No offence is meant by the title. The topic of bow torque, bow rotation and stabilisation is on a realistic level such a can of worms that anyone with any sense avoids the subject like the plague. I'm happy to be included among this group. What I'll try to do here , based on some simplistic examples, is put across some basic concepts about how objects ( e.g.. a bow) rotate when a force (torque) is applied. I shan't go through all the math just supply the relevant relationships.
The effect of a force on an object is to change it's momentum. In fact the definition of a force is the rate of change of momentum with time. It follows therefore that the change in momentum resulting from an applied force is the integral over time of the applied force i.e.put simply: change of momentum = force x time
In the following I'll use the symbol F for force with suffixes as appropriate.
The topic of Moment of Inertia has been covered on another page. The moment of inertia of an object around a particular axis can expressed as being equal to the object mass multiplied by the square of what is called the radius of gyration. The value of the radius of gyration of an object is dependant upon the position of the rotation axis. When you add stabilisers to a bow you to some extent increase the overall mass but what you are really trying to do is increase the value of the radius of gyration.
In the following I'll use the symbol M for object mass and K for the radius of gyration with suffixes as appropriate.
When you apply a torque to an object it rotates. The angle through which the bow rotates in a time T I'll represent by the symbol A.
When you apply a force to an object it changes it's momentum. There are two types of momentum, Linear Momentum and Angular Momentum. In general when you apply a force to an object both the linear and angular momentum are changed. There is a rule about momentum change called the 'Conservation of Momentum Principle' which is basically that momentum can't appear from nowhere or disappear down a black hole. The key to understanding how objects rotate under an applied force is that the changes in linear and angular momentum must simultaneously conform to the Conservation of Momentum Principle.
In the following discussion the bow is represented by a vertical black bar. It is assumed that the bow includes various appendages (sight, stabilisers etc.) so that the position of the bow centre of gravity is variable.
By Free Bow I mean that nothing other than a single applied force (which represents all the accelerating bits of the bow/arrow/string ) acts on the bow. i.e. there is no bow hand. Only bow rotation in the vertical plane is considered.
A force F acts on the bow at a distance b from
the centre of gravity at C and the assumed axis of the
consequent rotation is at R. The distance from R
to the bow centre of gravity is a. The centre of gravity
is assumed to lie in the plane of the bow. The force is assumed to
act for a time T, the time being short enough so we don't have
to worry about the value and position of F etc. changing.
When you apply the Conservation of Momentum principle as defined above to the bow it defines the point about which the bow rotates and the rate of rotation about this point.
The distance a is given by:
a = Kg 2/b ( further information)
Where Kg is the radius of gyration calculated at the centre of gravity C.
Several points follow from the definition of a above about the bow rotation axis. Firstly in this case (initially stationary riser and very small value of T) the 'natural' bow rotation axis can never be at the bow centre of gravity as this would contravene the Conservation of Momentum Principle. Secondly the closer the applied force is to bow centre of gravity the further away from C is the 'natural' rotation axis. Thirdly the higher the value of Kg the further away from C is the 'natural' rotation axis. (Here rotation axis is taken to mean a stationary point in the current reference frame around which the riser rotates.)
Assuming the force F is constant then the angle through which the bow rotates in time T is given by:
A = FT2(a+b)/2M( Kg 2 +a2)
No surprises here. The higher the bow mass and in particular the higher the value of Kg (lots of stabilisation) the less the bow will rotate in a given time. The closer the centre of gravity is to where the force F is applied then the less rotation you get. If the centre of mass is at the point where F is applied then there is no bow rotation.
This is the same situation as above but now a bow hand, located at a bow grip B, is introduced As is the usual situation the centre of gravity is located below the bow grip. The bow grip is located at a distance x above the bow centre of gravity. The bow grip is assumed to be located on the line between the applied force F and the centre of gravity C. This means that gravity has no effect on bow rotation.
When the arrow is shot and the force F is applied the bow
'tries' to rotate naturally in a clockwise direction with the rotation
axis at R as in the first example but the bow hand prevents
this. The bow 'leans' on the bow hand producing a force and reaction between
the grip and the hand of value F2. Assuming that the bow hand doesn't
move and the grip doesn't swivel about in the hand then the grip - hand
pressure point acts like a horizontal spindle and the rotation axis of
the bow is located at point B. The value of F2 is that required
to make B the rotation axis.
Applying the Conservation of Momentum Principle as defined above to the bow you get the expression for how much the bow rotates under the force F for a time T. i.e.
A = FbT2/2M(K g2+x2)
Agan the higher the bow mass and the radius of gyration the less the bow will rotate. As far as the geometry of the bow goes, to reduce the amount of bow rotation we want a low value of b and a high value of x. The rotation angle is directly proportional to b and so is very sensitive to it. In practice the distance b relates to the gap between the pressure button (F) and the bow grip (B). This is the basis for it being said that the smaller this gap the more 'torque forgiving' is the bow. How small you can make b is determined by needing to have enough clearance to shoot the arrow and fletchings over the bow hand and arrow shelf. The value of x is limited by the bow design/construction, there is only so much weight you can add below B to lower the centre of gravity and keep a reasonable overall bow weight. The above equation illustrates the basic requirements for the barebow where stabilisers are not allowed. The arrow ideally rests on the bowhand knuckles (minimising b) and weight is added somehow to the bottom of the bow to increase x . Note that it is not possible in this case to have a bow set-up with a zero value of A, it is not possible to completely 'torque stabilise' the bow.
