My Bedford and Fowler textbook (4th edition) has a chapter on numerical solutions. I'm having trouble understanding how the textbook splits up the components of force in the $x$ and $y$ directions to model the drag on a projectile:

The aerodynamic frag force on the projectile is of magnitude $C|{v^2}|$, where $C$ is a constant.

We must determine the $x$ and $y$ components of the total force on the projectile. Let $D$ be the drag force. Because $v/|v|$ is a unit vector in the direction of $v$, we can write the drag force as:

$$D = -C{|v|^2} \frac{v}{|v|} = C|v|v. $$

The external forces on the projectile are its weight and the drag so we have:

$$\sum F = -mgj - C|v|v,$$

and the total components of the force are:

$$\sum F_x = C\sqrt{v_x^2 + v_y^2}\quad v_x$$

$$\sum F_y = -mg- C\sqrt{v_x^2 + v_y^2}\quad v_y.$$

I don't understand how the individual components of the drag force $D = C|v|^2$ (acting in opposition to the projectile's movement) becomes something like this: $$-C\sqrt{v_x^2 + v_y^2}* v_x~ ?$$

I understand that the absolute value of something can be represented as: $$|a|=\sqrt{a^2}.$$

And the magnitude of velocity from its $x$ and $y$ components can be calculated as $$V=\sqrt{v_x^2+v_y^2}.$$

But I don't get how the non-absolute $v$ term magically becomes $v_x$ or $v_y$ while the absolute term doesn't become $|v_x|$ or $|v_y|$ too.

What about equations without absolute values in them like the drag force equation? $F_D = \frac{1}{2}pv^2AC_D$, where $p,A, C_D$ are constants. Would the horizontal and vertical components of forces then be:

$$F_x = -\frac{1}{2}pv_x^2AC_D$$ and $$F_y = -mg -\frac{1}{2}pv_y^2AC_D~ ?$$

  • $\begingroup$ The term is $\left| v \right| \vec v = \sqrt{v_x^2 + v_y^2} \big( \vec e_x v_x + \vec e_y v_y \big)$. Then you just split this up into components. The scalar multiplier will be the same for both components. $\endgroup$ Sep 6, 2015 at 12:48
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    $\begingroup$ @SebastianRiese You should post this as an answer. $\endgroup$
    – Bill N
    Sep 6, 2015 at 13:22
  • $\begingroup$ It's always better to write vector equations with vector symbols to avoid confusion like you have. You can't automatically do symbol substitution without thinking about what the concept of the symbol is. Also, you have a typo in the first equation with the $\Sigma$ sign. $\endgroup$
    – Bill N
    Sep 6, 2015 at 13:24
  • $\begingroup$ My confusion probably stems from the fact that the textbook page that I wrote this from does not have any vector symbols above the appropriate variables. $\endgroup$
    – user155876
    Sep 6, 2015 at 13:33
  • $\begingroup$ @user155876 Do they use bold-face for vectors? Normally this is explained in an introductory chapter. $\endgroup$
    – Bernhard
    Sep 6, 2015 at 13:45

2 Answers 2


I find this type of question is always easier if you draw a diagram. The drag force $F$ acts in the opposite direction to the velocity so it looks like:


and the components of the drag force are:

$$\begin{align} F_x &= F \cos\theta \\ F_y &= F \sin\theta \end{align}$$

$\cos\theta$ is $v_x/v$ and $v = \sqrt{v_x^2 + v_y^2}$ so we get:

$$ F_x = F \frac{v_x}{\sqrt{v_x^2 + v_y^2}} $$

Since $F = -Cv^2$ we get:

$$ F_x = -Cv^2 \frac{v_x}{\sqrt{v_x^2 + v_y^2}} $$

and because $v = \sqrt{v_x^2 + v_y^2}$ this becomes:

$$\begin{align} F_x &= -C \left(v_x^2 + v_y^2\right) \frac{v_x}{\sqrt{v_x^2 + v_y^2}} \\ &= -C v_x \sqrt{v_x^2 + v_y^2} \end{align}$$

And likewise for $F_y$.

