Abstract:
A method and apparatus for a transmission system selectively positioning sets of planet gear support bearings ( 50, 55, 80, 90 ) to achieve an optimized load distribution among a set of drive planet pinions ( 22 ) and a set of idler planet pinions ( 70 ) disposed in engagement between two reaction gears ( 30, 40 ) in the transmission system (A, A 1 ), for splitting an applied load between at least two pathways between an input and an output.

Description:
CROSS-REFERENCE TO RELATED APPLICATIONS 
       [0001]    The present application is related to, and claims priority from, U.S. Provisional Patent Application Ser. No. 61/323,648 filed on Apr. 13, 2010, and which is herein incorporated by reference. 
     
    
     STATEMENT REGARDING FEDERALLY SPONSORED RESEARCH 
       [0002]    Not Applicable. 
       BACKGROUND OF THE INVENTION 
       [0003]    The present invention relates, in general, to a load sharing mechanism for power transmissions and in particular to a load sharing mechanism for gear transmission with stepped gears. 
         [0004]    Rotary wing aircraft typically uses high-speed turbine engine to drive the rotor or propeller. A main gear transmission between the engine the rotor is necessary to transmit engine power while reducing the engine speed to the appropriate rotor speed. The main gear transmission is usually the heaviest subsystem in the drive train of the aircraft. Increasing power throughput and reducing the weight of the transmission is very desirable for modern rotary wing aircraft. 
         [0005]    One effective way to improve power density is to divide the input torque from the gas turbine engine into multiple paths. Each path uses a smaller individual gear member which leads to an overall transmission design that is lighter in weight, compact in size and has smaller gear face width due to the lower loads in each gear mesh. The smaller but numerous gears also require smaller bearings which have increased life span due to less applied torque. 
         [0006]    One embodiment of a power dense planetary gear transmission consists of a compound planetary gear-train having a set of stepped planet gears. Each stepped planet gear includes a large planet gear and a small drive planet pinion. The stepped planet gears may have flexible or pivot-able shaft. A set of small and simple idler planet pinions supported by a planet carrier are employed to share the torque, distributing load carried by the transmission among the drive planet pinions and the idler planet pinions. The idler planet pinions have non-floating shaft with respect to the planet carrier. 
         [0007]    Alternative transmission configurations such as split-torque face gear transmission may also be utilized, where a stepped gear is used to drive a primary face gears and two idler face gears that sandwich the primary face gear. The stepped gears have fixed shaft. Small idler gears were used as crossover gears to provide multiple power paths to share the load. Similarly, the idler gears also have fixed shafts. 
         [0008]    Load sharing mechanisms may be disposed within a power dense planetary gear transmission. These load sharing mechanisms may include a stepped gear cluster having a large gear and a small drive gear. The small drive gear meshes with two reaction gears, and the load sharing is achieved through a mechanical mechanism where a support structure of stepped gear cluster is devised utilizing a single pivoting support bearing which is selectively positioned between the large and small gears of the stepped gear cluster. The support structure and position of the pivoting support bearing allows the tangential forces of the small drive planet gear at mesh points with two reaction gears to be partitioned to achieve a desired load sharing ratio between the two reaction gears edge loading may occur if the gear teeth were not properly crowned. 
         [0009]    Accordingly, it would be advantageous to provide a flexible support structure in a gear transmission that allows for partitioning of a load between two reaction gears as desired, and for partitioning of a load between cluster support bearings to optimize the load distribution among drive planet pinions and idler planet pinions, and to maintain parallel gear engagement. In doing so, the maximum load capacity can be achieved. 
       BRIEF SUMMARY OF THE INVENTION 
       [0010]    Briefly stated, the present disclosure provides a power dense planetary gear transmission with a flexible support structure that allows a load to be split between two reaction gears as desired, and between a set of cluster support bearings to optimize the load distribution among drive planet pinions and idler planet pinions, maintaining proper parallel gear alignment for maximum load capacity. 
         [0011]    In an alternate embodiment, the present disclosure provides a face gear transmission with a flexible support structure that allows a load to be split between two reaction gears as desired, and between a set of cluster support bearings to optimize the load distribution among drive planet pinions and idler planet pinions, maintaining proper parallel gear alignment for maximum load capacity. 
         [0012]    As a method, the present disclosure provides a procedure for selectively positioning a set of support bearings to achieve an optimized load distribution among drive planet pinions and idler planet pinions in a transmission system incorporating a split gear assembly for splitting an applied load between two reaction gears or pathways. 
         [0013]    The foregoing features, and advantages set forth in the present disclosure as well as presently preferred embodiments will become more apparent from the reading of the following description in connection with the accompanying drawings. 
     
