Subject:
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RE: DC10 turning
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Newsgroups:
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lugnet.robotics.handyboard
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Date:
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Wed, 7 Aug 1996 17:52:12 GMT
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Original-From:
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Chuck McManis <cmcmanis@freegate&spamless&.net>
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Viewed:
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1504 times
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Keith wrote:
> I had the same problem. [d10 turning] I got some improvement by running the treads in
> opposite directions. Turns real quick then! And the drive tread does not
> have to carry the whole force of the turn plus "slip" or pivot the
> stationary tread.
A _long_ time ago, Dave Williams and I got into a discussion about tank treads and turning
etc. I ran some experiments on a Radio Shack Sherman Tank toy I was using (the original
"RoboTank" in my case)
In a nutshell we proposed a mathematical model for analyzing the power needed to turn a
treaded vehicle, in both the one tread drive turn, and the two tread drive turn. (another person
on the list from FMC wanted to help but all of his information on modelling the efficency of
turning treaded vehicles was classified :-) The model involved computing the dynamic friction
coefficient for the tread, F, and using that factor in an integration of the turning tread to compute
the power load. For those of you who would like to re-create this work we analyzed a turning
tank as having sections of tread described by a formula for the chord of a circle at a given
radius intersected with a function describing the rectangular shape of the tread in a radial
axis, computing the surface area of the "patch", the distance that patch travelled in a
90 degree turn times the dynamic friction coefficient.
The bottom line was that it took exactly the same amount of power to turn the tank, regardless
of the technique used. However, if you turned on both treads, then each motor needed to supply
only half the required power to achieve the turn so this was the reccomended way to turn.
As Jeff has noticed however, the force needed to move the tread is M*A*F where 'M' is the
mass of the tank, 'A' is the surface area of the tread, and 'F' is the friction coefficient.
Thus if you compute the power to turn the robot as Pm = MAF, then Mmax (maximum
mass of the robot) can be computed as Mmax = AF/Pmax (where Pmax is the most power
the motors can produce.) If you optimize for the lowest possible AF the formula reduces to
wheels. Also note that the analysis was done on flat surfaces with a fixed texture (in my
case experiments were carried out on a standard office chair roller mat.) For a complete robot
analysis you need to take into account whether the surface is constant or not, and if not
how widely can it vary.
--Chuck
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