Abstract:
An agricultural tractor having a rear suspension with draft compensating geometry to control the compression or extension of the rear suspension in response to a draft load. The draft compensating suspension enables the tractor hitch to be mounted on the suspended frame or chassis of the tractor rather than mounting the hitch to a non-suspended beam axle or locking out the suspension during draft work. The draft load is applied to the chassis by a ground engaging implement. The tractive load and torque at the rear wheels are reacted through the side view instantaneous center of the rear suspension. The placement of the instantaneous center will determine the suspension reaction to the wheel load and torque. Locating the suspension instantaneous center on a line of 100% draft-compensation will eliminate the motion of the suspension due to the draft load. If the instantaneous center is located below the 100% draft-compensation line, the suspension will compress under a draft load. If the instantaneous center is located above the 100% draft-compensation line, the suspension will extend under a draft load. The distance between the instantaneous center and the 100% draft compensation line will determine the amount of compression or extension of the suspension.

Description:
BACKGROUND OF THE INVENTION 
     1. Field of the Invention 
     The present invention relates to an agricultural tractor having a rear suspension and in particular a draft compensating rear suspension in which the suspension is designed to reduce or minimize the effect of draft loads on the tractor attitude by minimizing compression or extension of the rear suspension. 
     2. Description of the Related Art 
     An agricultural tractor is intended primarily for off-road usage and is designed primarily to supply power to agricultural implements. An agricultural tractor propels itself and provides a draft force in the direction of travel to enable an attached, soil engaging, implement to perform its intended function. Furthermore, an agricultural tractor may provide mechanical, hydraulic and/or electrical power to the implement. Agricultural tractors must be designed with sufficient normal force, i.e. a down force, acting on the drive wheels to produce the needed draft force. Agricultural tractors differ from cargo carrying vehicles, such as pickup trucks and semi-truck tractors, in that trucks are not designed to produce a continuous draft load. A truck only needs to produce a draft load during periods of acceleration and deceleration and relies on the weight of the cargo to produce the normal force on the drive wheels. 
     There is a continuous desire to increase the productivity of agricultural tractors. Productivity can be increased by tractor designs that maintain tire-soil contact force when traversing uneven terrain, both during field operations and during road transport. To maintain tire-soil contact force on an uneven terrain, it is necessary to provide a suspension system to allow the tires to follow the terrain. When a suspension is added to an agricultural tractor, between the chassis and the wheels, with the hitch on the suspended chassis, an undesirable interaction occurs between the suspension and draft load. The draft load reaction through the suspension tends to force the rear suspension into jounce, i.e. the suspension compresses, causing the wheels to move upward relative to the suspended chassis. As the suspension compresses, the pulled implement, such as a plow, runs deeper, increasing the draft load. The higher the draft load, the more the rear suspension compresses until it is fully compressed. Once fully compressed, the suspension no longer provides a benefit. If the suspension is only partially compressed, it will still reduce the amount of suspension travel available to maintain tire-soil contact force and to improve the tractor ride. Suspension compression also affects the clearance under the vehicle, the height of the drawbar and hitch above the ground plane, and the attitude of the tractor. 
     The interaction between the suspension and draft load can interfere with the proper operation of hitch controls. When the hitch control senses the need to reduce the depth of the implement in the ground, the suspension compresses, instead of raising the implement. The reverse is true if the hitch control senses the need to lower the implement. 
     One way to overcome the problem of suspension compression is to provide a system that compensates for the draft load by extending the suspension. This is accomplished by adding hydraulic fluid to the system to return the suspension to a center position, whereby suspension travel in each direction is still available. However, this significantly increases the spring rate of the suspension under a draft load, resulting in a harsher ride. Furthermore, adding and removing fluid to a circuit with an accumulator results in a large energy waste. 
     Another solution is to provide a rigid beam axle mounted to the chassis through the suspension and mount the hitch to the unsuspended beam axle. This avoids reacting of the draft load through the suspension. However, the implement will not receive the benefit of the suspension and will follow the motion of the axle. 
     Yet another alternative is to provide a means to lock out the suspension when performing draft work. This eliminates all benefits of the suspension. 
     All of the above solutions to the interaction of the suspension to the draft load reduce the effectiveness and benefit of the rear suspension. 
     SUMMARY OF THE INVENTION 
     The present invention relates to an agricultural tractor having a rear suspension with draft compensating geometry to control the compression of the rear suspension in response to a draft load. The draft load is applied to the chassis by a ground engaging implement. The tractive load and torque at the rear wheels are reacted through the side view instantaneous center of the rear suspension. The placement of the instantaneous center affects the chassis reaction to the wheel load and torque. Locating the suspension instantaneous center on a line of 100% draft-compensation will eliminate the motion of the suspension due to the draft load. If the instantaneous center is located below the 100% draft-compensation line, the suspension will compress under a draft load, whereas if the instantaneous center is located above the 100% draft-compensation line, the suspension will extend under a draft load. The distance between the instantaneous center and the 100% draft compensation line will determine the amount of compression or extension of the suspension. An expected response to the draft load is to compress the suspension. In a tractor without a suspension, the draft load typically compresses the rear tires. By properly locating the instantaneous center of the suspension, an opposing force is created to counteract, completely or partially, the suspension compression. 
     The draft compensating suspension geometry can be used to eliminate or reduce the amount of suspension travel used to react to the draft load. As a result, the suspension will have travel available to react to irregular terrain profiles, which provides the operator with better ride and control. It also permits the suspension to be less stiff to terrain inputs. A draft compensating suspension also reduces the interaction between the suspension and the hitch draft control system. A draft compensating geometry eliminates the need for complex control systems to control suspension height and vehicle attitude. 
    
