exception would be for wind effects, like you mentioned with the spin of the bullet "climbing" up the atmosphere in certain conditions. According to my understanding of Newtonian physics and notwithstanding any atmospheric effects, a bullet would fall directly from muzzle height immediately upon firing if fired level to the ground. (Excuse me; immediately upon leaving the barrel - not immediately upon "firing.") It would take the same number of seconds to impact the Earth regardless of muzzle velocity, although obviously it would land farther away at higher velocities.
As for stray bullets, let's do some quick analysis. This is clearly too simple, but I bet it accounts for 99% of the movements of a bullet:
When fired directly upward, the bullet would describe a ballistic trajectory, like any other thing shot or thrown. It would leave the muzzle of the gun at muzzle velocity. It would then lose velocity at the same rate that gravity accelerates everything; 9.81 m/s squared. At some point it will come to a stop at its highest point, then plummet back to Earth, experiencing the same acceleration. The same amount of time will elapse on the way down as on the way up. Discounting any atmospheric drag, the bullet would re-achieve muzzle velocity by the time it returns to the height from which it was fired. Yes, it is dangerous to fire a gun into the air!
The velocity reached by dropping it from the highest point should be the same as if it were fired up to that point. The result is that the bullet stops a pretty astounding way up (several miles, I'd guess,) and then accelerates all the way down. There is nothing stopping the bullet from reaching muzzle velocity, as the acceleration due to gravity does not "give up" part way down.
Where my calculations falter is with regard to air resistance. I don't know to what extent that will change the equation, but obviously a bullet has a pretty aerodynamic profile. I'd be willing to bet it doesn't change very much.
m121c, what were your findings on the subject?
No spin would cause tumble, and wind resistance would keep it from reaching more than 180 MPH compared to the 700MPH plus at muzzle velocity.
If my math is correct, it would take the bullet just over 20 seconds to reach the zenith of its flight and 20 seconds to fall back to Earth.
http://www.convertalot.com/ballistic_trajectory_calculator.html
Why 180 MPH, though? My guess is that a bullet is far more aerodynamic than that, even without spin.
https://web.archive.org/web/20080331192517/http://www.loadammo.com/Topi…
Except flip it so it only goes straight end over end instead of twisting 180 degrees and ending up upside down.
Good luck, nobody has done it yet.
The 180 is a loose terminal velocity speed figuring atmosphere.
In a vacuum, distance would be the only factor, and you might achieve the same muzzle speeds free falling.
Bullets, especially of the supersonic kind, must slow down pretty dramatically at first!
Pit Row
For the shots fired straight up, Falcon your engineering calculations are correct. But they all assume there is no atmosphere. The results are pretty staggering when you figure them that way. But knowing that wind resistance is such a big part of shot placement in the real world means they aren't very accurate for...... well, the real world. Terminal velocity on the way back down "might" be fudged slightly higher because of the density of the projectile but like was mentioned before, the bullet would likely start to tumble well before it apexed. Therefore, it should take considerably longer to return to the earth for a couple of reasons. First, it will be tumbling, which further increases the wind affects. Second, that wind resistance will not allow the item to go past it's relative terminal velocity. That terminal velocity will not be anywhere close to the actual muzzle speed of the fired projectile. Those bullets will still kill at terminal velocity although a smaller projectile might be less likely to kill.
If you have time, go watch this: https://youtu.be/jX7dcl_ERNs . It's a pretty good explanation of the Coriolis affect. There are other really cool vids out there showing how a spotter can actually see the vapor trail behind a high speed bullet and watch it's trajectory to impact. The things these guys do at distance is amazing to me and partly understanding the physics behind it makes it that much better.
Also, since we are talking about bullets, the trajectory of a bullet is always slightly angled up when trying to shoot a target in a horizonal orientation because of the way gun sights work. It's the reason guns have to be zeroed at certain distances and will be offset at any distance other than the zero distance. The sight of the gun is always above the gun bore. The bore has an axis and the sights, be it a magnified scope or fixed open sights, have their own axis. The gun will be "zeroed" where ever the sight axis is adjusted in intersect with the bore axis plus the amount of bullet drop. So the bullet will typically exit the end of the bore traveling slightly toward the sight axis. Like mentioned in an earlier post, at some point the bullet trajectory will cross the sight axis as it travels away and slightly up. Then, at some other point in time, after the bullet apexes, the bullet will cross the sight axis again traveling downward. So the gun can technically be zeroed at two distances. The Coriolis affect effectively means the target is moving relative to where it was when the bullet was fired. Some of the travel times are in excess of 5 seconds. Since the earth is round(ish), that means the target is traveling up and closer or down and farther away depending on the direction you've fired. It's most effective in shots due east or west. There are charts and computer programs developed for taking account for these things. Add to that the differences in bullet projectile mass, shape, and the amount of charge in the casing and you get a pretty complex calculation if you're trying to shoot a target the size of a watermelon (or a man) from a long way away.
Most of my physics relies on ignoring atmospheric effects.
Even in dead calm conditions, a bullet will drift right or left depending on barrel twist.
Kinda like when you duck hook or slice a tee shot.
https://youtu.be/tF_zv3TCT1U
Now you have something dropped from a high place. Nothing like leaving a muzzle.
Not even remotely like leaving a muzzle.
If I clamp a length of 6mm steel rod in a vice and push down on it it will bend. But if I pull upwards with the same force I *doubt* it would return to being a perfectly straight rod again. I imagine that at best it would have a kink by the vice. Why is that?
As I said, I haven’t tried this, so maybe if the forces and direction were *exactly* the same/opposite then it would. Or would it..?
Or maybe I'm talking out of my ass, I don't know.
By bending, you are breaking some of the molecular bonds in the metal. Those will not reconnect unless the steel is molten (I think.) You'll never again have the steel rod in the original condition. Imagine if you were to bend back and forth repeatedly. Eventually the rod would break at the spot where the bending occurs.
I do think you could get a perfectly straight rod again, with just the right application of forces. It just wouldn't be as strong.
Any metallurgists in the building??
https://en.m.wikipedia.org/wiki/Work_hardening
I was told this by Marzocchi, so who knows if it’s true, but it makes sense to me.
So when forks hit a bump, or land from a jump the forks try to bend/flex forward or backwards for due to the forces. (Everything flexes, even if only by a very small amount). They will flex the most at the weakest point/point of most leverage.
Conventional forks will flex where the chrome tube exits the lower triple clamp: that’s the point of greatest leverage. The lower part, where the outer tube is sliding over the chrome stanchion remains straighter.
With USD forks it’s different. The flex point is where the smaller chrome tube enters the outer tube. So the bent chrome tube is trying to go through the tight bushes in the outer tube. As a result it binds, whereas the conventional fork has a far smoother action because the maximum bend point is not at the point where the two tubes are sliding.
Not sure whether I explained that well enough. But as I said, it makes sense to me. I’m sure modern, larger diameter USD tubes may have mitigated that problem somewhat but it’s hard to argue against the logic.
To this day, the best forks I ever rode with were Marzocchi conventionals…
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