This article presents the method of determining the stability limit of the vehicle (robots or electric wheelchairs) that use tracks (chain, belt) for climbing and descending stairs. Research method was conducted by modeling vehicle which is working under the influence of some elements such as geometric, mass, velocity, acceleration, friction...Afterthat, we identify some conditions which lead to tipping.
Science & Technology Development, Vol 15, No.K1- 2012 DETERMINING THE STABILITY CONDITIONS OF THE TRACK-DRIVEN VEHICLE WHILE STAIR CLIMBING AND DESCENDING Pham Duc Khoi, Thai Thi Thu Ha DCSELAB, University of Technology, VNU-HCM (Manuscript Received on April 5th, 2012, Manuscript Revised November 20rd, 2012) ABSTRACT: This article presents the method of determining the stability limit of the vehicle (robots or electric wheelchairs) that use tracks (chain, belt) for climbing and descending stairs Research method was conducted by modeling vehicle which is working under the influence of some elements such as geometric, mass, velocity, acceleration, friction Afterthat, we identify some conditions which lead to tipping Finally, the results of research are constraint equations in order to ensure stable operation of vehicle Keywords: wheelchair An alternative solution consists of a cluster INTRODUCTION Nowadays, there are many kinds of mobile robots which have been used in different tasks One of the most interesting and useful mobile robots is mobiles robot capable of climbing and descending stairs They are powered wheelchair which greatly improve the mobility of people with disability, the robot security, reconnaissance, fire There were many types of vehicle have been developed and still are under development to make a mobile robot capable of climbing and descending steps, slopes and stairs The popular solutions make use of tracks, clusters of wheels, legged system The legged systems are mentioned in [1] Legged robots are versatile for obstacle overpassing and high mobility in difficult terrain or soil condition, but they are too complex, low speed, and low-load capacity Trang 36 of wheels that are attached to a rotating link A commercial available stairs-climbing wheelchair is shown in [2] where the wheel has a smaller radius of ladder height Each cluster is combined from two or more wheels, are arranged beam (star) Each wheel within a wheel spindle beam separately Indeed, there are several problems in using cluster of wheels A problem concerns with that each wheel of a cluster must have its own transmission system, and therefore a vehicle can be very heavy, large size, high energy consumption Track is a quite common solution A track is an endless belt or chain in self-propelled vehicle, and it helps the vehicle to distribute its weight more evenly over a larger surface area than wheels contacts only In obstacle climbing, tracks emulate a wheel with infinite radius so that an obstacle can be over passed as TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 15, SỐ K1- 2012 a slope by using track extension as bridge A high friction coefficient between step edge and uuur FD track is also needed to generate a proper uur N tangential force that allows the vehicle to climb an obstacle An example of using tracks in uur rcmc uur Ffc climbing and descending stairs is Top-chairs, uur Fa uuur FgΣ r the principles presented in [3], illustrated in l1 Figure This vehicle has two pairs of tracks uur r12c that can adapt their geometry to the initial and l2 final phases in climbing and descending stairs Figure Vehicle model The loaded vehicle can be modeled as illustrated in Figure Mass M, with a moment of inertia I cm about its center of mass The forces acting on M are the gravitational force uur r FgΣ = M g , the surface normal reaction ur force N , friction force uur Ffc , driving force uuur uur F D , inertial force F a , length of the contact Figure Top-chairs vehicle The next section will present the research methods of stability of vehicles that climb and descend steps by using tracks, reference to [4] surface lc = l1 + l2 2.1 The statistical stability MODELING Elements to be considered include the geometrical elements (dimensions and shape of the vehicle, the size of the stairs), the statistical elements (mass and inertia), the kinematical element (velocity and acceleration), dynamical elements (forces and moments) uuur N 3q uuur N 3p uuur N 2q the p uuur N 2p uur rcmc uuur FgΣ uuur N1q uuur N1p ln l2 hS lS Figure Stair climbing model Trang 37 Science & Technology Development, Vol 15, No.K1- 2012 When the vehicle is stationary or moving s without acceleration, system can be modeled as illustrated in Figure In case of vehicle upstairs, the forces acting on M are the uur s θ crit1 tan θ crit = at which the vehicle hc first starts to tip over depends on l1 : r Σ gravitational force Fg = mg , at each peak, l2 − p tan θ = S l1 − ln uur there are two perpendicular forces N iq and hc l1 − p < tan θ crit = s hc uur N i p , ( i = 1, 2, ) Angle of the ladder in this case is given by: tan θ S = hS , hS and lS are the lS uur rcmc height and width of each ladder, p is the uuur N1p distance between the nearest peaks of the ln The point of rotation will be about the downhill contact point O Torque equation can be written as: Mg ( hc sin θ − (l2 − ln ) cos θ ) − ∑N p i uur (i − 1) p = (1) r At the limit of stability, N i = as point lifts away from the stair, l2 < ln equation (1) can be simplified into: S tan θ crit = l2 − p < tan θ S = hc p l1 lS It is concluded that the vehicle moves uur p 2,3 just uuur N 2q uuur N 3q Figure Stair descending model uur S hS uuur N 2p uuur N1q hS + lS ladder and given by: p = S uuur FgΣ uuur N3p Σ stablity if the gravitational force Fg located in S S the space limited by the angle θcrit1 and θcrit , (illustrated in Figure 5) l − ln (2) hc Equation (2) gives the tipping stability s limit, the angle θ crit tan θ crit = s l2 − p hc which the vehicle first starts to tip over depends on l p s θ crit s θ crit l1 Similarly, in case of vehicle moves down the stairs (illustrated in Figure The tipping stability uur rcmc at limit, the p l2 hS lS angle Figure Limit of stable space Trang 38 TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 15, SOÁ K1- 2012 Subscript 2.2 The effect of the veclocity When the vehilce when moving down the ur stairs with veclocity v0 , certainly, the angular velocity is zero ω0 = If for any reason the vehicle stop suddenly, vehicles can be flipped refers to the situation immediately after impact The movement of vehicle include only translational motion with initial veclocity v0 and rotational motion with angular velocity ω0 At this time, center of forward, shown in Figure The next section gravity height : y1 = y Kinetic and potential will energy of the vehicle at this time, respectively analyze the conditions for vehicle overturned in this case To facilitate the calculation, assuming that no loss of energy when the vehicle brake suddenly Subscript refers to Subscript refers to the situation some time after impact Center of gravity is highest the situation immediately prior to impact, The movement of vehicle include only translational motion with initial T1 and V1 veclocity v0 Kinetic and potential energy of the vehicle at this time, T0 V0 positon at this time ( y2 ) Kinetic and potential energy of the vehicle at the time, respectively T2 and V2 The energy of the vehicle at is given by: T0 + V0 = 2 Mv0 + Mgy0 , The energy of the vehicle at includes T2 and V2 Applying ur r1 conservation uur v0 impact: of energy post- Mv0 + Mgy0 = T2 + V2 Applying the principle of conservation of angular momentum (O), ( I1 is inertia moment of system): r v r ( Mv ) h = ( Mv ) r + I1ω1 (3) Because the vehicle starts rotating about the instantaneous velocity of the vehicle O1 r when the center of gravity at the highest r position, we consider that: v1 = r1 ω1 Combining this expression with equation (3) Figure Vehicle collision leads to: Trang 39 Science & Technology Development, Vol 15, No.K1- 2012 r2 (M r ( Mv ) h = ) r2 (M r + I )ω r ( Mv h ) = ( M r + I ) ω r ( Mv h ) = (M r + I )ω r (M r + I ) (4) + I1 ω1 ( Mv ) h = Kinetic of system immediately prior to 1 impact is given by: 2 ⇔ 1 T0 = Mv is given by: 1 Combining this expression with equation (6) leads to : r2 2 T1 = Mv1 + I1ω1 = M r1 + I1 ω1 2 ( ) (6) subscript and 2: ( Mv h ) T1 = (8) r2 ( M r +I 1 Applying conservation of energy post-impact at ) Combining this expression with equation (5) leads to : (7) T1 + V1 = T2 + V2 The potential energy of the vehicle at and 2, Kinetic of system immediately after impact ⇔ (5) respectively ( Mv0 h )2 r ( M r 2+I 1 : V1 = Mgy1 = Mgy , V2 = Mgy2 , Kinetic energy of the vehicle at Solving for is zero: T2 = By solving equation (6), we can calculate the conditions required to induce unrecoverable tipping of the vehicle, where the vehicle's tilt angle in world co-ordinates has reached the S ) c vtilt = v0 gives us: r ( 2 g ( y2 − y1 ) M r + I1 c tilt (9) = Mg ( y2 − y1 ) v = v0 = Mh ) (10) In addition to increased stability, maximum which is generated upon amount of tip ∆y θcrit with zero tipping, which is a measure of how much time speed Tipping of the vehicle up to this point the wheelchair's wheels spend off the ground will be recoverable, as and out of play as control surfaces for the limit of statistical stability the vehicle will be statistically stable even at maximum tip, and wheelchair hence (2.25), we have: recover its initial position and orientation Equation (5) is re-written as: ∆ y = y − y1 Solving equation ∆y = Mv0 h ( r2 g M r1 + I1 = T0 h r2 ) g (M r + I1 ) (11) The ratio of change in potential energy ∆ V at maximum tilt to kinetic energy T0 on impact (illustrated in Figure 6) Trang 40 TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 15, SỐ K1- 2012 ∆V = T0 ⇔ Mh r2 (M r + I1 ∆V T0 ( Mh Adding safely factor k into equation, + I1 )