    
     BRIEF DESCRIPTION OF THE DRAWINGS 
     FIG. 1 is a side view of the agricultural tractor of the present invention having a draft compensation rear suspension. 
     FIG. 2 is a rear view of the left side of the rear suspension. 
     FIG. 3 is a perspective view of the left side rear suspension without the outboard final drive housing. 
     FIG. 4 is a side view of the left side chassis illustrating the connection of the control arms to the tractor frame. 
     FIG. 5 is a side view free body diagram of the dynamic loads acting on the tractor. 
     FIG. 6 is a side view schematic representing the suspension system. 
     FIGS. 7 and 8 are side schematics of tractors illustrating two different degrees of draft compensation. 
    
    
     DETAILED DESCRIPTION OF THE INVENTION 
     An agricultural tractor  10  according to the present invention is shown in FIG.  1 . The tractor  10  includes a frame  12 , front wheels and tires  14 , rear wheels and tires  16  and a cab  18  forming an operator station. A front hood  22  covers an engine  24 . 
     The frame  12  includes a pair of channel members  28  between which a transmission  26  is housed. Only one of the channels is shown in FIG. 1. A rear drive differential case  30  is attached to the rear ends  32  of the steel channels  28  and thus forms a part of the tractor frame structure. The term “frame structure” is used broadly to mean both separate frame elements such as the channels  28  and other structural members mounted thereto such as the differential case and transmission, etc. Some tractors may be designed without separate frame members and form a frame structure entirely of drive train components such as the engine, transmission, front and rear drive differential cases, etc. The differential case  30  houses a rear differential drive that is driven by the transmission  26 . The rear differential has left and right outputs  34  to drive the left and right rear wheels  16 . 
     A three-point hitch  36  is mounted to the rear of the differential case  30 . The three-point hitch includes a pair of lower draft links  38  and an upper link  40 . Lift links  42  are attached to the draft links  38  and extend upwardly therefrom. The lift arms and the rock shaft of a conventional three point hitch are not shown. In addition to the three-point hitch, a drawbar  44  extends rearwardly from the differential case  30 . A PTO housing  46  with a PTO shaft  48  is also mounted to the differential case  30 . 
     The rear wheels  16  are mounted to outboard planetary final drives  50 , only the left side final drive  50  is shown in FIG.  2 . The left and right sides of the rear axle and suspension are substantially identical. The final drive  50  is attached to the frame  12  by upper and lower suspension control arms  52 ,  54 . The upper control arm  52  is generally V-shaped, having a front leg  56  and a rear leg  58 . The two legs are joined together at an outboard apex  60  that forms a ball joint  61  having a ball stud  62  that is coupled to the housing of the final drive  50  at the top of the housing. The inboard ends of the legs  56 ,  58  are joined to the frame  12  by ball joints  64  and  66  respectively. The ball joints  64 ,  66  include ball studs  68  and  70 , respectively, shown in FIG.  4 . 
     The lower control arm  54  is similar to the upper control arm  52 , having a front leg  72  and a rear leg  74  joined together at an outboard apex  76 . The apex  76  forms a ball joint  78  having a ball stud  79  for attachment to the final drive housing at the bottom of the housing. The legs  72 ,  74  of the lower control arm are joined to the frame  12  by ball joints  90 ,  92  having ball studs  94 ,  96 . The ball joints at the inboard and outboard ends of the control arms  52 ,  54  enable the outboard final drives  50  to move up and down relative to the frame  12 , as shown by the double arrow  98  in FIG.  2 . One or both of the ball joints at the inner end of each control arm can be replaced with pivot pins. 
     Vertical loads are transmitted between the differential case  30  and the outboard final drives  50  by front and rear hydraulic cylinders  100 , shown only schematically by a phantom line. The front hydraulic cylinder extends between front mounting brackets  102  on the frame  12  and a stud (not shown) on the front of the final drive housing. The rear hydraulic cylinder  100  extends between rear mounting brackets  104  on the differential case and a rear stud  106  on the final drive housing. The hydraulic cylinders  100  are part of a hydraulic circuit that includes a pressure accumulator in a known manner to provide the suspension system with a hydro-pneumatic spring system. Other spring devices can be used in place of the hydraulic cylinders such as metal springs, air bags, etc. 
