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Old 11-07-2009, 01:17 PM   #31 (permalink)
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what does the vertical force on the paintball have to do with its X direction velocity?
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Old 11-07-2009, 02:08 PM   #32 (permalink)
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With a 50 FS round, you would give up a lot of the advantages available to each. They won't fit in a hopper, so you can't fit many more of them. And you'd give up putting them in conventional guns. Including the .50 guns available today, if you elongate the round.
Which is why both "finned" paintballs and "elongated" paintballs all failed in the marketplace during the 90s.

It is an issue of usability. I remember trying the similar Sniperballs. They seemed more accurate, but for every sniperball fired, you could fire 20-30 regular shots which had a far greater chance of hitting someone, and ended up costing about the same.

The only real attraction comes from the milsim aspects. Players might like how it can increase the accuracy of the simulation.
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Old 11-07-2009, 03:13 PM   #33 (permalink)
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what does the vertical force on the paintball have to do with its X direction velocity?
I'll give you an opportunity to think that through and say.. "nevermind, I made a mistake". I think you can figure that one out vector boy.

Actually.. no.. I'll explain it. The ball moves along the x plane.... but because we have gravity we have introduced a y plane component. Gravity acts on the ball... it pulls it down... which means the ball now has a y component to the velocity in addition to the x component. Which gives you an angular velocity vector. The more it moves down (closer to the ground), increase in y component you have a reduction in the x component.

In space.. because there is no gravity, there will be no y/perpendicular component to the flight path.

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Old 11-07-2009, 04:57 PM   #34 (permalink)
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I'll give you an opportunity to think that through and say.. "nevermind, I made a mistake". I think you can figure that one out vector boy.

Actually.. no.. I'll explain it. The ball moves along the x plane.... but because we have gravity we have introduced a y plane component. Gravity acts on the ball... it pulls it down... which means the ball now has a y component to the velocity in addition to the x component. Which gives you an angular velocity vector. The more it moves down (closer to the ground), increase in y component you have a reduction in the x component.

In space.. because there is no gravity, there will be no y/perpendicular component to the flight path.

E
dot.

The x and y components of motion are absolutely independent. Gravity near the surface of the earth pulls on all objects at 9.81m/s/s, so both the .50 cal and .68 cal projectiles will fall at the same rate (neglecting air resistance of course).
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Old 11-07-2009, 10:11 PM   #35 (permalink)
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Cody... go back to taking photos.

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Old 11-08-2009, 09:57 AM   #36 (permalink)
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hahah, will do. My understanding of flight of projectiles is minimal right now. Centripetal acceleration and forces on the other hand....
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Old 11-08-2009, 07:39 PM   #38 (permalink)
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I'll give you an opportunity to think that through and say.. "nevermind, I made a mistake". I think you can figure that one out vector boy.

Actually.. no.. I'll explain it. The ball moves along the x plane.... but because we have gravity we have introduced a y plane component. Gravity acts on the ball... it pulls it down... which means the ball now has a y component to the velocity in addition to the x component. Which gives you an angular velocity vector. The more it moves down (closer to the ground), increase in y component you have a reduction in the x component.

In space.. because there is no gravity, there will be no y/perpendicular component to the flight path.

E
the Y component is independent of the X component however.

without drag, the ball would simple fall at the same rate downward as a ball not flying 300 fps. this is one of those junior high science experiments with tennis balls. gravity is a state function, it wont "bend" the initial X velocity downward, the X velocity will be unchanged by the Y force of gravity. only drag will slow down the ball in the X direction.



perpendicular forces do not effect each other, and there cross product does not equal a moment. the only reduction to the velocity will be from drag, in a level fired shot.
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Last edited by Cockerpunk; 11-08-2009 at 08:01 PM.
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Old 11-09-2009, 09:28 AM   #39 (permalink)
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Wow.. Thanks for the explanation!

It's amazing how I've almost forgotten everything from Middle school ('88-'91) and High school ('91-'95) and freshmen year of college ('95-'96). I guess I've found more important information to store in the limited space up there. Oh well. Having a young mind like yours around here that can remember this stuff is refreshing. Saves me from trying to mine it out of some dark corner of my brain.

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