Drive-train Basics Team 1640 Clem McKown - mentor October 2009 (r3) Topics What’s a Drive-train? Basics Components Propulsion Drivetrain Model Center of Mass Considerations Automobile versus robot tank drive 4wd versus 6wd robot tank drive Some Conclusions & Good Practices Unconventional Drive-trains Introducing Concept Pivot Chassis The mechanism that makes the robot move Comprising: What’s a Drive-train? Motors Transmissions Gearboxes Power Gearboxes transmission Wheels Axles Bearings Bearing blocks Wheels Motors Power transmission Note: this is an unrealistic chain run. We would always run individual chain circuits for each wheel. This way, if one chain fails, side drive is preserved. Basics - Components Motors Transmission Gear Reduction (optional shifting) Power transmission to wheels Wheels Axles Bearings Bearing blocks Electrical Power (W) 12 V DC Current per Motor performance Basics - Motors Controlled via Pulse Width Modulation (PWM) Motors convert electrical power (W) to rotational power (W) Power output is controlled via Pulse Width Modulation of the input 12 V DC CCL Industrial Motors (CIM) Rotational Speed FR801-001 Torque Basics - Motors CIM Performance Curves Speed ½ Motor curve @ 12 V Effic DC 100 350 Current Power Allowed a max of 90 300 (4) CIM Motors on 80 the Robot 70 250 y Motors provide 60 power at too low 200 50 torque and too high 150 speed to be directly 40 useful for driving Speed (1/s); Efficienc 30 100 Current (Amps); Power (W) robot wheels 20 50 Each CIM weighs 10 2.88 lb 0 0 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 Torque, ft lb Basics – Transmission Transmission Reduces motor rotational speed and increases torque to useful levels to drive wheels Transmits the power to the wheels Optional – it may allow shifting gears to provide more than one effective operating range High gear for speed Low gear for fine control Generally consists of two parts Gearbox for gear reduction & shifting Power transmission to the wheels – which may include additional gear reduction as well Gearbox examples •AndyMark 2-Speed •AndyMark Toughbox •Bainbots planetary gearbox •10.67:1 and 4.17:1 •5.95:1 or 8.45:1 •9:1; 12:1 or 16:1 (2-stage) •Output: 12 tooth sprocket •Output: ½” keyed shaft •Output – ½” keyed shaft •1 or 2 CIM motors •1 or 2 CIM motors •1 CIM motor (2 available) •4.14 lb •2.5 lb •2.56 lb •Used on our previous 2 robots •Can drive wheel directly •3:1 or 4:1 reduction/stage •1 to 4 stages available •3:1 to 256:1 available 1640 Custom gearbox Modified AndyMark 2-Speed Sprocket output replaced w/ 20-tooth gear & additional 45:20 (9:4) reduction added Direct-Drive ½” shaft output 9.4:1 & 24:1 1 or 2 CIM motors Used successfully on Dewbot V Power Transmission Chains & Sprockets Traditional Allows further reduction (via sprocket sizing) 3/8” pitch chain Steel – 0.21 lb/ft Polymer – 0.13 lb/ft Direct (w/ Bainbots gearbox) Gears (Team 25) Shafts Use your imagination Basics – Wheels - examples Kit Wheel Performance Wheel Omni Wheel Mecanum Wheel 6” diameter 8” diameter 8” diameter 8” diameter High-traction tread Circumferential rollers Angled rollers μt,s = 1.07 μt,s = 0.70 μt,k = 0.90 μt,k = 0.60 μx,s = 0.20 μx,s = 0.70 μx,k = 0.16 μx,k = 0.60 m = 0.48 lb m = 1.41 lb m = 1.13 lb m = 2.50 lb There are left & right mecanums Drive Basics - Propulsion Fn = normal force Ff = Friction Force between frictive surfaces Ff = μ Fn For a 120 lbm robot with Fn weight equally distributed over four wheels, F would μ = coefficient of friction n be 30 lbf at each wheel. The same robot with six For objects not sliding relative wheels would have Fn to each other of 20 lbf at each wheel (at equal loading). μ = μs (static coefficient of friction) τ For objects sliding relative to each other r τ= torque μ = μ (kinetic coefficient of friction) k r = wheel radius as a rule, μs > μk (this is why anti-lock brakes are such a good idea) Fp = Propulsive Force Fd = Drive Force μ For wheels not sliding on drive surface: Fd = τ/r s Fp = -Fd; Fp ≤ Ff/s μk For wheels slipping on drive surface: Fp = Ff/k dv Gnτ MS gc ⎛ τ ML G ⎞ = ⎜1− − v⎟ dt mrW ⎝ τ MS 2πrWν MU ⎠ Drive-train Model Excel-based model calculates acceleration, velocity & position versus time for a full-power start Predicts and accounts for wheel slippage Allows “what if?” scenarios A tool for drive-train design Center of Mass considerations Help! I’ve fallen and I can’t score anymore Center of Mass The point in space about which an object (robot) balances If the projection of the CoM falls outside the wheelbase, the robot will tip over Center of Mass Projection Straight down if robot is not accelerating Straight down on a ramp also (but the projected point shifts) Projection shifted by inverse of acceleration vector (see 2 diagram at right) a = 32.2 ft/s g = 32.2 ft/s2 Remember that stopping and turning are also accelerations The CoM can move as the robot deploys… …it can move a lot! How Robots