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Drivetrain Lessons Learned Summer 2008

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)

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Drivetrain Lessons Learned Summer 2008

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  1. Drivetrain Lessons LearnedSummer 2008 Team 1640 Clem McKown - mentor August 2008

  2. 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

  3. 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

  4. t Drive Basics - Propulsion Fn = normal force between frictive surfaces Ff = Friction Force Ff = mFn m = coefficient of friction For objects not sliding relative to each other m = ms (static coefficient of friction) For objects sliding relative to each other m = mk (kinetic coefficient of friction) as a rule, ms > mk (this is why anti-lock brakes are such a good idea) For a 120 lbm robot with weight equally distributed over four wheels, Fn would be 30 lbf at each wheel. The same robot with six wheels would have Fn of 20 lbf at each wheel (at equal loading). Fn • = torque r = wheel radius r Fp = Propulsive Force Fd = Drive Force Fd = t/r ms mk For wheels not sliding on drive surface: Fp = -Fd; Fp ≤ Ff/s For wheels slipping on drive surface: Fp = Ff/k

  5. Robot Propulsion (Pushing) • Symmetric 4wd Robot • Symmetric 6wd Robot • Conclusions

  6. Propulsion Force (Fp) – Symmetric 4wd Propulsion Force per wheel Rolling without slipping: Fp/w = t/rw - up to a maximum of Fp/w = ms Fn Pushing with slipping: Fp/w = mk Fn Assumptions / Variables: t = torque available at each axle m = mass of robot Fn = Normal force per wheel = ¼ m g/gc (SI Fn = ¼ m g) – evenly weighted wheels rw = wheel radius Robot Propulsion Force Fp/R = S Fp/w Rolling without slipping: Fp/R = 4t/rw Pushing with slipping: Fp/R = 4mk Fn Fp/R = mk m g/gc (SI):Fp/R = mk m g

  7. Fp – Symmetric 6wd Propulsion Force per wheel Rolling without slipping: Fp/w = 2/3t/rw - up to a maximum of Fp/w = ms Fn Pushing with slipping: Fp/w = mk Fn Assumptions / Variables: 2/3t = torque available at each axle same gearing as 4wd w/ more axles m = mass of robot Fn = Normal force per wheel = 1/6 m g/gc (SI Fn = 1/6 m g) – evenly weighted wheels rw = wheel radius Robot Propulsion Force Fp/R = S Fp/w Rolling without slipping: Fp/R = 6 2/3t/rw = 4t/rw Pushing with slipping: Fp/R = 6mk Fn Fp/R = mk m g/gc (SI):Fp/R = mk m g Conclusion Would not expect 6wd to provide any benefit in propulsion (or pushing) vis-à-vis 4wd.

  8. 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 (Fn) 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.

  9. Stationary turning ofsymmetric robot • Assume equal loading of all wheels • Assume turn axis is center of wheelbase • Some new terms need an introduction: • mt – wheel/floor coefficient of friction in wheel tangent direction (kinetic unless otherwise noted) • mx – wheel/floor coefficient of friction in wheel axis direction (kinetic unless otherwise noted) – note that omni-wheels provide mx significantly lower than mt • Ft – wheel propulsive force in turn tangent direction • Fx – wheel drag force in wheel axis direction • Fr – wheel resistance to turn (force) in turn tangent direction • A Premise: Stationary turning demands wheel slippage, therefore drive force (Fd) must be capable of exceeding static friction (msFn) as a prerequisite for turning.

  10. Fr = Fx sin a = Fx = drag force against turn in the direction of the turning tangent Ft = Fp cos a = Fp Ft = Fp cos a = Fp = propulsion force for turn in the direction of the turning tangent l √(w²+l²) w √(w²+l²) w √(w²+l²) Stationary turning – 4wd a l a Fp = mtFn a Ft Fp = Propulsion force in direction of wheel tangent Fr Fx = mx Fn = axial direction drag (force) resisting turning a rturn = √(w²+l²) w a = tan-1(l/w) tturn = 4[Ft–min(Ft,Fr)]rturn = 4[Ft-min(Ft,Fr)]√(w²+l²) = 4[Fpw – min(Fpw,Fxl )] = m[mtw – min(mtw,mxl )]g/gc turning resistance Turning is possible if mtw > mxl a Fp = mtFn a propulsion propulsion a Ft a turning resistance

  11. Fr = Fx sin a = Fx = drag force against turn in the direction of the turning tangent Ft = Fp cos a = Fp = propulsion force for turn in the direction of the turning tangent w √(w²+l²) l √(w²+l²) Stationary turning – 6wd Fx = mx Fn = axial direction drag (force) resisting turning a l Fr Fp = mtFn Fp = mtFn a Ft Fp = Propulsion force in direction of wheel tangent tturn = 4(Ft–Fr)rturn + 2Fpw = 4(Ft-Fr)√(w²+l²) + 2Fpw = 6Fpw – 4Fxl = m(mtw – 2/3mxl )g/gc (SI)= mg(mtw – 2/3mxl ) rturn = √(w²+l²) w a = tan-1(l/w) Turning is possible if mtw > 2/3mxl All other factors being equal, 6wd reduces resistance to turning by 1/3rd Additional benefit: center wheels could turn w/out slippage, therefore use ms rather than mk (increased propulsion)

  12. Twitch drive testing – steering overview • These observations are consistent with this analysis where: tturn = m(mtw – mxl )g/gc • 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

  13. 6wd Geometry l = 6 w = 4.455 l/w = 1.35 steerable w/ 4” wheels (but not w/ 5” knobbies shown – it’s a power thing) r = 7.473 r = 4.455 α = 53.4°

  14. Connection to observations6wd • 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.

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