Dot Physics

Throwing a football, Part II

In part I of this post, I talked about the basics of projectile motion with no air resistance. Also in that post, I showed that (without air resistance) the angle to throw a ball for maximum range is 45 degrees. When throwing a football, there is some air resistance this means that 45 degree is not necessarily the angle for the greatest range. Well, can’t I just do the same thing as before? It turns out that it is a significantly different problem when air resistance is added. Without air resistance, the acceleration was constant. Not so now, my friend.

The problem is that air resistance depends on the velocity of the object. Search your feelings, you know this to be true. When you are driving (or riding) in a car and you stick your hand out the window, you can feel the air pushing against your hand. The faster the car moves, the greater this force. The air resistance force depends on:

  • Velocity of the object. The typical model used for objects like a football would depend on the direction and the square of the magnitude of the velocity.
  • The density of air.
  • The cross sectional area of the object. Compare putting an open hand out the car window to a closed fist out the car window.
  • Some air drag coefficient. Imagine a cone and a flat disk, both with the same radius (and thus same cross sectional area). These two objects would have different air resistances due to the shape, this is the coefficient of drag (also called other things I am sure).

So, since the air force depends on the velocity, it will not be a constant acceleration. Kinematic equations won’t really work. To easily solve this problem, I will use numerical methods. The basic idea in numerical calculations is to break the problem into a whole bunch of little steps. During these small steps, the velocity does not change much so that I can “pretend” like the acceleration is constant. Here is a diagram of the forces on the ball while in the air.

i-01b8e367a57ba319724893cfb20b36bf-air-resistance-diagram-1.jpg

Before I go any further, I would like to say that there has been some “stuff” done on throwing a football before – and they probably do a better job than this post. Here are a few references (especially with more detailed discussion about the coefficient of drag for a spinning football):

And now for some assumptions:

  • I hereby assume that the air resistance is proportional to the square of the magnitude of the velocity of the object.
  • The orientation of the football is such that the coefficient of drag is constant. This may not actually be true. Imagine if the ball were thrown and spinning with the axis parallel to the ground. If the axis stayed parallel to the ground, for part of the motion the direction of motion would not be along the axis. Get it?
  • Ignore aerodynamic lift effects.
  • Mass of the ball is .42 kg.
  • The density of air is 1.2 kg/m3.
  • The coefficient of drag for the football is 0.05 to 0.14
  • Typical initial speed of a thrown football is around 20 m/s.

And finally, here is the recipie for my numerical calculation (in vpython of course):

  • Set up initial conditions
  • Set the angle of the throw
  • Calculate the new position assuming a constant velocity.
  • Calculate the new momentum (and thus velocity) assuming a constant force.
  • Calculate the force (it changes when the velocity changes)
  • Increase the time.
  • Keep doing the above until the ball gets back to y=0 m.
  • Change the angle and do all the above again.

The answer

First, I ran the program with an initial velocity of 20 m/s. Here is the data:

Air Resistance Diagram 1

At 35 degrees, this gives a distance of 23 meters (25 yards). This doesn’t seem right. I know a quarterback can throw farther than that. What if I change the coefficient to 0.05? Then the greatest angle is closer to 40 degrees and it goes 28 meters. Still seems low (think Doug Flutie). What about with no air resistance? Then it goes 41 meters (at 45 degrees). So, here is the Doug Flutie throw.

From the video, it looks like he threw the ball from the 36ish yard line to about the 2 yard line. This would be 62 yards (56.7 meters). I am going to assume a coefficient of 0.07 (randomly). So, what initial speed will get this far? If I put in an initial velocity of 33 m/s, the ball will go 55.7 meters at an angle of 35 degrees.

Really the thing that amazes me is that someone (not me) can throw a ball that far and essentially get it where they want it. Even if they are only sometimes successful, it is still amazing. How is it that humans can throw things somewhat accurately? We obviously do not do projectile motion calculations in our head – or maybe we do?

Comments

  1. #1 Uncle Al
    December 29, 2008

    The football is launched, spinning, with its long axis at 35-45 degrees elevation from the ground. Does it real world impact with its leading pole still up in the air or pointing toward the ground?

  2. #2 Pablo Sanches
    October 9, 2009

    where is the video i want to watch it

  3. #3 Rhett
    October 9, 2009

    @Pablo,

    Is the video not showing up in the above post? If not, here is the direct url from youtube:

    http://www.youtube.com/watch?v=q3ykWbu2Gl0

  4. #4 Patrice Vezeau
    July 3, 2010

    human don’t calculate we use Memory,muscle memory and intuitive deduction from those memory and previous experience to estimate how to throw the ball.

    Personalty i am playing tennis and I know every time I start to play i must warm up hitting many ball to adjust depending on the court condition like heat, wind,humidity, type of ball,and if I play with a used string bed or new one.

    But I start to hit from my previous experience and do little adjustment with trial and error then the confidence grow and I can hit harder or very sharp angle shot without missing because I know how the ball will react.