For an assignment of mine I am creating a very basic model of an airplane, where I choose to consider a fixed amount of parameters that affect flight and to disregard others which I find too difficult to model due to lack of experience. I am modelling the acceleration to start, and I've gotten this far:

$Acceleration = Force * Mass$

$Force = TotalThrust - Drag$

I have the total thrust for my model of airplane, the C-130 Hercules, however for drag is where complications arise. I wish to plot acceleration vs. time in a graph of mine to derive an equation of it, however I don't know how I would express drag as a function of time if somehow possible.

Being that the equation for Drag is the following:

$Drag = (Drag Coefficient) * (Density of Air) * (Terminal Velocity)^2 * (Area of Wing)$,

I was wondering if I could express either the density of air or terminal velocity as a function of time if somehow possible, so that I could create an acceleration vs. time graph. Is there any way at all to manipulate variables such that the expression becomes as a function of time?

Keep in mind I am making the primitive assumption that everything else stays fixed. I.e., if my terminal velocity becomes a function of time, I'll keep my drag coefficient, and density of air as fixed values just for the sake of making it less complicated.

  • $\begingroup$ Your question isn't very clear. Your model doesn't seem to include lift either. $\endgroup$ – Gert Mar 26 '16 at 2:16
  • $\begingroup$ The density of air is usually considered to be constant. Where I have seen drag calculations, the velocity is the current velocity, not the terminal velocity. You can keep the state as a three position matrix, location, velocity, and acceleration. Then you update each one every timestep based on your acceleration and drag equations. $\endgroup$ – Ross Millikan Mar 26 '16 at 4:37

Force is the product of mass and acceleration. Therefore, acceleration is not force times mass, but force divided by mass. By Newton's Second Law of Motion:

$$ \vec F = m\vec{a} \Rightarrow \vec a = \frac{\vec F}{m} $$

Therefore, the sum of forces on the aircraft in the horizontal direction $\hat i$

$$ \Sigma \vec F = \vec F_t - \vec F_d = m\vec a$$

where $F_t$ is the thrust and $F_d$ is the force due to drag. Since you said you know the force due to thrust, we can write this as

$$ \vec F_d = \vec F_t - m\vec a$$

And therefore

$$ \frac{1}{2} \rho u^2 C_d A = \vec F_t - m\vec a$$

where $u$ is the flow velocity relative to the plane's motion. Using this, you should be able to solve for whichever variable you want, depending on your circumstances.

Unless this is a theoretical problem/derivation we are dealing with, the air density $\rho$ is not going to be changing significantly with time. Therefore defining it with time will yield functions such that

$$\frac{d\rho}{dt} = 0$$

That said, if you wanted to define this in terms of time, you could do so implicitly by letting acceleration be a function of time. This would require that you have a vector-valued function to define position $\vec r(t)$. Then

$$ \vec a(t) = \frac{d^2\vec r}{dt^2} $$

In this setup, you would be able to plot the coefficient of drag, air density, etc., implicitly using the values for acceleration. You could calculate the values in excel, then plot them.


Don't go into the project cold. Read John Denker's on-line book about aviation. I will try to explain the basics.

For ordinary aircraft, in ordinary flight:

  • the airplane has a bank angle, which it uses to tilt the lift vector. When the lift vector is tilted, the horizontal component of the lift serves to move the aircraft in a horizontal arc - this is how it turns.

  • lift and drag are both proportional to two things: angle of attack, and velocity squared. Since these two factors are in trade-off, that is how the pilot controls speed - by adjusting the angle of attack.

  • the airplane is loaded so that its center of gravity is forward of its center of lift. This means that the tail planes are flying inverted - they create a downward push at the tail. This is crucial for maintaining a stable speed, because it results in a self-correcting feedback. If the plane slows down, the downward push at the tail decreases, so the nose drops, so the plane speeds up "down-hill". If the plane speeds up, the opposite happens. This also determines how the pilot controls vertical motion. Increasing power results in increasing speed, which the aircraft corrects by going up. Decreasing power brings it down.

These are Flying 101. You need at least this to simulate flight.


Both thrust and drag will change over speed, and the easiest will be to cut the process into small time slices in which you assume a constant speed. Basically, model the airplane with a simple finite difference method.

Next, cut the process into steps. A take-off consists of

  1. Ground run: The aircraft accelerates while rolling on the ground.
  2. Rotation: The angle of attack is increased when close to the lift-off speed until the aircraft lifts off the ground.
  3. Initial climb: The aircraft gains altitude and accelerates further until the conditions for the take-off (usually 1.3 times minimum speed and 35 or 50 ft of altitude) have been reached.

If you want to express the acceleration as a function of time, you can model the energy of the aircraft during each of the three steps and assume a constant thrust and drag during each, but at the price of lower accuracy. The total energy gain $E$ is $$E = \frac{m}{2}\cdot (1.3\cdot v_{min})^2 + m\cdot g\cdot h = \int_{t_0}^{t_1}Pdt$$ with $m$ the mass of the aircraft, $v_{min}$ its minimum speed in take-off configuration, $h$ the height at which the take-off is considered to be concluded and $P$ the effective power of its engines.

To model drag, use the quadratic polar: $$c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$

To model thrust, make it inverse to speed: $$T \varpropto v^{n_v}$$ with $n_v$ around -0.8 for turboprop engines. The equation for static thrust $T_0$ is: $$T_0 = \sqrt[3]{P^2\cdot\eta_{Prop}^2\cdot\pi\cdot \frac{d_P^2}{2}\cdot\rho}$$

$c_L \:\:\:$ lift coefficient
$n_v \:\:\:$ thrust exponent
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing. This is 10.1 in case of the C-130
$\epsilon \:\:\:\:\:$ the wing's Oswald factor. The C-130 should reach 0.9 here.
$c_{D0} \:$ zero-lift drag coefficient. Use 0.02 for a clean aircraft and 0.03 in take-off configuration if you don't have better values.
$\eta_{Prop}$ propeller efficiency. Use 80% for the C-130
$d_P \:\:$ propeller diameter
$\rho \:\:\:\:$ air density


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.