Built on Facts

Challenge Problem!

This one’s from Young and Freedman, and I pick it out because it’s both from the chapter I’m teaching and it’s a great conceptual problem as well. (I’ve modified it slightly.)

A shotgun fires a large number of pellets upward, with some pellets traveling vertically and some as much a 1 degree from the vertical. Ignore air resistance and assume the pellets leave the gun at 150 m/s. Within what radius from the point of firing will the pellets land? Will air resistance tend to increase or decrease this number?

The book gives the range equation directly even though it can be derived easily, so since we’re still in the “mostly conceptual” week here at BoF, I’m going to just give the equation. Wikipedia follows the derivation I’d use, for the curious.


It’s a short and sweet equation, but make sure to keep track of units and the distinction between degrees and radians.


Fig. 1: Marines prefer experimental physics to theoretical physics.

The interesting part is of course the second part. For instance, air resistance will tend to slow the horizontal motion which should shrink the circle. But it will also limit the downward speed to some comparatively small terminal velocity, giving the pellets more time to come apart – which would tend to increase the circle. Think about this stuff long enough and the answer is clear – so fire away! Mathematically, anyway.


  1. #1 Russell
    February 9, 2009

    The notion of “more time to to come apart” makes no sense in this context. That would make sense, if the increased time weren’t because of a less velocity. Increased air resistance means dampened velocity means smaller spread.

  2. #2 Uncle Al
    February 9, 2009

    Firing birdshot requires an appropriate choke or all that obtains is Darth Cheney decorating a lawyer. Around #4 buckshot unchoked you get about 1″ spread/yard horizontal flight. If you want your ounce of lead to impact a tight CEP, fire a deer slug (though for the same recoil you can fire quite the rifle round).

  3. #3 Brando
    February 9, 2009

    Ballistics for shotguns is a bit tricky as there are more variables involved. Shooting a projectile through a rifled barrel is one thing, but having a wad accelerate a bundle of lead shot (.00 Buck, for example) down a smoothbore and then discarding after leaving the muzzle imparts many multi-directional forces on the projectiles from firing and upon exiting the barrel. This is why it’s easier to compute trajectories on rifle bullets; there are still many, many variables, but almost all can be accounted for and predicted while the process of shooting a shotgun has substantial “noise” in the variables.

  4. #4 DA
    February 9, 2009

    I’ve gotta disagree with Russell. The x and y velocities are independent. The time increase is based on the vertical terminal velocity. Air resistance is proportional to velocity (or velocity squared), and since the horizontal velocity is on the order of 2 m/s and the vertical velocity is on the order of 150 m/s, there will be a negligible effect on the horizontal motion and a significant effect on the vertical motion. Meaning: the pellets will travel the same speed horizontally over a longer period of time. The radius will increase.

  5. #5 MattXIV
    February 9, 2009

    Smaller spread. Higher air resistance equals more time in the air from the same height, but the apex will be lowered by the air resistance on the way up.

  6. #6 Jon Claerbout
    February 9, 2009

    Interesting you could put equation derivations at Wikipedia. When I derived the world oil peak equations there, they kicked it out — not suitable for an encyclopedia. So it’s on my web site in PDF, namely
    (just in case you are interested in the world’s financial “readjustment”)

  7. #7 Alex
    February 9, 2009

    The Wikipedia derivation only applies to cases where air resistance is proportional to v, not v^2 (i.e. they assume low Reynolds number). Since most projectile problems are high Reynolds number, this assumption is inaccurate. At high Reynolds number, the x and y components of velocity are coupled. I don’t know if these coupled non-linear equations can be solved analytically.

  8. #8 Tim Gaede
    February 10, 2009

    It will be a smaller spread. I’ve simulated enough trajectories using quadratic wind resistances on computer to know this.

    One simple way would loop:

    force = gravity + air_resistance
    acc = force / mass
    vel += acc * time_increment
    pos += vel * time_increment

    for each dimension (or just use a vector class for everything except mass, which would be a scalar).

    It seems that increasing air thickness moves the trajectory completely within the old trajectory. I’m not sure how to elegantly prove that this will always be the case. I would have to think about this more.

  9. #9 Tony
    February 10, 2009

    Smaller spread, because energy is lost due to drag. You can ignore height, pellet spread, the fact that it is a shotgun, and everything else. Just look at one pellet with some horizontal trajectory, just as you would look at one bullet being fire from a rifle at some angle. Without air drag the bullet travels the maximum range. With drag, it travels a shorter distance (and it doesn’t go as high).

  10. #10 rob
    February 10, 2009

    the first part of the question is what is the radius of the spread?

    the range equation is given. using v=150 m/s, theta=1° and g=9.8 m/s/s

    radius= (150)^2 sin(2°)/9.8 = 80 meters.

    i will think some more about the second part. since i assume the question is for an intro physics course and is in the section about 2-d motion, any discussion of air resistance beyond that the air resistance is proportional to velocity is out of the scope of the problem. even assuming that the resistance is proportional to the velocity is probably out of the scope of the problem.

  11. #11 Brando
    February 10, 2009

    Interesting thread. The increased drag with velocity bit is a huge part of tuning a precision long range load. When shooting into the 2000+ meter arena with precision, we often struggle to find the right balance of projectile mass and velocity. Lower mass bullets moving at high velocities give you flatter trajectories with very low max ordinate values. This helps tremendously when dealing with elevation adjustments on optics. However those very fast moving projectiles (3100+ fps) slow down much more rapidly than the heavier ones. Conversely a higher mass projectile tends to require more energy to get to high velocity and stays supersonic much longer (ie is more stable) and often it’s just not practical to attempt to reach such velocities as you hit a pressure safety wall (approx 60,000 psi), let alone the increased wear on an expensive precision barrel. This is why extreme range interdiction shooters tend to use CNC lathed solid (copper/nickel alloy or spun brass) projectiles that when shot at relatively high velocities maintain a very high ballistic coefficient and are much more consistent and predictable in the 2km range.

  12. #12 rob
    February 10, 2009

    here is what the radius does with air resistance:

    the range equation is independent of time, so you can’t argue about how air resistance increases or decreases the time of flight. the angle and g are constant, so the only thing that can change is velocity.

    since air resistance adds another force that decreases velocity, the v in the range equation is not the initial velocity. it should be some velocity between zero and v initial. this means that the radius for the pellets is smaller than without air resistance.

  13. #13 JK
    February 11, 2009

    All you are asked for is the direction of the change.

    Consider the limit of very high resistance.

    Radius decreases.

New comments have been disabled.