There are a couple of other considerations that may be worth a mention.
The reaction force F2 at the bowhand is given by:
F2 = F (1-xb/(Kg2+x 2))
If the centre of gravity is at the bow grip then F2 = F. As the centre of gravity is lowered then the value of F2 decreases. The friction between the hand and the grip depends on the force between them so as you lower the centre of gravity hand - grip friction decreases. As this friction opposes any movement of the grip in the hand during the shot it suggests that there is a benefit in having a low value of x i.e. having the centre of gravity near or at the bow grip. Many archers use shooting gloves to increase the hand - grip friction. There are probably also physiological and psychological effects on how the value of F2 ultimately effects arrow groups.(On which I'm not qualified to comment).
Another consideration with respect to the centre of gravity position issue is energy. The energy in our revolving bow comes from the energy stored in the bow when drawing the arrow. So the more energy that ends up in the bow means less energy that goes into the arrow (less arrow speed). The energy E that ends up in bow due to the force F acting on it is given by:
E = FAb
The value of b is defined by the bow manufacturer. The smaller we can make A (maximise torque stability) gives a side benefit of increasing the bow energy efficiency.
The tentative conclusions that can be made from the simple model above are that to make the bow 'torque forgiving' the archer should increase the bow mass and radius of gyration (moment of inertia) - well you all knew that anyway. The model also suggests that overall the stabilisers should be arranged so that the centre of gravity is located at the bow grip - hand pressure point, a suggestion that archers would invariably disagree with.
The big simplification in this model is that the bow is assumed to be vertical with F, B, C and R all in the plane of the riser. This results in gravity having no effect on the bow rotation. In practice when shooting any distance the bow has to be rotated in the vertical plane so this condition no longer holds and gravity does have an effect. Another consideration is that in general archers locate the centre of gravity vertically below the grip and in front of the riser. This raises the questions why do this and how far down and forward should the centre of gravity be. A model including the effect of gravity is required.
In this example the bow is elevated from the horizontal by an angle p. The distance from the button to the bow - hand pressure point is b. The centre of gravity C of the bow is located a vertical distance x below the pressure point and a horizontal distance forward w . In this case gravity has an effect on bow rotation. For convenience the viewpoint is rotated so that the bow appears to be vertical.
As in the previous example the bow is assumed to naturally rotate
into the bow hand making B the rotation axis, which as previously
acts like a horizontal spindle. In the earlier examples the fact that
the bow hand had to support the weight of the bow was ignored as it had
no effect on the bow rotation. In this case the weight does effect rotation
so the value of F2 is the total force on the bow hand resulting
from the mechanical action of the bow (F) and the gravitational force
(F3) on the bow.
There are now two torques, in opposite directions, acting to rotate the bow, the torque from the bow mechanical action and the torque from the bow mass. In principle then, if with the bow geometry these torques can be balanced we end up with a bow that doesn't rotate. We get a 'torque stabilised' bow.
Applying the Conservation of Momentum Principle as defined above to the bow and using g for the gravitational acceleration you get the expression for how much the bow rotates under the force F for a time T. i.e.
A = T2(Fb+Mg( wCos(p)+xSin(p))/(2M( Kg2+w2+x2 ))
For a 'torque stabilised' bow A would be zero i.e. the terms in brackets in the numerator in the above equation
(Fb+Mg(wCos(p)+xSin(p)) would equate to zero.
The term Fb is the (clockwise) torque from the action of the bow and is constant because the length of the lever arm doesn't depend on on the bow angle p. The term Mg(wCos(p)+xSin(p)) is the gravitational (anticlockwise) torque, Mg being the force and (wCos(p)+xSin(p) being the length of the lever arm which does depend on the bow angle. For a specific bow angle p and value of x the value of w can be adusted to give a 'torque stabilised bow'. You can rearrange the net torque equals zero relationship to indicate how w depends on x. i.e. w = - xTan(p)-Fb/MgCos(p)
This is just an equation of a straight line with slope -Tan(p). In other words for a given bow angle there a lots of positions of the centre gravity that will result in a 'torque stabilised' bow it's just a matter of getting the right combination of x and w.
The diagram illustrates the relationship between x and w
for torque stabilisation for bow angles of 10 and 3 degrees (typical
Fita round say).
The assumed properties of a 'typical' recurve bow are:
Mass M = 2.5 Kg
gap b = 5.5 cms
average F = 60 Newtons (guesstimated from typical draw weight)
x is assumed to vary from level with to 10 cms below position B
As x increases the required value of w decreases related to the bow angle and vice versa.
