Abstract:
A belt transmission comprising a housing, a first belt trained between a first shaft and a first intermediate shaft, a second belt trained between the first intermediate shaft and a second intermediate shaft, a third belt trained between the second intermediate shaft and a second shaft, a first tensioner and second tensioner each engaged with the housing and each engaged about the first intermediate shaft whereby each tensioner exerts a force upon the first intermediate shaft which thereby imparts a tension to the first belt and to the second belt, and a third tensioner and fourth tensioner each engaged with the housing and each engaged about the second intermediate shaft whereby each tensioner exerts a force upon the second intermediate shaft which thereby imparts a tension to the second belt and to the third belt.

Description:
FIELD OF THE INVENTION 
       [0001]    The invention relates to a belt transmission, and more particularly, to a belt transmission comprising a first tensioner and second tensioner engaged about an intermediate shaft, the first tensioner and second tensioner each comprising a first arm and second arm and a torsion spring engaged therebetween, the first arm and second arm each bearing upon a surface, the first arm and second arm rotatable by operation of the torsion spring to exert a force upon the intermediate shaft whereby a tension is imparted to a first belt and a second belt engaged with the intermediate shaft. 
       BACKGROUND OF THE INVENTION 
       [0002]    Electric power assist steering systems (EPAS) have been around since the 1960&#39;s. Hydraulic power assist steering has traditionally dominated the market. Hydraulic systems have high parasitic energy loss when the hydraulic pump is pumping, but power assist is not required. Early attempts to eliminate this parasitic loss involved fitting an electric motor to the pump and only driving the pump when necessary. 
         [0003]    Electric hydraulic assisted power steering systems use an electric motor to drive a hydraulic pump to feed a hydraulic power steering system. These systems are an intermediate step by the industry and their use will likely fade with the increased use of EPAS. EPAS systems allow realization of reduced noise, reduced energy use, active safety features, and adjustability to meet driving conditions. However, the use of these systems has remained limited until recent C.A.F.E. requirements became more difficult to meet. This is driving automotive manufactures to turn to EPAS systems more and more in an effort to improve vehicle fuel economy. EPAS systems eliminate the parasitic losses typically found in hydraulic assist power steering systems. 
         [0004]    For example, one difficulty that slowed implementation of EPAS systems was meeting the power requirement with a 12 volt electric motor. Recently systems have been developed that successfully solve this problem. Further, all EPAS systems require a control module to sense driver input and control the electric motor to provide the desired assist. The control module measures driver input torque and uses this to determine the amount of assist required. Assist can be tuned to meet the drivers need depending on driving conditions. The system can even have a tunable “feel” available to the driver. 
         [0005]    Even though the main driver for automotive EPAS is fuel economy improvement, EPAS has additional benefits. The system can make steering assist available even when the vehicle&#39;s engine is not running. It also enables the use of the automatic parallel parking systems available today. 
         [0006]    There are two main types of EPAS systems; column assist and rack assist. Rack assist EPAS systems have an electric motor that is connected to the steering rack. The electric motor assists the rack movement usually through driving a lead screw mechanism. Column assist EPAS systems have an electric motor connected to the steering column. The electric motor assists the movement of the column shaft usually through a worm gear type arrangement. One advantage of these types of systems is the electric motor can be placed in the passenger compartment freeing up valuable space under the hood. This also keeps any sensitive electrical components out of the harsh under hood environment. 
         [0007]    Worm drive column assist systems are usually used in small cars where the assist power requirements are lower than what would be needed in a large heavy vehicle. These systems are limited by the speed of the steering wheel and the ratio of the worm drive. The steering wheel at its fastest speed rotates relatively slowly at approximately 60 rpm. With a 60 rpm speed of the steering wheel and a worm drive ratio of 15:1, the max speed of the electric motor would only be 900 rpm. Worm drives are limited to ratios under 20:1 because ratios higher than that cannot be back-driven. 
         [0008]    The steering system must be able to be operated with no power. This requires the worm drive be able to operate with the gear driving the worm (back-driven). Having a low motor speed and limited ratio worm drive causes the need for high torque motor. Even with a high torque motor, these types of systems have not been made successful on heavy vehicles. Small vehicles are light and require less steering effort thus enabling the use of these systems. Worm drive column assist SPAS systems are the lowest cost systems and thus also lend themselves to smaller less expensive vehicles. 
