##
To the editors:

The many-brained and almost-all knowing Cecil Adams, who recently discussed correctly a question involving a somewhat subtle point in the general theory of relativity [June 30], has now flubbed what he (a bit hastily) describes as a problem “straight out of Physics 101” [Letters, July 21; Straight Dope, December 16]. I refer to the matter of the comparative times before hitting the ground of a plummeting and a (horizontally) fired bullet.

In the first place, Cecil blunders in his description of the factors involved: “The arc described by the bullet shot from the gun is a function of two forces, the exploding gunpowder and gravity.” The exploding gunpowder (which, properly speaking, is not a “force,” although of course it exerts a force upon the bullet) has done its work by the time the bullet leaves the barrel of the gun–at which time the problem proper really begins. The relevant forces on the projected bullet, from this time on, are that of gravity (the bullet’s weight), the buoyant effect of the air (which can be regarded simply as reducing the weight, and which acts equally upon the projected and the falling bullet), and the resistance of the air to motion through it.

That criticism, however, only concerns the correct use of technical terms. What is more important is a serious error in the mathematical analysis of the problem. Here Cecil says that “wind resistance affects the two”–in the context, the “two forces” he has referred to; but I take him to mean the two (relevant) components of the bullet’s velocity: horizontal and vertical–“independently”; and he goes on to conclude that the vertical motion is slowed by the same amount in the case of the fired and the falling bullet, so that, were it not for the earth’s curvature, the two would take equal times to hit the ground (starting from the same height).

Now, that’s plumb wrong. The air is not smart enough to take separate account, directly, of the horizontal and vertical components of the motion. What it “knows” is merely that it is being penetrated, at a certain rate, in a certain direction. Its resistance is a force, in the direction opposite to that of the moving body, whose magnitude is some function of the speed of that motion.

An easy bit of elementary vector algebra (or geometry) leads to the conclusion that the actual effect produced by such a force will be the same as “independent action upon the two components” (that is, considering only the vertical component as the one of interest, that the vertical resistance will be the same as if only the vertical component of the total motion were in fact present), if and only if the law of dependence of the magnitude of the resistance upon the speed is that of direct proportionality. A more delicate (but still fairly elementary) analytical argument shows that if the resistance increases more rapidly, with increasing speed, than in direct proportion to the speed, the effect of the resistance will increase the time of fall more for a projected object moving obliquely downwards than for the same object plummeting straight down.

Now, the actual facts about the drag exerted by a fluid (whether liquid or gas) upon a body moving relatively to it are extremely complicated: far from being a matter for physics 101, they are today far from being completely understood theoretically at all, since at high speeds–like that of a fired bullet–they involve the phenomenon of turbulence (Cecil might find it interesting to look at vol. ii, ch. 41, of The Feynman Lectures on Physics for a general picture of the situation). That the drag is not just directly proportional to the speed is also evident from the fact (surely Cecil knows this) that a singularity (“shock wave”) occurs at the speed of sound: the famous “sound barrier.” A bullet from a high-powered gun has a muzzle-velocity considerably greater than the speed of sound.

Your correspondent Alfred Magnus, to whom Cecil’s remarks are directed, is therefore also in error when he says that the resistance of the air “is roughly proportional to the square of the speed of the object.” Newton developed a theory of projectiles in resisting media based on this assumption; but that theory neglects factors that play a role in the constitution of real media. According to a famous formula of Sir George Stokes, for small velocities the resistance is approximately proportional to the speed. Perhaps this is the assumption that was used in Cecil’s Physics 101 class: for a plummet and, say, a bean shot horizontally from a bean-shooter, the times of fall will be practically the same. But Stokes himself knew that this law leads to paradoxical results, and cannot be strictly correct. A more precise analysis, due to the Danish physicist C. Oseen, shows that a better approximation is given by a formula that adds to the Stokes resistance a second term, small at low velocity, proportional to the square of the speed–so that even at low velocity, the drag increases (a little) faster than in direct porportion to the speed. In any case, at the speed of a fired bullet, where turbulence and the ratio to the speed of sound (the Mach number) are of great importance, the increase over direct proportionality is great (that is, over most of the range of speed–there is an exceptional part of this range in which the drag actually falls as the speed increases); and Mr. Magnus is right in saying that the time of fall is increased much more by air resistance than by the curvature of the earth.