    
     
       BRIEF DESCRIPTION OF THE SEVERAL VIEWS OF THE DRAWINGS 
         [0014]    In the accompanying drawings which form part of the specification: 
           [0015]      FIG. 1  is a cut-away perspective view of power dense planetary gear transmission incorporating a flexible support structure of the present disclosure in the form of a set of cluster support bearings; 
           [0016]      FIG. 2  is a cut-away perspective view similar to  FIG. 1 , illustrating vectors of interacting forces within the power dense planetary gear transmission; 
           [0017]      FIG. 3  is a perspective load diagram for a stepped cluster gear in the power dense planetary gear transmission of  FIG. 1 ; 
           [0018]      FIG. 4  is a tangential force balance diagram illustrating bearing loads in the tangential direction for first and second support bearings of a stepped cluster gear; 
           [0019]      FIG. 5  is top plan view tangential force diagram fro the stepped cluster gear and idler gear of the power dense planetary gear transmission of  FIG. 1 ; 
           [0020]      FIG. 6  is a perspective load diagram for an idler gear in the power dense planetary gear transmission of  FIG. 1 ; 
           [0021]      FIG. 7  is a sectional view of a helicopter main gear box incorporating the load sharing mechanisms of the present disclosure; 
           [0022]      FIGS. 8A and 8B  are perspective and cut-away perspective views of a planet carrier structure for use in the gear box of  FIG. 7 ; 
           [0023]      FIG. 9  is a sectional view of a face gear transmission incorporating the load sharing mechanisms of the present disclosure on the stepped cluster gear and idler gears; 
           [0024]      FIG. 10  is a perspective view of an alternate embodiment planet carrier structure; and 
           [0025]      FIG. 11  is an enlarged view of a portion of  FIG. 10 . 
       
    
    