     Since the final drive  50  is attached to the upper and lower control arms through ball joints, it is possible to rotate the final drive  50  about a generally upright axis passing through the ball joints  61  and  78 . To prevent this rotation, a fixed length link  108  extends between the differential case  30  and the outboard final drive housing  50 , rearward of the lower control arm  54 . 
     A drive shaft  110  shown schematically by a phantom line connects the left output  34  from the differential to the final drive  50 . 
     With reference to FIG. 4, the ball joints  64 ,  66  that attach the upper control arm  52  to the frame  12  define an upper control arm pivot axis  84 . Likewise, the lower ball joints  90 ,  92  define a lower control arm pivot axis  86 . The axes  84 ,  86  converge toward one another, forward of the rear wheels  16  at the side view instantaneous center of the rear suspension. The location of the side view instantaneous center is a critical factor in designing the suspension geometry so that it provides the desired degree of draft compensation as described below. 
     The various loads on a tractor are shown in reference to the free body diagram in FIG.  5 . The forces shown in FIG. 5 are as follows: 
     D A  Aerodynamic drag 
     R xf  Rolling resistance of the front tires 
     R xr  Rolling resistance of the rear tires 
     F xf  Longitudinal force at the front tires (traction) 
     F xr  Longitudinal force at the rear tires (traction) 
     W f  Load on the front wheel 
     W r  Load on the rear wheel 
     W Total Vehicle weight 
     a x  Longitudinal acceleration 
     R dbx  Longitudinal drawbar load 
     R dbz  Vertical drawbar load 
     θ Angle of slope 
     Assuming the tractor does not accelerate in pitch, the load on the front axle can be found by summing moments about the point O r , with clockwise moments being positive:                    W   f        L     +       D   A          h   a       +         Wa   x        h     g     +       R   dbx          h   db       +       R   dbz          d   db       +                Wh                   sin      θ       -     Wc                 cos                 θ       =   0           1   )                                
     Solving for the load on the front axle:                W   f     =       1   L          (       Wc                 cos                 θ     -       D   A          h   a       -         Wa   x        h     g     -       R   dbx          h   db       -     Wh                 sin                 θ       )               2   )                                
     The load on the rear axle can be found by summing moments about the point O f .                    D   A          h   a       +         Wa   x        h     g     +       R   dbx          h   db       +       R   dbz          (       d   db     +   L     )       -       W   r        L     +     Wh                 sin                 θ     +     Wb                 cos                 θ       =   0           3   )                                
     Solving for load on rear axle:                W   r     =       1   L          (       Wb                 cos                 θ     +       D   A          h   a       +       R   dbx          h   db       +       R   dbz          (       d   db     +   L     )       +     Wh                 sin                 θ     +         Wa   x        h     g       )               4   )                                
     For static loads on a level ground, θ, D A , R dbx , R dbz  and a x  all equal zero, thus:                W   f     =       W   fs     =     W        c   L                 5   )                 W   r     =       W   rs     =     W        b   L                 6   )                                
     For low speed accelerations on level ground θ and D A  equal zero then:                W   f     =       W        c   L       -       R   dbx            h   db     L       -       R   dbz            d   db     L       -         Wa   x     g          h   L                 7   )                 W   r     =       W                   b   L       +       R   dbx            h   db     L       +         R   dbz          (       d   db     +   L     )       L     +         Wa   x     g          h   L                 8   )                                
     Since any suspension is equivalent to a trailing arm, the pitch control performance can be quantified by analyzing the free body diagram of the suspension shown in FIG.  6 . In FIG. 6, points A f  and A r  are the virtual pivot points of the front and rear suspensions on the vehicle body. Since the arm is rigidly attached to the axle (resisting axle wind up), it has the ability to transmit a vertical force to the sprung mass which can be designed to counteract draft loads that compress the rear suspension. 
     The sum of the moments about A f  or A r  must be zero when the system is in equilibrium. Note that the rear load is characterized as a static component, W rs , plus a dynamic component, ΔW r , rising from longitudinal load transfer. For simplicity, axle weights are neglected. Counter-clockwise torque&#39;s are positive. 
     
