FAQ

Questions

• At which angle of departure does a bullet achieve its maximum range?

 
• Why is the rotation of the bullet, after leaving the muzzle, clockwise

and why not counterclockwise?
 
• Is it true that if a bullet and its shell are released simultaneously, they

will both hit the ground at the same time? Why?
 
• If a bullet is fired horizontally from a barrel that is perfectly level, will

the bullet, at some point, rise?
 
• Centrifugal force is a ficticious force, it does not exist! There is only a force radially inward which is the centripetal force.

 
• Can a conventional gun fire an ordinary bullet in the vacuum of space?

 
• Will a bullet stabilize in space (absense of atmosphere), or tumble?

 
• How fast does a bullet lose its spin velocity?

Answers

A: 1. If one neglects the atmosphere and considers bullet motion in vacuum, the
maximum range will be reached for a 45° departure angle.

2. For bullets fired from handguns through the atmosphere, the maximum range (typically in the range of a few kilometres or less) is usually reached for a 30°- 35° departure angle. This is a consequence of bullet retardation by the drag force.

3. Artillery shells may reach maximum ranges of a few dozens of kilometres. When fired at higher departure angles, these projectiles are capable to reach much higher altitudes than handgun bullets. However in those altitudes the air density is considerably smaller than the ground air density. Lower air density goes along with lower drag and this is the reason why artillery shells reach their maximum range for higher angles of departure (typically at 45°).

A: Of course - in the real world - there exists clockwise (right handed) as well as counterclockwise (left handed) twist. For unknown reasons, at least for handguns, the majority of barrels have clockwise twist. Just for the purpose of simplification this article restricts only to clockwise twist. In fact, for counterclockwise twist, some forces (e.g. Magnus) change their orientation.

A: It would be exactly true if there was no drag - or with other words if those objects were dropped in vacuum. In vacuum the only active force is gravity F_G= mg; m=mass g= gravitational constant. As a consequence the fall time t is independent of the mass (s = falling distance): If both objects are dropped in air, their fall times to reach the ground will not be exactly the same, because they experience different retardations, depending on their shape, the way they tumble while they fall, and other factors.
However, while falling, the drag of both objects will be very small, and they will "approximately" hit the ground at the same time.

A: There is no way a bullet can rise over the axis of bore of the gun, because there is no "upward" directed force:
1. The drag acts opposite to the direction of movement and simply retards the bullet
2. the force of gravity is directed downward.
Some kind of confusion may arise, because the bullet  normally rises over the line of sight of a gun. The line of sight however is usually inclined with respect to the axis of bore.

A: For a full answer please see  .

A: There is no reason why a conventional gun shouldn´t fire in vacuum.
First, the primer in the cartridge (which contains explosive material) is mechanically ignited. Hot particles are produced which ignite the powder charge.
The powder however already contains the oxygen which is needed for "burning".
If this wouldn´t be the case (if e.g. the powder had to take oxygen from the surrounding atmosphere), the burning process would be too slow.

It would be interesting to ask NASA for a verification!

A: No, the bullet will not tumble. Except the (weak) gravitational forces of the planets, which apply at the bullet´s CG and do not cause any moment, there will be no other forces or moments, especially no destabilizing moments. However, any bullet precession induced at muzzle exit will continue.

This would be a second experiment for NASA!

A:This question cannot be answered in general. As a rule, spin is much less reduced than velocity: An estimate for the M80 bullet (7.62 x 51 Nato) fired vertically up gives the following values:
• all of the velocity has been lost at the summit
• only approx. 36% of the angular velocity has been lost at the summit.