WO2015176169A1

WO2015176169A1 - Synchronous endless drive member tensioner with tooth skip protection

Info

Publication number: WO2015176169A1
Application number: PCT/CA2015/000339
Authority: WO
Inventors: Marek Frankowski; Nghia H PHAM
Original assignee: Litens Automotive Partnership
Priority date: 2014-05-23
Filing date: 2015-05-25
Publication date: 2015-11-26
Also published as: DE112015002425T5

Abstract

In an aspect, a drive arrangement includes a synchronous belt and a tensioner. The belt is driven by a crankshaft of the engine in a first direction, and, in turn, drives a driven component. The tensioner includes a base, an arm pivotable about the base, a biasing member biasing the arm in a first arm direction, and an engagement member connected to the arm for engaging the belt. As the belt moves in the first direction, it urges the arm via friction in a second arm direction. As the belt moves in a second direction, it urges the arm via friction in the first arm direction. A sum of at least: moments generated by the arm biasing member, by friction during movement of the belt, and by a hubload from the belt, prevents movement of the arm in the second arm direction.

Description

SYNCHRONOUS ENDLESS DRIVE MEMBER TENSIONER WITH TOOTH SKIP

PROTECTION

CROSS-REFERENCE TO RELATED APPLICATIONS [0001] This application claims the benefit of US Provisional Application No. 62/002,680, filed May 23, 2014, the contents of which are incorporated herein by reference in their entirety.

FIELD OF INVENTION [0002] This disclosure relates generally to the field of tensioners for synchronous endless drive members and more particularly to timing belt tensioners where the timing belt is lubricated with oil.

BACKGROUND OF INVENTION [0003] It is known to provide a tensioner for a timing drive on an engine. Such tensioners are important for reducing the likelihood of tooth skip on the timing belt. However, in some circumstances it has been found that the timing belt can skip a tooth, leading to loss of synchronization between the components driven by the timing belt and the crankshaft, which can be very harmful to the engine. It would be advantageous to provide a tensioner and endless drive arrangement that reduces the likelihood of tooth skip.

SUMMARY

[0004] In an aspect, an endless drive arrangement for an engine is provided, and includes a synchronous endless drive member and a tensioner. The synchronous endless drive member is driven by a crankshaft of the engine in a first drive member direction, and, in turn, drives at least one driven component. The tensioner includes a base that is mounted to a stationary member, a tensioner arm that is pivotable relative to the base about a tensioner arm axis, an arm biasing member that biases the tensioner arm in a first arm direction, and an engagement member that is connected to the tensioner arm and engages the endless drive member. During movement of the endless drive member in the first drive member direction, the endless drive member urges the tensioner arm via friction in a second arm direction that is opposite the first arm direction. During movement of the endless drive member in a second drive member direction, the endless drive member urges the tensioner arm via a first frictional force in the first arm direction. For this endless drive arrangement:

Hi ^*M_g ^*d + Ms - H|^*R is greater than or equal to zero,

wherein

Hi = a hub load on the engagement member by the endless drive member, μ₉ = a coefficient of friction through which the endless drive member generates the first frictional force to urge the tensioner arm in the first arm direction,

d = a moment arm between the first frictional force and the tensioner arm axis,

Ms = a moment applied by the arm biasing member on the tensioner arm in the first arm direction, and

R = a moment arm between the hub load and the tensioner arm axis.