       
         Σ M   Ar   =W   r   d   r   −W   rs   d   r   −ΔW   r   d   r   −F   xr ( e   r   −r   r )=0  9) 
       
     
     
       
         Σ M   Af   =−W   f   d   f   +W   fs   d   f   +ΔW   f   d   f   −F   xf ( e   f   −r   f )=0  10) 
       
     
     Where: 
     W fs  and W rs  are the static loads on the front and rear axles and suspensions respectively. 
     ΔW f  and ΔW r  are the dynamic changes in front and rear suspension loads respectively. 
     r f  and r r  are the rolling radii of the front and rear tires respectively. 
     Solving equation 9 for ΔW r :                Δ                   W   r       =         W   r     -     W   rs     -       F   xr          (         e   r     -     r   r         d   r       )         =       k   r          δ   r                 11   )                                
     Where: 
     k r  is the rear suspension spring rate. 
     δ r  is the rear suspension deflection (positive jounce). 
     Then substituting        W                   b   L                            
     for W rs  from equation 6 and          W   r     =       W        b   L       +       R   dbx            h   db     L       +       R   dbz            (       d   db     +   L     )     L       +         Wa   x     g          h   L                                              Δ                   W   r       =         R   dbx            h   db     L       +       R   dbz                       (       d   db     +   L     )     L       +         Wa   x     g          h   L       -       F   xr          (         e   r     -     r   r         d   r       )                 12   )                                 
     Since:                F   xr     =         (     1   -   ξ     )          (       R   dbx     +       Wa   x     g       )       =       (     1   -   ξ     )          F   x                 13   )                                
     Where: 
     ξ is the fraction of the total tractive force developed on the front tires. 
     F x  is the total tractive force developed by the tractor. 
     Therefore:                Δ                   W   r       =           R   dbx            h   db     L       +       R   dbz            (       d   db     +   L     )     L       +         Wa   x     L          h   L       -       (     1   -   ξ     )            F   x          (         e   r     -     r   r         d   r       )           =       k   r            δ   r     .                 14   )                                
     Solving equation 10 for ΔW f :                Δ                   W   f       =         W   f     -     W   fs     +       F   xf          (         e   f     -     r   f         d   f       )         =       k   f          δ   f                 15   )                                
     Where: 
     k f  equals the front suspension spring rate 
     δ f  equals the front suspension deflection (positive jounce). 
     Substituting          W   fs     =     W        c   L                              
     from equation 5 and          W   f     =         W   c     L     -       R   dbx            h   db     L       -       R   dbz            d   db     L       -         Wa   x     g          h   L                                
     from equation 7, results in:                Δ                   W   f       =         -     R   dbx              h   db     L       -       R   dbz            d   db     L       -         Wa   x     g          h   L       +       F   xf          (         e   f     -     r   f         d   f       )                 16   )                                
     Since:                F   xf     =       ξ                   F   x       =     ξ        (       R   dbx     +       Wa   x     g       )                 17   )                                