Drive Automobile driving Robot (Tank) driving How an automobile drives Motor Power source Steering Front wheels change angle to direct line Differential of travel Provides equal Transmission drive torque Reduces rpm while to Left & Right increasing torque drive wheels to useful levels Suspension Maintains wheel contact on uneven surface How a (typical) robot drives Transmission Reduces rpm while increasing torque Steering to useful levels Robots steer like tanks - not like cars - Motor Power source by differential left & right Dual left & right drives side speeds or directions Unlike a car, robot (tank) steering requires wheel sliding Suspension Most FRC robots lack a suspension Car - Robot Comparison Automobile Drive Robot (Tank) Drive + Efficient steering - High energy steering + Smooth steering - Steering hysterisis + Avoids wheel sliding - Wheels slide to turn + Low wheel wear - High wheel wear - Large turn radius + Zero turning radius - Cannot turn in place + Turns in place - Limited traction + Improved traction 4wd – 6wd Comparison Propulsion Force (Fp) – Symmetric 4wd Propulsion Force per wheel Rolling without slipping: Fp/w = τ/rw - up to a maximum of Fp/w = μs Fn Assumptions / Variables: τ = torque available at each axle Pushing with slipping: Fp/w = μk Fn m = mass of robot Fn = Normal force per wheel = ¼ m g/gc (SI Fn = ¼ m g) Robot Propulsion Force – evenly weighted wheels rw = wheel radius Fp/R = Σ Fp/w Rolling without slipping: Fp/R = 4τ/rw Pushing with slipping: Fp/R = 4μk Fn Does not depend on evenly weighted wheels Fp/R = μk m g/gc (SI): Fp/R = μk m g Fp – Symmetric 6wd Propulsion Force per wheel Rolling without slipping: 2 Fp/w = /3τ/rw - up to a maximum of Fp/w = μs Fn Assumptions / Variables: 2 /3τ = torque available at each axle Pushing with slipping: Fp/w = μk Fn same gearing as 4wd w/ more axles m = mass of robot Fn = Normal force per wheel Robot Propulsion Force 1 1 = /6 m g/gc (SI Fn = /6 m g) – evenly weighted wheels Fp/R = Σ Fp/w rw = wheel radius 2 Rolling without slipping: Fp/R = 6 /3τ/rw = 4τ/rw Conclusion Pushing with slipping: Fp/R = 6μk Fn Fp/R = μk m g/gc Would not expect 6wd (SI): Fp/R = μk m g to provide any benefit in propulsion (or pushing) vis-à-vis 4wd (all other factors being equal) Stationary turning of symmetric robot Assume center of mass and turn axis is center of wheelbase Some new terms need an introduction: μt – wheel/floor coefficient of friction in wheel tangent direction wheel μx – wheel/floor coefficient wheel of friction in wheel axial axial tangent direction (omni-wheels direction provide μ << μ ) direction x t (x) (t) Fx – wheel drag force in wheel axis direction Stationary turning – 4wd α l α F = μ F F = F sin α p t n r x α l F F t = Fx r ) Fp = √(w²+l²) Fx = μx Fn ² = axial direction +l Propulsion = drag force α ² Ft = Fp cos α drag (force) w force in against turn √( w direction resisting turning = F in the direction w = p √(w²+l²) of wheel of the turning n r = propulsion tangent tangent r tu force for turn -1 l α = tan ( /w) in the direction of the turning τturn = 4(Ft –Fr)rturn tangent = 4(Ft -Fr)√(w²+l²) = 4(Fpw –Fxl ) Turning is possible if μtw > μxl turning resistance = m(μtw – μxl )g/gc Chris Hibner – Team 308 shows that turning resistance is reduced by shifting the center α of mass forward or back from the center of wheelbase. F = μ F propulsion α p t n propulsion α F F = F cos α t t p w = F α p √(w²+l²) turning resistance Stationary turning – 6wd Fx = μx Fn = axial direction α l drag (force) F r resisting turning Fp = μtFn Fp = μtFn α F t F = ²) p Propulsion τ = 4(F –F )r + 2F w ²+l F = F cos α turn t r turn p w t p force in √( w direction = 4(Ft-Fr)√(w²+l²) + 2Fpw = F w = p √(w²+l²) of wheel = 6F w –4Fl n p x r = propulsion tangent 2 r tu = m(μtw – /3μxl )g/gc force for turn α = tan-1(l/ ) (SI) = mg(μ w – 2/ μ l ) w in the direction t 3 x of the turning But this is based on tangent Equal weight distribution 2 Turning is possible if μtw > /3μxl Analysis indicates F = F sin α center wheels support r x l All other factors being equal, 6wd = F disproportionate weight: x √(w²+l²) 1 rd 40-60% of total - @ 40%: reduces resistance to turning by /3 = drag force 2 τturn = m(μtw – (1-.4) /3μxl)g/gc against turn = m(μtw –0.4μxl)g/gc in the direction of the turning ∴ turning benefit of 6wd is tangent considerable Additional benefit: center wheels could turn w/out slippage, therefore use μs rather than μk (increased propulsion) 4wd – 6wd Tank Drive Comparison 4wd Tank Drive 6wd Tank Drive + Simplicity - More complex + Weight - Weight (2 wheels) - Constrains design o Traction o Traction o Stability o Stability - Turning + Turning - Steering hysterisis + Less hysterisis - Wheel wear + Reduced wear + Ramp climbing Conclusions & Good Practices Provided that all wheels are driven, all other factors being equal, the number of drive wheels does not influence propulsion or pushing force available.