[0005] In another aspect, an endless drive arrangement for an engine is provided, and includes a synchronous endless drive member and a tensioner. The synchronous endless drive member is driven by a crankshaft of the engine in a first drive member direction, and, in turn, drives at least one driven component. The tensioner includes a base that is mounted to a stationary member, a tensioner arm that is pivotable relative to the base about a tensioner arm axis, an arm biasing member that biases the tensioner arm in a first arm direction, and an engagement member that is connected to the tensioner arm and engages the endless drive member. During movement of the endless drive member in the first drive member direction, the endless drive member urges the tensioner arm via friction in a second arm direction that is opposite the first arm direction. During movement of the endless drive member in a second drive member direction, the endless drive member urges the tensioner arm via a first frictional force in the first arm direction. For this endless drive arrangement, the absolute value of (H ^*p_g ^*d + Ms - H ^*R) is less than the absolute value of (M_a + M_r ), wherein

Hi = a hub load on the engagement member by the endless drive member, p_g = a coefficient of friction through which the endless drive member generates the first frictional force to urge the tensioner arm in the first arm direction,

d = a moment arm between the first frictional force and the tensioner arm axis,

Ms = a moment applied by the arm biasing member on the tensioner arm in the first arm direction,

R = a moment arm between the hub load and the tensioner arm axis, Ma = a moment generated by a second frictional force which is resistive to movement of the tensioner arm about the tensioner arm axis resulting from any axial support of the tensioner arm, and Mr = a moment generated by a third frictional force which is resistive to movement of the tensioner arm about the tensioner arm axis resulting from any radial support of the tensioner arm.

[0006] In yet another aspect, an endless drive arrangement for an engine is provided, and includes a synchronous endless drive member and a tensioner. The synchronous endless drive member is driven by a crankshaft of the engine in a first drive member direction, and, in turn, drives at least one driven component. The tensioner includes a base that is mounted to a stationary member, a tensioner arm that is pivotable relative to the base about a tensioner arm axis, an arm biasing member that biases the tensioner arm in a first arm direction, and an engagement member that is connected to the tensioner arm and engages the endless drive member. During movement of the endless drive member in the first drive member direction, the endless drive member urges the tensioner arm via a first frictional force in a second arm direction that is opposite the first arm direction. During movement of the endless drive member in a second drive member direction, the endless drive member urges the tensioner arm via the first frictional force in the first arm direction, such that a net moment on the tensioner arm, which is the sum of at least: a moment generated by the arm biasing member, a moment generated by the first frictional force and a moment generated by a hub load on the engagement member due to engagement of the engagement member with the endless drive member, prevents movement of the tensioner arm in the second arm direction.

BRIEF DESCRIPTION OF THE DRAWINGS

[0007] The foregoing and other aspects of the disclosure will be more readily appreciated by reference to the accompanying drawings, wherein: [0008] Figure 1 is an elevation view of an engine with an endless drive arrangement in accordance with an embodiment;

[0009] Figure 2 is a plan view of the tensioner shown in Figure 1 ;

[0010] Figure 3 is a section elevation view of the tensioner taken along section line 3-3 in Figure 2; [0011] Figure 4A is a plan view of the tensioner shown in Figure 1 , showing the forces that act on an arm of the tensioner during movement of the endless drive member in a first direction;

[0012] Figure 4B is a plan view of the tensioner shown in Figure 1 , showing the forces that act on an arm of the tensioner during movement of the endless drive member in a second direction;

[0013] Figure 5 is a plan view of a tensioner in accordance with another embodiment; and [0014] Figure 6 is a sectional elevation view of the tensioner shown in Figure 5.

DETAILED DESCRIPTION OF EMBODIMENTS

[0015] Reference is made to Figure 1 , which shows an engine 10 for a vehicle with an endless drive member arrangement 11 thereon. The endless drive member arrangement 11 includes a crankshaft 12, which drives a synchronous (e.g. toothed) endless drive member 14 via a crankshaft pulley 16 (which may be toothed, as shown). The endless drive member 14 may be referred to as a timing belt 14 or simply as belt 14 for convenience, with the understanding that any other suitable synchronous endless drive member could instead be used. Via the timing belt 14, the crankshaft 12 drives at least one driven component 18. In the embodiment shown, the crankshaft 12 drives a pair of camshafts 18 via camshaft pulleys 20. The camshafts 18 control the operation of valves (not shown) for the cylinders (not shown) of the engine 10.

