Drivetrain Lessons Learned Summer 2008 Team 1640 Clem McKown - mentor August 2008
Observations – Bi-axial (Twitch) Twitch feature worked beautifully – effortless, radiusless 90° turns Robot was basically unsteerable in “x” orientation (drive aligned w/ long axis) Robot steered easily in “y” orientation (drive aligned w/ short axis) w/ 4 omni-wheels, robot steered easily, but was also easily pushed about w/ 2 catter-corner omni-wheels, robot turned one direction, but not the other
Observations – 6wd 6wd w/ 5” knobbies could not turn. 6wd w/ 4” wheels turned well Quite okay w/ 6 std wheels – not easily pushed Easier w/ 4 std & 2 catter-corner omni-wheels - #2 in terms of being pushed Similar turning w/ 4 std & 2 aft omni-wheels - #3 in terms of being pushed Easiest turning w/ 2 mid std & 4 corner omni- wheels – and a real easy pushover
Drive Basics - Propulsion r FnFn F d = Drive Force F d = /r F n = normal force between frictive surfaces For a 120 lb m robot with weight equally distributed over four wheels, F n would be 30 lb f at each wheel. The same robot with six wheels would have F n of 20 lb f at each wheel (at equal loading). F p = Propulsive Force For wheels not sliding on drive surface: F p = -F d ; F p ≤ F f/s For wheels slipping on drive surface: F p = F f/k = torque r = wheel radius F f = Friction Force F f = F n = coefficient of friction For objects not sliding relative to each other = s (static coefficient of friction) For objects sliding relative to each other = k (kinetic coefficient of friction) as a rule, s > k (this is why anti-lock brakes are such a good idea) sksk
Robot Propulsion (Pushing) Symmetric 4wd Robot Symmetric 6wd Robot Conclusions
Propulsion Force (F p ) – Symmetric 4wd Assumptions / Variables: = torque available at each axle m = mass of robot F n = Normal force per wheel = ¼ m g/g c (SI F n = ¼ m g) – evenly weighted wheels r w = wheel radius Rolling without slipping: F p/w = /r w - up to a maximum of F p/w = s F n Pushing with slipping: F p/w = k F n Propulsion Force per wheel Robot Propulsion Force F p/R = F p/w Rolling without slipping: F p/R = 4 /r w Pushing with slipping: F p/R = 4 k F n F p/R = k m g/g c (SI): F p/R = k m g
F p – Symmetric 6wd Assumptions / Variables: 2 / 3 = torque available at each axle same gearing as 4wd w/ more axles m = mass of robot F n = Normal force per wheel = 1 / 6 m g/g c (SI F n = 1 / 6 m g) – evenly weighted wheels r w = wheel radius Propulsion Force per wheel Rolling without slipping: F p/w = 2 / 3 /r w - up to a maximum of F p/w = s F n Pushing with slipping: F p/w = k F n Robot Propulsion Force F p/R = F p/w Rolling without slipping: F p/R = 6 2 / 3 /r w = 4 /r w Pushing with slipping: F p/R = 6 k F n F p/R = k m g/g c (SI): F p/R = k m g Conclusion Would not expect 6wd to provide any benefit in propulsion (or pushing) vis-à-vis 4wd.
Propulsion Conclusions Provided that all wheels are driven, for a robot of a given mass and fixed total driving force, the number of drive wheels does not influence propulsion or pushing force available. The existence of undriven wheels, which support weight but do not contribute to propulsion, necessarily reduce the available pushing force as long as those undriven wheels are weighted. For a robot with a rectangular envelope, given wheelbase, mass and center of gravity, (4) wheels (driven or not) provide the maximum stability. Additional wheels neither help nor hurt. A common (l-r) side drive-train (linked via chains or gears) has the following propulsion advantage over a drive-train having individual motors for each wheel: As wheel loading (F n ) changes and becomes non-uniform, a common drive-train makes more torque available to the loaded wheels. Motor stalling (and unproductive spinning) are therefore less likely under dynamic (competition) conditions.
Stationary turning of symmetric robot Assume equal loading of all wheels Assume turn axis is center of wheelbase Some new terms need an introduction: t – wheel/floor coefficient of friction in wheel tangent direction (kinetic unless otherwise noted) x – wheel/floor coefficient of friction in wheel axis direction (kinetic unless otherwise noted) – note that omni-wheels provide x significantly lower than t F t – wheel propulsive force in turn tangent direction F x – wheel drag force in wheel axis direction F r – wheel resistance to turn (force) in turn tangent direction A Premise: Stationary turning demands wheel slippage, therefore drive force (F d ) must be capable of exceeding static friction ( s F n ) as a prerequisite for turning.
Stationary turning – 4wd w l F p = t F n = tan -1 ( l / w ) F t = F p cos = F p = propulsion force for turn in the direction of the turning tangent w √(w²+l²) FtFt F p = t F n F t = F p cos = F p w √(w²+l²) FtFt turning resistance turn = 4[F t –min(F t,F r )]r turn = 4[F t -min(F t,F r )]√(w²+l²) = 4[F p w – min(F p w,F x l )] = m[ t w – min( t w, x l )]g/g c r turn = √(w²+l²) F x = x F n = axial direction drag (force) resisting turning FrFr F r = F x sin = F x = drag force against turn in the direction of the turning tangent l √(w²+l²) F p = Propulsion force in direction of wheel tangent Turning is possible if t w > x l propulsion
Stationary turning – 6wd F p = t F n F p = Propulsion force in direction of wheel tangent w FtFt F t = F p cos = F p = propulsion force for turn in the direction of the turning tangent w √(w²+l²) = tan -1 ( l / w ) l F x = x F n = axial direction drag (force) resisting turning FrFr F r = F x sin = F x = drag force against turn in the direction of the turning tangent l √(w²+l²) F p = t F n turn = 4(F t –F r )r turn + 2F p w = 4(F t -F r )√(w²+l²) + 2F p w = 6F p w – 4F x l = m( t w – 2 / 3 x l )g/g c (SI) = mg( t w – 2 / 3 x l ) r turn = √(w²+l²) Turning is possible if t w > 2 / 3 x l All other factors being equal, 6wd reduces resistance to turning by 1 / 3 rd Additional benefit: center wheels could turn w/out slippage, therefore use s rather than k (increased propulsion)
Twitch drive testing – steering overview These observations are consistent with this analysis where: turn = m( t w – x l )g/g c Would expect FRC bot to be steerable in y mode, but not in x mode w/out omni-wheels Model does not explain catter-corner omni-wheel steering asymmetry
6wd Geometry r = r = α = 53.4° l = 6 w = l/w = 1.35 steerable w/ 4” wheels (but not w/ 5” knobbies shown – it’s a power thing)
Connection to observations 6wd The 6wd prototype w/ 5” diam knobby wheels could not turn. It was clearly underpowered. The 6wd prototype w/ 4” diam wheels turned satisfactorily in all tested configurations with/without omni- wheels.