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 [quote="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: [i][b]a[/b][/i] = - ([i]GM[/i]e[i][b]r[/b][/i])/[i]r[/i]3 - [i][b][/i]w[i][/b][/i] x ([i][b][/i]w[i][/b][/i] x [i][b]r[/b][/i]) - 2[b]w[/b] x [i][b]v[/b] - b[b]v[/b]/m[/i] [b]w[/b] is the vector angular velocity of the earth's rotation (2p/day, pointed along the earth's axis). The x represents the cross product, [i][b]v[/b][/i] is the object's velocity, [i]b[/i] is the drag coefficient, [i]m[/i] is its mass, and [i][b]r[/b][/i] is its vector position from the center of the Earth. The drag term ( [i]- b[b]v[/b]/m[/i]) opposes the object's motion (obviously), while the Coriolis term ( - 2[i][b]w[/b][/i] x [i][b]v[/b][/i]) 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 (- [i][b]w[/b][/i] x ([i][b]w[/b][/i] x [i][b]r[/b][/i])) pushes the object away from the Earth's axis of rotation, or up and south (in the northern hemisphere). ------------------ You will respect my philosophai. [This message has been edited by Dr. Borodog (edited 02-01-2004 12:54 PM).][/quote]
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Buzzsaw
Posted: Mon Feb 02, 2004 4:56 pm    Post subject: 1

Quote:
If you really want to know the answer go out side and fire a gun.

At first, I didn't think this was possible to do this sober, but I just fired a 22 rifle from my front porch.(that chickadee had had enough to eat and was hogging the feeder, RIP)

I did learn somethin and that's that learning never taught me nothin and I certainly didn't learn no math from shoting off that gun just now.

I think the neighbor lady is calling the police.
Dr. Borodog
Posted: Mon Feb 02, 2004 4:27 pm    Post subject: 0

".005 mililmetters"

*cough BULLSHIT! cough*

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You will respect my philosophai.
Will
Posted: Mon Feb 02, 2004 4:35 am    Post subject: -1

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.
Lepton
Posted: Mon Feb 02, 2004 3:51 am    Post subject: -2

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).]
Courk
Posted: Mon Feb 02, 2004 3:11 am    Post subject: -3

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)
Courk
Posted: Mon Feb 02, 2004 3:04 am    Post subject: -4

You last took physics when I was born.
Posted: Sun Feb 01, 2004 11:01 pm    Post subject: -5

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.
Dr. Borodog
Posted: Sun Feb 01, 2004 5:53 pm    Post subject: -6

Thanks DP!

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You will respect my philosophai.
Dr. Borodog
Posted: Sun Feb 01, 2004 5:50 pm    Post subject: -7

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 = - (GMer)/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).

------------------
You will respect my philosophai.

[This message has been edited by Dr. Borodog (edited 02-01-2004 12:54 PM).]
DP
Posted: Sun Feb 01, 2004 5:25 pm    Post subject: -8

abcdefghijklmnopqrstuvwxyz
Dr. Borodog
Posted: Sun Feb 01, 2004 5:23 pm    Post subject: -9

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

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You will respect my philosophai.
Posted: Sun Feb 01, 2004 5:14 am    Post subject: -10

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).]
Chuck
Posted: Sun Feb 01, 2004 4:37 am    Post subject: -11

It looks like it's firing backwards. Doesn't that cause accidents?
Courk
Posted: Sun Feb 01, 2004 4:32 am    Post subject: -12

You'll notice in my picture that I already solved this problem. The gun 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.
Termital
Posted: Sun Feb 01, 2004 4:12 am    Post subject: -13

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?
Dr. Borodog
Posted: Sun Feb 01, 2004 3:46 am    Post subject: -14

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

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You will respect my philosophai.
The Ktulu
Posted: Sun Feb 01, 2004 3:35 am    Post subject: -15

Barometric pressure. Don't forget barometric pressure.
wordcross
Posted: Sun Feb 01, 2004 2:49 am    Post subject: -16

if we're going to take natural phenomenon into considerateion, then you have to factor in wind speed, gravitational fluctuations, magnetic interference, spin, etc.
Pablo
Posted: Sun Feb 01, 2004 1:48 am    Post subject: -17

We were one-upping. THAT was more like five or six-upping.
Courk
Posted: Sun Feb 01, 2004 12:18 am    Post subject: -18

Since we're one-upping each other:
Marvin
Posted: Sat Jan 31, 2004 9:19 pm    Post subject: -19

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).]
i_h8_evil_stuff
Posted: Sat Jan 31, 2004 9:19 pm    Post subject: -20

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).]
Pablo
Posted: Sat Jan 31, 2004 9:17 pm    Post subject: -21

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.
firemeboy
Posted: Sat Jan 31, 2004 9:08 pm    Post subject: -22

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.