Slippage Down on Rolling Mobile Robots While Overcoming Inclined Obstacles

For this work, Lázaro [43] was used, which is a skid steer robot weighing W=255.1 N, with an attached arm that has two joints: a rotational one (θ₁) that moves the first link of the arm, and a prismatic one (d₂) that moves the second link of the arm (Fig. 2a). This arm is used in maneuvers to overcome obstacles and prevent tipping by acting as a counterweight or by placing its wheeled endeffector in contact with the ground. Additionally, it serves as a platform for installing sensors like a 2D rangefinder This allows the system to perceive its surroundings and more accurately detect and estimate obstacles. Subsequently, a frame fixed to the robot (X_C Y_C Z_C) was defined, where the X_C Y_C plane coincides with the plane containing the four contact points of the wheels with the ground, the X_C axis coincides with the longitudinal direction of the robot (forward motion), the Z_C axis is perpendicular to this plane, and its origin O_C is located at the intersection of the axis of the first joint in the robot's arm with the X_C Y_C plane (Fig. 2a). The main dimensions of this robot are shown in Tab. 1.

Fig. 2.

Lázaro robot showing dimensions and X_c Y_c Z_c frame a) Axono metric view; b) Side view

Additionally, it is important to note that in Fig. 2b, Δx_f = R_rsin(φ_c – φ_s2) and Δx_r = R_rsin(φ_c – φ_s1), which correspond to variations in the position of the front and rear wheel contact points respectively (from the ideal position), occur when the wheels are positioned on surfaces that are not parallel to the X_C Y_C plane. In this case, R_r is the radius of the wheels (the other variables will be defined later). Although this change in the wheel contact point position with the ground also induces a variation in the Z_c direction called Δz, it will be assumed negligible due to its small magnitude.

On the other hand, α_c and φ_c must be considered, which are the roll and pitch angles of the X_c Y_c Z_c frame with respect to a global frame X_M Y_M Z_M (to be defined later) and define the robot's inclination relative to a horizontal plane; these angles are obtained from an Inertial Measurement Unit (IMU) installed on the robot. With all the described variables, the position of the center of gravity and the weight vector (both defined according to the X_c Y_c Z_c frame) can be obtained for any inclination or position of the arm, through (3) and (4) respectively (Fig. 2a), where a, b, c, d, and r are constants specific to the robot. 3[ WxWyWz ]=W[ sinϕccosϕcsinαccosϕccosαc ]\left[ {\matrix{ {{W_x}} \hfill \cr {{W_y}} \hfill \cr {{W_z}} \hfill \cr } } \right] = W\left[ {\matrix{ {\sin {\phi _c}} \cr {\cos {\phi _c}\sin {\alpha _c}} \cr {\cos {\phi _c}\cos {\alpha _c}} \cr } } \right] 4[ xcogycogzcog ]=[ a+rcosθ1b+rsinθ1cd2+d ]\left[ {\matrix{ {{x_{cog}}} \hfill \cr {{y_{cog}}} \hfill \cr {{z_{cog}}} \hfill \cr } } \right] = \left[ {\matrix{ {a + r\;\cos {\theta _1}} \cr {b + r\;\sin {\theta _1}} \cr {c\;{d_2} + d} \cr } } \right]

According to the established premises, a simplified obstacle was defined as two flat surfaces: Surface 1, where the robot is located before overcoming the obstacle, and Surface 2, where the robot is located after overcoming the obstacle (Fig. 3). These surfaces are delimited by two edges considered straight lines: Edge 1 delimits Surface 1, and Edge 2 delimits Surface 2, so that the space between these edges defines the obstacle. Now, based on this, three frames were defined (Fig. 3):

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  A fixed frame (X_s1 Y_s1 Z_s1), attached to Surface 1, where the X_s1 Y_s1 plane is contained within Surface 1, the Z_s1 axis is perpendicular to this surface, and the origin of this frame O_s1 coincides with point O_c at an initial time instant (t = 0), when the robot has not yet started the obstacle overcoming process. At this instant, X_s1 Y_s1 Z_s1 coincides with X_c Y_c Z_c.

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  A fixed global frame (X_M Y_M Z_M) where X_M Y_M is located on a horizontal plane, the X_M axis coincides with the horizontal projection of X_s1, Z_M is vertical, and the origin O_M coincides with O_s1.

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  A fixed frame (X_s2 Y_s2 Z_s2), attached to Surface 2, where the X_s2 Y_s2 plane is contained within Surface 2, the Z_s2 axis is perpendicular to this surface, the origin of this frame O_s2 coincides with the intersection of Edge 2 with the X_s1 Z_s1 plane, and the X_s2 axis coincides with the intersection of Surface 2 with the X_s1 Z_s1 plane.

Finally, the orientation of the X_s1 Y_s1 Z_s1 and X_s2 Y_s2 Z_s2 frames relative to the fixed frame X_M Y_MZ_M must be established using the Roll (α) and Pitch (φ) angles. Therefore, the following angles are defined:

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  α_s1, φ_s1 : Roll and pitch angles of the X_s1 Y_s1 Z_s1 frame relative to the X_MY_MZ_M frame (define the inclination of Surface 1 relative to horizontal plane).

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  α_s2, φ_s2 : Roll and pitch angles of the X_s2 Y_s2 Z_s2 frame relative to the X_M Y_M Z_M frame (define the inclination of Surface 2 relative to horizontal plane).

Fig. 3.

Geometric parameters used to characterize the obstacle

Lastly, an angle called γ_b1 is added, which measures the inclination of Edge 1 relative to the X_s1 axis (or direction of robot advancement). All of these angles can be obtained using the IMU and a LiDAR (or arm with 2D rangefinder) installed on the robot.

Let s be a surface where a wheel is positioned; this surface is defined through a frame X_s Y_s Z_s with its origin located at the wheel's contact point n (n = 1 …4) with the ground, the X_s Y_s plane contained in surface s, the Z_s axis perpendicular to the surface, and the X_s axis in the rolling direction. With this, we can establish its pitch (φ_s) and roll (α_s) angles relative to a fixed frame X_M Y_M Z_M with the same origin as the X_s Y_s Z_s frame (Fig. 4a).

Fig. 4.