       [0026]    Corresponding reference numerals indicate corresponding parts throughout the several figures of the drawings. It is to be understood that the drawings are for illustrating the concepts set forth in the present disclosure and are not to scale. 
         [0027]    Before any embodiments of the invention are explained in detail, it is to be understood that the invention is not limited in its application to the details of construction and the arrangement of components set forth in the following description or illustrated in the drawings. 
       DETAILED DESCRIPTION 
       [0028]    The following detailed description illustrates the invention by way of example and not by way of limitation. The description enables one skilled in the art to make and use the present disclosure, and describes several embodiments, adaptations, variations, alternatives, and uses of the present disclosure, including what is presently believed to be the best mode of carrying out the present disclosure. 
         [0029]    Turning to the figures, and to  FIGS. 1 and 2  in particular, an embodiment of present disclosure is shown generally incorporated into a power dense gear transmission (A). The gear transmission (A) includes a drive gear  10 , a set of stepped cluster gears  20 , each having a large gear  21  and a small drive pinion gear  22 , a first reaction gear  30 , a second reaction gear  40 , and a set of cluster gear support bearings  50  and  55 . 
         [0030]    The drive gear  10 , driven via an associated bevel gear  1  and ring gear  5 , defines a first axis of rotation AR 1 , while each of the stepped cluster gears  20  defines an associated second axis of rotation AR 2 . The two axis of rotation AR 1  and AR 2  define a plane S. The two gears in each stepped cluster gear  20  are spaced apart axially along the axis AR 2  by a distance L, as seen in  FIG. 2 . The first support bearing  55  is disposed between the large gear  21  and the small drive pinion  22  at a position that is at an axial distance L 1  from the large gear  21  along the axis AR 2 . The second support bearing  50  is disposed between the large gear  21  and the small drive pinion  22  at an axial position which is at a distance L 2  along the axis AR 2  from the small drive pinion gear  22 . The large gear  21  in the cluster gear has a pitch diameter of D 0 , while the small drive pinion gear  22  has a pitch diameter of D 1 . At the first bearing position, the first bearing  55  along with an associated housing structure provides an effective tangential support stiffness KDt 1  and an effective radial support stiffness KDr 1 . At the second bearing position, the second bearing  50  along with an associated housing structure provides an effective tangential support stiffness KDt 2  and an effective radial support stiffness KDr 2 . 
         [0031]    The power dense gear transmission further includes a set of idler gears  70 . Each idler gear  70  is straddle mounted on a common support structure (planet carrier)  60  through a third bearing  80  and a fourth bearing  90 . The third bearing  80  is located axially at a distance L 3  from the center of the idler gear  70 . The fourth bearing  90  is located axially at a distance L 4  from the center of the idler bearing  70 . At the third bearing position, the third bearing  80 , along with an associated housing structure in the planet carrier, provides effective tangential support stiffness Klt 3 . At the fourth bearing position, the fourth bearing  90 , along with an associated housing structure in the planet carrier, provides effective tangential support stiffness Klt 4    
         [0032]    During operation of the transmission (A), the drive gear  10  meshes with the large gear  21  of each cluster gear  20 , exerting a meshing force that has a tangential component F 0  perpendicular to the plane S defined by the axis AR 1  and AR 2 . The first reaction gear  30  meshes with the small drive pinion gear  22  at the same side to axis AR 1  as does the drive gear  10 . The mesh between the reaction gear  30  and the small drive pinion gear  22  generates a meshing force that has a tangential component F perpendicular to the plane S. The second reaction gear  40  meshes with the small drive pinion gear  22  on the opposite side from the first reaction gear  30 . The mesh between the second reaction gear  40  and the small drive pinion gear generates a meshing force which has a tangential component F+ΔF, similarly perpendicular to the plane S. 
         [0033]    The meshing forces acting on the stepped cluster gear  20  are balanced by resulting forces in support bearings  50  and  55 , as shown in the load diagram of  FIG. 3 . Similarly,  FIG. 4  illustrates a tangential force balance diagram for the stepped cluster gear  20 , where bearing loads in a tangential direction are Ft t  for the first support bearing  55 , and Ft 2  for the second support bearing  50 . The effective bearing and housing support stiffness in tangential direction are KDt 1  for the first bearing position, and KDt 2  for the second bearing position. 
         [0034]    The first reaction gear  30  also meshes with the idler gears  70 , transferring the tangential meshing force F to the idler gears  70 . In doing so, the reaction gear  30  is rotationally balanced. Each idler gear  70 , in turn further meshes with the second reaction gear  40 , generating a matching tangential meshing force F to rotationally balance the tangential meshing force F received from the first reaction gear  30 . In doing so, the third bearing  80  receives a tangential load Ft 3 , and the fourth bearing  90  receives a tangential load of Ft 4 , best seen in  FIG. 5  which illustrates a tangential force diagram for a stepped gear  20  and an idler gear  70 , as well as in  FIG. 6  which is specific to an idler gear  70 . 
         [0035]    To partition the tangential meshing forces at the mesh with two reaction gears  30  and  40 , while maintaining the various gears in parallel engagement, bearing positions L 1  and L 2  supporting the stepped gear  20  are carefully selected. Assuming that the desired ratio of the tangential meshing force generated between the first reaction gear  30  and the small drive pinion gear  22 , relative to the tangential meshing force generated between the second reaction gear  40  and the small drive pinion gear  22  is: 
         [0000]    
       