         [0009]    Typical steering systems with worm drive assists are limited in their efficiency. EPAS systems must be designed to operate when there is no power available. Due to the nature of worm drive&#39;s tendency to lock up during back driving when ratios exceed approximately 20:1, worm drive EPAS systems efficiency is not greater than approximately 85% and nearer to 65% during back-driving conditions. 
         [0010]    Representative of the art is U.S. Pat. No. 8,327,972 which discloses a vehicle steering system transmission comprising a housing, an input shaft journalled to the housing, an electric motor connected to the housing and coupled to the input shaft, an output shaft journalled to the housing, the input shaft and the output shaft coupled by a first pair of sprockets having a first belt trained therebetween and having a first ratio, the first belt and first pair of sprockets comprising a helical tooth configuration, the input shaft and the output shaft coupled by a second pair of sprockets having a second belt trained therebetween and having a second ratio, and the input shaft and the output shaft coupled by a third pair of sprockets having a third belt trained therebetween and having a third ratio. 
         [0011]    What is needed is a belt transmission comprising a first tensioner and second tensioner engaged about an intermediate shaft, the first tensioner and second tensioner each comprising a first arm and second arm and a torsion spring engaged therebetween, the first arm and second arm each bearing upon a surface, the first arm and second arm rotatable by operation of the torsion spring to exert a force upon the intermediate shaft whereby a tension is imparted to a first belt and a second belt engaged with the intermediate shaft. The present invention meets this need. 
       SUMMARY OF THE INVENTION 
       [0012]    The primary aspect of the invention is to provide a belt transmission comprising a first tensioner and second tensioner engaged about an intermediate shaft, the first tensioner and second tensioner each comprising a first arm and second arm and a torsion spring engaged therebetween, the first arm and second arm each bearing upon a surface, the first arm and second arm rotatable by operation of the torsion spring to exert a force upon the intermediate shaft whereby a tension is imparted to a first belt and a second belt engaged with the intermediate shaft. 
         [0013]    Other aspects of the invention will be pointed out or made obvious by the following description of the invention and the accompanying drawings. 
         [0014]    The invention comprises a belt transmission comprising a housing, a first belt trained between a first shaft and a first intermediate shaft, a second belt trained between the first intermediate shaft and a second intermediate shaft, a third belt trained between the second intermediate shaft and a second shaft, a first tensioner and second tensioner each engaged with the housing and each engaged about the first intermediate shaft whereby each tensioner exerts a force upon the first intermediate shaft which thereby imparts a tension to the first belt and to the second belt, and a third tensioner and fourth tensioner each engaged with the housing and each engaged about the second intermediate shaft whereby each tensioner exerts a force upon the second intermediate shaft which thereby imparts a tension to the second belt and to the third belt. 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         [0015]    The accompanying drawings, which are incorporated in and form a part of the specification, illustrate preferred embodiments of the present invention, and together with a description, serve to explain the principles of the invention. 
           [0016]      FIG. 1  is an exploded view of the device. 
           [0017]      FIG. 2  is a front perspective view of the device. 
           [0018]      FIG. 3  is a back perspective view of the device. 
           [0019]      FIG. 4  is a view of the tensioner assembly. 
           [0020]      FIG. 5  is an exploded view of a tensioner assembly. 
           [0021]      FIG. 6  is a detail of the tensioner assembly in the device. 
           [0022]      FIG. 7  is a detail of the input pulley and belt. 
           [0023]      FIG. 8  is a detail of the compound pulley sprocket. 
           [0024]      FIG. 9  is a diagram of the forces on the shaft of the first compound pulley/sprocket. 
           [0025]      FIG. 10  is a diagram of the position of the forces along the input shaft. 
           [0026]      FIG. 11  is a diagram of the angular positions of the forces on the input shaft. 
           [0027]      FIG. 12  is a detail of the tensioner. 
           [0028]      FIG. 13  is a detail of the tensioner. 
           [0029]      FIG. 14  is a detail of the force components in the tensioner arms. 
           [0030]      FIG. 15  is a perspective view of housing  9 . 
           [0031]      FIG. 16  is a perspective view of housing  10 . 
           [0032]      FIG. 17  is a perspective view of the assembled housing parts. 
           [0033]      FIG. 18  is a plan view of the assembled housing parts. 
           [0034]      FIG. 19  is a detail of a tensioner arm. 
       
