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To: Pikachu_Dad; kjam22

>> “Bullet begins to drop as soon as it leaves the muzzle.” <<

.
Not if it is aimed upward on a ballistic trajectory.

What do you think the “elevation” settings on a sight or scope are for?

A projectile aimed upward continues upward for a considerable distance. Brush up on your trigonometry, and ballistics.
.


36 posted on 08/05/2014 7:56:46 PM PDT by editor-surveyor (Freepers: Not as smart as I'd hoped they'd be)
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To: editor-surveyor
“Bullet begins to drop as soon as it leaves the muzzle.”

It does begin to drop as soon as it leaves the muzzle, when compared to the mathematical line that includes the axis of the barrel and which extends out from the muzzle to infinity.

If the muzzle is aimed at a point five degrees above the horizon, the bullet starts to fall below that line the instant it leaves the muzzle. The bullet is still climbing with respect to a horizontal, but it is falling with respect to the muzzle axis.

When he said he aimed at a point 75 yards above the target, that does not mean that the round ever got 75 yards above the target, not even close.

44 posted on 08/05/2014 8:04:13 PM PDT by Steely Tom (How do you feel about robbing Peter's robot?)
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To: editor-surveyor

Sorry, had to drive home before I tackled the next step.


54 posted on 08/05/2014 9:48:55 PM PDT by Pikachu_Dad (Impeach Sen Quinn)
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To: editor-surveyor

If we add in the angle.

horizontal terms
Sin(ang)=vert/fps. cos(ang)=dist/fps

So. Fps*Cos(ang)= dist (ft/sec)
Time= distance/dist = distance/(fps)*Cos(ang)

Vertical terms
Initial vertical velocity (fps) = fps*Sin(ang)
Less acceleration due to gravity = a = -9.8 m/s

Velocity = integral ( a dt) from t to to = at + Vi. = fps*Sin(ang) - 9.8 m/s * t

Then height is. Integral of above = Vi t - 9.8/2 * t^2. = fps*Sin(ang) - 9.8/2 * t


56 posted on 08/05/2014 10:04:42 PM PDT by Pikachu_Dad (Impeach Sen Quinn)
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To: editor-surveyor

>> “Bullet begins to drop as soon as it leaves the muzzle.” <<

.
Not if it is aimed upward on a ballistic trajectory.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

The statement “...begins to drop as soon as it leaves the muzzle.” is correct. The drop manifests as a reduction in the vertical component of the projectile velocity over time.
If one dropped an object from near the surface of earth, it would accelerate at a nearly constant rate, in the downward direction (toward center of mass of the planet.) Similarly, the projectile in this example does the same thing, as described above. Effects of aerodynamics, and relativistic effects, and the location of the center of mass of the universe, and possibly others, have been dismissed from this discussion.


81 posted on 08/06/2014 6:16:05 PM PDT by loungitude (The truth hurts.)
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