     Where ξ is the fraction of the total tractive force developed on the front tires. Therefore:                Δ                   W   f       =           -     R   dbx              h   db     L       -       R   dbz            d   db     L       -         Wa   x     g          h   L       +     ξ                     F   x          (         e   f     -     r   f         d   f       )           =       k   f          δ   f                 18   )                                
     The pitch angle of the tractor, θ p , is simply the sum of the suspension deflection divided by the wheel base.                θ   p     =         δ   r     -     δ   f       L             19   )                                
     Substituting into equation 19, equations 14 and 18 for δ r  and δ f  results in:                θ   p     =         1   L            R   dbx       k   r              h   db     L       +       1   L            R   dbz       k   r              (       d   db     +   L     )     L       +       1   L            Wa   x       gk   r            h   L       -       1   L          (     1   -   ξ     )            F   x       k   r            (         e   r     -     r   r         d   r       )       +       1   L            R   dbx       k   f              h   db     L       +       1   L            R   dbz       k   f              d   db     L       +       1   L            Wa   x       gk   f            h   L       -       1   L        ξ                     F   x       k   f            (         e   f     -     r   f         d   f       )                 20   )                                
     For field operations, quasi-steady state conditions exist, and a x =0. Then θ p  becomes:                θ   p     =         1   L            R   dbx       k   r              h   db     L       +       1   L            R   dbz       k   r              (       d   db     +   L     )     L       -       1   L          (     1   -   ξ     )            F   x       k   r            (         e   r     -     r   r         d   r       )       +       1   L            R   dbx       k   f              h   db     L       +       1   L            R   dbz       k   f              d   db     L       -       1   L        ξ          F   x       k   r            (         e   f     -     r   f         d   f       )                 21   )                                
     Since F x =R dbx :                θ   p     =         1   L            R   dbx       k   r              h   db     L       +       1   L            R   dbz       k   r              (       d   db     +   L     )     L       -       1   L          (     1   -   ξ     )            R   dbx       k   r            (         e   r     -     r   r         d   r       )       +       1   L            R   dbx       k   f              h   db     L       +       1   L            R   dbz       k   f              d   db     L       -       1   L        ξ          R   dbx       k   f            (         e   f     -     r   f         d   f       )                 22   )                                
     Since the suspension reacts the torque of the outboard planetary final drive  50  of the tractor  10 , r r =r f =0. θ p  becomes:                θ   p     =         1   L            R   dbx       k   r              h   db     L       +       1   L            R   dbz       k   r              (       d   dbz     +   L     )     L       -       1   L          (     1   -   ξ     )            R   dbx       k   r              e   r       d   r         +       1   L            R   dbx       k   f              h   db     L       +       1   L            R   dbz       k   f              d   db     L       -       1   L        ξ                     R   dbx       k   f              e   f       d   f                   23   )                                
     Since: 
     
       
           R   dbx   =R   dh  cos β 
       
     
     
