$$ a_{xd} = k({v_x}^2+{v_y}^2)cos\beta $$ $$ a_{yd} = k({v_x}^2+{v_y}^2)sin\beta $$

According to the above equations (relating to a projectile experiencing drag), the drag acceleration components are dependent on both the parallel AND the perpendicular velocity components. I'm struggling to understand how, say, the horizontal drag acceleration could depend on the vertical velocity.


For reference, the equations have been derived for a projectile as follows: $$ a_d=\frac{1}{2m}C_d\rho A V^2 $$ where $C_d$ is the drag coefficient, $\rho$ is the medium density, A is the projected area of the body and V is the velocity.

For the purpose of these calculations: $$ k = \frac{1}{2m}C_d\rho A $$ Therefore the magnitude of the drag acceleration can be expressed as: $$ |a_d| = k ({v_x}^2+ {v_y}^2) $$ If the angle of elevation beta is expressed as: $$ \beta = \arctan{\frac{v_y}{v_x}} $$ Then the x and y components for the drag acceleration can be expressed as in the image above.

  • $\begingroup$ Please define the symbols in your equations in your post and describe the situation for which the formulas are intended - otherwise, writing an answer to your question will result in quite a bit of guesswork ;) $\endgroup$
    – Sanya
    Dec 23, 2016 at 17:34
  • $\begingroup$ Physics.SE is a MathJax-enabled site; so use that facility to write your equations. For a quick look, check this Meta Math.SE post. $\endgroup$
    – user36790
    Dec 23, 2016 at 17:35
  • $\begingroup$ Apologies, the derivations for the equations have been added. $\endgroup$
    – user120568
    Dec 23, 2016 at 17:59

1 Answer 1


Well, with the expansion of your original question you have pretty much answered it yourself: You can see that the drag depends on the square of the absolute value of the velocity, which includes both velocity components. One should add, however, that the drag coefficient $C_d$ will also be a function of the angle of attack $\beta$, so the way the components of drag depend on the velocity components is more complex than what your equations suggest.

The reason for this complex dependence of the drag components on the velocity components is due to the way the velocity field of the flow around the projectile changes as a function of angle of attack. The velocity field ultimately appears as the solution of a certain set of nonlinear partial differential equations, and the way that this solution changes as a function of $\beta$ is highly non-trivial in general.

As another example, consider the wing of an airplane. At zero effective angle of attack, the wing will only produce a drag force that is parallel to the velocity of the oncoming flow. However, if you increase the angle of attack, the wing will start producing lift, which is a force normal to the flow direction that can quickly be much larger (by one to two orders of magnitude) than the drag force.

If your projectile moves at a non-zero angle of attack, it will likewise produce some lift force. The lift will not be as large as the one of a wing, but it can be substantial, and roughly reach the same magnitude as the drag force.


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