    
    
     DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT 
       [0035]      FIG. 1  is an exploded view of the device. The inventive device comprises a three stage belt drive transmission. Drive stage one comprises a multi-ribbed belt  100  with a drive ratio of 2.4:1. Stage two comprises a toothed or synchronous belt  101  with a ratio of 3.8:1. Stage  3  comprises a toothed or synchrounous belt  102  with a ratio of 3.5:1. The overall drive ratio of the transmission is 31.9:1. Of course, a desired drive ratio can be selected by altering the diameter of the pulleys and sprockets as described herein. 
         [0036]    The inventive device comprises input shaft  2 , input pulley  1 , multi-ribbed belt  100 , compound pulley/sprocket  3 , a first intermediate shaft  4 , automatic tensioner assemblies  5 ,  6 ,  7 , and  8 , a compound sprocket  11 , a second intermediate shaft  12 , a 3 mm pitch toothed or synchronous belt  101 , a 5 mm toothed or synchrounous belt  102 , an synchronous sprocket  13 , housing portion  9 , housing portion  10 , a plurality of bearings ( 50 ,  51 ,  52 ,  53 ,  54 ,  55 ,  56 ), a motor mount  14 , and a plurality of fasteners  15 . A synchronous or toothed belt comprises teeth which extend across a width of the belt. 
         [0037]    Input shaft  2  is mounted on bearings  50 ,  51 . Input pulley  1  is press fit to input shaft  2 . Bearings  51  and  50  are mounted in each housing  9  and housing  10  respectively, thereby supporting input shaft  2 . 
         [0038]    Compound pulley/sprocket  3  is mounted on first intermediate shaft  4 . First intermediate shaft  4  is mounted on bearings  52 ,  53 . In turn, bearings  53 ,  52  are each mounted within an automatic tensioner  5  and automatic tensioner  6  respectively. Automatic tensioner  5  and automatic tensioner  6  along with bearings  53 ,  52  are each in contact with housing  9  and housing  10 , respectively. Compound pulley/sprocket  3  comprises pulley  32  for engaging belt  100  and sprocket  31  for engaging belt  101 . 
         [0039]    Second intermediate shaft  12  is mounted on a pair of bearings  54 ,  55 . Compound sprocket  11  is mounted to intermediate shaft  12 . Bearings  54 ,  55  are each mounted in automatic tensioner  7  and automatic tensioner  8  respectively. Automatic tensioner  7  and  8  with bearings  54 ,  55  respectively are each in contact with housing  9  and housing  10 . Compound sprocket  11  comprises sprocket  110  for engaging belt  101  and sprocket  111  for engaging belt  102 . 
         [0040]    Output sprocket  13  is mounted on a bearing  56 . Bearing  56  is mounted in housing  10 . Housing portion  9  and housing portion  10  are bolted together using fasteners  15 . Motor mount  14  is bolted to housing  10 . A motor or other driver (not shown) can be mounted to motor mount  14 . Sprocket  13  engages belt  102 . 
         [0041]    Multi-ribbed belt  100  transmits power from input pulley  1  to pulley  32 . A multi-ribbed belt comprises ribs that extend in the endless of longitudinal direction of the belt. Belt  101  transmits power from sprocket  31  to sprocket  110 . Belt  102  transmits power from sprocket  111  to output sprocket  13 . 
         [0042]    Output sprocket hub  130  is configured to enable connection to a vehicle steering shaft (not shown). Input shaft  2  is configured to allow connection to an electric motor or other power source (not shown). Housing  10  further comprises a bracket  82 , see  FIG. 18 , for mounting the inventive device to a vehicle (not shown). 
         [0043]    Known tensioners typically comprise a rigidly mounted base and a moveable arm assembly with an idler pulley journalled to the moveable arm. The idler pulley is forceably engaged with a belt by a torsion spring which tensions a belt. Each automatic tensioner  5 ,  6 ,  7 , and  8  differs from the prior art wherein the prior art tensioner base is replaced by an arm which acts as a second tensioner arm in the inventive device, see  FIG. 5 . 
         [0044]    Automatic tensioner  5  and  6  act cooperatively to position shaft  4  thereby tensioning belt  100  and belt  101 . Automatic tensioner  7  and  8  act cooperatively to position shaft  12  thereby tensioning belt  101  and belt  102 . 