       
           R   dbz   =R   db  sin β 
       
     
     Where: 
     R db  is the drawbar pull 
     β is the angle of the drawbar pull with respect to the vehicle horizontal axis. 
     For small angles of β, sin β0 and the cos β1.                θ   p     =         1   L            R   db       k   r              h   db     L       -       1   L          (     1   -   ξ     )            R   db       k   r              e   r       d   r         +       1   L            R   db       k   f              h   db     L       -       1   L        ξ                       R   db         k   f              e   f       d   f                   24   )                 θ   p     =       1   L            R   db          [         1     k   r              h   db     L       -       1   L          (     1   -   ξ     )          1     k   r              e   r       d   r         +       1     k   f              h   db     L       -     ξ                     1       k   f              e   f       d   f           ]                 25   )                                
     For the pitch angle to be zero, the term in brackets must equal zero.              0   =         h   db     L     -       (     1   -   ξ     )            e   r       d   r         +         k   r       k   f              h   db     L       -     ξ                     k   r       k   f              e   f       d   f                   26   )                   e   r       d   r       =       1     (     1   -   ξ     )            [         h   db     L     +         k   r       k   f              h   db     L       -     ξ          k   r       k   f              e   f       d   f           ]               27   )                                
     The equation 27 defines a line  80  shown in FIG. 7 that extends from the center of the rear tire patch upward and forward. If the side view instantaneous center of the rear suspension,  82 , is located on this line, the suspension geometry will produce a 100% compensation of the draft forces, resulting in no rear suspension compression or extension when the tractor produces a draft force. As a result, the line  80  is referred to as the “100% draft compensation line”. 
     The side view instantaneous center of the rear suspension is the point of intersection  82  of the pivot axis  84  of the upper control arm and the pivot axis  86  of the lower control arm. In FIG. 7, the instantaneous center  82  is located above the line  80 . This is the lift zone where the draft force will produce an extension of the rear suspension. In FIG. 8, the instantaneous center  82  is located below the 100% draft compensation line and is in the compression zone. A draft force will result in compression of the rear suspension. The further the instantaneous center  82  is from the 100% draft compensation line, the greater the effect of the draft load on the suspension system. 
     The amount of draft compensation can be quantified by:                        e   r       d   r          actual           e   r       d   r          100      %                 draft                 compensation       ×   100     =     %                 draft                 compensation             28   )                                
     If the location of the instantaneous center produces, for example, 60% draft compensation, then 60% of the draft load is compensated by an opposing force that counteracts compression or extension of the suspension. The remaining 40% of the draft force will result in suspension compression and can be partially or totally reacted by the suspension load leveling system, depending on the leveling system design. Some compression is desired as it is the expected response of a tractor to a draft load, resulting from tire compression in non-suspended tractors. Load leveling will increase the suspension spring rate, but not nearly as much as would be required if the load leveling system reacted to the entire draft force. With a draft compensating suspension geometry, a less complex load leveling system is needed and requires less power to operate. Load leveling suspension systems are generally known. 
     The line  80  defined by the equation 27 above provides draft compensation for the horizontal component of the draft force. If the angle β (the draft force angle to horizontal) is a value other than zero, there will be both horizontal and vertical components to the draft force. Only the horizontal component is compensated for by the suspension geometry. As noted above, the compensation may be full or partial, depending on the location of the suspension instantaneous center relative to the line of 100% compensation. The vertical component of the draft force will act to compress or extend the rear suspension, depending on the direction of the vertical draft component. The effect of the vertical component on the tractor attitude may be reduced, or eliminated, by a load leveling system. Since the suspension instantaneous center is fixed on the tractor, the percentage of the total draft [compensation] that is compensated by the suspension geometry varies as a function of the angle β. The exact amount of draft compensation will vary depending on the particular implement and angle β of the draft force produced by the implement. Accordingly, the amount of draft compensation is usually expressed as a range. 
     If the suspension did not have a load leveling system, the suspension geometry could be designed with the instantaneous center above the line of 100% draft compensation to produce a suspension extension to counter a vertical draft component to control the vehicle attitude. This can be expressed as a draft compensation value that is greater than 100%. 
     The equation 27 above results from equation 22 where r r  and r f =0 in the case of an outboard planetary final drive. For a tractor having an inboard final drive, equation 22 would resolve differently for the slope of the 100% draft compensation line since the dimensions r r  and r f  are the rolling radii of the tires. 
     The desired percentage of draft compensation produced by the suspension geometry will vary depending upon the tractor design, the total amount of tire motion available through the suspension system, the suspension spring rate and the load leveling system. Where greater suspension travel is available, a lower amount of draft compensation may be needed to still have sufficient suspension travel to traverse an uneven terrain. Where a relatively large amount of suspension travel is available, a suspension geometry providing 30% draft compensation may be adequate. The tractor  10  above has draft compensation between 40-60%. However, where the suspension travel is smaller, draft compensation between 60-80% may be required to insure adequate suspension travel remains for traversing a rough terrain. 
     The draft compensating suspension of the present invention on an agricultural tractor enables the implement hitch to be mounted on the suspended frame of the tractor and avoid the negative consequences of the draft load reacting through the suspension to either compress or extend the suspension. The suspension allows the implement to be suspended as well, providing greater control over the implement operation. 
     The invention should not be limited to the above-described embodiment, but should be limited solely by the claims that follow.