Forces at the wheel-ground contact point: a) robot wheel; b) wheel at the end effector

If wheel n is in contact with surface s, a normal force F_n is produced along the Z_s axis, and a friction force F_rn is contained within the X_s Y_s plane at an angle of inclination (δ_n) relative to the X_s axis. If the friction force F_rn = μ_iF_n, where μ_i is the instantaneous coefficient of friction between the wheel and the ground, then the resulting force sFTn→\overrightarrow {^s{F_{Tn}}} at contact point n measured in the X_s Y_s Z_s frame is defined by: 5sFTn→=Fn[ μicδnμisδn1 ]\overrightarrow {^s{F_{Tn}}} = {F_n}\left[ {\matrix{ {{\mu _i}c{\delta _n}} \cr {{\mu _i}s{\delta _n}} \cr 1 \cr } } \right]

This force can be expressed in terms of X_M Y_M Z_M through (6): 6MFTn→=[ cϕssϕssαssϕscαs0cαs−sαs−sϕscϕssαscϕscαs ]sFTn→\overrightarrow {^M{F_{Tn}}} = \left[ {\matrix{ {c{\phi _s}} & {s{\phi _s}s{\alpha _s}} & {s{\phi _s}c{\alpha _s}} \cr 0 & {c{\alpha _s}} & { - s{\alpha _s}} \cr { - s{\phi _s}} & {c{\phi _s}s{\alpha _s}} & {c{\phi _s}c{\alpha _s}} \cr } } \right]\overrightarrow {^s{F_{Tn}}}

Finally, MFTn→\overrightarrow {^M{F_{Tn}}} can be transformed using a frame parallel to the fixed frame of the robot X_c Y_c Z_c (with origin at the wheel-ground contact point), to obtain cFTn→\overrightarrow {^c{F_{Tn}}} through (7) with its components [ FnxFnyFnz ]T{\left[ {\matrix{ {{F_{nx}}} \hfill & {{F_{ny}}} \hfill & {{F_{nz}}} \hfill \cr } } \right]^T} defined by (8)-(10) in terms of the normal force F_n. 7cFTn→=[ FnxFnyFnz ]=[ cϕc0−sϕcsϕcsαccαccϕcsαcsϕccαc−sαccϕccαc ]MFTn→\overrightarrow {^c{F_{Tn}}} = \left[ {\matrix{ {{F_{nx}}} \hfill \cr {{F_{ny}}} \hfill \cr {{F_{nz}}} \hfill \cr } } \right] = \left[ {\matrix{ {c{\phi _c}} & 0 & { - s{\phi _c}} \cr {s{\phi _c}s{\alpha _c}} & {c{\alpha _c}} & {c{\phi _c}s{\alpha _c}} \cr {s{\phi _c}c{\alpha _c}} & { - s{\alpha _c}} & {c{\phi _c}c{\alpha _c}} \cr } } \right]\overrightarrow {^M{F_{Tn}}} 8Fnx=Fn [ cϕc(μicϕscδn+μisϕssαssδn+sϕscαs)− sϕc(−μisϕscδn+μicϕssαssδn+cϕscαs) ]{F_{nx}} = {F_n}\left[ {c{\phi _c}\left({{\mu _i}c{\phi _s}c{\delta _n} + {\mu _i}s{\phi _s}s{\alpha _s}s{\delta _n} + s{\phi _s}c{\alpha _s}} \right) - } \right.\left. {s{\phi _c}\left({ - {\mu _i}s{\phi _s}c{\delta _n} + {\mu _i}c{\phi _s}s{\alpha _s}s{\delta _n} + c{\phi _s}c{\alpha _s}} \right)} \right] 9Fny=Fn [ sϕcsαc(μicϕscδn+μisϕssαssδn+sϕscαs)+ cαc(μicαssδn−sαs)+cϕcsαc( −μisϕscδn+ μicϕssαssδn+cϕscαs ) ]{F_{ny}} = {F_n}\left[ {s{\phi _c}s{\alpha _c}\left({{\mu _i}c{\phi _s}c{\delta _n} + {\mu _i}s{\phi _s}s{\alpha _s}s{\delta _n} + s{\phi _s}c{\alpha _s}} \right) + } \right.c{\alpha _c}\left({{\mu _i}c{\alpha _s}s{\delta _n} - s{\alpha _s}} \right) + c{\phi _c}s{\alpha _c}\left({ - {\mu _i}s{\phi _s}c{\delta _n} + } \right.\left. {\left. {{\mu _i}c{\phi _s}s{\alpha _s}s{\delta _n} + c{\phi _s}c{\alpha _s}} \right)} \right] 10Fnz=Fn [ sϕccαc(μicϕscδn+μisϕssαssδn+sϕscαs)− sαc(μicαssδn−sαs)+cϕccαc( −μisϕscδn+ μicϕssαssδn+cϕscαs ) ]{F_{nz}} = {F_n}\left[ {s{\phi _c}c{\alpha _c}\left({{\mu _i}c{\phi _s}c{\delta _n} + {\mu _i}s{\phi _s}s{\alpha _s}s{\delta _n} + s{\phi _s}c{\alpha _s}} \right) - } \right.s{\alpha _c}\left({{\mu _i}c{\alpha _s}s{\delta _n} - s{\alpha _s}} \right) + c{\phi _c}c{\alpha _c}\left({ - {\mu _i}s{\phi _s}c{\delta _n} + } \right.\left. {\left. {{\mu _i}c{\phi _s}s{\alpha _s}s{\delta _n} + c{\phi _s}c{\alpha _s}} \right)} \right]

The end effector of the arm installed on the Lázaro robot is a steerable wheel that, due to its characteristics, is considered omnidirectional. Therefore, the friction present occurs due to rolling effects; the magnitude of this friction is much lower than the sliding friction present in the other wheels of the robot. Therefore, for this wheel, friction is considered negligible, which implies that when this wheel touches the ground, it is only subjected to a normal force F_n (Fig. 4b). Thus, if we follow the procedure defined in the previous section but consider that μ_i=0, we can find equations (11)–(13) that define the projections of the normal force F₅ of this wheel with respect to the X_c Y_c Z_c frame. Finally, it is worth noting that the Lázaro robot has a resistive force sensor attached to the end effector, which allows estimating the magnitude of F_5z, from which the rest of the components can be estimated. 11F5x=F5(cϕcsϕscαs−sϕccϕscαs){F_{5x}} = {F_5}\left({c{\phi _c}s{\phi _s}c{\alpha _s} - s{\phi _c}c{\phi _s}c{\alpha _s}} \right) 12F5y=F5(sϕcsαcsϕscαs−cαcsαs+cϕcsαccϕscαs){F_{5y}} = {F_5}\left({s{\phi _c}s{\alpha _c}s{\phi _s}c{\alpha _s} - c{\alpha _c}s{\alpha _s} + c{\phi _c}s{\alpha _c}c{\phi _s}c{\alpha _s}} \right) 13F5z=F5(sϕccαcsϕscαs+sαcsαs+cϕccαccϕscαs){F_{5z}} = {F_5}\left({s{\phi _c}c{\alpha _c}s{\phi _s}c{\alpha _s} + s{\alpha _c}s{\alpha _s} + c{\phi _c}c{\alpha _c}c{\phi _s}c{\alpha _s}} \right)

