According to § (2.4), due to the inelastic collision of the trailing leg with the floor and due to the pendular motion of the mass and the standing leg, part of the energy is lost when transitioning into the next step. This energy loss associated to the projection of the velocity from the previous step, to
the polar versor of the next step, is instantly restored so that the mechanical energyEdoesn’t change value. Since the floor was horizontal and flat, both the feet had the same height. This means that both the feet of the robot had the same potential energy at any point of the surface. In the case of a non-horizontal surface, this specific scenario may not happen. In order to reset the energy associated to the standing leg and the mass, we must account for the potential energy gained or lost through the height difference between the respective feet and the mass. The energy we are interested in conserving is determined at the beginning of each step. Figure (3.3) represents the situation described where there is a height difference between the foot of the the standing leg associated to stepn,(xSF(n),ySF(n))and
Figure 3.3:Height difference between the standing leg in stepnand the standing leg in stepn+1. The energy of the stepn is determined by measuring the velocity and height of massm at the final instant of stepn. To determine the energy of stepn+1, the velocity and height are measured in a frame located at the initial instant of stepn+1. If there is no control on the energy received or taken by this height difference, our model may be unable increment a step. Since the heighthi(n+1), from the foot of the standing leg in stepn+1 to the massm, is lower than the heighthf(n), from foot of the standing leg in the previous step to the mass, a bigger angular velocity is needed to be added in order for the system to conserve the energyEassociated to the step.
Upon reaching the instant when there is a step transition, it is necessary to know which velocityv0(n+1) to add so the system does not lose any energy. It is possible to determinev0(n+1)as well as the initial velocity of the next step,vi(n+1) = vf(n)cosβ+v0(n+1). The illustration in Fig. (3.3) helps identify the
The standing leg from stepnis rotated of an angleβwith the standing leg in stepn+1 at the step transition instant, we can eliminateφi(n+1),
φi(n+1)−φf(n)=β. (3.14)
where the positive sign is taken. The instantaneous velocity added in the beginning of the step,v0, can be expressed as,
The energy that is added due to the control mechanism,E(n+1)0 , in the case of an irregular surface, is given by
Equations (3.17) and (3.22) correspond respectively to the velocity added along the direction of the versor−eˆp(ti(n+1))and the energy added to the system instantaneously when the new step starts for irregular surfaces. Not every value associated toφ
tf(n)
can be attributed since it is not possible for the mass to be below the floor, or one of the legs to collide internally with the floor. In order to address this limitation, a variable substitution was applied. Withσ(n)being the angle that the segment which connects the two feet makes with an horizontal floor on the final instant of stepn. It can be defined, taking into consideration Figs. (2.1) and (3.3),
σ(n)=φ meaning the the mass is below the floor, or, there is an internal collision with the floor, rendering an invalid step. We can thus, express Eqs. (3.17) and (3.22) in terms ofσ(n),
v0(n+1) where the factor E/mlgis an adimensional quantity greater than 1 in order to assure that the mass can cross its vertical position. We will restrict the type of floors available to those that have a fixed slope, meaning that, the floor f can be defined by a general function of the type
f(x) =tan(ψ)x, (3.26)
whereψcorresponds to the angle the floor makes with a horizontal line. In this case, the value ofψ coincides withσ(n),
σ(n)=ψ. (3.27)
Figure (3.4) represents the robot walking on a flat but inclined surface. In this situationσ(n)=ψ, with ψbeing the angle associated to the slope of the floor.
ψ σ(1)
Figure 3.4: Robot walking down the first step on a flat inclined surface. Generally, the angle the segment that connects both feet at the step transition makes with a horizontal line,σ(n), is different from the slope of the floor, ψ, but in the case of a flat surface they coincide.
Figures (3.5) and (3.6) correspond respectively to the contour plots associated tov0(n+1)andE0(n+1) in Eqs. (3.24) and (3.25) for three different values of energyE, along the angle of the slope,ψ, and the angle between the legs, β. There are some similarities regarding the flat horizontal case. For example, the relationship with β remains the same, meaning that the higher theβ, the bigger are v0(n+1)andE0(n+1). However, there are also some major differences and one of these differences is the introduction of the variableψ, the fixed slope of the floor. Whenψ>0, the difference of the heights in step transitions is negative,∆h(n)=hi(n+1)−hf(n)<0, which means that the final velocity in the step before,vf(n), will be lower than the velocity at the start of the next step,vi(n+1), due to the reduction of potential energy at the start of the next step. Analogously, whenψ< 0 the height difference due to the step transition is positive∆h(n)>0 which means that the velocity on the final instant of the step before,vf(n)will be higher than the velocity at the start of the next step,vi(n+1), because of the increase of potential energy at the start of the next step. This has different implications whetherβis bigger or smaller thanπ/2.
Assumingβ > π/2, the projected final velocity has a negative component along the direction of the versor−eˆp(ti(n+1)), which means that the control mechanism must stop the speed from the last step and add the necessary velocity in order to maintain the energy of the system constant. With ψ=0, the case is identical to the horizontal floor case. By looking at Figs. (3.5(a)) and (3.6(a)), when ψ < 0, but close to 0, E0(n+1) and v0(n+1) drop. This happens because, even though the projected velocity is negative, which means that more energy and speed is required to sustain the movement, the initial velocityvi(n+1)is also lower due to the increase in potential energy in the next step according to Eq. (3.15). Whenψ<0 but close to−π/2 the values ofE0(n+1)andv0(n+1)raise since the potential energy increase passes its maximum value, mlg, and starts to be smaller, which increases the value ofvi(n+1). Whenψ>0, this potential energy is decreased, increasing the value ofvi(n+1), but the final velocity,vf(n), also decreases, which has the result of lowering the velocity that needs to be stopped, this has the effect of lowering E0(n+1) andv0(n+1). When ψ > 0 but closer toπ/2, the final velocity becomes greater than its minimum since the standing leg associated to the final instant of the step is not close to the vertical position, therefore,E0(n+1)andv0(n+1)are raised untilψ=π/2.
