Oops again. I thought the revised figure of 2.5 acres was too good to be
true---and it was! I put the ``e'' in the wrong place. As efficiency (e)
goes up, the area (a) should go up. So I should have said:
a=ecvw/g
and with the revised estimates of growth g=1E-7m/s and longest cutting
length c=0.02m we get the depressingly realistic
a=0.2*0.02*0.1*0.1/1E-7
=400 m^2
Where can we get a real value for g? Whoever mows their lawn next must
measure a typical cutting and work it out for us!
If this latest calculation is right, then we are right on the border of
acceptable performance. Since I'd like to have a generous margin on
this, it looks as though we have no choice but to have a wider cutter (w)
and move faster (v). We might also have to consider how we can be more
efficient (e) and if we can put up with longer cuttings (c).
What's the value of area (a) we must allow for? In my case, it's
about 500 m^2, but Monty Stein has already claimed a massive 2 acres
(8093 m^2). To achieve this, we would have to improve the product of w,
v, c and e by a factor of 20. I can live with width w=0.3m maybe (*3)
and velocity v=0.2m/s at a push (*2); perhaps working 24 hours lets us
get efficiency e up to 0.6 (*3) but c=0.02m is surely near the acceptable
limit already. But *18 is almost good enough. Or is this is pushing
too hard? Should just resign ourselves to tackling a more modest area,
with the expectation that two or more mowbots would be needed to cope
with bigger spaces. Opinions?
Robin.
-- R.M.O'Leary <robin nospam at acm.org> +44 973 310035 P.O. Box 20, Swansea SA2 8YB, U.K.