         
           
             
               
                 
                   
                     LR 
                     T 
                   
                   ≡ 
                   
                     F 
                     
                       F 
                       + 
                       
                         Δ 
                          
                         
                             
                         
                          
                         F 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     1 
                   
                   ) 
                 
               
             
           
         
       
     
         [0036]    The bearing tangential force ratio of the second bearing  50  to the first bearing  55  is: 
         [0000]    
       
         
           
             
               
                 
                   
                     φ 
                     21 
                   
                   ≡ 
                   
                     
                       Ft 
                       2 
                     
                     
                       Ft 
                       1 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     2 
                   
                   ) 
                 
               
             
           
         
       
     
         [0037]    Correspondingly, for the idler gears, the bearing tangential force ratio of the fourth bearing  90  to the third bearing  80  is: 
         [0000]    
       
         
           
             
               
                 
                   
                     φ 
                     43 
                   
                   ≡ 
                   
                     
                       Ft 
                       4 
                     
                     
                       Ft 
                       3 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     3 
                   
                   ) 
                 
               
             
           
         
       
     
         [0038]    To achieve equilibrium or rotational balance, the following relationships have to be met 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         
                           L 
                           1 
                         
                         + 
                         
                           
                             φ 
                             21 
                           
                            
                           
                             ( 
                             
                               L 
                               - 
                               
                                 L 
                                 2 
                               
                             
                             ) 
                           
                         
                       
                       
                         L 
                         - 
                         
                           L 
                           1 
                         
                         + 
                         
                           
                             φ 
                             21 
                           
                            
                           
                             L 
                             2 
                           
                         
                       
                     
                     = 
                     
                       
                         ( 
                         
                           
                             1 
                             - 
                             
                               LR 
                               T 
                             
                           
                           
                             1 
                             + 
                             
                               LR 
                               T 
                             
                           
                         
                         ) 
                       
                        
                       
                         
                           D 
                           0 
                         
                         
                           D 
                           1 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   and 
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     4 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       L 
                       3 
                     
                     
                       L 
                       4 
                     
                   
                   = 
                   
                     φ 
                     43 
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     5 
                   
                   ) 
                 
               
             
           
         
       
     
         [0039]    where: 
         [0040]    D 0  is the radius of the mesh circle to the large gear  21 ; 
         [0041]    D 1  is the radius of the mesh circle of the small drive pinion gear  22 ; 
         [0042]    L 1  is the center distance from large gear  21  to the first support bearing  55 ; 
         [0043]    L 2  is the center distance from the small drive pinion gear  22  to the second support bearing  50 ; 
         [0044]    L 3  is the center distance from the idler gear  70  to the third bearing  80 ; and 
         [0045]    L 4  is the center distance from the idler gear  70  to the fourth bearing  90  (see  FIG. 9 ). 
         [0046]    Under tangential load, the center of the first bearing  55  experiences tangential displacement by amount equal to: 
         [0000]    
       
         
           
             
               
                 
                   
                     SDt 
                     1 
                   
                   = 
                   
                     
                       Ft 
                       1 
                     
                     
                       KDt 
                       1 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     6 
                   
                   ) 
                 
               
             
           
         
       
     
         [0047]    Similarly, the center of the second bearing  50  experiences a tangential displacement of an amount equal to: 
         [0000]    
       
         
           
             
               
                 
                   
                     SDt 
                     2 
                   
                   = 
                   
                     
                       Ft 
                       2 
                     
                     
                       KDt 
                       2 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     7 
                   
                   ) 
                 
               
             
           
         
       
     
         [0048]    To keep the cluster gear  20  properly aligned, minimizing both gear mesh misalignment, and edge loading, it is desirable to have SDt 1 =SDt 2 . This leads to the following relationship: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       KDt 
                       2 
                     
                     
                       KDt 
                       1 
                     
                   
                   = 
                   
                     
                       
