alex33 said:
Can I ask you to keep the name of the variables other than those already adopted? Otherwise I'm afraid to get myself confused... until now we've talked about L to indicate the length of the Rover (actually it would only be the length of the wheelbase, the Rover is 80 cm longer). As for h, I had defined the height of the chassis
As a matter of course, I generally define length variables as half lengths of symmetric objects, otherwise fractions keep appearing in the algebra.
I start with these constants:
| r | 2.982245324 | L | 1.019987973 | I/m | 0.394925155 |
| v | 2.6 | omega | 0.4638891017 | J/m | 0.68 |
| alf | 0.3490658504 | del t | 0.1666666667 | s | |
| h | 0.38 | | | | |
| g | 1.65 | | | | |
Note that my L (half car length) is ##r\sin(\alpha)##. I know now it should be a bit longer, but I don't think it changes much.
I/m is the moment of inertia of the car per unit mass: ##\frac 13(h^2+L^2)##.
J/m is the impulse on the rear axle on landing, per unit mass. The impulse is normal to the ground.
del t (Δt) is the time between the front wheels exiting the dip and the rear wheels reaching it.
Omega is the angular velocity when the front wheels reach the dip. As discussed, the vertical velocity of the mass centre then is ##\frac 12v\tan(\alpha)##, giving a rotation rate of ##\frac 1{2L}v\tan(\alpha)##.
As origin, I use the point halfway across the dip but at the surrounding ground level.
Velocities and displacements are positive to the left and up; rotations are clockwise from horizontal.
The simulation during a free fall tracks the following:
t: time since start of phase
xc, yc: coordinates of mass centre
xf, yf: coordinates of front wheels
xr, yr: etc.
theta: chassis orientation
These are the initial values of the velocity and position of the mass centre and the orientation at the start of the current free fall phase. During such a phase, we can compute the current values from those and the time since the start of the phase.
I also track total mechanical energy, E= kinetic+potential as a check on the equations and to ensure J is not excessive.
During del t, the rear wheels are still on the ground but the front wheels are airborne.
Taking moments about the rear wheels, gravity on the mass centre exerts a torque -mgL. The moment of inertia about that axis is ##\frac 13(h^2+4L^2)m##, so the angular acceleration during del t is ##-3\frac{gL}{(h^2+4L^2)}##. (Its magnitude will reduce slightly during del t, but not much.) Hence, when the rear wheels reach the dip the rotation rate is ##\frac 1{2L}v\tan(\alpha)-3\frac{gL}{(h^2+4L^2)}\Delta t##. This is ##w_0 ## for the first free fall phase.
From that we can deduce ##v_{0c}=w_0*L## and for ##y_{0c}## I use the average rotation rate during del t: ##y_{0c}=(w_0+\frac 1{2L}v\tan(\alpha))/2*L*\Delta t##. (Yes, these are approximations based on theta still being quite small.)
That sets up the conditions at the start of this free fall phase. At time t into the phase we can calculate xc, yc and theta easily: ##x_c=x_{0c}+v_{0x}*t##, ##\theta=\theta_0+w_0*t##, ##y_c=y_{0c}+v_{0y}*t-\frac 12gt^2##.
The position of the rear wheels is trickier.
##x_r=x_c-L\cos(\theta)+h\sin(\theta)##
##y_r=y_c-L\sin(\theta)-h\cos(\theta)##
Similarly the front wheels:
##x_f=x_c+L\cos(\theta)+h\sin(\theta)##
##y_f=y_c+L\sin(\theta)-h\cos(\theta)##
Finally, the "touch" column tracks whether the rear wheels have landed:
##r^2-x_r^2-(r\cos(\alpha)-y_r)^2##.
When that value goes negative, we've hit regolith.
For the second free fall phase, we have to set up a new set of initial conditions and start t at 0 again. For this we need a value for J/m. Something around 0.6 to 0.7 m/s seems to be reasonable. More on that later.