When a robot moves on an inclined surface, it can experience downhill slipping that may cause a deviation from its trajectory. In [23], a metric was developed to quantitatively estimate this risk in such cases. However, this metric is not useful when the robot is overcoming an obstacle because, in this scenario, the robot is positioned on two surfaces that may have different inclinations and properties. Therefore, it became imperative to develop a new metric in this case, which was defined based on the following premises:

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  Although uncommon, the general case was considered where each surface may have its own static coefficient of friction (μ_s) when it is in contact with the robot's wheels. Therefore, μ_s1 and μ_s2 are the static coefficients of friction on surfaces 1 and 2, respectively.

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  Each wheel has its own instant coefficient of friction (μ_i) with the ground. However, considering that the wheels on the same surface touch the same material (with identical properties), a single average instant coefficient of friction was assumed for the wheels touching a surface. Therefore, when the robot is overcoming an obstacle and its wheels touch two surfaces, μ₁ and μ₂ appear as the instant coefficients of friction of the wheels on surfaces 1 and 2, respectively.

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  During the obstacle traversal, if a pair of wheels positioned on one surface starts to slip, they will immediately induce slippage in the other two wheels (positioned on the other surface); this indicates that slippage is not independent on each surface. Operationally, if slippage occurs when the instantaneous coefficient of friction (μ_i) equals the static coefficient of friction (μ_i = μ_s), both pairs of wheels simultaneously reach their respective μ_s if: 14μ1μ2=μS1μS2{{{\mu _1}} \over {{\mu _2}}} = {{{\mu _{S1}}} \over {{\mu _{S2}}}}

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  When an individual element positioned on a surface starts to slip, it tends to move in the direction of maximum slope of the surface. However, this premise does not necessarily hold true in a complex system like a robot with multiple wheels, whose contacts are simultaneously distributed between two surfaces.

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  The friction force on each wheel has an orientation angle δ_n with respect to the X_s axis. This angle depends mainly on the slope of the surface where the wheel is positioned and on the overall robot inclination; although it can also be affected by additional forces from other wheels and external disturbances. However, in this work, these additional influences were neglected, so it was considered that the wheels on the same surface will have the same δ_n, resulting in δ₁ and δ₂ corresponding to friction forces on surfaces 1 and 2, respectively.

Now, there is no slipping between two surfaces when F_{s max} ≥ F_s, where F_s is the static friction force and F_{s max} is the maximum static friction force. In the case of a wheel n touching a surface, F_s = μ_iF_n, where F_n is the normal force; furthermore, F_{s max} = μ_sF_{n s}, where F_{n s} is the normal force calculated for the limit condition before slipping occurs. Therefore, the relationship F_{s max} ≥ F_s can be rewritten as: 15μsFns≥μiFn{\mu _s}{F_{n\;s}} \ge {\mu _i}{F_n}

Additionally, if it is considered that when two wheels n and m are on the same surface, they have the same μ_s and μ_i, it can be stated according to (15) that: 16μs(Fm+Fn)s≥μi(Fm+Fn){\mu _s}{\left({{F_m} + {F_n}} \right)_s} \ge {\mu _i}\left({{F_m} + {F_n}} \right)

Where (F_m + F_n)_s is the sum of the normal forces belonging to wheels m and n calculated for the limit condition before slipping, while (F_m + F_n) corresponds to the sum of these forces for an instantaneous condition with coefficient of friction μ_i. Based on this, a normalized slipping index I_s was defined by (17) whose magnitude is bounded between 1 and 0:1 for maximum grip with the ground and 0 for imminent slipping of the pair of wheels under study. 17Is=μs(Fm+Fn)s−μi(Fm+Fn)μs(Fm+Fn)s{I_s} = {{{\mu _s}{{\left({{F_m} + {F_n}} \right)}_s} - {\mu _i}\left({{F_m} + {F_n}} \right)} \over {{\mu _s}{{\left({{F_m} + {F_n}} \right)}_s}}}

Finally, if we consider that during obstacle overcoming the robot positions two wheels on each surface (Fig. 2b), an I_s can be defined for each surface (I_s1, I_s2), and a general index I_sg given by (18). This index allows evaluating the robot's propensity for slipping as a single system. 18Isg=min(Is1,Is2){I_{sg}} = \min \left({{I_{s1}},{I_{s2}}} \right)

Now, an analysis of the normal forces of wheels m and n was conducted in section 2.2.1, but the equations found include μ_i and δ_n, which are variables depending on the robot's instantaneous position at a given moment. These parameters are not directly obtained; therefore, in the following sections, an attempt is made to estimate these parameters.

As previously mentioned, the orientation angle of the friction force on a wheel δ_n with respect to the X_s axis primarily depends on the inclination of the surface where the wheel is positioned, the inclination of the robot, and other factors already described that make its exact determination difficult, considering the robot as a system that has multiple δ_n but simplified to two in our case: one for each pair of wheels touching the same surface. Therefore, to initiate this study, the following angles shown in Fig. 5a were determined, which subsequently allow estimating δ_n depending on the system's conditions. These angles are:

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  β_s : slope angle of surface s (maximum slope) relative to the X_s axis, defined in the X_sY_sZ_s frame. 19βs=tan−1(−sinαScosϕSsinϕS){\beta _s} = {\tan ^{ - 1}}\left({{{ - \sin {\alpha _S}\cos {\phi _S}} \over {\sin {\phi _S}}}} \right)

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  ε_s : slope angle opposite to the inclination of surface s (opposite to the maximum slope) relative to the X_s axis, defined in the X_sY_sZ_s frame. 20εs=tan−1[ sinαS(sin2ϕS+1)−cosϕSsinϕS ]{\varepsilon _s} = {\tan ^{ - 1}}\left[ {{{\sin {\alpha _S}\left({{{\sin }^2}{\phi _S} + 1} \right)} \over { - \cos {\phi _S}\sin {\phi _S}}}} \right]

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  β_c : director angle of the robot's inclination (maximum slope of the X_cY_c plane) with respect to the X_M axis, defined in the X_MY_MZ_M frame.

