FAQ

Questions and remarks of readers

• 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?
 
• If a bullet is fired horizontally from a barrel and another bullet is dropped from the same altitude at the same instant, will they both hit the ground at the same time?

 
• 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?
  
  
• How fast do bullets travel through the air?

• What is the unit of the drag coefficient c_D and what is the connection between c_D and the projectile caliber?
  
  
• Where might I find more information about estimating C_M (overturning moment coefficient) for various shaped bullets, primarily boat tail and flat base match bullets. 
  
  
• If a bullet is fired vertically from a rifle, what will its terminal velocity be if it strikes the top of someones head on its way back down?
  
  
• How can bullet drift can be calculated from spin?
  
  
• I would like to photograph a bullet in flight with the shadowgraph technique. What equipment do I need?

Answers of the author

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, theseprojectiles 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, atleast 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: This is an interesting question and the answer is not trivial.
It is true that the horizontally fired bullet and the dropped bullet would hit the surface at the same moment, if the experiment happens in vacuum. In vacuum there is only the force of gravity which affects both objects in the same way.
However if the shooting occurs in air there is the additional force of drag.
Both objects - let us assume spheres - experience drag. The difference however is, that the horizontally fired bullet has a much higher velocity. Only the "downward" velocity components v_y at t=0 are the same (v_y=0) for both bullets.
The force of drag is (roughly) proportional to the square of the velocity v (v = sqrt(v_x² + v_y²)) and not only to the v_y component!  Thus, the drag experienced by the fired bullet is much higher than the drag experienced by the dropped bullet. As a consequence the fired bullet will reach the surface later.
Example: Sphere of 10 mm diameter, 10 g mass, fired at 500 m/s from 10 m height
1. Horizontally fired: flight time 1.649 s; terminal velocity 160.2 m/s; point of impact at approx. X =  400 m (range)
2. Dropped: fall time 1.432 s, terminal velocity 13.9 m/s

Q: Centrifugal force is a ficticious force, it does not exist! There is only a force radially inward which is the centripetal force.

A: For a discussion of that subject please see this source .

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!

Please also read that Internet source! It seems that the Soviets already thought about using guns in outer space!

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.

A: The answer depends on the type of gun.
Typically bullets fired from pistols and revolvers travel at 300 - 500 m/s.
Hunting or miltary bullets are faster (approx. 800 - 1000 m/s).
Air rifles are in the 100 - 200 m/s range.

Q: What is the unit of the drag coefficient c_D and what is the connection between c_D and the projectile caliber?

A:The drag coefficient is a dimensionless number (in the area of 0.1 - 1) and
depends on Mach number, Reynolds Number, Froude number,...
The drag coefficient is usually measured by Doppler Radar or other velocity
loss measurements. There is no simple relationship between bullet geometry (length, diameter,
shape) and c_D.

A: All the aerodynamic coefficients are usually hard to obtain. Military research institutes measure them but only for military bullets. Almost no data is available for bullets from the civilian market. There is some (expensive) software available which estimates the aerodynamic coefficients from bullet geometry.

A: This question is hard to answer in general. The best I can give is a "worst-case" estimation.
When a gun is fired vertically, the bullet after some time reaches a summit where the velocity  is zero, and then falls back. The bullet will fall back base first which is hard to calculate. I can estimate the velocity if it would fall nose first, that is the normal flying position for which drag is well known - so the real terminal velocity will actually be smaller than the following prediction.
 

• For a .22 lr bullet (m=40 grain, v0 = 1150 ft/s) 
  the summit will be at 1164 ft, the total flight time 30 seconds and the terminal velocity 270 ft/s

• For a SS109 military bullet (m= 55 grain, v0=3200 ft/s)
  the summit will be at 2650 ft, the total flight time 44 seconds and the terminal velocity 404 ft/s.
  For this bullet are indications that it will become unstable. This will further reduce summit height and terminal velocity considerably.

A: This is not an easy task and can only be done with some accuracy by applying exterior ballistics software. There is a simple formula for estimation purposes.
 
z = k1*T²

T is the total flight time in seconds
k1 is a factor in the area of 0,1 ... 0,12 m/s² which depends on spin, muzzle velocity and bullet parameters
z is the side deviation in meters

Generally bullet drift at short distances (100 - 300 yd) is by far smaller than the normal scatter.
Drift is only of some practical importance for artillery shells, at ranges of several miles.

A: The most important (and expensive part) of the equipment will be a light source -
a spark flash of very small flash time (in the area of 1 millionth of a second).
This is something you do not get in the supermarket.