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If you shot a 270 straight up using a 130 grain bullet(bc of .4 and mv of 3,000 fps)....how fast would the bullet be traveling when it hit the ground?

In SA, we still have people shooting pistols in the air to celebrate.

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If you shot a 270 straight up using a 130 grain bullet(bc of .4 and mv of 3,000 fps)....how fast would the bullet be traveling when it hit the ground?

In SA, we still have people shooting pistols in the air to celebrate.

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Fast enough to get a speeding ticket.

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same speed as a 500 gr bullet or a henweigh

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Roger was the bus driver and the bus was red

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I may be wrong, but I seem to remember falling velocity as

200 ft. per second, increasing by 100 fps every 200 ft.

Terminal velocity (the fastest an object will fall without outside influence is about 210 MPH., or 308 mps. Therefore an object will reach terminal velocity after falling about 350 ft. Wind resistance will pay a part, as will sectional density (think cannon ball and soft ball).

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You start with the velocity equation: V(f) = V(o) + at

V(f) = 0fps this is at the bullet's peak and V(o) = 3200fps. You solve this for a time of 99.3788 seconds. So it takes over a MINUTE AND A HALF just for the bullet to reach its peak.

Now you use the quadratic distance formula: x = x(o) +V(o)t + (.5)(a)(t^2)

I used an initial height of 6ft assuming the bullet leaves the barrel 6ft off the ground. And an initial velocity of 3200fps & the acceleration due to gravity a= -32.2 f/(s^2)

And you just solved for t=

With this you get a maximum height of

Now using your kinetic (E) and potential (U) energy equations: E = (.5)(m)(V^2) & U = m*a*h

By the way...I looked up the actual weight of a 130grain bullet...it is

You can solve for your final velocty when the bullet hits the ground with the energy equations. Just to let you know the maximum energy is

But ultimately the equations yield a final velocity of

***REMEMBER THIS IS IN A VACUUM WITH NO WIND RESISTANCE!***

In the real world it will be less than this as the wind slows it down and the bullet will not fall straight, but will tumble in the air.

This is just an engineer being bored at work...take it as you will

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AFishinMan14- That puts the bullet falling at a rate of 2,182 Miles per hour. This seems a little fast, even in a vacuum. That puts it about the same as muzzle velocity. I have never jumped out of an airplane, but those who have told me the human body (on average) falls at 210 mph at terminal velocity. I am sure a bullet would go faster, having less wind resistance, but 2182 seems a bit high. Oh, and my previous post should read fps, not mps.

Bored at work too, need to go hunt! :slimer:

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In a vacuum there is no wind resistance which is basically friction on the bullet slowing it down. The only thing slowing it down is gravity (32.2 ft/(s^2)). Therefore the distance traveled and velocities reached in a vacuum are MUCH greater than on Earth...My equations are correct, for an unrealistic situation. If only gravity is applied, there is no "terminal velocity" or a maximum velocity. As long as an object is falling towards the Earth...it is accelerating. It never stops accelerating and gainging velocity.

All said an done...these calculations have no practical relevance to the question, because the fact is that we do have wind resistance. Which wouldn't surprise me if such a factor cut my numbers in half.

But in an ideal environment, the math is correct.

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Your analysis does however show how important wind resistance is on ballistics.

EDIT: Sorry, didn't read your statement in the 2nd post.

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The problem is,

Once the bullet starts to fall, that BC goes out the window as the bullet is likely to be falling/flipping every which way.

The ballistic coefficient is a relation of the "drag coefficient". The drag coefficient itself is a function of the area of the object normal (perpendicular) to the direction of travel. If the bullet isn't spinning like a football in a tight spiral, the drag coefficient is different.

The "drag coefficient" is used for calcuating the drag force (which is a function of velocity squared)

Terminal velocity is the velocity where this drag force equals the force of gravity.

So without knowing exactly how the bullet is "spinning" as it falls, it's tough to say.

Also, a BC of 0.4 doesn't mean the drag coefficient is 60% from ideal. Drag coefficients can be greater than 1. It means that is is 60% from some "standard bullet" which has a BC of 1 (not a drag coefficient of 1).

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Fd=Cd/2 * Rho * V^2 * A

where

Fd= Drag force

Cd=Drag Coefficient

Rho=Density of Air

V=Velocity

A=Cross Sectional Area

By definition, terminal velocity is where the drag force equals the force of gravity (Fg)

So if:

Fg=m*ag

where

m=mass

ag=acceleration due to gravity

Then if:

Fd=Fg

Cd/2 * Rho * V^2 * A = m * ag

and

V^2=(2*m*ag)/(Cd*Rho*A)

and finally

V= Sqrt((2*m*ag)/(Cd*Rho*A))

So, if you want to know, plug and chug with this equation. You'll need to know:

the diameter of the bullet

the length of the bullet

the mass of the bullet

the density of air

the BC for a cylinder falling sideways

the BC for a cylinder falling longways

The bullet falling sidways would be your ''minimum" velocity. The bullet falling longways would be your "maximum" velocity. The true number will be somewhere between the two.

Unless I've done something wrong....it is getting late over here!!!

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I did a little research on the internet and found this....

In 1920 the U.S. Army Ordnance conducted a series of experiments to try and determine the velocity of falling bullets. The tests were performed from a platform in the middle of a lake near Miami, Florida. The platform was ten feet square and a thin sheet of armor plate was placed over the men firing the gun. The gun was held in a fixture that would allow the gun to be adjusted to bring the shots close to the platform. It was surmised that the sound of the falling bullets could be heard when they hit the water or the platform. They fired .30 caliber, 150 gr., Spitzer point bullets, at a velocity of 2,700 f.p.s. Using the bullet ballistic coefficient and elapsed time from firing until the bullet struck the water, they calculated that the bullet traveled 9,000 feet in 18 seconds and fell to earth in 31 seconds for a total time of 49 seconds.

As a comparison, the .30 caliber bullet fired in a vacuum at 2,700 f.p.s. would rise nearly 21.5 miles and require 84 seconds to make the ascent and another 84 seconds to make its descent. It would return with the same velocity that it left the gun. This gives you some idea of what air resistance or drag does to a bullet in flight

Based on the results of these tests it was concluded that the bullet return velocity was about 300 f.p.s. For the 150 gr. bullet this corresponds to an energy of 30 foot pounds. Earlier the Army had determined that, on the average, it required 60 foot pounds of energy to produce a disabling wound. Based on this information, a falling 150 gr. service bullet would not be lethal, although it could produce a serious wound.

Many other experiments have been made to find the amount of air drag on a .30 caliber bullet at various velocities and it was found that the drag at 320 f.p.s. balances the weight of the .021 lb. (150 gr.) bullet and terminal velocity is achieved. For larger calibers the bullet terminal velocity is higher since the bullet weight is greater in relation to the diameter. Major Julian Hatcher in his book

So if you're hit by a bullet falling straight down and it's from a firearm, you'll probably live. If you're hit by a bullet that was allowed to stabilize...not straight up and down...it's pretty deadly.

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I saw a girl in New Orleans during New Years some years ago that had a bullet hole in the top of her head.......don't know if the bullet went straight up or at angle but it did kill her!!!

I also saw the Mythbuster show where they disproved it could happen......but I saw what I saw and the girl was dead...........

Catfish48

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