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  ε_c : angle opposite to the inclination of the robot (opposite to the maximum slope of the X_cY_c plane) with respect to the X_M axis, defined in the X_MY_MZ_M frame.

Although β_c and ε_c are defined in the X_M Y_M Z_M frame, their projection can be found in the X_s Y_s Z_s frame corresponding to the surface that the wheel touches (see Fig. 5b) through: 21sβc=tan−1(sϕssαssϕccαc−cαssαccϕssϕccαc)^s{\beta _c} = {\tan ^{ - 1}}\left({{{s{\phi _s}s{\alpha _s}s{\phi _c}c{\alpha _c} - c{\alpha _s}s{\alpha _c}} \over {c{\phi _s}s{\phi _c}c{\alpha _c}}}} \right) 22sεc=tan−1(sϕssαSsϕccαc+cαSsαc−cϕssϕccαc)^s{\varepsilon _c} = {\tan ^{ - 1}}\left({{{s{\phi _s}s{\alpha _S}s{\phi _c}c{\alpha _c} + c{\alpha _S}s{\alpha _c}} \over { - c{\phi _s}s{\phi _c}c{\alpha _c}}}} \right)

Once these angles were defined, a study was conducted based on multiple simulations carried out on a robot model in the MSC. ADAMS software that allowed studying δ_n and drawing some conclusions about it (associated with the angles defined previously). The results of these simulations are shown in Fig. 6, and from this, it can be stated that:

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  When both surfaces are inclined in the same direction (β₁ = β₂), the robot slides in the direction of the maximum slope of these surfaces (see Fig. 6b, d, f, h), therefore, the friction force takes the opposite direction, such that δ₁ = δ₂ = ε_s.

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  When one surface is horizontal, the friction of the wheels on that surface tends to take the direction of the friction of the other surface (see Fig. 6a, c, e, g), therefore δ₁ = δ₂.

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  When both surfaces have a similar inclination (|β₁ – β₂| ≤ 90°), the friction on the wheels on both surfaces tends to take the opposite direction to that shown by the maximum slope on each surface (see Fig. 6i, j, q, r). Therefore, it is assumed that δ₁ = ε₁ and δ₂ = ε₂.

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  When both surfaces have a non-analogous inclination (|β₁ – β₂| > 90°, see Fig. 6k - p), the direction of the friction force tends to mostly follow the opposite direction of the maximum slope of the X_c Y_c plane that defines the robot's inclination (δ₁ = ^s1 ε_c and δ₂ = ^s2 ε_c), in situations where slipping is possible:

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    On surfaces resembling a concavity (see Fig. 6m, o), this occurs if the surface facing the robot's natural sliding direction has a high inclination (HIS) while the other surface has a low inclination (LIS).

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    In the case of convex surfaces (see Fig. 6 k, n, p), this happens if the surface facing the robot’s natural sliding direction has a low inclination (LIS) while the other surface has a higher inclination (HIS).

Fig. 5.

Direct angles of the maximum slope on Surface s and on the X_C Y_C plane

Fig. 6.

Results of simulations to estimate δ_n

This rule is not absolute, as increasing the inclination of the surfaces, this configuration can change (see Fig. 6 l), and there can even be a wedge effect that limits the possibility of sliding.

Once the value of δ_n has been estimated in the previous section, this segment records a procedure for obtaining the instantaneous friction coefficient μ_i at the moment when the robot is overcoming the obstacle and has a pair of wheels positioned on each surface (see Fig. 7). This procedure is described separately for the case of overcoming the obstacle with the elevation of the front wheels (Fig. 7a) and overcoming with the side wheels (Fig. 7b).

Fig. 7.

Robot overcoming obstacle with two wheels on each surface

Overcoming obstacles with front wheels: In this case, during the obstacle overcoming, the two front wheels are on surface 2 while the two rear wheels are positioned on surface 1. The reaction forces of the 4 wheels represented in the X_c Y_c Z_c frame through their components F_nx, F_ny and F_nz (Fig. 7a) are obtained from (8)-(10). 23{ a=cϕccϕs2cδ2+cϕcsϕs2sαs2sδ2+sϕcsϕs2cδ2−sϕccϕs2sαs2sδ2b=cϕcsϕs2cαs2−sϕccϕs2cαs2c=sϕcsαccϕs2cδ2+sϕcsαcsϕs2sαs2sδ2+cαccαs2sδ2−cϕcsαcsϕs2cδ2+cϕcsαccϕs2sαs2sδ2d=sϕcsαcsϕs2cαs2−cαcsαs2+cϕcsαccϕs2cαs2e=sϕccαccϕs2cδ2+sϕccαcsϕs2sαs2sδ2−sαccαs2sδ2−cϕccαcsϕs2cδ2+cϕccαccϕs2sαs2sδ2f=sϕccαcsϕs2cαs2+sαcsαs2+cϕccαccϕs2cαs2\left\{ {\matrix{ {a = c{\phi _c}c{\phi _{s2}}c{\delta _2} + c{\phi _c}s{\phi _{s2}}s{\alpha _{s2}}s{\delta _2} + s{\phi _c}s{\phi _{s2}}c{\delta _2} - s{\phi _c}c{\phi _{s2}}s{\alpha _{s2}}s{\delta _2}} \hfill \cr {b = c{\phi _c}s{\phi _{s2}}c{\alpha _{s2}} - s{\phi _c}c{\phi _{s2}}c{\alpha _{s2}}} \hfill \cr {c = s{\phi _c}s{\alpha _c}c{\phi _{s2}}c{\delta _2} + s{\phi _c}s{\alpha _c}s{\phi _{s2}}s{\alpha _{s2}}s{\delta _2} + c{\alpha _c}c{\alpha _{s2}}s{\delta _2} - c{\phi _c}s{\alpha _c}s{\phi _{s2}}c{\delta _2} + c{\phi _c}s{\alpha _c}c{\phi _{s2}}s{\alpha _{s2}}s{\delta _2}} \hfill \cr {d = s{\phi _c}s{\alpha _c}s{\phi _{s2}}c{\alpha _{s2}} - c{\alpha _c}s{\alpha _{s2}} + c{\phi _c}s{\alpha _c}c{\phi _{s2}}c{\alpha _{s2}}} \hfill \cr {e = s{\phi _c}c{\alpha _c}c{\phi _{s2}}c{\delta _2} + s{\phi _c}c{\alpha _c}s{\phi _{s2}}s{\alpha _{s2}}s{\delta _2} - s{\alpha _c}c{\alpha _{s2}}s{\delta _2} - c{\phi _c}c{\alpha _c}s{\phi _{s2}}c{\delta _2} + c{\phi _c}c{\alpha _c}c{\phi _{s2}}s{\alpha _{s2}}s{\delta _2}} \hfill \cr {f = s{\phi _c}c{\alpha _c}s{\phi _{s2}}c{\alpha _{s2}} + s{\alpha _c}s{\alpha _{s2}} + c{\phi _c}c{\alpha _c}c{\phi _{s2}}c{\alpha _{s2}}} \hfill \cr } } \right.

