Another question for the math people...

[firemeboy]

Hey, if I were to fire a rifle into the air at a 45 degree angle, what does the trajectory look like? Is it a mirror image from the apex? In other words, does it hit the ground (when it hits the ground), at a 45 degree angle, or does gravity take over and it hits pretty much straight down? I assume that air resitance has somethign to do with it.

[Pablo]

Without wind resistance it is a perfect parabola, mirror image as you describe. With wind resistance, the horizontal velocity in the first half of the trajectory is greater than in the second half, so as it slows down horizontally, it will impact the ground at a steeper angle than 45 degrees. Also, the apex will be farther from you than from the point of impact.

[i_h8_evil_stuff]

Neglecting air resistance, the bullet should have an angle of -45 degrees at the moment that it is at the same height as what it was fired at. Also, it will have the same velocity as when fired at that moment.

Edit: Damn you, Pablo.



[This message has been edited by i_h8_evil_stuff (edited 01-31-2004 04:19 PM).]

[Marvin]

I'd expect it would be a mirror image but for air resistance, like you said. That, and the curvature of the earth.
[Bah at both of you. But I mentioned curvature.]

[This message has been edited by Marvin (edited 01-31-2004 04:20 PM).]

[Courk]

Since we're one-upping each other:

[Pablo]

We were one-upping. THAT was more like five or six-upping.

[wordcross]

if we're going to take natural phenomenon into considerateion, then you have to factor in wind speed, gravitational fluctuations, magnetic interference, spin, etc.

[The Ktulu]

Barometric pressure. Don't forget barometric pressure.

[Dr. Borodog]

Damn. A physics question and it's already pretty much answered.

I could write down the equations for you if you wanted.


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You will respect my philosophai.

[Termital]

When the explosives go, isn't part of the force expended on the gun? They do kick back don't they? So doesn't the moving barrel skew the bullet's path so long as the bullet is in?

[Courk]

You'll notice in my picture that I already solved this problem. The gun the robot is holding is a Synergy Technologies Laser Propulsion Firearm 3000-X2 with built in silencer and an explosive force resistor system. They're standard on all STLPFs versions 2056-X3 and up.

[Chuck]

It looks like it's firing backwards. Doesn't that cause accidents?

[Dread Pirate Westley]

It may look like that, but actually the oversized butt holds all the Synergistic Technology (tm) and Laser Propulsion gizmos, along with the explosive force resistors. They're working on miniturizing these systems for the 3100 series for a weapon that's 20% lighter.

[This message has been edited by Dread Pirate Westley (edited 02-01-2004 12:14 AM).]

[Dr. Borodog]

Is there an easy way to get greek letters into your posts?

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You will respect my philosophai.

[DP]

abcdefghijklmnopqrstuvwxyz

[Dr. Borodog]

The acceleration of a moving object near the surface of the Earth, subject to the Earth's gravity, the centrifugal and Coriolis forces, and wind resistance (assuming the air is motionless relative to the Earth's surface) is given by:

a = - (GM_er)/r³ - w x (w x r) - 2w x v - bv/m

w is the vector angular velocity of the earth's rotation (2p/day, pointed along the earth's axis). The x represents the cross product, v is the object's velocity, b is the drag coefficient, m is its mass, and r is its vector position from the center of the Earth.

The drag term ( - bv/m) opposes the object's motion (obviously), while the Coriolis term ( - 2w x v) will push a north-moving object east (in the northern hemisphere), a west-moving object towards the ground (and north), a south-moving object west, and an east moving object up and south. The centrifugal term (- w x (w x r)) pushes the object away from the Earth's axis of rotation, or up and south (in the northern hemisphere).



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You will respect my philosophai.



[This message has been edited by Dr. Borodog (edited 02-01-2004 12:54 PM).]

[Dr. Borodog]

Thanks DP!

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You will respect my philosophai.

[Samadhi]

Bah. I need to get back to school and take some physics (last class was in High School, 1986).

Could you give the units of each term? I would be better able to understand it at that point.

[Courk]

You last took physics when I was born.

[Courk]

G doesn't have units (I don't think), it's a constant. M sub e is the mass of the earth in kilograms. r is the radius of the earth, in meters, omega (the little w looking thing) is, as Boro said, 2 pi radians per day (or per many seconds, or minutes, etc). v is measured in meters per second. Drag coeficient (b) might be a constant with no units, but don't quote me. a, acceleration, is measured in meters per square seconds, m/(s^2)

[Lepton]

Constants aren't necessarily unitless, Courk...

I guess we're in MKS, so:
G = 6.6726E-11 Nm^2/kg^2
omega has units of s^-1
the drag coefficient is given in units of kg/s, and is somewhat interesting, because it depends on things like humidity and that sort of thing.

[This message has been edited by Lepton (edited 02-01-2004 10:51 PM).]

[Will]

the guy that one an olympics gold metal for shotting does this.

He has to shot a target at 55 yards away the size of a quarter when he gets up to shot he has to get everything to stop. He gets his digestive system to stop by not eating in the last 12 hours. he stops breathing at that moment to slow up his pulse. He then feels his pulse through his body and he then shoots at the exact moment that is between each pulse. then there is all the mathmatical stuff to go through. The wind speed. The gravatational speed (the drop the bullet will take). And then he says if the tip of the barral is close to .005 mililmetters off the bullet will drop to the next circle and cost him 30 points. If you really want to know the answer go out side and fire a gun.

[Dr. Borodog]

".005 mililmetters"

*cough BULLSHIT! cough*

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You will respect my philosophai.

[Buzzsaw]