[0016] The engine 10 is shown as a simple rectangle, for purposes of representing an engine, however, it will be understood that the engine 10 may have any suitable form.

[0017] In operation, the crankshaft 12 and crankshaft pulley 16 rotate clockwise in the view shown in Figure 1 , so as to drive the camshafts 18 as needed for the operation of the valves. As a result of the rotation of the crankshaft 12, a first span 14a of the timing belt 14 on a first side of the crankshaft 12 will typically be relatively slack and may be referred to as a normally slack side span 14a of the timing belt 14, and a second span 14b of the timing belt 14 on a second side of the crankshaft 12 will typically be relatively tight and may be referred to as a normally tight side span of the timing belt 14. [0018] An idler pulley shown at 22 may be provided on the normally tight side span 14b. The idler pulley 22 engages the belt 14 and helps to maintain selected a wrap angle of the timing belt 14 on the camshaft pulley 20 immediately preceding it and helps to maintain a selected wrap angle on the crankshaft pulley 16. [0019] A tensioner 24 may be provided on the normally slack side span 14a. The tensioner 24 engages the timing belt 14 to maintain tension in the belt 14, and helps to maintain a selected wrap angle on the camshaft pulley 20 immediately proceeding it and on the crankshaft pulley 16. With reference to Figure 3 in particular, the tensioner 24 includes a base, a tensioner arm 28, an engagement member 30, and an arm biasing member 32.

[0020] The base 26 may include a base plate 26a and a shaft 26b that is fixedly connected to the base plate 26a. The base 26 is mounted to a stationary member, such as the engine block shown at 10a in Figure 1 , by means of a tensioner mounting fastener 34, such as a bolt that is received in a threaded aperture in the engine block 10a.

[0021] Referring to Figure 3, the tensioner arm 28 is pivotably connected to the base 26, for pivoting movement about an arm pivot axis AA. A pivot bushing 36 may be provided between the arm 28 and the base 26 to facilitate the pivotal movement of the arm 28 on the base 26. The pivot bushing 36 may be made from any suitable material, such as a polymeric material. A washer 38 and a polymeric thrust bushing 40 may be provided between the arm 28 and the head of the tensioner mounting fastener 34 (not shown in Figure 3) to facilitate the pivotal movement the arm 28 relative to the fastener 34. [0022] The arm biasing member 32 biases the tensioner arm 28 in a first arm direction shown at DA1 in Figure 2, which may be referred to as a free arm direction (which is clockwise in the view shown in Figure 2). In the embodiment shown, the arm biasing member 32 is a helical torsion spring, however it will be understood that the arm biasing member 32 may alternatively be any other suitable type of biasing member. [0023] The engagement member 30 is connected to the tensioner arm 28 and is positioned to engage the belt 14 in order to maintain tension in the belt 14. In the embodiment shown in Figures 2 and 3, the engagement member 30 is a shoe 41 that is fixedly connected to, and is therefore integral with, the tensioner arm 28 (e.g. by way of a press-fit between the outer surface 47 of the tensioner arm 28 and the inner surface 49 of the shoe 41). The shoe 41 has an engagement surface 42 that engages a rear surface 43 of the belt 14. The shoe 41 has a profile configured such that, pivoting of the arm 28 about the axis AA changes how far the engagement surface 42 presses into the belt 14. [0024] In the embodiment shown in Figures 5 and 6, the engagement member 30 is a pulley 44 that is rotatably connected to the arm 28 via a pulley support bushing 46.

