# Understanding equation for force acting on stones skipping over water

I think I have a decent conceptual understanding of the forces at work when stones are skipped over water. My question pertains to this equation, $$F = C_L\rho U^2S\sin({\alpha + \beta})$$ which appears without derivation in Water-skipping stones and spheres (Tadd Truscott, Jesse Belden and Randy Hurd, Physics Today, December 2014, page 70), where

• $C_L$ is the coefficient of lift, given as 1/2 for a disk
• $F$ is the force
• $\rho$ is the density of the water
• $U$ is the impact velocity
• $S$ is the wetted area
• $\alpha$ is instantaneous attack angle
• $\beta$ is the instantaneous course angle

The article also contains a useful image.

I suppose my question comes down to the final sine factor. I suspect deriving this equation comes somehow from the lift co-efficient formula: $$L = \frac12 \rho^2 v^2 A C_L.$$

Here is my barrage of questions

1. Where does the sine factor come from?
2. Why doesn't the fact that the stone experiences different densities above and below change the equation?
3. Why does an equation that deals with a single fluid apply to this situation at the boundary of two fluid?
• Single fluid probably applies because they consider the lift from the air negligible compared to the lift from the water. At most it would give a small correction factor to the parabolic flight path, but the dominant lift force in the system is lift due to water. – MonkeysUncle Mar 22 '15 at 1:32

1) The $\sin$ term appears to be resolving the velocity of the water relative to the face of the disk. $U$ is the just the speed, it's necessary to determine what part of that velocity will produce force on the object. It might help to visualize extreme cases, setting $\alpha$ and $\beta$ to 0, $\pi /2$ etc. For $\alpha = \beta =0$ you have a perfectly level stone sliding across the surface of the water (no lift). For $\alpha = 0$ and $\beta = \pi/2$ you have a level stone falling straight down (pure lift).
2/3) As @MonkeysUncle notes, the forces due to air are negligible in this case. The aero/hydrodynamic forces scale linearly with the density of the fluids. So the forces due to air ($\rho = 1.27 \,~\text{kg/m^3}$) will be about 0.1 % as large as those due to water ($\rho = 1000\, ~\text{kg/m^3}$).
• @ReidErdwien a more common meaning of 'lift' is "to raise or direct upwards" as is appropriate in this case. In fact, I strongly suspect that the more technical definition arises from the fact that, in most applications (wings etc.), force in the the cross-stream direction happens to be upwards. Further, in the equation given $C_L$ is used, rather than $C_d$, implying that the resultant force is defined as lift in the context of this problem. Really, it's just a matter of nomenclature (semantics, if you will). – user3823992 Mar 23 '15 at 5:48