I apologize for the length of this letter, but the subject seemed to demand it. I daresay the letter is too long for you to print; but perhaps Cecil might find it worth reading.

Howard Stein

Professor in the Department of Philosophy and the Committee on the Conceptual Foundations of Science

University of Chicago

##
Cecil Adams replies:

I am not going to abuse you about this, doc, though you richly deserve it, because your mistake is a common one. Even Jearl Walker, author of the Amateur Scientist column in Scientific American and a professor of physics at Cleveland State University, initially agreed with you when I called to discuss the matter with him. I am pleased to report, however, that he eventually came around. With luck, so will you.

Drag is of course a unitary phenomenon, but it may be divided into vertical and horizontal components according to the following formulas, taken from the article on ballistics in the McGraw-Hill Encyclopedia of Science and Technology:

m(d2x/dt2)= -Dcos0

m(d2y/dt2)= -Dsin0 – mg

The first equation says the horizontal component of drag D is proportionate to the cosine of the projectile’s line of travel relative to the ground (angle O). The second says the vertical drag is porportionate to the sine. (The term mg indicates the contribution of gravity to the vertical vector of velocity.)

Let’s now consider an extreme example. Suppose we simultaneously shoot a gun and drop a bullet from an elevation of 300 meters above the ground. Suppose further that the fired bullet has an initial horizontal velocity of 300 meters per second. No one disputes that in a vacuum both bullets would hit the ground at the same time. Using other formulas in McGraw, we calculate that the time of flight for both bullets in a vacuum would be 7.8 seconds, that they would achieve a maximum vertical vector of velocity of 76.68 meters per second, and that the shot bullet would strike the ground at an angle of slightly more than 14 degrees relative to the horizon.

Now let’s add drag to the picture. Using the formulas above, we determine that the horizontal component of drag for the shot bullet at the moment of impact is .969 D and the vertical component is .247 D. (The situation is artificial, since we’re bringing drag in at the last moment of flight, but this is a defensible shortcut.)

Now note the following interesting relation:

DH/DV = cos 14 +0/sin 14 +0 = .967/.247 = 300/76.68 = VH/VV

In other words, the ratio of the horizontal component of drag to the vertical component is equal to the ratio of the horizontal vector of velocity to the vertical vector. This suggests that (1) the vertical drag is entirely a function of the velocity imparted to the bullet by gravity (I’m being a little imprecise here, but you get the idea); (2) the vertical drag is therefore the same as it would be on a bullet that was simply dropped through the air from 200 meters; and therefore (3) a shot bullet and a dropped bullet would hit the ground at the same time, even taking air drag into account.

It should now be clear that the fact that drag increases with the square of velocity is irrelevant to the problem. The dropped bullet and the shot bullet are traveling downward at identical speeds; the vertical component of drag on them is the same. Your argument is tantamount to saying that some of the horizontal component of drag slops over into the vertical. It is tempting to believe this occurs, but the equations above indicate otherwise.

Having said that, I will concede there are certain circumstances that would make a trivial difference in the time required for each bullet to hit the ground. Jearl Walker points out that the shot bullet presumably would curve in flight so that its long dimension was always coincident with its axis of travel. (I know there may be a certain amount of precession due to longitudinal spinning imparted by rifling in the gun barrel, but never mind that.) Therefore the shot bullet would present minimal cross section to the wind and so experience minimal drag. The dropped bullet, on the other hand, would present maximum cross section to the wind, at least at the outset, presuming it was horizontal when dropped. It would thus experience greater drag and so hit the ground after the fired bullet.

Another possibility is that the fired bullet might generate positive or negative lift as a result of the Bernoulli effect due to topspin or backspin while in flight. If the bullet were spherical and had spin of this kind, it would sink or sail in flight just like a golf ball or baseball does. In this case it would land before or after the dropped bullet.

If there’s anything in the above that isn’t clear Cecil will be glad to explain. I’d also be happy to go over any class notes, book manuscripts, etc that you may have. By God, I’ll get this country straightened out yet.