question about velocity

[KillerCharlie]

Originally posted by: presidentender

Originally posted by: tcsenter
What if you fired a bullet from a barrel inside a rail gun enclosed in a rocket....cripe!

There is already a pretty good model of what you are talking about; the cannons mounted on fighter planes. e.g.

If muzzle velocity of the bullet fired while the fighter is not moving = 3000 ft/s

Then yes, muzzle velocity of the bullet fired while fighter is moving = fighter speed + 3000 ft/s

This is also why you can pee in the bathroom of an airliner traveling 500MPH.

Heck, we could stack these all day, although you probably wouldn't want rail guns within rail guns.

Drag really starts to bite you though...

 

[presidentender]

Originally posted by: dkozloski
FWIW. There have been a couple of documented instances where a jet fighter has shot itself down by encountering projectals from it's own weapons.

The F-11 Tiger is noted for being the first jet aircraft to shoot itself down. On 21 September 1956, during a test firing of its 20 mm cannons, pilot Tom Attridge fired two bursts mid-way through a shallow dive. As the velocity and trajectory of the cannon rounds decayed, they ultimately crossed paths with the Tiger as it continued its descent, disabling it and forcing Attridge to crash land the aircraft. The pilot survived.[1][2]

The plane, though, have constant thrust. The rounds from the cannon don't; they slow down, the plane doesn't.

 

[futuristicmonkey]

There are three cases for which the bullet's kinetic energy will change once the charge is detonated. As I see them, they are:

i) If the casing is heavier than the bullet the bullet will have the largest gain in KE. Think about shooting a gun. The gun and you do not gain a significant amount of KE because the bullet is much more easily accelerated. |delta-v| of the casing << |delta-v| of the bullet.

ii) If the mass of the casing equals the mass of the bullet the bullet's KE will only moderately increase. |delta-v| of the casing ~= |delta-v| of the bullet.

iii) If the mass of the casing is much less than that of the bullet, the bullet will experience only a very slight gain in KE. |delta-v| of the casing >> |delta-v| of the bullet.


Edit: I forgot to state the foundation for what I've said. Newton's third law (I think it's the third) ensures that the force the bullet experiences will be equal in magnitude and opposite in direction, compared with the force on the casing. Newton's second law states that the acceleration of an object is proportional to the force it endures and inversely proportional to its mass.

-ben

 

[DrPizza]

Originally posted by: KIAman
????? Where the heck are YOU getting your formula from?

Kinetic Energy = 1/2M * V^2, I don't even know where you pulled (mv^2)/2 from, that doesn't even derivate into any other energy equations.

EDIT: The formula is one and the same, haha. Wow, I am really tired.

And no, it is NOT appropriate to measure a bullet's power in Force. Force has a deceleration component that you absolutely cannot know or predict. Is the bullet going to magically come to a complete stop once it hits its target in.... 1 milli second? in 1/10 of a second? How do you know? What is the object and what is the property of that object? You cannot know.

Joules is a good metric to describe a bullet's power because it is independent of the object it is hitting.

Let's make an example to help clear this up.

Situation 1: Let's say you shoot a .22 hunting rifle at a distance of 100y at a piece of meat 2 feet thick.

Situation 2: Let's say you shoot a .22 hunting rifle at a distance of 100y at a piece of depleted uranium 2 feet thick.

Using logic, you would think that the bullet has the same impact power on either target, because you are firing the exact same bullet with the exact same speed and distance. Using F=MA, situation 2 will create MUCH more force then situation 1. I'll let you let that sink in as to why.

Damnit, now I have to think and my brain is hurting recovering all my servers. Thanks! Jerk! 🙂

You toss around physics terms - force, power, etc., as if you pull them out of a hat.
Power: watts
Energy: joules
Force: newtons
Momentum: kgm/s
impulse: N*s = kgm/s

And, in your depleted uranium example, in both cases (assuming the bullet doesn't pass through the meat), both objects will receive the same impulse (more or less*), but since the time is greater in the meat, it experiences a smaller force & vice versa for the uranium.
*more or less because I assume the meat is less dense than the uranium, thus it will be moving faster than the uranium after the impact. This difference is, for all practical purposes, negligible.

Originally posted by: tcsenter
This is also why you can pee in the bathroom of an airliner traveling 500MPH.

However, when you take drag into account, this is why if you pee in the direction you're traveling while strapped to the roof of a plane like a stunt rider, you're going to end up with wet clothing.


OP: since we're going with the assumption of 50 feet per second when fired in the open, while at rest, then:
If we ignore air resistance, final speed = 5050 feet per second
If we don't ignore air resistance, 5000 < final speed < 5050 feet per second.

What the heck is so hard about that?!

Reason it will be less than 5050: when the gunpowder ignites, it imparts an impulse to the bullet - which has a component of time. At rest, a bullet doesn't go from 0 to 50 ft per second instantaneously; there's a very small amount of time involved. During that time, drag forces act on the bullet; if there were no air, then the final speed would be slightly greater than 50 ft per second. Now, when it's already going 5000 ft per second, assuming that the gunpowder were able to provide the same impulse - over the same time (I can't see what's wrong with that assumption), then drag forces are going to act on the bullet for that very brief amount of time. However, as you increase from 5000 ft per second to 5050 ft the amount of drag is significantly more than the amount encountered during 0 to 50.

 

[PlasmaBomb]

Originally posted by: KIAman
^ Wrong.

The impact energy of the bullet can be measured in kinetic energy.

E = 1/2M * V^2

Speed has MUCH more impact to the kinetic energy as mass does. I'm too tired to do math right now (I'm at work and trying to recover some broken servers) but there is a certain falling off point where enough mass needs to be loss to equal an overall loss of kinetic energy while speed is gained.

Multiple "bullets" fired from a single ejected mass is the simplest form of a anti personnel weapon, aka artillery shells, fragmentation grenades, etc.

As we are manly men, let us use the .50 cal as an example. The bullet weighs about 50 g (depending on manufacturer - wiki has some weights if you are interested). The round (inc. casing and propellant) weighs in about 150 g. Thus causing the bullet to be fired during flight (remember no barrel) would result in a small increase in bullet speed and a substantial drop in mass.

The problem would be that the casing now has a big open hole at the front (O) which will increase drag and likely mean the casing won't hit the target. Another problem no one has mentioned is destabilization. If the bullet is fired without a barrel to guide it, there is no guarantee that it will stay on course.

Another interesting though is that it would only take the railgun slug ~250 microseconds to overtake the round once it has caught up with it. So it may well just squidge the round...