  • $\begingroup$ I fully understand now thanks to your step by step guide through the algebra. It's so simple and I can't believe I didn't realise it. Although it's against the rules to do so in a comment, I cannot thank you enough. $\endgroup$
    – user155876
    Sep 6, 2015 at 14:06
  • $\begingroup$ Wow, that's a great answer to a seemingly trivial but ultimately confusing question. Thank you! :) $\endgroup$
    – user273231
    Aug 26, 2020 at 21:27

Review of vectors

A vector is a quantity with a magnitude and a direction, which makes it somewhat different from how normal quantities, which are just magnitudes. One huge difference is that vectors "transform" a certain way when we rotate our coordinates, for example. They "transform" just the same way that an arrow in space transforms, if we do not care about where the arrow is located but only how the arrow is pointing. So when you get really technical, this will become the definition of a vector; we'll just say that a vector is anything which "transforms like a vector" and give the example case of displacement vectors in space to define how we want them to transform.

We have a couple of different notations for vectors, but a very common one is to write a vector as its "coordinates" in some basis, defining some $x,y,z$ axes which are perpendicular to each other and then writing down $$\vec v = \begin{bmatrix} v_x \\ v_y \\ v_z\end{bmatrix}.$$ The only thing I don't like about this notation is writing it in-line in a paragraph like this one, which is where I sometimes switch to semicolons to divide the components: $\vec v = [v_x;\, v_y;\, v_z].$ Some people instead prefer $\vec v = \hat e_x v_x + \hat e_y v_y + \hat e_z v_z$ or $\mathbf v = v_x \mathbf i + v_y \mathbf j + v_z \mathbf k;$$ these are the same numbers written in slightly different ways. I prefer the first one above.

There are some things which are invariant under coordinate rotations. This means that it does not matter what coordinates you use when you are these things in a calculation, and since your choice of coordinates shouldn't matter, this is very important for writing correct expressions in general.

The first one is the vector scaling, where we multiply the magnitude of the vector by some parameter k. In coordinates, this looks like:$$k \vec v = k ~ \begin{bmatrix}v_x\\ v_y\\ v_z\end{bmatrix} = \begin{bmatrix}k v_x\\ k v_y\\ k v_z\end{bmatrix}.$$ The point is, no matter how your coordinates are rotated, everyone can agree that this vector is the other vector scaled by this particular parameter $k.$

The next is vector sums, $$\vec u + \vec v = \begin{bmatrix}u_x\\ u_y\\ u_z\end{bmatrix} + \begin{bmatrix}v_x\\ v_y\\ v_z\end{bmatrix} = \begin{bmatrix}u_x + v_x\\ u_y + v_y\\ u_z + v_z\end{bmatrix}.$$ This has a nice geometrical interpretation as placing the "base" of one arrow on the "tip" of the other, then drawing an arrow from the base of the latter to the tip of the former.

The third one is about coordinate rotations preserving the angles between different vectors, and the cleverest way to phrase it involves the dot product, which turns two vectors into a scalar (a pure "magnitude" with no direction beyond a possible $\pm$ sign). This scalar is: $$\vec u \cdot \vec v = |\vec u| |\vec v| \cos\theta = \begin{bmatrix}u_x\\ u_y\\ u_z\end{bmatrix} \cdot \begin{bmatrix}v_x\\ v_y\\ v_z\end{bmatrix} = u_x~v_x + u_y~v_y + u_z~v_z.$$ As I mention in the equation, this has a nice interpretation as the magnitude of $\vec u$ times the magnitude of $\vec v$ times the cosine of the angle that lies between them, but it also has this super-simple algebraic form on the right.