                         Ft 
                         2 
                       
                       
                         Ft 
                         1 
                       
                     
                     = 
                     
                       φ 
                       21 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     8 
                   
                   ) 
                 
               
             
           
         
       
     
         [0049]    It is further desirable to have both Ft 1  and Ft 2  in the same direction to reduce bearing load. This is to say: 
         [0000]      φ 21 ≧0  (Eqn. 9)
 
         [0050]    Under a tangential load Ft 3 , the center of the third bearing  70  experiences tangential displacement of the amount: 
         [0000]    
       
         
           
             
               
                 
                   
                     SIt 
                     3 
                   
                   = 
                   
                     
                       Ft 
                       3 
                     
                     
                       KIt 
                       3 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
               
             
           
         
       
     
         [0051]    Under tangential load Ft 4 , the center of the forth bearing  80 , experiences tangential displacement of the amount: 
         [0000]    
       
         
           
             
               
                 
                   
                     SIt 
                     4 
                   
                   = 
                   
                     
                       Ft 
                       4 
                     
                     
                       KIt 
                       4 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     11 
                   
                   ) 
                 
               
             
           
         
       
     
         [0052]    To keep the idler gear  70  properly aligned, minimizing both gear mesh misalignment and edge loading, it is desirable to have Slt 3 =Slt 4 . This leads to: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       KIt 
                       4 
                     
                     
                       KIt 
                       3 
                     
                   
                   = 
                   
                     
                       
                         Ft 
                         4 
                       
                       
                         Ft 
                         3 
                       
                     
                     = 
                     
                       φ 
                       43 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     12 
                   
                   ) 
                 
               
             
           
         
       
     
         [0053]    To maintain integrity of gearing system SDt 1 =SDt 2 =Slt 3 =Slt 4  (Eqn. 13) must hold true, thus: 
         [0000]    
       
         
           
             
               
                 
                   
                     KR 
                     ≡ 
                     
                       
                         KIt 
                         3 
                       
                       
                         KDt 
                         1 
                       
                     
                   
                   = 
                   
                     
                       
                         2 
                          
                         
                           ( 
                           
                             1 
                             + 
                             
                               φ 
                               21 
                             
                           
                           ) 
                         
                       
                       
                         1 
                         + 
                         
                           φ 
                           43 
                         
                       
                     
                     · 
                     
                       
                         
                           LR 
                           T 
                         
                         · 
                         DR 
                       
                       
                         
                           1 
                           + 
                           
                             LR 
                             T 
                           
                           + 
                           
                             
                               ( 
                               
                                 1 
                                 - 
                                 
                                   LR 
                                   T 
                                 
                               
                               ) 
                             
                             · 
                             DR 
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     14 
                   
                   ) 
                 
               
             
           
         
       
     
         [0054]    or: 
         [0000]    
       
         
           
             
               
                 
                   
                     LR 
                     T 
                   
                   = 
                   
                     
                       KR 
                       · 
                       
                         ( 
                         
                           1 
                           + 
                           
                             φ 
                             43 
                           
                         
                         ) 
                       
                       · 
                       
                         ( 
                         
                           DR 
                           + 
                           1 
                         
                         ) 
                       
                     
                     
                       
                         2 
                          
                         
                           
                             ( 
                             
                               1 
                               + 
                               
                                 φ 
                                 21 
                               
                             
                             ) 
                           
                           · 
                           DR 
                         
                       
                       + 
                       
                         KR 
                         · 
                         
                           ( 
                           
                             1 
                             + 
                             
                               φ 
                               43 
                             
                           
                           ) 
                         
                         · 
                         
                           ( 
                           
                             DR 
                             - 
                             1 
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     15 
                   
                   ) 
                 
               
             
           
         
       
     
         [0055]    where: 
         [0000]    
       
         
           
             
               
                 
                   DR 
                   ≡ 
                   
                     
                       D 
                       0 
                     
                     
                       D 
                       
                         1 
                          
                         
                             