Assumingβ < π/2, the projected final velocity has a positive component along the direction of the versor−eˆp(ti(n+1)), which means that the control mechanism uses the speed from the last step and adds the remaining amount in order for the energy of the system to maintain constant. Whenψ>0, not only is the final velocity smaller, but also, the potential energy of the system is lowered, which causes a sharp raise inE0(n+1)andv0(n+1). Whenψapproachesπ/2 andβis small, the differences in potential energy are not so significant and the projection along−eˆp(ti(n+1))becomes relevant. When ψ< 0 andβ< π/2, the slope of the floor helps the movement, hence, the final velocity is increased and the potential energy associated to the next step is increased. In this region, E0(n+1) and v0(n+1) decrease, reaching, in some cases, 0 or negative values. When E(n+1)0 and v0(n+1) are negative, it means that the control mechanism must take energy or speed from the robot in order to continue the movement with the same value of mechanical energyE.
-π2 -π4 0 π
(a)Projectionv0n+1added instantaneously when the new step starts along with−eˆp(ti(n+1))as a func-tion of the slopeψand the angleβfor the energy E/mlg=1
(b)Projection v0n+1 added instantaneously when the new step starts along with−eˆp(ti(n+1))as a function of the slopeψand the angleβfor the energyE/mlg=2
(c)Projection v0n+1 added instantaneously when the new step starts along with−eˆp(ti(n+1))as a function of the slopeψand the angleβfor the energyE/mlg=3
Figure 3.5:Projectionv0n+1as a function of the slope of the floorψin Eq. (3.27) and the angleβbetween the legs at the transition instants for three different energies. As the energyE/mlgincreases, the associated valuesv0(n+1) also increase. As in the case of a flat surface,v0is strongly related toβ, meaning that the higher theβ, the bigger the projectionv0(n+1). WhenE/mlg=1,ψ>0 andβ>π/2, the projected final velocity is lower because of the height difference, which implies that less velocity is needed to stop the movement of the mass from the previous step, which causesv0n+1to drop slightly.
-π2 -π4 0 π4 π2
(a)Energy E0n+1 added instantaneously when the new step starts due to the control mechanism as a function of the slopeψand the angleβfor the energyE/mlg=1
(b)Energy E0n+1 added instantaneously when the new step starts due to the control mechanism as a function of the slopeψand the angleβfor the energyE/mlg=2
(c)Energy E0n+1 added instantaneously when the new step starts due to the control mechanism as a function of the slopeψand the angleβfor the energyE/mlg=3
Figure 3.6: EnergyE0(n+1) as a function of the slope the floorψand the angle between the legsβin transition instants between steps. This quantity shares the same relationship withβasv0(n+1)in Fig. (3.5), however, in the caseE/mlg=1 there is a strict increase whenβ>π/2 andψ>0 becauseE0(n+1)must account from the drop in potential energy from stepnto stepn+1.
If the control mechanism is removed, E0(n+1) = 0, the energy Ewill change from step to step, and according to Eqs. (2.5), (3.20), (3.23) and (3.27), the following expression can be obtained for the
energy associated to the next step, E(n+1)
mlg =
E(n) mlg−cos
ψ− β
2
cos2β+cos
ψ+β 2
, (3.28)
which is exactly the numerical application of the fixed point method,E(n+1) =ρ E(n)
[19]. There-fore, if no control mechanism is applied, andE/mlg >1, the energy of the step will converge to its fixed point, given that|cos2β| < 1 for 0 < β ≤ π/2. Since the energy converges to its fixed point from step to step, the robot will enter a limit cycle in which it was proven its existence for bipedal walking models going down a slope,ψ < 0 [14, 20]. Figure (3.7) is a plot of the phase space(φ, ˙φ) resultant of the iteration of 5 steps of the robot walking down a slope with no control mechanism, in which the energy converges from iteration to iteration to its fixed point. It can be obtained with the following parameters,
ψ=−π/10 rad, β=1.205 rad, l=1 m, m=80 Kg, g=9.8 m/s2,E0=1123.5 J. (3.29)
t [ s ]
0 1 2 3 4
Figure 3.7: Phase space,(φ, ˙φ), of the robot walking down a slope for 5 steps with no control mechanism. We can see that the energy associated to the movement, from step to step, is converging since the differences become smaller.
The fixed point associated to the slope of floor,ψand the angle between the legsβ, can be obtained by Eqs. (3.25) and (3.27), withE0(n+1)at 0, the result is the following solution
E
We can thus expand the number of steps, and by calculating the difference between the fixed point at Eq. (3.30) and the numerical energy at the 40th step,
E(40)= 1 which is of the order of the numerical error. Figure (3.8) is a density plot which relates the energy necessary to maintain a limit cycle trajectoryE/mlgwith a given angle of the slopeψ, and an aperture between the legs at the start of the stepβaccording to Eq. (3.30).
E
Figure 3.8: Regions of parameter space associated to the existence of limit cycles. The colours correspond to E/mlgand represent the energy necessary to maintain a limit cycle trajectory with a givenψandβ. Highlighted at red are the solutions in whichE/mlg=1, which is the energetic sufficient condition in order for the movement to take place. Due to the singularity associated to the term sin2β, the energyE/mlghas a big variation whenβ is small. Whenβis big, the variations associated toE/mlgare small. Generally, whenβ<π/4 and, the lower is ψ, the larger isE/mlg.