Elsewhere , we have looked at how
fast you can go around a turn. For the purposes of this section, assume
that your best possible speed is 0.75m/s. Now that is the average speed
of the mouse. The outer wheel must be going faster than the inner wheel.
However, when we approach the turn, both wheels are going the same speed.
You can’t instantaneously change the wheel speeds so, if you approach
the turn already doing your maximum turn speed, the outer wheel will need
to speed up and the inner wheel will need to slow down. During this period,
the mouse will not be describing a constant radius turn. Rather, the turn
radius will get smaller, down to some minimum and then get larger until
we are going straight again. If you can complete a turn entirely within
a cell, then it will be a relatively simple matter to perform any combination
of turns.

Calculating the velocity of the wheels during the turn is a bit tricky.
In fact, so far, I have not been able to do it.

I do know that it is a goal worth chasing. Best case for a 90 degree
smooth turn is about 0.18 seconds.

To stop and turn in place will take much longer.

Assume you have a best acceleration (and thus deceleration) of 0.7g,
say 6.8m/s/s. You need to be stationary at the centre of the cell where
you are going to turn. You can enter the cell at a higher speed of 1.1m/s
and come to a halt in 0.163 seconds. Ignore the rotational inertia of
the mouse for a moment and limit ourselves only by friction. Given a wheel
track of 9cm, a quarter turn in place at maximum acceleration (6.8m/s/s)
would take 0.2 seconds. You then need another 0.163 seconds to get to
the edge of the cell, again doing about 1.1m/s.

OK, so you can enter the cell faster and leave it going faster. The entire
operation took 0.526 seconds – nearly three times as long as the smooth
turn. In a maze with a lot of turns (there will be), the overall effect
will be disastrous. No competitive mice turn
in place on a speed run.

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