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     16 
                   
                   ) 
                 
               
             
           
         
       
     
         [0056]    The following procedure implements the above relationships into the design process for a gear transmission (A) that incorporates a load sharing mechanism as presented in the current disclosure: 
         [0057]    Step 1—Determine a load partitioning ratio LR T  of the two reaction gears  30  and  40  using Eqn. (1), and determine bearing load partitioning ratios using Eqn. (2) and Eqn. (3); determine a cluster gear pitch diameter ratio using Eqn. (16). 
         [0058]    Step 2—Select the bearings and their initial positions (L 1 , L 2 , L 3 , and L 4 ) using Eqn. (4) and Eqn. (5); design and engineer bearings and supporting structures such as the planet carrier to achieve predetermined support stiffness (force) ratios; calculate and check actual support stiffness (force) ratios using Eqn. (8), Eqn. (12) and Eqn. (14). Iteratively repeat the design and engineering process, if necessary until actual support stiffness (force) ratios for the bearings which are within an acceptable tolerance of the predetermined support stiffness ratios are achieved. 
         [0059]    Step 3—Calculate and check the actual gear load partitioning ratios using 
         [0060]    Eqn. (15), iteratively, and adjust using the previous step or step(s), if necessary, until a gear loading partitioning ratio which is within an acceptable tolerance of a predetermined gear load partitioning ratio is achieved. 
         [0061]    Step 4—Chose and adjust support bearing positions (L 1 , L 2 , L 3 , and L 4 ) using Eqn. (4) and Eqn. (5) together with the actual support stiffness ratios and gear load partitioning ratio(s). 
         [0062]    The above process may be used alone, or along with other procedures, such as specifically configured planet carrier structures, to yield desirable solutions and design configurations having the required support stiffness. 
         [0063]    The load sharing concepts of the present disclosure may be utilized, for example, in a helicopter main gear box as seen at (A 1 ) in  FIG. 7 . The gear system (A 1 ) is a compound planetary gear train, coupled to a source of driving power (not shown) via a drive shaft  100  which is engaged with a ring gear  105  associated with drive gear  110 . The drive gear  110  in turn is engaged with, and drives one or more planet cluster gears  120 . Each planet cluster gear  120  consists of a large planet gear  121  and a drive planet gear  122  coaxially coupled thereto. Drive planet gears  122  each function as small drive pinions disposed between first and second reaction gears. The first reaction gear takes the form of an idler sun gear  130 , and the second reaction gear taking the form of a fixed ring gear  140 . A set of idler planet gears  170  are used in addition to the drive planet gears  122  to improve the load carrying capacity of the main gear box compound planetary gear train (A 1 ). Within the compound planetary gear train (A 1 ), a planet carrier  160  is utilized to support the various planet gears. Each planet cluster gear  120  is supported on the planet carrier  160  by a pair of bearings  150  and  155  mounted in associated housings having a degree of flexibility in the tangential direction. The first support bearing  155 , together with an associated housing structure carried by the planet carrier  160  is configured to provide a relatively soft support in the tangential direction and a rigid support in the radial direction. Likewise, the second support bearing  150  provides relatively soft support in the tangential direction and a rigid support in the radial direction. 
         [0064]    For each of the idler planet gears, the third support bearing  180 , and the fourth support bearing  190 , with their respective housing structures in the planet carrier  160 , provide rigid supports in the tangential direction. This allows the drive planet gears  122  to float more easily than the idler planet gears  170  in the annular space between idler sun gear  130  and fixed ring gear  140 , facilitating a transfer of a portion of the applied loads through a second power path. That is to say, the third and fourth bearing positions provide stiffer support for the idler planet gears than the first and second bearing positions provide for the drive planet gears. 
         [0065]    As can be appreciated, the gear teeth of each drive planet gear  122  are subjected to uni-directional bending, while the gear teeth of the idler planet gears  170  are subjected to bi-directional bending. The maximum tangential force for the drive planet  122  is F+ΔF and the maximum tangential force for idler planet gear  170  is F. To maintain equal safety margin against gear tooth bending failure, it is highly desirable to have the idler planet gears  170  transmitting less tangential force than the drive planet gears  122 . That is to say: 
         [0000]    
       