These equations include both the normal force F_n and the friction force (represented through μ_i and δ_n), of which F_n for each wheel and μ_i for each pair of wheels positioned on a surface are unknown. Therefore, these equations can be rewritten for the front wheels as: 23F1x=F1(μ2a+b)F1y=F1(μ2c+d)F1z=F1(μ2e+f)\matrix{ {{F_{1x}}} \hfill & { = {F_1}\left({{\mu _2}a + b} \right)} \hfill & {{F_{1y}}} \hfill & { = {F_1}\left({{\mu _2}c + d} \right)} \hfill \cr {{F_{1z}}} \hfill & { = {F_1}\left({{\mu _2}e + f} \right)} \hfill & {} \hfill & {} \hfill \cr } 24F2x=F2(μ2a+b)F2y=F2(μ2c+d)F2z=F2(μ2e+f)\matrix{ {{F_{2x}}} \hfill & { = {F_2}\left({{\mu _2}a + b} \right)} \hfill & {{F_{2y}}} \hfill & { = {F_2}\left({{\mu _2}c + d} \right)} \hfill \cr {{F_{2z}}} \hfill & { = {F_2}\left({{\mu _2}e + f} \right)} \hfill & {} \hfill & {} \hfill \cr }

Where μ₂ is the instantaneous friction coefficient of wheels 1 and 2 positioned on surface 2, and a, b, c, d, e, and f are process variables defined by (23).

In the case of the rear wheels positioned on surface 1, the reaction forces are defined by: 26F3x=F3(μ1g+h)F3y=F3(μ1i+j)F3z=F3(μ1k+l)F4x=F4(μ1g+h)F4y=F4(μ1i+j)\matrix{ {\matrix{ {{F_{3x}}} \hfill & { = {F_3}\left({{\mu _1}g + h} \right)} \hfill & {{F_{3y}}} \hfill & { = {F_3}\left({{\mu _1}i + j} \right)} \hfill \cr {{F_{3z}}} \hfill & { = {F_3}\left({{\mu _1}k + l} \right)} \hfill & {} \hfill & {} \hfill \cr } } \hfill \cr {\matrix{ {{F_{4x}}} \hfill & { = {F_4}\left({{\mu _1}g + h} \right)} \hfill & {{F_{4y}}} \hfill & { = {F_4}\left({{\mu _1}i + j} \right)} \hfill \cr } } \hfill \cr } 27F4z=F4(μ1k+l){F_{4z}} = {F_4}\left({{\mu _1}k + l} \right)

Where μ₁ is the instantaneous friction coefficient of wheels 3 and 4 positioned on surface 1, and g, h, i, j, k, and l are process variables defined by (28).

28{ g=cϕccϕs1cδ1+cϕcsϕs1sαs1sδ1+sϕcsϕs1cδ1−sϕccϕs1sαs1sδ1h=cϕcsϕs1cαs1−sϕccϕs1cαs1i=sϕcsαccϕs1cδ1+sϕcsαcsϕs1sαs1sδ1+cαccαs1sδ1−cϕcsαcsϕs1cδ1+cϕcsαccϕs1sαs1sδ1j=sϕcsαcsϕs1cαs1−cαcsαs1+cϕcsαccϕs1cαs1k=sϕccαccϕs1cδ1+sϕccαcsϕs1sαs1sδ1−sαccαs1sδ1−cϕccαcsϕs1cδ1+cϕccαccϕs1sαs1sδ1l=sϕccαcsϕs1cαs1+sαcsαs1+cϕccαccϕs1cαs1\left\{ {\matrix{ {g = c{\phi _c}c{\phi _{s1}}c{\delta _1} + c{\phi _c}s{\phi _{s1}}s{\alpha _{s1}}s{\delta _1} + s{\phi _c}s{\phi _{s1}}c{\delta _1} - s{\phi _c}c{\phi _{s1}}s{\alpha _{s1}}s{\delta _1}} \hfill \cr {h = c{\phi _c}s{\phi _{s1}}c{\alpha _{s1}} - s{\phi _c}c{\phi _{s1}}c{\alpha _{s1}}} \hfill \cr {i = s{\phi _c}s{\alpha _c}c{\phi _{s1}}c{\delta _1} + s{\phi _c}s{\alpha _c}s{\phi _{s1}}s{\alpha _{s1}}s{\delta _1} + c{\alpha _c}c{\alpha _{s1}}s{\delta _1} - c{\phi _c}s{\alpha _c}s{\phi _{s1}}c{\delta _1} + c{\phi _c}s{\alpha _c}c{\phi _{s1}}s{\alpha _{s1}}s{\delta _1}} \hfill \cr {j = s{\phi _c}s{\alpha _c}s{\phi _{s1}}c{\alpha _{s1}} - c{\alpha _c}s{\alpha _{s1}} + c{\phi _c}s{\alpha _c}c{\phi _{s1}}c{\alpha _{s1}}} \hfill \cr {k = s{\phi _c}c{\alpha _c}c{\phi _{s1}}c{\delta _1} + s{\phi _c}c{\alpha _c}s{\phi _{s1}}s{\alpha _{s1}}s{\delta _1} - s{\alpha _c}c{\alpha _{s1}}s{\delta _1} - c{\phi _c}c{\alpha _c}s{\phi _{s1}}c{\delta _1} + c{\phi _c}c{\alpha _c}c{\phi _{s1}}s{\alpha _{s1}}s{\delta _1}} \hfill \cr {l = s{\phi _c}c{\alpha _c}s{\phi _{s1}}c{\alpha _{s1}} + s{\alpha _c}s{\alpha _{s1}} + c{\phi _c}c{\alpha _c}c{\phi _{s1}}c{\alpha _{s1}}} \hfill \cr } } \right.