[0025] During operation of the endless drive member arrangement 11 , the crankshaft 2 rotates clockwise to move of the timing belt 14 in a first drive member direction (shown in Figure 1 at D1) so as to drive the rotation of the camshafts 18. However, when the engine 10 is turned off, as the engine components come to a stop, they can end up moving backwards briefly, which in turn causes the crankshaft to turn backwards (i.e. counterclockwise). During backwards rotation of the crankshaft 12, the belt 14 is driven in the drive member direction shown at D2, and the slack and tight spans of the belt 14 switch places with one another, as compared to when the crankshaft 12 rotates in direction D1. As a result, the first span 14a becomes the tight side span and the second span 14b becomes the slack side span. At all times during operation of the engine 10 it is important to prevent all spans of the belt 14 from becoming so slack that there is the risk of disengagement of the belt 14 from some of the teeth of the pulleys of the driven components (e.g. the camshaft pulleys 20) and/or the crankshaft pulley 16 (referred to as tooth skip). If such disengagement were to occur, the timing between the crankshaft 12 and the driven components (e.g. the camshafts 18) would no longer be correct, which can lead to potentially significant damage to the engine 10. Examples of the damage that can occur include collisions between the engine pistons with the valves in engines that are referred to as having an interference design, which include valves that open into the swept area of the pistons.

[0026] If, when the crankshaft 12 turns backwards (i.e. moves in the second drive member direction D2), the tensioner 24 moves into the belt 14, or at least does not move from where it was prior to the crankshaft 12 turning backwards, the second span 14b of the belt 14 would be prevented from becoming so slack that there is a risk of tooth skip. In other words, if the net moment on the tensioner arm 28, which is the sum of the torques acting on the tensioner arm 28 ensures that the tensioner arm 28 moves into the belt 14 or at least does not move from the position it was in prior to the crankshaft 12 turning backwards then the risk of tooth skip would be mitigated.

[0027] For the purpose of achieving such a net moment, the tensioner 24 is configured such that, during movement of the endless drive member 14 in the second drive member direction D2 (as shown in Figure 4B), the endless drive member D2 urges the tensioner arm 28 in the first arm direction DA1 , via a first frictional force Fg.

[0028] The sum of the torques acting on the tensioner arm 28 includes at least Ms, which is the torque applied by the arm biasing member 32, Mh, which is the torque applied by the hub load, and Mg, which is the torque applied due to the frictional force Fg, which is the frictional force caused by the movement of the belt 14 during engagement with the engagement member 30. Additional torques that can act on the tensioner arm 28 are frictional torques that, in general, are resistive to movement of the arm 28 about the tensioner arm axis A_A regardless of the direction of movement of the arm 28. For example, such additional torques include Ma, which is the frictional torque resulting from any axial support of the tensioner arm (e.g. the frictional torque applied by the axial thrust bushing 40), and Mr, which is (e.g. the frictional torque applied by the pivot bushing 36).

[0029] The torque Ms applied by the biasing member 32 drives the tensioner arm 24 in the first (free arm) direction DA1.

[0030] The torque Mh applied by the hub load is equal to the Hi multiplied by the working eccentric R (i.e. the perpendicular distance between the direction of the hub load and the arm pivot axis AA). In other words, Mh = H ^*R. The effective centre of the shoe 41 , through which the hub load Hi acts, is shown at 45. The torque Mh drives the tensioner arm 28 in a second arm direction DA2, which may be referred to as a load stop direction, and which is the opposite direction to DA1.