Torque stabilistation is related to the distance shot. You can only ever pick a compromise value for a range of distances.
The slope is flatter the lower the bow angle which suggests that cranking up the bow draw weight gives some benefit with respect to overall bow torque stabilisation.
Perhaps the most relevent feature of the graph is that there is a specific value of x and w at which the bow is torque stabilised at both angles (both distances).
This diagram illustrates how much bow rotation (A) you get as
w (or x) vary around the 'torque balance' situation. (the
bow actually rotates in opposite directions either side of the optimum value
but I've shown A as positive both sides)
What you get is a 'valley' shape. The closer you are to the optimum position of the bow centre of gravity (A=0) the less rotation you get.
If you plot the w-x curves for all the range of bow angles
(distances) you shoot then you get a bunch of tramlines.
There is no single set of values of x and w which will give you a 'torque stabilsed' bow at all distances. All you can do is pick an area in which all the lines are relatively close (the black circle).
In reality for an actual bow the values of x and w (along with just about everything else) vary during the shooting process, so the best you can ever do is choose a compromise position which seems to work well overall.
The idea put forward in the previous example that reducing the value of b (the button to grip distance) makes the bow more torque forgiving in this case is no longer necessarily true because there are two torques affecting bow rotation. It's rather like a tug of war with the bow action at one end and gravity at the other. Reducing b means that at any given moment the 'bow' team is pulling less hard. The 'gravity' team's pull variation (x and w changing) can therefore result in more movement of the middle of the rope. i.e. reducing b can make the bow less torque forgiving.
The denominator in the formula for A includes the term M(Kg2+w2+x2) which is the Moment of Inertia of the bow with respect to rotation around the hand - grip pressure point. The bigger this term is then the less the bow rotates during the shot. The benefit of having a high bow mass is clear. The term Kg relates to the radius of gyration at the centre of gravity. The x and w values relate to how far the bow rotation axis is from the centre of gravity. The value of Kg will depend on the values of x and w.
The basic conclusion is that for a given bow mass the further the grip - bow hand rotation axis is from the centre of gravity and/or the higher the value of Kg the more 'torque forgiving' the bow will be. (And at the same time the values of x and w are selected to best obtain 'torque stabilisation').
How far the bow centre of gravity is from B is limited by mechanical effects on the bow. There are 3 principal limitations on this. The first obvious one is the overall bow weight. Too much weight will result in archer fatigue and the inability to hold the bow steady. (Bow Hand Loading) The second limitation is the stiffness of the stabiliser rods and the connections between rods and bow. As the length of a rod increases the stiffness drops rapidly (as the cube of the length). The more the rod bends when the riser twists then the less effective it's going to be in reducing the amount of riser twist. The third limitation is the effect on the bow tiller. If you can imagine pulling down on the end of a long rod of a bow at full draw then the effect will be to increase the bending of the top limb and reduce the bending of the bottom limb. When the string is released this differential bending disappears. So the further forward the bow centre of gravity is in front of the grip the more the bow tiller will require adjustment to compensate. As you can never completely compensate for bow tiller effects the higher the tiller the worse off you're going to end up.
To summarise a stabiliser system has two main functions
The above two criteria are to be met while keeping a 'manageable' force for the archer on the bowhand.
The first of the above criteria is met by use of a long rod pushing the bow centre of gravity forward of the grip. This property is enhanced by use of a very rigid extender which effectively increases the rod length.
The second of the above criteria is met by the extender, vbar and twin rod combination. As discussed in the section on bow hand loading the vertical bow mass load on the bowhand has to be adjusted to give the best biomechanical result. The V bar arrangement allows (twin rod) mass to be added at or close to a horizontal axis through the bow grip. This is the most efficient method of adding bow mass load to the bow hand bearing in mind the overall bow hand force needs to be within manageable limits.
Because the bow centre of gravity is in front of the grip, after the shot the bow rolls forward. It is sometimes suggested that this roll forward can affect the arrow while it's being shot. The arguments vary between this roll forward being a good thing or a bad thing. If you actually quantify the combined effect of the roll forward and the bow recoil backward it is clear that bow roll has no significant effect on the arrow.
The above models relate to bow rotation around a horizontal axis. Bow rotation around a vertical axis is more or less described by the 'constrained bow' model above. In this case in principle we can have both b and x equal to zero as F, B and C can be in the same vertical plane. There is only a torque problem if the bow hand centre of pressure (B) is horizontally shifted by incorrect hand placement. The archers paradox effect itself generates bow torque as it results in the horizontal direction of F changing (which results in a non zero varying value of b). This effect is essentially catered for in the 'bow tuning' process but as the buckling of the arrow on release depends on the loose a bad release can generate torque affecting where the arrow ends up.
These simple models do, I hope, put across some of the basic concepts about bow rotation and stabilisation. A real bow of course rotates in three dimensions, the bow hand - grip interface isn't fixed and the values of F, x, w etc. all vary during the shooting process so the first law of archery comes into effect. Suck it and see!
Last Revision 1 July 2009