Speaking of magnitudes, since there is no angle between a vector and itself, and $\cos 0 = 1$, this lets us finally define what a magnitude is as a coordinate-invariant property, $|\vec v| = \sqrt{v_x^2 + v_y^2 + v_z^2} = \sqrt{\vec v \cdot \vec v}.$ In turn we can define what a direction is as a "unit vector", a vector scaled by its inverse magnitude, $\hat v = \vec v / | \vec v |$ so that $|\hat v| = 1.$

Now the meaning of $\hat e_x$ or $\mathbf i$ are apparent: they are unit vectors $[1;\,0;\,0]$ pointing in the coordinate directions, and you can recover the full vector simply by scaling these directions with the components and summing them back together.

In your specific case

The aerodynamic drag force for a particle moving with velocity $\vec v$ through a fluid is defined to have a magnitude $k v^2$ for some $k$, usually $\frac 12 C_D A \rho,$ but the key point is that it is quadratic in velocity. Here $v = |\vec v|$ is the magnitude of the vector-velocity, if that wasn't 100% clear.

Of course a force is a vectorial quantity and so it also has a direction; for aerodynamic drag its direction is opposite to $\vec v$. So we scale the unit vector $\hat v = \vec v / |\vec v|$ by $-k |v|^2$ to find the vector $$\vec F_d = \frac{-k |\vec v|^2}{|\vec v|} ~ \vec v = \frac{-k\left(v_x^2 + v_y^2 + v_z^2\right)}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \begin{bmatrix} v_x\\v_y\\v_z\end{bmatrix}.$$ If we want we can divide by $|\vec v|$ and multiply through all of the coordinates to simply get:$$ \vec F_d = \begin{bmatrix} - k v_x \sqrt{v_x^2 + v_y^2 + v_z^2}\\-k v_y \sqrt{v_x^2 + v_y^2 + v_z^2}\\-k v_z \sqrt{v_x^2 + v_y^2 + v_z^2}\end{bmatrix},$$ so those are the components of the aerodynamic drag force.

The principles are really simple, but it can definitely look a little hairy when you are using the scaling rule, which multiplies all of the components by some $\lambda$, with some particular $\lambda = -k (\vec v \cdot \vec v)$ where the scaling parameter itself comes from a dot product! It can very much look like this is supposed to secretly be $v_x^2$ or so, but the rules tell us it is not!

This has a very important consequence. Normally if you are bicycling North and a wind starts blowing from the East, you have to work against the a drag force. Bacteria and paramecia, who live in a world with linear drag forces $|\vec F_d| = -k |\vec v|$, do not have this problem! The work done per unit time for a force $\vec F$ is always $\vec F \cdot \vec v$ and for linear drag these forces are hopelessly perpendicular; but for quadratic drag they do mix a little bit. If the expression had instead been $[v_x^2;\,v_y^2;\,v_z^2]$ then you would not have seen this effect, and we would just be like those bacteria -- but then the force would not have had magnitude proportional to $v^2$ but rather $\sqrt{v_x^4 + v_y^4 + v_z^4}$, which is not coordinate-invariant: for this expression it really depends which coordinates you use. An example is $[1;\,0;\,0]$ yielding $1$ but its rotation by $\pi/4$ into the $y$-direction, $[\sqrt{1/2};\,\sqrt{1/2};\,0],$ yielding $\sqrt{1/4 + 1/4} = \sqrt{1/2},$ which is obviously not $1$.

  • $\begingroup$ Heh, took me so long to write this that @JohnRennie swooped in underneath me. It happens. ^_^ $\endgroup$
    – CR Drost
    Sep 6, 2015 at 15:01
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    $\begingroup$ My name in Dnepropetrovsk is cursed, When he finds out I publish first! with apologies to Tom Lehrer. $\endgroup$ Sep 6, 2015 at 15:43
  • $\begingroup$ Another great answer to a good question :) $\endgroup$
    – user273231
    Aug 26, 2020 at 21:28

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