Game Development Reference
In-Depth Information
directions will become initial velocities in each direction, and they will be included in
the equations of motion once they've been integrated. The initial velocities will show
up in the velocity and displacement equations just like they did in the example in
Chapter 2 . You'll see this in the following sections.
X Components
The first step is to make the appropriate substitutions for the force terms in the equation
of motion, and then integrate to find an equation for velocity.
-F wx - F dx = m(dv x /dt)
-(C w v w cos γ) - C d v x = m dv x /dt
dt = m dv x / [-(C w v w cos γ) - C d v x ]
(0 to t) dt = ∫ (vx1 to vx2) -m / [(C w v w cos γ) + C d v x ] dv x
t = -(m/C d ) ln((C w v w cos γ) + C d v x )| (vx1 to vx2)
t = -(m/C d ) ln((C w v w cos γ) + C d v x2 ) + (m/C d ) ln((C w v w cos γ)
+ C d v x1 )
(C d /m) t = ln[((C w v w cos γ) + C d v x1 ) / ((C w v w cos γ) + C d v x2 )]
e (C d /m) t = e ln[((C w v w cos γ) + Cd v x1 ) / ((Cw vw cos γ) + C d v x2 )]
e (Cd/m) t = ((C w v w cos γ) + C d v x1 ) / ((C w v w cos γ) + C d v x2 )
((C w v w cos γ) + C d v x2 ) = ((C w v w cos γ) + C d v x1 ) e -(C d /m) t
v x2 = (1/C d ) [ e (-C d /m) t (c w v w cos γ + C d v x1 ) - (C w v w cos γ)]
To get an equation for displacement as a function of time, you need to recall the equation
v dt = ds , make the substitution for v (using the preceding equation) and then integrate
one more time.
v x2 dt = ds x
(1/C d ) [e (-C d /m) t (c w v w cos γ + C d v x1 ) - (c w v w cos γ)] dt = ds x
(0 to t) (1/C d ) [e (-C d /m) t (c w v w cos γ + C d v x1 ) - (c w v w cos γ)] dt
=
= ∫ (sx1 to sx2) ds x
s x2 = [(m/C d ) e (-C d /m) t (-(C w v w cos γ) / C d - v x1 ) - ((C w v w cos
γ) / C d ) t] -
[(m/C d ) (-(C w v w cos γ) / C d - v x1 )] + s x1
Yes, these equations are ugly. Just imagine if we hadn't made the simplifying assumption
that drag is proportional to speed and not speed squared! You would have ended up
with some really nice equations with an arctan term or two thrown in.