         [0045]    Automatic tensioner  5  and  7  act upon housing  9 . Automatic tensioner  6  and  8  act upon housing  10 , which in turn the combination creates a reaction force upon the movable intermediate shaft  4 . The reaction force exterted on the moveable intermediate shaft  4  positions the shaft to a position of equilibrium based upon the tension in belt  100  and belt  101 . Shaft  4  and pulley  3  move into a position where the belt tension is equal to the combined force pf tensioners  5  and  6 . The same operating principle is realized by tensioners  7  and  8  acting on intermediate shaft  12  and thereby pulley  11 . In this Figure tensioner  5  and tensioner  7  are shown in exploded view. Tensioner  6  and tensioner  8  are not shown in exploded view. Tensioners  5 ,  6 ,  7 ,  8  are of the same design and description. 
         [0046]      FIG. 2  is a front perspective view of the device. Housing  9  is omitted from this drawing. 
         [0047]      FIG. 3  is a back perspective view of the device. Shaft  2  engages an electric motor or other suitable driver (not shown). Member  82  mounts the device to a suitable mounting surface (not shown). Bearing  52  supports tensioner  6 . Bearing  55  supports tensioner  8 . 
         [0048]      FIG. 4  is a view of the tensioner assembly. The inventive automatic tensioner comprises an arm  500 , a bushing  502 , a torsion spring  504 , and an arm  501 . Arm  500  is rotatably connected to arm  501  with bushing  502  providing a low friction surface to facilitate movement. One end  509  of the torsion spring  504  rests against a face  510  on arm  500 . The opposite end  507  of the torsion spring  504  rests against a face  508  on arm  501 . Spring  504  is loaded in the unwinding direction. Arm  500  comprises tangs  503  which hold the tensioner assembly together. Arm  501  comprises tangs  511  which hold the tensioner assembly together. Arm  501  comprises arcuate tensioner surface  506 . Surface  506  contacts a bracket surface  92  on housing  9 , see  FIG. 15 . Arm  500  comprises arcuate tensioner surface  505 . Surface  505  contacts a bracket surface  91  on housing  9 , see  FIG. 15 . This description is also applicable to automatic tensioners  6 ,  7  and  8 . 
         [0049]      FIG. 5  is an exploded view of a tensioner assembly. Tensioner  5  receives a bearing  52  which in turn engages shaft  4 . Tensioner arms  500 ,  501  are cam like in configuration. The cam like arms rotate around the center of the bearing  52 , namely, the rotation center, see  FIG. 19 . Arms  500 ,  501  are each configured similarly, that is, a circle within a circle having offset centers and different radii, see  FIG. 19 . 
         [0050]    Arms  500 ,  501  comprise surfaces  505 ,  506  respectively which rest on bracket surface  91  and bracket surface  92 , see  FIG. 15 . Torsion spring  504  provides a moment to each arm in opposing directions. End  507  bears against tab  508 . End  509  bears against tab  510 . A spring force forcibly rotates the arm surfaces  505 ,  506  against surfaces  91 ,  92  of the housing  91 ,  92 . Since the arms are cam like in operation this causes the rotation center of bearing  52  and thus shaft  4  and pulley  3  to move in a direction which properly tensions belts  100  and  101 . The movement stops when the belt tension is equal to the force of tensioners  5  and  6 . This description is also applicable to operation of automatic tensioners  6 ,  7  and  8  as well. 
         [0051]      FIG. 6  is a detail of the tensioner assembly in the device. The tensioners operate in pairs, namely, tensioners  5  and  6  act cooperatively to support shaft  4 . Tensioners  7  and  8  act cooperatively to support shaft  12 . Each pair of tensioners forcibly position shaft  4  and shaft  12  which provides the force necessary to properly tension the belts. 
         [0052]    For the two belts ( 101 ,  102 ) engaged with each compound pulley/sprocket ( 3 ,  11 ) the tensioning force is preferably oriented such that the proper force in the proper direction is applied to create the desired tension in each belt. 
         [0053]    Proper belt tension depends on the diameter of the pulley and the desired torque in the system. For example, a torque input to input pulley  1  is 1.88 Nm and the pulley diameter is 30 mm. This yields a force of 125.3 N (or ΔT=125.3 N) applied to belt  100  by pulley  1 . This is the difference in tension in belt  100  due to torque regardless of the installed tension in the belt. 
         [0000]      Torque=Force×distance
 