[0031] The torque Mg applied due to the frictional force Fg is equal to Fg multiplied by the guide radius d, which is the perpendicular distance between the direction of the frictional force Fg and the arm pivot axis AA. The friction force Fg is equal to the hub load Hi multiplied by the coefficient of friction p_g between the endless drive member 14 and the engagement surface 42 of the engagement member that is engaged with the endless drive member 14. Thus, the torque Mg = H ^* _g ^*d. The direction of the torque Mg depends on the direction of travel of the belt 14. [0032] As can be seen from Figures 4A, during forward travel of the belt 14 (i.e. travel in direction D1 ), the torque generated from frictional force Fg generated by the movement of the belt 14 on the engagement member 30 acts against the biasing member torque Ms. However, as can be seen in Figure 4B, during backward travel of the belt 14 the torque generated from frictional force Fg acts in the same direction as the biasing member torque Ms. In Figures 4A and 4B, the teeth on the belt 14 are not shown, solely for convenience. Additionally, the illustration of the belt wrapped around the engagement member 30 appears to differ when comparing Figure 1 to Figures 2, and 4A and 4B, however such differences are to be disregarded, as the drawings are not to scale. [0033] To prevent tooth skip during backwards movement of the belt 14 (i.e. movement in the second drive member direction D2), the tensioner 24 may be configured to have a torque condition such that: Ms + Mg - Mh is greater then or equal to zero (i.e. such that Hi ^*p_g ^*d + M_s - H ^*R is greater than or equal to zero). As noted above, the additional frictional torques Ma and Mr that are applied to the tensioner arm 28 would only drive the arm 28 in the second arm direction DA1 in a case where the net result of the aforementioned torque condition is greater than zero. Since the torque Mr and Ma are only resistive torques and thus can never actually drive the arm 28 in any particular direction, it is sufficient to meet the aforementioned torque condition without regard to the torques Mr and Ma, in order to ensure that the tensioner arm 28 does not move in the second arm direction during backward movement of the belt 14.

[0034] It will be noted that it would be possible for the torques Ms + Mg - Mh to sum to less than zero (which would urge the arm 28 to move in the second arm direction DA2) as long as they were less than the resistive torques Mr and Ma, which would mean that the tensioner arm 28 would not move. Thus a second torque condition that could alternatively be met would be for the absolute value of (Ms + Mg - Mh) to be less than the absolute value of (Mr + Ma).

[0035] The hub load Hi is correlated to the belt tension by the following formula: Hi = 2^*T ^*sin (PHI/2), where T is the belt tension differential between in the belt spans 14a and 14b needed to drive the driven components and PHI is the wrap angle of the belt 14 on the engagement member 30. In an non-limiting example, it has been determined that a belt tension differential (i.e. a difference in the belt tensions between spans 14a and 14b) of at least 500N is required to overcome the sources of internal resistance to movement of the belt 14. Thus, if there is a belt tension of 500N in the tight span (which is span 14a when the belt 14 is moving backwards) and a belt tension of zero in the slack span (i.e. span 14b when the belt 14 is moving backwards), and if there is a 60 degree wrap angle (i.e. PHI = 60 degrees) then the hub load Hi may be calculated as being 2^*500^*sin(60/2) = 500N.

[0036] In some instances, it will be desired to solve for a minimum value of d using values that have been established for parameters involved in the first torque condition above such as the peak belt tension differential T, the wrap angle PHI, and the coefficient of friction p_g. To solve for d: d is greater than or equal to (2*T^*R^*sin(PHI/2) - Ms) / (p_g ^*2^*T^*sin(PHI/2)

[0037] Some example values for parameters that can be used in a non-limiting example of an endless drive member arrangement for an engine are:

Friction coefficient for pivot bushing 36 - p_b = 0.1

Guide/belt friction coefficient - p_g = 0.1

Friction coefficient for thrust bushing - p_w = 0.1

Working eccentric - R =7mm Guide radius - d = 50mm bushing radius - r = 26mm Washer radius - r_w= 15mm

Hub load acting on the tensioner guide necessary to break resistance of the camshafts (to rotate camshafts) H, = 500N