Next, static equilibrium operations were carried out using the second law of Newton (summation of forces and moments) with respect to the X_c Y_c Z_c frame, i.e., ∑F_Xc = 0, ∑F_Yc = 0, ∑F_Zc = 0 and ∑M_3–4 = 0, to find the following system of equations with 4 unknowns [(F₁ + F₂), (F₃ + F₄), μ₁, μ₂]: 29{ (F1+F2)(μ2a+b)+(F3+F4)(μ1g+h)+m=0(F1+F2)(μ2c+d)+(F3+F4)(μ1i+j)+n=0(F1+F2)(μ2e+f)+(F3+F4)(μ1k+l)+o=0(F1+F2)(μ2e+f)p+q=0\left\{ {\matrix{ {\left({{F_1} + {F_2}} \right)\left({{\mu _2}a + b} \right) + \left({{F_3} + {F_4}} \right)\left({{\mu _1}g + h} \right) + m = 0} \hfill \cr {\left({{F_1} + {F_2}} \right)\left({{\mu _2}c + d} \right) + \left({{F_3} + {F_4}} \right)\left({{\mu _1}i + j} \right) + n = 0} \hfill \cr {\left({{F_1} + {F_2}} \right)\left({{\mu _2}e + f} \right) + \left({{F_3} + {F_4}} \right)\left({{\mu _1}k + l} \right) + o = 0} \hfill \cr {\left({{F_1} + {F_2}} \right)\left({{\mu _2}e + f} \right)p + q = 0} \hfill \cr } } \right.

Where m, n, o, p, and q are process variables defined by: 30{ m=Wx−F5xn=F5y−Wyo=F5z−Wzp=xt−Δxr+Δxfq=F5z(xr−Δxr+a1cθ1)−F5x(d2−d2c)−Wz(xr−Δxr+xcog)−Wxzcog\left\{ {\matrix{ {m = {W_x} - {F_{5x}}} \hfill \cr {n = {F_{5y}} - {W_y}} \hfill \cr {o = {F_{5z}} - {W_z}} \hfill \cr {p = {x_t} - \Delta {x_r} + \Delta {x_f}} \hfill \cr {q = {F_{5z}}\left({{x_r} - \Delta {x_r} + {a_1}c{\theta _1}} \right)} \hfill \cr { - {F_{5x}}\left({{d_2} - {d_{2c}}} \right) - {W_z}\left({{x_r} - \Delta {x_r} + {x_{cog}}} \right) - {W_x}{z_{cog}}} \hfill \cr } } \right.

Then, by algebraically manipulating the system given in (29), the following can be found: 31F1+F2=−qp(μ2e+f){F_1} + {F_2} = {{ - q} \over {p\left({{\mu _2}e + f} \right)}} 32F3+F4=q−opp(μ1k+l){F_3} + {F_4} = {{q - op} \over {p\left({{\mu _1}k + l} \right)}} 33rμ1+sμ1μ2+tμ2+v=0r{\mu _1} + s{\mu _1}{\mu _2} + t{\mu _2} + v = 0

Where r, s, t, and v are process variables such that: 34{ r=−qk(b+d)+f(q−op)(g+i)+pfk(m+n)s=−qk(a+c)+e(q−op)(g+i)+pek(m+n)t=−ql(a+c)+e(q−op)(h+j)+pel(m+n)v=−ql(b+d)+f(q−op)(h+j)+pfl(m+n)\left\{ {\matrix{ {r = - qk(b + d) + f(q - op)(g + i) + pfk(m + n)} \hfill \cr {s = - qk(a + c) + e(q - op)(g + i) + pek(m + n)} \hfill \cr {t = - ql(a + c) + e(q - op)(h + j) + pel(m + n)} \hfill \cr {v = - ql(b + d) + f(q - op)(h + j) + pfl(m + n)} \hfill \cr } } \right.

Finally, knowing that μ₁ and μ₂ are theoretically related through (14), we can solve for μ2 to find that: 35μ2=−(rμs1+tμs2)±(rμs1+tμs2)2−4svμs1μs22sμs1{\mu _2} = {{ - \left({r{\mu _{s1}} + t{\mu _{s2}}} \right) \pm \sqrt {{{\left({r{\mu _{s1}} + t{\mu _{s2}}} \right)}^2} - 4sv{\mu _{s1}}{\mu _{s2}}} } \over {2s{\mu _{s1}}}}

If μ₂ ≥ μ_s2, then μ₂ = μ_s2 so that slipping is imminent (I_s = 0). Finally, knowing μ₂, we can find μ₁ through (14), F₁ + F₂ from (31) and F₃ + F₄ from (32), which are necessary to determine the slipping index in each wheel pair given by (17). Additionally, ((F₁ + F₂)_s and (F₃ + F₄)_s can be obtained from (31) and (32) respectively by substituting μ₁ with μ_1s and μ₂ with μ_2s.

For obstacle traversal with lateral wheels: In this case, during the obstacle traversal, the two wheels on the same side are on surface 2 while the remaining two wheels are positioned on surface 1 (Fig. 7b). The process to obtain the corresponding instantaneous friction coefficients for each surface (μ₁ and μ₂) along with the normal reaction forces (F_n + F_m) of each wheel pair is similar to that described for obstacle traversal with front wheels using equations (23)–(35), except that now the variables F₁ and F₂ (with their components) are replaced by the numbers of wheels on surface 2, while F₃ and F₄ (with their components) are replaced by the numbers of wheels on surface 1. After this, the entire process remains the same, including the process variables, except for p and q, which now take the following values: 36{ p=−ytq=−F5z(yt2−a1sθ1)−F5y(d2−d2c)+Wz(yt2−ycog)+Wyzcog\left\{ {\matrix{ {p = - {y_t}} \hfill \cr {q = - {F_{5z}}\left({{{{y_t}} \over 2} - {a_1}s{\theta _1}} \right) - {F_{5y}}\left({{d_2} - {d_{2c}}} \right) + {W_z}\left({{{{y_t}} \over 2} - {y_{cog}}} \right) + {W_y}{z_{cog}}} \hfill \cr } } \right.

