Drivetrain Lessons Learned

Summer 2008

Team 1640Clem 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

Fn

Fd = Drive ForceFd = /r

Fn = normal forcebetween frictive surfaces

For a 120 lbm robot withweight equally distributedover four wheels, Fn wouldbe 30 lbf at each wheel.

The same robot with sixwheels would have Fn

of 20 lbf at each wheel(at equal loading).

Fp = Propulsive ForceFor wheels not sliding on drive surface: Fp = -Fd; Fp ≤ Ff/s

For wheels slipping on drive surface: Fp = Ff/k

= torquer = wheel radius

Ff = Friction ForceFf = Fn

= coefficient of friction

For objects not sliding relativeto each other = s (static coefficient of friction)

For objects sliding relative toeach other = k (kinetic coefficient of friction)

as a rule, s > k

(this is why anti-lock brakes are such agood idea)

s

k

Robot Propulsion (Pushing)

Symmetric 4wd Robot Symmetric 6wd Robot Conclusions

Propulsion Force (Fp) – Symmetric 4wd

Assumptions / Variables: = 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

Rolling without slipping: Fp/w = /rw - up to a maximum of Fp/w = s Fn

Pushing with slipping: Fp/w = k Fn

Propulsion Force per wheel

Robot Propulsion Force

Fp/R = Fp/w

Rolling without slipping: Fp/R = 4/rw

Pushing with slipping: Fp/R = 4k Fn

Fp/R = k m g/gc

(SI): Fp/R = k m g

Fp – Symmetric 6wd

Assumptions / Variables: 2/3 = 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

Propulsion Force per wheel

Rolling without slipping: Fp/w = 2/3/rw - up to a maximum of Fp/w = s Fn

Pushing with slipping: Fp/w = k Fn

Robot Propulsion Force

Fp/R = Fp/w

Rolling without slipping: Fp/R = 6 2/3/rw = 4/rw

Pushing with slipping: Fp/R = 6k Fn

Fp/R = k m g/gc

(SI): Fp/R = k m g

Conclusion

Would not expect 6wdto provide any benefitin 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 (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.

Stationary turning ofsymmetric 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

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 (sFn) as a prerequisite for turning.

Stationary turning – 4wd

w

lFp = tFn

= tan-1(l/w)

Ft = Fp cos = Fp

= propulsionforce for turnin the directionof the turningtangent

w√(w²+l²)

F t

Fp = tFn

Ft = Fp cos = Fp

w√(w²+l²)

Ft

turning resistance

turning resistance

turn = 4[Ft–min(Ft,Fr)]rturn

= 4[Ft-min(Ft,Fr)]√(w²+l²) = 4[Fpw – min(Fpw,Fxl )] = m[tw – min(tw,xl )]g/gc

r turn =

√(w

²+l²)

Fx = x Fn

= axial directiondrag (force)resisting turning

Fr

Fr = Fx sin = Fx

= drag forceagainst turnin the directionof the turningtangent

l√(w²+l²) Fp =

Propulsionforce indirectionof wheeltangent

Turning is possible if tw > xl

propulsion propulsion

Stationary turning – 6wd

Fp = tFn

Fp = Propulsionforce indirectionof wheeltangent

w

F t

Ft = Fp cos = Fp

= propulsionforce for turnin the directionof the turningtangent

w√(w²+l²)

= tan-1(l/w)

l

Fx = x Fn

= axial directiondrag (force)resisting turningF r

Fr = Fx sin = Fx

= drag forceagainst turnin the directionof the turningtangent

l√(w²+l²)

Fp = tFn

turn = 4(Ft–Fr)rturn + 2Fpw = 4(Ft-Fr)√(w²+l²) + 2Fpw = 6Fpw – 4Fxl = m(tw – 2/3xl )g/gc (SI) = mg(tw – 2/3xl )

r turn =

√(w

²+l²)

Turning is possible if tw > 2/3xl

All other factors being equal, 6wdreduces resistance to turning by 1/3

rd

Additional benefit: center wheels could turn w/outslippage, therefore use s rather than k (increased propulsion)

Twitch drive testing – steering overview

These observations are consistent with this analysis where: turn = m(tw – xl )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

Drive Orientation x x y yScale ½ (VEX) Full (FRC) ½ (VEX) Full (FRC)l 7.41 14.85 5.00 9.88w 6.27 11.45 8.67 16.42l/w 1.18 1.30 0.58 0.60

Steerability Not Steerable Very Good

6wd Geometry

r =

4.4

55

r = 7.473

α = 53.4°

l = 6w = 4.455

l/w = 1.35

steerable w/4” wheels (but not w/5” knobbiesshown – it’sa power thing)

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.