         [0000]      Torque=1.88 Nm 
         [0000]      Distance=Diameter/2=0.030 m/2=0.015m 
         [0000]      Force=Torque/distance 
         [0000]      Force=1.88 Nm/0.015m 
         [0000]      Force=125.3 N 
         [0054]      FIG. 7  is a detail of the input pulley and belt. The difference between the tight side tension and the slack side tension of belt  100  is 125.3 N. The slack side tension cannot drop below a certain value without the drive slipping. This value is determined with the calculation of the minimum tension as follows: 
         [0000]    
       
         
           
             
               
                 T 
                  
                 
                     
                 
                  
                 2 
               
               
                 T 
                  
                 
                     
                 
                  
                 1 
               
             
             = 
             
                
               μθ 
             
           
         
       
     
         [0055]    Where T 2 =tight side tension
       T 1 =slack side tension   μ=friction=1   Θ=wrap angle on pulley=139.7 degrees       
 
         [0059]    Solving for T 2 : 
         [0000]        T 2 =T 1 e   μθ   
         [0060]    Additionally the torque is equal to the radius of the pulley times the difference between the tight side tension and the slack side tension: 
         [0000]      Torque= r*ΔT=r ( T 2 −T 1) 
         [0000]    Substituting for T 2  and solving for T 1 : 
         [0000]    
       
         
           
             
               T 
                
               
                   
               
                
               1 
             
             = 
             
               Torque 
               
                 R 
                  
                 
                   ( 
                   
                     
                        
                       μθ 
                     
                     - 
                     1 
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               T 
                
               
                   
               
                
               1 
             
             = 
             
               1.88 
               
                 0.015 
                  
                 
                   ( 
                   
                     
                        
                       
                         1 
                         * 
                         2.72 
                       
                     
                     - 
                     1 
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               T 
                
               
                   
               
                
               1 
             
             = 
             
               12.0 
                
               
                   
               
                
               N 
             
           
         
       
     
         [0061]    Since ΔT=125 N we get T 2 =137.3 N 
         [0062]    The value calculated above for T 1  is the minimum value so a factor of safety is added to the system, for example, this value is doubled to 24 N which gives a tight side tension of 149.3 N for belt  100 . When there is no torque in the drive, the tight side and slack side tensions equalize to become the installed tension. The magnitude of that is one half the total tension: 
         [0000]    
       
         
           
             
               
                 
                   
                     Installed 
                      
                     
                         
                     
                      
                     tension 
                   
                   = 
                     
                    
                   
                     T 
                      
                     
                         
                     
                      
                     1 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     T 
                      
                     
                         
                     
                      
                     2 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       1 
                       2 
                     
                      
                     
                         
                     
                      
                     total 
                      
                     
                         
                     
                      
                     tension 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       1 
                       2 
                     
                      
                     
                       ( 
                       
                         149.3 
                         + 
                         24 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     86.6 
                      
                     
                         
                     
                      
                     N 
                   
                 
               
             
           
         
       
     
         [0063]    The hubload is then the resultant of the sum of these tension forces applied at the angle of the belt. To determine the angle of the belt we need to know the wrap angle of the belt around the pulley. Simple geometry yields the following formula for wrap angle: 
         [0000]      WA=π−2 sin( R 2 −R 1/center distance)
 
         [0064]    Where:
       R 2 =radius of opposing pulley=36 mm   R 1 =radius of subject pulley=15 mm   Center distance=the distance between the centers of the pulleys=61 mm       
 
         [0068]    This results in a wrap angle of 139.7 degrees. 
         [0069]    The angle of the tension force is: 
         [0000]      Tension force angle=TFA=(180−WA)/2
 