Spring torque - Ms =1.5Nm Spring axial force - Fa =30N

Radial, (for example pivot bushing 36) frictional torque - Mr = Mb

Axial (for example thrust bushing 40) frictional torque - Ma =Mw

[0038] Upon verifying the value of the sum of the three torques, Ms, Mh and Mg, (i.e. Ms+Mg-Mh) it can be seen that the net torque on the tensioner arm 28 is: [0039] The sum of the three torques (Ms, Mg, Mh) equals 1.5Nm + 500N*0.1 *0.05m - 500N*0.007m, which equals + 0.5Nm. Thus, the net torque is positive, which indicates that the tensioner arm 28 is urged towards the belt 14. Whether the arm 28 moves towards the belt 14 would be determined by comparing the sum of the three torques, Ms, Mh and Mg and to the resistive torque applied by the frictional elements, (i.e. Mr and Ma which result from the pivot bushing 36 and the thrust bushing 40). If the net torque from the three torques is greater than the resistive torque, then the arm 28 will move towards the belt. If the resistive torque is greater, then no movement of the tensioner arm 28 occurs. In either case however, the belt 14 is safe from tooth skip. Other values for the parameters above can be selected for which the net torque from the three torques would be negative, but less than the resistive torque from the friction sources, (i.e. the pivot bushing 36 and the thrust bushing 40) in which case, the belt 14 would still be safe from tooth skip.

[0040] For the example parameters above with a belt tension differential of 500N, if one uses a working eccentricity (shown as R) of 7mm, a wrap angle of 60 degrees, a spring torque of 1 .5Nm (i.e. the value for Ms) and a friction coefficient μ₉ of 0.1 , the distance d, based on the first torque condition above, would be greater than or equal to (H|*R - Ms)/(H * g). Thus, d may be greater than or equal to about 40mm. Based on such a value it can be seen that, if one controls the other parameters provided (such as using a sufficiently small value for the working eccentric, R), a tensioner having reasonable dimensions for use on engines today can be designed to meet the torque conditions described above.

[0041] By contrast, if the working eccentric were 25mm, d would be about 143mm, which would result in a tensioner that is potentially too large to fit on some engines, (although it could fit on a large enough engine).

[0042] A working eccentric of somewhere between about 5mm and about 25mm has been found to result in tensioners that can fit on a useful range of engine sizes while protecting against tooth skip. [0043] If it is desired to simplify the first torque relationship in order to facilitate designing the tensioner 24, one can amend the first torque condition so as to eliminate the biasing member torque from it. In other words one can provide a torque condition where Mg - Mh is greater than or equal to zero. Expressed differently, H ^* _g ^*d is greater than or equal to H ^*R. This third torque condition may be simplified to: d is greater than or equal to R/ _g. If this third torque condition is respected, tooth skip will be prevented. It will be noted, however, that this third torque condition is merely a subset of the first torque condition above and results in a more conservative minimum sizing of the shoe 41 relative to the size of the working eccentric R as compared to applying the first torque condition above. [0044] Reference is made to Figures 5 and 6, which show a tensioner 124 that could be used in place of the tensioner 24, to similar effect. Instead of using a shoe that is fixedly connected to the arm, the tensioner 124 differs from the tensioner 24 in that the tensioner 124 includes a pulley shown at 170 as the engagement member 30, and is rotatably connected to the tensioner arm 128 tensioner arm (shown at 128) by means of a pulley bushing 172. For the tensioner 124, the pulley 170 rotates about a pulley axis Ap on the arm 128 that is offset from the arm pivot axis AA.

[0045] The tensioner 124 further includes a base 126 which includes a base plate 126a and a shaft 126b that is fixedly connected to the base plate 126a. The tensioner arm 128 is pivotally connected to the base 126 (i.e. to the shaft 126b) by means of a pivot bushing 136, which may be similar to the pivot bushing 36. A thrust bushing 138 and a washer 140 are provided to support the arm 128 in the axial direction. A fastener that is similar to the fastener 34 shown in Figure 1 passes through the fastener aperture shown at 174 in the shaft 126b and has a head that abuts the washer 140 to hold the tensioner 124 to a stationary member such as the engine block 10a.

[0046] As the belt 14 moves either forwards or backwards, the pulley 170 rotates with the movement of the belt 14. The friction force Fg in this embodiment is not the friction force between the belt 14 and the pulley 170; it is the friction force at the pulley bushing 172 generated by rotation of the pulley 170 relative to the arm 128. The friction force Fg here is equal to the hub load Hi multiplied by the coefficient of friction associated with the rotation of the pulley 170 relative to the arm 128. For example, p_g in the example shown in Figures 5 and 6 is the coefficient of friction associated with the rotation of the pulley 170 relative to the arm 128 via the pulley pushing 172.