         
           
             
               
                 
                   
                     LR 
                     T 
                   
                   = 
                   
                     
                       F 
                       
                         F 
                         + 
                         
                           Δ 
                            
                           
                               
                           
                            
                           F 
                         
                       
                     
                     ≤ 
                     1 
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     17 
                   
                   ) 
                 
               
             
           
         
       
     
         [0066]    In practical application, it is recommended to have a load partitioning ratio LR T  between 0.5 and 1.0. The endurance limiting stress for a reverse bending gear tooth is roughly 70% of the endurance limiting stress for a unidirectional bending gear tooth. Thus, the load partitioning ratio LR T =0.7 is suggested. For practical considerations, it is desirable to adopt an equal tangential load partition between the first bearing  155  and the second bearing  150 , and between the third bearing  180  and the fourth bearing  190 . That is φ 21 =1 and φ 43 =1. 
         [0067]    Consequently, the stiffness relationships at the first, second, third and the fourth bearing positions are determined as: 
         [0000]    
       
         
           
             
               
                 
                   
                     KDt 
                     1 
                   
                   = 
                   
                     KDt 
                     2 
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     18 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     KIt 
                     3 
                   
                   = 
                   
                     KIt 
                     4 
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     19 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       KIt 
                       3 
                     
                     
                       KDt 
                       1 
                     
                   
                   = 
                   
                     
                       1.4 
                        
                       DR 
                     
                     
                       1.7 
                       + 
                       
                         0.3 
                          
                         DR 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     20 
                   
                   ) 
                 
               
             
           
         
       
     
         [0068]    The locations for the first and second bearings and for the third and fourth bearings along the respective axis about which they rotate are then determined as: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         L 
                         + 
                         
                           L 
                           1 
                         
                         - 
                         
                           L 
                           2 
                         
                       
                       
                         L 
                         - 
                         
                           L 
                           1 
                         
                         + 
                         
                           L 
                           2 
                         
                       
                     
                     = 
                     
                       0.1765 
                        
                       DR 
                     
                   
                    
                   
                     
 
                   
                    
                   and 
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     21 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     L 
                     3 
                   
                   = 
                   
                     
                       L 
                       4 
                     
                     . 
                   
                 
               
               
                 
                   ( 
                   
                     Eqn 
                     . 
                     
                         
                     
                      
                     22 
                   
                   ) 
                 
               
             
           
         
       
     