         [0000]      TFA=(180−139.7)/2
 
         [0070]    The belt tension forces are at angles of +/−20.15 degrees from the line formed between pulley centers. 
         [0071]    The hubload (HL) is then: 
         [0000]      Hubload=2(installed tension*cos(TFA)) 
         [0000]      HL=2*(86.6*cos(20.15))=162.6 N 
         [0072]    The hubload is applied along a line formed through the centers of each pully pair at the mid width of the belt. 
         [0073]    The force on the output pulley is equal and opposite the force on the input pulley. 
         [0074]    When the pulley is a compound pulley or sprocket, see  FIG. 8 , the hubload must be determined for both belts and applied in the appropriate direction and location along the shaft.  FIG. 8  is a detail of the compound pulley sprocket  3 . Since the forces on each shaft cancel, it is possible to calculate the forces necessary from each tensioner to balance the hubloads on the shaft. 
         [0075]      FIG. 9  is a diagram of the forces acting on the shaft  4  of the first compound pulley/sprocket  3 .  FIG. 10  is a diagram of the position of the forces along the input shaft  4 .  FIG. 11  is a diagram of the angular positions of the forces on the input shaft  4 . 
         [0076]    FH 1  is the force of hubload from belt  100 . 
         [0077]    FH 2  is the force of hubload from belt  102 . 
         [0078]    FT 1  is the force from tensioner  1 . 
         [0079]    FT 2  is the force from tensioner  2 . 
         [0080]    In order to determine the forces required in each tensioner, the calculation is simplified by separating the calculations into the forces from each belt and then adding them together. The forces are resolved into an x component and a y component. The x axis is normal to a line formed between the centers of the pulleys of the input drive (z-axis). Considering the x direction from FH 1  we get: 
         [0081]    Given:
       FH 1 =157.2 N   FH 2 =600 N   β=85 deg   FH 1  is in the positive X direction   z 1 =33.5 mm   Z 2 =48.0 mm   Z 3 =13.5 mm       
 
         [0089]    Summing the forces in the X direction (see  FIGS. 7 ,  8 , and  9 ): 
         [0000]      0=FH1−FT2 x −FT1 x  
 
         [0090]    Where: 
         [0091]    FT 2   x  is the force from tensioner  2  in the x direction. 
         [0092]    FT 1   x  is the force from tensioner  1  in the x direction. 
         [0093]    Summing the moments about point A ( FIG. 10 ): 
         [0000]      0=−FH1* z 1+FT1 x*z 2
 
         [0000]      FT1 x =( z 1/ z 2)*FH1 
         [0000]      Then: 
         [0000]      FT1 x= 109.7 N 
         [0000]      Substituting: 
         [0000]      FT2 x =FH1−FH1 x  
 
         [0000]      FT2 x= 47.5 N 
         [0094]    Repeating the calculations for the x direction from FH 2 : 
         [0000]      FH2 cos β=FT2 x ′+FT1 x′ 
 
         [0000]      FT2 x ′=FH2 cos β−FT1 x′ 
 
         [0095]    Summing moments about A: 
         [0000]      0=−FH2 cos β* z 3+FT1 x′*z 2
 
         [0000]      FT1 x ′=FH2 cos β*( z 3/ z 2)
 
         [0000]      FT1 x′= 14.7 N 
         [0000]      Substituting: 
         [0000]      FT2 x ′=FH2 cos β−FT1 x′ 
 
         [0000]      FT2 x′= 37.6 N 
         [0096]    Adding the respective forces in the x direction for the tensioners gives: 
         [0000]      FT1 x ″=FT1 x +FT1 x′ 
 
         [0000]      FT1 x″= 109.7 N+14.7 N=124.4 N 
         [0000]      And 
         [0000]      FT2 x ″=FT2 x +FT2 x′ 
 
         [0000]      FT2 x″= 47.5 N+37.6 N=85.1 N 
         [0097]    Repeating these calculations for forces in the Y direction yields: 
         [0000]      FT1 y″= 168.1 N 
         [0000]      FT2 y″= 583.0 N 
         [0098]    Geometry informs the magnetude of FT 1  and FT 2  by: 
         [0000]      FT1=√{square root over (FT1 x″   2 +FT1 y″   2 )}
 
         [0000]      FT2=√{square root over (FT2 x″   2 +FT2 y″   2 )}
 
         [0000]      FT1=209.1 N 
         [0000]      FT2=589.2 N 
         [0099]    From this, simple geometry gives us the angles of these forces: 
         [0000]      Θ= a  sin(FT2 y ″/FT2)=81.7 deg
 
