Dennis responds:
<snip>
>If we assume that both cars have the same level of grip (f),
There's where you've gotten. Available grip =~ Cf*W, where Cf =
tire coefficent of friction, assumed constant to a first
approximation, and W = force on the tire normal to the ground
plane. Now, if Cf is relatively constant wrt vertical loading -
only approximately true and only within a range of vertical
loading values - then the car that's twice as heavy will have
twice the available cornering force, which is exactly how much it
needs to maintain the same lateral acceleration as the lighter
car.
> then Car B must
>slow down to make the turn.
See above.
>Solving for v, we get
>
>sqrt(rf/m) = v
Correct, but f is directly proportional to the mass, as shown in
the above approximate equation.
>Meaning that v varies as the square root of one over the mass.
(uhh, right?)
No, since f varies as the first power of the mass, v =~ constant.
Now, a tire's Cf isn't really perfectly constant wrt vertical
loading: it tends to decrease slightly as vertical loading
increases outside some optimal range. If this were not the case,
the weight of a car could be increased without limit with no
reduction in its steady-state cornering capability (transient
behavior is a whole 'nother subject).
Now, does it make more sense? A heavier car needs more absolute
grip to corner at the same speed as a lighter one, but its
additional weight creates (more or less) the required added grip,
which is a function of the component of tire loading normal to
the pavement.
Jay
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