# [mowbot] motor power, induction charging

robin nospam at acm.org
Thu, 26 Sep 1996 01:10:37 +0100

Dave Everett wrote:
> If we can get 6v motors for all the drive components then we will get
> almost double the amp/hour capacity for the same size and weight of 12v
> batteries.
Unfortunately, you only get to run for twice as long by moving half as fast:-)

At this point we need another experiment: just how powerful a
motor do we need to move the (unknown) weight of the Mowbot over rough
grass at (say) 0.2m/s against the (unknown) drag of the cutting device?
There are a few imponderables, but it should be easy enough to get
an approximate idea. Assuming two motors, my guess is no more than 1 amp
per motor.

I haven't done the experiment, but I have looked at suitably chunky 12V
d.c. motors in the RS catalogue and they have ratings like: 300mA (60rpm,
0.3Nm), 493mA (65rpm, 0.3Nm), 1.2A (20rpm, 4Nm). The latter is perhaps
outside our budget, being 88 UKP (\$132) for two, but it is all metal,
126x42x64mm and looks *really* heavy duty. I can do the experiment on
paper at least.

The radius of the drive wheel, r (m), is fixed by the rotation speed
of the motor, S (/s), and the linear velocity we want v (m/s) by
2*pi*r*S = v
The force f (N) the motor can apply is just its torque, t (Nm), divided
by the distance to its application, in this case to the ground at r.

Taking the cheapest motor above costing 21 UKP (\$31) each,
S = 60rpm = 1rev/s. From earlier calculations about area and speed of
coverage, v=0.2m/s, so the radius of the wheel is
r = v/(2*pi*S) = 0.2/2pi = 0.031m (1.2in)
Then force is
f = t/r = 0.3/0.031 = 9.7N
So assuming perfect conditions (i.e. no other drag), this motor could
just manage to move forward while lifting vertically a mass of
9.7/g ~= 1 kg (2.2lb).
If Mowbot were powered with two such motors, it could exert
twice this force. If we guess the whole assemblage is likely to
mass around 6kg, this would mean that in perfect conditions, it
could just climb a slope of arcsin 2/6 = 17 degrees (1 in 3).
Hmm. Perhaps we need a just a bit more torque than that.

Repeating the calculation for the heftiest motor above gives
r = 0.2/(2*pi*20/60) = 0.095m (3.7in)
That's satisfyingly chunky! And the force it exerts is
f = 4/0.095 = 42N
Again neglecting drag, this could climb arcsin 4.3/6 = 45 degrees.
I think that's more than good enough. So the motor of choice
lies somewhere in this league.

> We don't need contacts, we could use inductive charging. The robot only
> needs to be in close proximity to the charger (maybe a plate on the
> ground).
Unless you can arrange to have the two parts of the induction
system interlocking so they make a pretty good magnetic circuit, the
efficiency is terrible, so a simple plate is out. But something
like this would work pretty well: (this is supposed to be a cut-away
drawing of a cylindrical object)
.-------------------.
| |<--ferrite core
| .---. .---. |
| | =| |= | |
| | =| |=<-+---+---first winding
| | `---' | |
| | .---. | |
| | =| |=<-+---+---second winding
`---' =| |= `---'
.-------' `-------.
| |
`-------------------'

This is exactly the charging mechanism in my electric toothbrush!
I haven't measured its efficiency, but it should be close to that
of an ordinary 1:1 transformer.

Robin.

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