         [0000]      α= a  sin(FT1 y ″/FT1)=53.5 deg
 
         [0100]    Similar determinations of tensioner force can be made for each tensioner position and then each tensioner can be configured to create the required force. Table 1 below is a summary of the required tensioner forces. The values in Table 1 are provided only as examples and are not intended to limit the scope of the invention. 
         [0000]    
       
         
               
               
               
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
               
             
           
               
                   
                 TABLE 1 
               
               
                   
                   
               
               
                   
                   
                   
                 Belt 
                   
                 angle 
                   
                 Tens 
                   
                 Tens. 
               
               
                   
                 Input 
                 Output 
                 inst. 
                 Hub 
                 between 
                 Tens. 
                 force 1 
                 Tens. 
                 force 2 
               
               
                   
                 Torque 
                 Torque 
                 Tension 
                 load 
                 stages 
                 force 1 
                 angle 
                 force 2 
                 angle 
               
               
                   
                 (Nm) 
                 (Nm) 
                 (N) 
                 (N) 
                 (deg) 
                 (N) 
                 (deg) 
                 (N) 
                 (deg) 
               
               
                   
                   
               
             
             
               
                   
               
             
          
           
               
                 Interm. 
                 Stage 1 
                 1.88 
                 4.51 
                 86.64 
                 162.68 
                 85.00 
                 137.19 
                 19.40 
                 635.18 
                 81.88 
               
               
                 Shaft 4 
                 Stage 2 
                 4.51 
                 17.14 
                 368.59 
                 647.14 
               
               
                 Interm. 
                 Stage 2 
                 4.51 
                 17.14 
                 368.59 
                 647.14 
                 88.00 
                 967.18 
                 80.70 
                 1633.23 
                 71.68 
               
               
                 Shaft 
                 Stage 3 
                 17.14 
                 62.13 
                 942.03 
                 1669.02 
               
               
                 12 
               
               
                   
               
             
          
         
       
     
         [0101]      FIG. 12  is a detail of the tensioner. Again turning to tensioner  5 , each arm  500 ,  510  has a rotation center about the center of shaft  4 , also see  FIG. 19 . Torsion spring  504  simultaneously applies a rotational force to each arm  500 ,  501 . The arms function as an opposing pair with the same torque being applied to each arm. Each arm surface  505 ,  506  rests against a surface of the housing  9 , namely  91 ,  92  respectively. The torque applied to the arms by the torsion spring  504  causes them to rotate. The resulting rotation causes the tensioner center of rotation, and thereby the center of shaft  4 , to move. The center of rotation moves until an opposing force prevents it, namely belt tension. The opposing force which equilibrates the system is the desired belt tension force. 
         [0102]    Each arm  500 ,  501  has a circular profile at the contact surface  505 ,  506  respectively. The distance between the tensioner rotation center (shaft  4  center) and a line perpendicular to the bracket surface  91  at the point of contact with the arm surface  505 , see  FIG. 12  and  FIG. 19 , is the effective tensioner arm length “E”. The effective arm length E changes with the rotation of the tensioner arms. 
         [0103]      FIG. 13  is a detail of the tensioner. In  FIG. 13  line A is perpendicular to the bracket surface  91  and is perpendicular to line a. Line B is perpendicular to line b. The effective arm length E is the distance from line A to the rotation center along line a. 
         [0104]    The center of curvature of the arm surface is offset a fixed distance from its center of rotation. The effective arm length is equal to the offset only when lines A and B are coincident with one another. When lines A and B are not coincident, the effective arm length is less than the center offset (CO) as a function of the angle formed between the lines. 
         [0000]        E =Effective arm length=CO*cos(ω)
 