[0047] In Figure 5, the teeth on the belt 14 are not shown, solely for convenience. [0048] The tensioner 124 further includes an arm biasing member 132 that biases the arm 128 in a free arm direction (first arm direction DA1 ) which may be similar to the biasing member 32. The biasing member torque Ms for the tensioner 124 is similar to the biasing member torque Ms for the tensioner 24. In other words, it is the torque applied by the biasing member 132. [0049] As is the case with the tensioner 24, the formula for determining the hub load Hi for the tensioner 124 is the same and uses the belt tension differential T and the wrap angle of the belt 14 on the pulley 170. The torque applied by the hub load Th is the hub load multiplied by the working eccentric between the hub load and the arm pivot axis AA, shown at R. [0050] Irrespective of the difference in where the friction force Fg is generated (i.e. between the pulley 170 and the arm 128, as opposed to between the shoe 41 and the belt 14), the first and second torque conditions described above in relation to the tensioner 24 remain applicable for preventing tooth skip for the tensioner 124. [0051] Those skilled in the art will understand that a variety of modifications may be effected to the embodiments described herein without departing from the scope of the appended claims.

Claims

1 . An endless drive arrangement for an engine, comprising:

a synchronous endless drive member that is driven by a crankshaft of the engine in a first drive member direction, and which, in turn, drives at least one driven component;

a tensioner, including:

a base that is mounted to a stationary member;

a tensioner arm that is pivotable relative to the base about a tensioner arm axis;

an arm biasing member that biases the tensioner arm in a first arm direction; and

an engagement member that is connected to the tensioner arm and engages the endless drive member,

wherein during movement of the endless drive member in the first drive member direction, the endless drive member urges the tensioner arm via a first frictional force in a second arm direction that is opposite the first arm direction, wherein during movement of the endless drive member in a second drive member direction, the endless drive member urges the tensioner arm via the first frictional force in the first arm direction,

wherein

Hi *p_g*d + Ms - H|*R is greater than or equal to zero,

wherein

Hi = a hub load on the engagement member by the endless drive member, Pg = a coefficient of friction through which the endless drive member generates the first frictional force to urge the tensioner arm in the first arm direction,

d = a moment arm between the first frictional force and the tensioner arm axis,

R = a moment arm between the hub load and the tensioner arm axis.

2. An endless drive arrangement as claimed in claim 1 , wherein the engagement member is integral with the tensioner arm and μ₉ = a coefficient of friction between the endless drive member and a surface of the engagement member that is engaged with the endless drive member.

3. An endless drive arrangement as claimed in claim 1 , wherein the engagement member is a pulley that is rotatably mounted to the tensioner arm and μ₉ = the coefficient of friction associated with the movement of the pulley relative to the tensioner arm.

4. An endless drive arrangement as claimed in claim 1 , wherein the engagement member is a pulley that is rotatably mounted to the tensioner arm via a pulley bushing and p_g = the coefficient of friction associated with the movement of the pulley relative to the tensioner arm through the pulley pushing.

5. An endless drive arrangement as claimed in any one of claims 1-4, wherein R is in the range of about 5 mm to about 25 mm.

6. An endless drive arrangement as claimed in any one of claims 1-5, wherein d is greater than or equal to R/p_g.

7. An endless drive arrangement as claimed in any one of claims 1-6, wherein Hi *Pg^*d + Ms - Hi ^*R - (Ma +Mr ) is greater than zero, wherein, Ma = a moment generated by a second frictional force which is resistive to movement of the tensioner arm about the tensioner arm axis resulting from any axial support of the tensioner arm, and

Mr = a moment generated by a third frictional force which is resistive to movement of the tensioner arm about the tensioner arm axis resulting from any radial support of the tensioner arm.