This case occurs during certain stages of obstacle traversal when the robot is only supported by two wheels plus the end effector (Fig. 8). In this scenario, only two wheels provide sufficient grip to prevent slipping, so it is necessary to analyze them to determine their slip index (I_s). If the robot is supported by two wheels, m and n, and the weight components (W_x, W_y, W_z) are known, the normal force F₅ on the end effector can be determined according to the following cases:

• –

  The front wheels are lifted (F₁ + F₂ = 0): F₅ is obtained through (37),

• –

  The rear wheels are lifted ((F₃ + F₄ = 0): F₅ is obtained through (38),

• –

  The left-side wheels are lifted (F₁ + F₄ = 0): F₅ is obtained through (39),

• –

  The right-side wheels are lifted (F₂ + F₃ = 0): F₅ is obtained through (40). 37F5=[ Wz(xr−Δxr+xcog)+Wxzcog ]/ [ (xr−Δxr+a1cθ1)(sϕccαcsϕscαs+sαcsαs+cϕccαccϕscαs)+( cϕcsϕscαs− sϕccϕscαs )(d2−d2c) ]{F_5} = \left[ {{W_z}\left({{x_r} - \Delta {x_r} + {x_{cog}}} \right) + {W_x}{z_{cog}}} \right]/\left[ {\left({{x_r} - \Delta {x_r} + {a_1}c{\theta _1}} \right)\left({s{\phi _c}c{\alpha _c}s{\phi _s}c{\alpha _s} + s{\alpha _c}s{\alpha _s} + c{\phi _c}c{\alpha _c}c{\phi _s}c{\alpha _s}} \right) + \left({c{\phi _c}s{\phi _s}c{\alpha _s} - } \right.} \right.\left. {\left. {s{\phi _c}c{\phi _s}c{\alpha _s}} \right)\left({{d_2} - {d_{2c}}} \right)} \right] 38F5=[ Wz(xf+Δxf−xcog)−Wxzcog ]/ [ (xf+Δxf−a1cθ1)( sϕccαcsϕsicαsi+sαcsαsi+ cϕccαccϕsicαsi )+(cϕcsϕsicαsi−sϕccϕsicαsi)(d2−d2c) ]{F_5} = \left[ {{W_z}\left({{x_f} + \Delta {x_f} - {x_{cog}}} \right) - {W_x}{z_{cog}}} \right]/\left[ {({x_f} + \Delta {x_f} - {a_1}c{\theta _1})\left({s{\phi _c}c{\alpha _c}s{\phi _{si}}c{\alpha _{si}} + s{\alpha _c}s{\alpha _{si}} + } \right.} \right.\left. {\left. {c{\phi _c}c{\alpha _c}c{\phi _{si}}c{\alpha _{si}}} \right) + \left({c{\phi _c}s{\phi _{si}}c{\alpha _{si}} - s{\phi _c}c{\phi _{si}}c{\alpha _{si}}} \right)\left({{d_2} - {d_{2c}}} \right)} \right] 39F5=[ Wz(yt2+ycog)−Wyzcog ]/ [ (yt2+a1sθ1)(sϕccαcsϕsicαsi+sαcsαsi+cϕccαccϕsicαsi)−( sϕcsαcsϕsicαsi− cαcsαsi+cϕcsαccϕsicαsi )(d2−d2c) ]{F_5} = \left[ {{W_z}\left({{{{y_t}} \over 2} + {y_{cog}}} \right) - {W_y}{z_{cog}}} \right]/\left[ {\left({{{{y_t}} \over 2} + {a_1}s{\theta _1}} \right)\left({s{\phi _c}c{\alpha _c}s{\phi _{si}}c{\alpha _{si}} + s{\alpha _c}s{\alpha _{si}} + c{\phi _c}c{\alpha _c}c{\phi _{si}}c{\alpha _{si}}} \right) - \left({s{\phi _c}s{\alpha _c}s{\phi _{si}}c{\alpha _{si}} - } \right.} \right.\left. {\left. {c{\alpha _c}s{\alpha _{si}} + c{\phi _c}s{\alpha _c}c{\phi _{si}}c{\alpha _{si}}} \right)\left({{d_2} - {d_{2c}}} \right)} \right] 40F5=[ Wz(yt2−ycog)+Wyzcog ]/ [ (yt2−a1sθ1)(sϕccαcsϕsicαsi+sαcsαsi+cϕccαccϕsicαsi)+( sϕcsαcsϕsicαsi− cαcsαsi+cϕcsαccϕsicαsi )(d2−d2c) ]{F_5} = \left[ {{W_z}\left({{{{y_t}} \over 2} - {y_{cog}}} \right) + {W_y}{z_{cog}}} \right]/\left[ {\left({{{{y_t}} \over 2} - {a_1}s{\theta _1}} \right)\left({s{\phi _c}c{\alpha _c}s{\phi _{si}}c{\alpha _{si}} + s{\alpha _c}s{\alpha _{si}} + c{\phi _c}c{\alpha _c}c{\phi _{si}}c{\alpha _{si}}} \right) + \left({s{\phi _c}s{\alpha _c}s{\phi _{si}}c{\alpha _{si}} - } \right.} \right.\left. {\left. {c{\alpha _c}s{\alpha _{si}} + c{\phi _c}s{\alpha _c}c{\phi _{si}}c{\alpha _{si}}} \right)\left({{d_2} - {d_{2c}}} \right)} \right]

Fig. 8.

