How can I properly model paper airplane physics? [closed]

I am working on a game involving flying and steering a paper airplane for WP7. I want the plane to fly just like how normal paper airplanes (online game) but I can't seem to find an equation for how paper airplanes fly.

Anyone have any experience with this? In my game now, it just follows the usual motion for an object in a vacuum, which makes for some flight, but it doesn't feel perfect, and traveling at a slight downward angle makes you lose speed, which isn't right.

I am asking do paper air planes have a maximum velocity brought on by its paper design even if control and thrust was possible maybe using composite cardboard material?

On Physics.SE: Paper airplane physics

• This honenstly isn't as simple of a question as you may think. You might try looking into third party physics engines to help you model the flight of the paper airplane... – Jay Carr Mar 22 '16 at 3:07
• A paper airplane flies exactly the same as any other heavier than air machine. In a vaccum, it cannot fly. It will travel some distance when momentum is imparted to it but it will hit the ground at exactly the same time as if you just dropped it. – Simon Mar 22 '16 at 6:37
• What is WP7? ... – Ward - Reinstate Monica Mar 22 '16 at 6:57
• please do not copy questions from other SEs (and from other users). If you think they would be more suited here, flag them in the other SE for migration. – Federico Mar 22 '16 at 7:37
• @Ward WP7 = Windows Phone 7 – Ron Beyer Mar 22 '16 at 14:13

I would model the energy flows between kinetic and potential energy. Start with a full bucket of potential energy plus the appropriate speed and drain kinetic energy away over time depending on flight speed.

To know how much energy is lost to drag, you need to model the drag using two components:

1. Friction drag, which grows with the square of airspeed, and
2. Induced drag, which drops with the square of airspeed.

If your drag is the sum of both, its minimum will be at some moderate speed. I plotted the drag components for a glider below, but since the physics are the same for a paper airplane, this plot should do for now.

The nonlinear behavior of the induced drag curve at low speed is due to flow separation, and something very similar will happen for a paper airplane. The important thing is: The drag curve has a minimum.

The energy loss over time is drag $D$ times speed $v$. This energy $E$ has to come from the reduction of height $h$ over time $t$: $$\frac{\delta E}{\delta t} = \frac{\delta(m\cdot g\cdot h)}{\delta t} = D\cdot v = m\cdot g\cdot v_z$$ with $m$ the mass of the paper airplane, $g$ gravitational acceleration and $h$ its height above ground. Only $h$ changes over time, so derivation is easy and the derivative of $h$ is $v_z$, the vertical speed.

For picking a realistic drag it helps to rephrase the equation above by introducing lift $L = m\cdot g\cdot n_z$. $n_z$ is the load factor and is approximately one in straight flight. The ratio between lift and drag for a paper airplane is somewhere between 4 and 10 - just pick a number which results in a realistic simulation. To calculate the sink speed $v_z$ as a function of flight speed $v$ use this formula: $$v_z = \frac{c_{D0}\cdot S\cdot v^3\cdot\rho}{m\cdot g\cdot 2} + \frac{m\cdot g\cdot 2}{v\cdot\pi\cdot b^2\cdot\rho}$$ S is the wing area of the paper airplane and $b$ its wing span. $\rho$ is air density, and for the zero-lift drag coefficient $c_{D0}$ you should pick a number which makes the paper airplane look realistic. Start maybe with 0.05.

• Your answers never cease to astound me... Of course, I'm pretending I understand more than about half of them. /math wasn't my strongest suit – FreeMan Mar 22 '16 at 13:05
• @FreeMan: I also struggled with math, especially when it was not related to airplanes. Over time I assembled a collection of formulas and equations, which comes handy for questions like this one. The last equation comes from this answer, only $c_L$ was replaced by $\frac{m\cdot g\cdot 2}{S\cdot v^2\cdot\rho}$. And yes, I noted a mistake in it when typing this comment. – Peter Kämpf Mar 22 '16 at 15:36