8. An endless drive arrangement as claimed in claim 1 , wherein d is greater than or equal to (R^*2^*T^*sin (PHI/2) -Ms)/ (M_g*2^*T^*sin (PHI/2)), wherein,

T = a peak tension differential in the endless drive member in order to initiate movement of the endless drive member, and

PHI = a wrap angle of the endless drive member on the engagement member.

9. An endless drive arrangement for an engine, comprising: a synchronous endless drive member that is driven by a crankshaft of the engine in a first drive member direction, and which, in turn, drives at least one driven component;

a tensioner, including:

a base that is mounted to a stationary member;

an engagement member that is connected to the tensioner arm and engages the endless drive member, wherein during movement of the endless drive member in the first drive member direction, the endless drive member urges the tensioner arm via friction in a second arm direction that is opposite the first arm direction,

wherein during movement of the endless drive member in a second drive member direction, the endless drive member urges the tensioner arm via a first frictional force in the first arm direction,

wherein the absolute value of (H ^* _g ^*d + Ms - H ^*R) is less than the absolute value of (M_a + M_r), wherein

d = a moment arm between the first frictional force and the tensioner arm axis,

R = a moment arm between the hub load and the tensioner arm axis, Ma = a moment generated by a second frictional force which is resistive to movement of the tensioner arm about the tensioner arm axis resulting from any axial support of the tensioner arm, and

10. An endless drive arrangement as claimed in claim 9, wherein the engagement member is integral with the tensioner arm and μ₉ = a coefficient of friction between the endless drive member and a surface of the engagement member that is engaged with the endless drive member.

11. An endless drive arrangement as claimed in claim 9, wherein the engagement member is a pulley that is rotatably mounted to the tensioner arm and p_g = the coefficient of friction associated with the movement of the pulley relative to the tensioner arm.

12. An endless drive arrangement as claimed in claim 9, wherein the engagement member is a pulley that is rotatably mounted to the tensioner arm via a pulley bushing and pg = the coefficient of friction associated with the movement of the pulley relative to the tensioner arm through the pulley pushing.

13. An endless drive arrangement as claimed in any one of claims 9-12, wherein R is in the range of about 5 mm to about 25 mm.

14. An endless drive arrangement as claimed in any one of claims 9-13, wherein d greater than or equal to R/p_g.

15. An endless drive arrangement as claimed in any one of claims 9-14, wherein the tensioner arm is pivotably connected to the base via a pivot bushing, and is at least partially supported by a thrust bushing, wherein Mr is based on friction associated with the pivot bushing, and the wherein Ma is based on friction associated with the thrust bushing.

An endless drive arrangement for an engine, comprising: a synchronous endless drive member that is driven by a crankshaft of the engine in a first drive member direction, and which, in turn, drives at least one driven component; a tensioner, including:

a base that is mounted to a stationary member;

wherein, during movement of the endless drive member in the first drive member direction, the endless drive member urges the tensioner arm via friction in a second arm direction that is opposite the first arm direction,

wherein, during movement of the endless drive member in a second drive member direction, the endless drive member urges the tensioner arm via a first frictional force in the first arm direction, such that a net moment on the tensioner arm, which is the sum of at least: a moment generated by the arm biasing member, a moment generated by the first frictional force and a moment generated by a hub load on the engagement member due to engagement of the engagement member with the endless drive member, prevents movement of the tensioner arm in the second arm direction.

17. An endless drive arrangement as claimed in claim 16, wherein the engagement member is integral with the tensioner arm and the first frictional force is between the endless drive member and a surface of the engagement member that is engaged with the endless drive member.

18. An endless drive arrangement as claimed in claim 16, wherein the engagement member is a pulley that is rotatably mounted to the tensioner arm and the first frictional force is associated with the movement of the pulley relative to the tensioner arm.