Robot overcoming obstacle supported by 2 wheels and the end effector

After obtaining F₅, its components (F_5x, F_5y, F_5z) defined in the X_c Y_c Z_c frame (see Fig. 8) can be derived using equations (11)-(13). Subsequently, the principles of the second law of Newton (sum of forces equals zero) according to the X_c Y_c Z_c frame can be used to find the sum of reaction forces belonging to the two wheels touching the ground on the same surface: 41cF→mn=[ FmnxFmnyFmnz ]=[ Fmx+FnxFmy+FnyFmz+Fnz ]=[ −(Wx+F5x)Wy−F5yWz−F5z ]\matrix{ {^c{{\vec F}_{mn}}} \hfill & { = \left[ {\matrix{ {{F_{mnx}}} \hfill \cr {{F_{mny}}} \hfill \cr {{F_{mnz}}} \hfill \cr } } \right]} \hfill & { = \left[ {\matrix{ {{F_{mx}} + {F_{nx}}} \hfill \cr {{F_{my}} + {F_{ny}}} \hfill \cr {{F_{mz}} + {F_{nz}}} \hfill \cr } } \right]} \hfill & { = \left[ {\matrix{ { - \left({{W_x} + {F_{5x}}} \right)} \cr {{W_y} - {F_{5y}}} \cr {{W_z} - {F_{5z}}} \cr } } \right]} \hfill \cr }

Then, F_mn can be expressed in terms of the X_M Y_M Z_M frame through (42) and subsequently, with respect to the X_s Y_s Z_s frame corresponding to the surface where the two wheels are positioned, through (43): 42MF→mn=[ cϕcsϕcsαcsϕccαc0cαc−sαc−sϕccϕcsαccϕccαc ]ccF→mn\matrix{ {^M{{\vec F}_{mn}}} \hfill & = \hfill & {{{\left[ {\matrix{ {c{\phi _c}} \hfill & {s{\phi _c}s{\alpha _c}} \hfill & {s{\phi _c}c{\alpha _c}} \hfill \cr 0 \hfill & {c{\alpha _c}} \hfill & { - s{\alpha _c}} \hfill \cr { - s{\phi _c}} \hfill & {c{\phi _c}s{\alpha _c}} \hfill & {c{\phi _c}c{\alpha _c}} \hfill \cr } } \right]}^c}} \hfill & {^c{{\vec F}_{mn}}} \hfill \cr } 43sF→mn=[ sFmnxsFmnysFmnz ]=[ cϕs0−sϕssϕssαscαscϕssαssϕscαs−sαscϕscαs ]mF→mn\matrix{ {^s{{\vec F}_{mn}}} \hfill & { = \left[ {\matrix{ {^s{F_{mnx}}} \cr {^s{F_{mny}}} \cr {^s{F_{mnz}}} \cr } } \right]} \hfill & { = \left[ {\matrix{ {c{\phi _s}} & 0 & { - s{\phi _s}} \cr {s{\phi _s}s{\alpha _s}} & {c{\alpha _s}} & {c{\phi _s}s{\alpha _s}} \cr {s{\phi _s}c{\alpha _s}} & { - s{\alpha _s}} & {c{\phi _s}c{\alpha _s}} \cr } } \right]\;{\;^m}{{\vec F}_{mn}}\;} \hfill \cr }

Finally, considering the assumptions made in this work: both wheels on the same surface have a single μ_i and δ_n, we can consider that the combined normal force between the two wheels is ^sF_mnz, while the combined friction force is sFmnx2+sFmny2\sqrt {^sF_{mnx}^2{ + ^s}F_{mny}^2} . Therefore, μ_i can be determined using (44) and δ_n using (45). With this information known, the sliding index I_s can be determined through (17) and it is also considered that the general index I_sg = I_s since there is only sliding contact with one surface. 44μi=sFmnx2+sFmny2sFmnz{\mu _i} = {{\sqrt {^sF_{mnx}^2{ + ^s}F_{mny}^2} } \over {^s{F_{mnz}}}} 45δn=tan−1sFmnysFmnx{\delta _n} = {\tan ^{ - 1}}{{^s{F_{mny}}} \over {^s{F_{mnx}}}}

Although the effectiveness of I_s was verified using MSC. ADAMS while the robot moved on some test surfaces in the previous section, this section describes other tests where the robot is overcoming an obstacle and sliding completely, losing the ability to successfully complete the operation. The purpose of these tests is to verify if in these cases, the previously defined I_s is able to predict such a situation. For this, three simulations (A, B, and C) were performed where the robot executes a strategy to overcome an obstacle, but in this case, the static friction coefficient between the wheels and the surfaces (μ_s1 and μ_s2) was reduced to promote sliding during the simulation. Tab. 4 shows, for each simulation, the geometry of the obstacles and the coefficients μ_s1 and μ_s2.

The results for Simulation A are shown in Fig. 15: the robot begins the obstacle overcoming process and reaches Stage f where the front wheels are positioned on Surface 2, and the final elevation of the robot must start due to the action of the second link of the arm that is deployed. In the figure, it can be observed that, at the start of this stage, I_s undergoes a sharp decrease until reaching 0, at the same instant, there are oscillations in the longitudinal (V_x) and transversal (V_y) velocities of the robot since it effectively begins to slide at the time predicted by I_s and moves back, losing its position on the obstacle (as shown in the thumbnail image on Stage F in Fig. 15). The abrupt decrease in I_s is related to the μ_s2 of Surface 2, which is very low (μ_s2 = 0.2), so when the robot is supported only by the front wheels positioned on this surface, there is not enough grip (no support from the rear wheels), and it begins to slide, a situation predicted by I_s ≈ 0 at that moment.

Fig. 15.

Analysis of stages for overcoming obstacles until slipping (Simulation A)

Similar results are obtained when studying Simulation B (Fig. 16), where the robot begins the obstacle overcoming process until reaching Stage C where the right-side wheels must be lifted while the second arm link is deployed. At the start of this stage, it is observed that when F_5z > 0 due to the contact of the arm's end effector with the ground, I_s undergoes a sharp decrease, this time below 0.1, while simultaneously, visible variations occur in the longitudinal velocity (V_x) of the robot as it starts to slide forward. Once again, the abrupt decrease in I_s occurs when the robot is supported only by two wheels on Surface 1, which in this case has a μ_s2 = 0.4, which is not sufficient to provide the necessary grip to two wheels. It is worth noting that in this simulation, I_s did not reach the value of 0 during the moment of sliding, so it is important to avoid I_s < 0.2 as a safety condition, which was already explained in the previous section. Next, the results of Simulation C are not shown because they are very similar to those experienced by the robot in Simulation B (the robot also slides in Stage C).

Fig. 16.

Analysis of stages for overcoming obstacles until slipping (Simulation B)