         [0105]    Given:
       CO=6 mm   ω=8 deg       
 
         [0000]      Effective arm length=6 cos(8)=5.94 mm 
         [0108]    The force from each tensioner arm is equal to the torque on the arm divided by the effective arm length. 
         [0109]    Knowing the force required of the tensioner acts against the angular surfaces of the housing, for exmaple,  91 ,  92 , at the point of contact of the tensioner arm and the surface, one can determine the force required at these surfaces and from that, the torque required in the tensioner arm. 
         [0110]    Given: 
         [0111]    SA=Surface  91  angle=30 deg 
         [0112]    TF=Tensioner force=635 N 
         [0113]    Then: 
         [0114]    Arm force=(TF/2)*cos(SA) 
         [0115]    AF=(635/2)*cos(30) 
         [0116]    AF=275 N 
         [0117]    The torque required in the arm is simply the arm force (AF) times the effective arm length (EAL). 
         [0000]      Torque=AF*EAL 
         [0000]        T= 275 N*0.00594 m=1.63 Nm 
         [0118]    Tensioners  5 ,  6 ,  7 ,  8  are designed such that as the arms rotate, the effective arm length is reduced. Each respective torsion spring ( 504 ,  604 ,  704 ,  804 ) also provides less torque as the tensioner arms rotate. If the torsion spring has a spring rate of 0.01 Nm/deg and the arms rotate 20 degrees, then the torque from the spring drops by 0.2 Nm. The effective arm length changes from the above 5.94 mm to 5.30 mm. The resulting tensioner arm force remains nearly constant at 270 N. 
         [0119]    The included angle of the faces of the housing surfaces  91 ,  92  can range between 180 deg to 90 deg giving a surface angle of 0 deg to 45 deg as described above, see  FIG. 14 ,  FIG. 15  and  FIG. 19 . 
         [0120]    If the angle between surfaces  91 ,  92  is 0 degrees, there is no horizontal force component from the tensioner arms. Surface angles greater than zero causes the tensioner to self center due to the horizontal component of the force being equal and opposite from each tensioner arm. If the surface angle exceeds 45 degrees, these horizontal components exceed the tensioning force. This creates a condition of “diminishing returns” on the spring torque. As the spring torque is increased, the horizontal component of tensioner force grows more than the tensioning force. 
         [0121]      FIG. 14  is a detail of the force components in the tensioner arms. Vector “A” indicates the force on surface  505  exerted by surface “TS” at the point of contact between  505  and TS. Vector “B” indicates the force on surface  506  exerted by surface “TS” at the point of contact between  506  and TS. Surface TS is analogous to surface  91  and surface  92 . Surface TS depicts the 180 degree condition between surfaces  91 ,  92 . Given the offset of each tensioner arm  500 ,  501 , see  FIG. 5  and  FIG. 19 , vectors A and B are not co-axial. 
         [0122]      FIG. 15  is a perspective view of housing  9 . Housing  9  comprises bracket surface  91  and bracket surface  92 . Tensioner surface  505  and tensioner surface  506  engage surfaces  91  and  92  respectively. 
         [0123]      FIG. 16  is a perspective view of housing  10 . Bearing  50  engages receiving portion  80 . Bearing  56  engages receiving portion  81 . Tensioner surface  805  engages surface  91   c . Tensioner surface  806  engages surface  92   c . Tensioner surface  605  engages surface  91   a . Tensioner surface  606  engages surface  92   a.    
         [0124]      FIG. 17  is a perspective view of the assembled housing parts. Bracket  82  on housing  10  provides means to attach the device to a mounting surface (not shown). Tensioner  5  engages surfaces  91  and  92 . For tensioner  6 , arcuate surfaces  605  and  606  engage surfaces  91   a  and  92   a  respectively. For tensioner  7 , arcuate surfaces  705  and  706  engage surfaces  91   b  and  92   b  respectively. For tensioner  8 , arcuate surfaces  805  and  806  engage surfaces  91   c  and  92   c  respectively. 
         [0125]      FIG. 18  is a plan view of the assembled housing parts. 
         [0126]      FIG. 19  is a detail of a tensioner arm. Rotation center (RC) is the point about which the arm  500  rotates during operation. RC also coincides with the axis of rotation of shaft  4 . The arm profile center (PC) is the center of curvature of surface  505 , see  FIG. 13 . The distance between the two points is the offset. The rotation center radius (R 1 ) is less than the radius of curvature (R 2 ) of surface  505 . This description is also applicable to arm  501 . This description for  FIG. 19  also applies to each of the arms for tensioners  6 ,  7  and  8 . 
         [0127]    Although a form of the invention has been described herein, it will be obvious to those skilled in the art that variations may be made in the construction and relation of parts and method without departing from the spirit and scope of the invention described herein.