         [0069]    Turning to the  FIGS. 8A and 8B  illustrate an exemplary planet carrier for obtaining the desired stiffness relationships. The carrier structure includes a base hub  162  for supporting the drive planet gears  122  at equidistantly spaced housings, and a upper plate  164  which, in conjunction with the base hub  162 , supports the idler planet gears  170 , at equidistantly spaced and axially aligned reinforced housings. The base hub includes a tapered portion  168  and a splined boss  169 . The upper plate  164  is supported in the axial direction by a set of support posts  166  disposed between the upper plate  164  and the base hub  162 , about the circumference of the base hub  162 . Those of ordinary skill in the art will recognize that are other carrier configurations that, in conjunction with suitable supporting bearings, will provide adequate support stiffness. 
         [0070]    For example, an alternate embodiment of the planet carrier structure for obtaining the desired support stiffness relationship is shown at  400  in  FIGS. 10 and 11 . The planet carrier  400  has a base hub  420  and an upper plate  410 . The upper plate  410  has extrusion posts  460  for connecting with, and supporting relative to, the carrier base hub  420 . Stiffened housings  470  and  471  are formed in the upper plate and base hub of the planet carrier for hosting the idle planet support bearings ( 180  and  190 ). Flexible housings  480  and  481  are formed from the base hub  420  for hosting the stepped planet gear support bearings ( 150  and  155 ). The flexible housings are constructed within a common hub  440  which is supported by offset horizontal walls  494  and  495  and vertical drop down walls  491 ,  492 , and  493 , as best seen in  FIG. 11 . Multiple relief channels ( 431 ,  432 , and  433 ) are disposed in the structures of the planet carrier  400  to further provide flexibility for the stepped planet gear support in a circumferential direction, and for improved stress distribution. The drop down wall  491  (or  492 ), the hub body  440 , and the channel  431  form a double “U” structure with the openings arranged in opposite directions. This helps to maintain the alignment as the axis of stepped gear moves circumferentially. The planet carrier structure thus created is capable of providing the required gear support stiffness ratios as defined by (Eqn. 8), (Eqn. 12) and (Eqn. 14) while maintaining the supported gears in proper alignment. 
         [0071]    Those of ordinary skill in the art will recognize that the load sharing mechanisms of the present disclosure are not limited to use in the planetary gear systems shown in  FIGS. 1 and 7 , but may be adapted for use in other types of gear transmissions, such as a split-torque face gear transmission as shown in  FIG. 9 . In a split-torque face gear transmission application, the first reaction gear takes the form of an idler face gear  330 ; the second reaction gear is a primary face gear  340  coupled to an output shaft  365 . A drive cluster gear  320  defines the stepped gear, including a driven large gear  321  and a small drive gear  322 . The split-torque face gear transmission further includes a set of idler pinions  370 . The drive cluster gear  320  is supported by a pair of bearings, including a first bearing  355  and a second bearing  350 , relative to a housing or planet carrier of the gear transmission. Each of the idler pinions  370  is straddle-mounted to the housing by a pair of bearings, including a third bearing  380  and a fourth bearing  390 . 
         [0072]    The large gear  321  of the drive cluster gear  320  meshes with, and is driven by an input drive gear  310 , which in turn is coupled to an input shaft  305  and driving engine (not shown). The small drive gear  322  of the drive cluster gear  320  is sandwiched between, and meshes with, both the idler face gear  330 , supported on bearings  395  relative to the housing, and the primary face gear  340 . Similarly, the idler pinion  370  is sandwiched between, and meshed with, both the idler face gear  330  and the primary face gear  340 . The rotational axis of the input gear AR 1 , the rotational axis of the drive cluster gear  320  AR 2 , and the rotational axis of the idler and primary face gears AR 3  each lie in a common plane S. 
         [0073]    During operation, input power is transmitted from the input shaft  305  to the output shaft  365  through the split-torque face gear transmission. The power is split at the small drive gear  322 , with portion being delivered directly to the primary face gear  340 , and portion being delivered to the idler face gear  330 . The idler pinion  370  then acts as a crossover gear, passing the power back from the idler face gear  330  to the primary gear  340 . In doing so, the driving power is re-combined at the primary face gear  340  to drive the output shaft  365 . 
         [0074]    The amount of power split between the idler face gear  330  and the primary face gear  340  is determined, among other factors, by the positions of the bearings ( 350 ,  355 ,  380 , and  390 ) along with the associated tangential support stiffness at the bearing positions. The relationships set forth in above Equations (1)-(22) are applicable to this embodiment, and may be utilized to selectively position the bearings to achieve the desired power split. Those of ordinary skill in the art will recognize that when applied to a split-torque face gear transmission, any radial dimensions referred to previously in Equations (1)-(22) should interpreted as axial directions. 
         [0075]    Other variations and applications of the current disclosure are possible without deviating from the sprit of the disclosure. The embodiments and application disclosed herein should be considered as ways of explaining and implementing, not as ways of limiting the scope of the current disclosure. As various changes could be made in the above constructions without departing from the scope of the disclosure, it is intended that all matter contained in the above description or shown in the accompanying drawings shall be interpreted as illustrative and not in a limiting sense.