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Supination resistance and orthoses prescription

Discussion in 'Biomechanics, Sports and Foot orthoses' started by Simon Spooner, Apr 12, 2011.

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    All, the supination resistance test seems to have become quite popular among colleagues since my good friend Kevin Kirby first described it in 1992. My question is this: the assumption seems to be that the higher the supination resistance then the higher the supination force required by the orthosis- right? But, have there ever been any studies which show changes in supination resistance with foot orthoses in situ? And has the relationship between supination resistance and say, rearfoot posting angle been evaluated?
  2. I guess Ian G your Guy on this one, I could not find anything - good paper idea

    But if we look at this study -
    Gives a good starting point maybe ?
  3. Presenting the pilot at biomechanics summer school. Just wanted to know if anyone else had already done this.
  4. Craig Payne

    Craig Payne Moderator

    Look forward to seeing it. I got some data ... half analysed ... but ...'God put me ...'
  5. I guess it is time to set the record straight here. I don't believe I ever said that if the supination resistance is high then the orthosis should always be designed to apply a greater amount of supination force to the foot. I have seen others suggest this and teach this concept, but I don't believe this is the best way to use the supination resistance test that I invented in the late 1980s.

    For example, if the patient has a peroneal tendinitis but is found to have a relatively high supination resistance, then I will still try to evert (i.e. pronate) the foot with the orthosis in order to decrease the tensile force on the peroneal tendons during weightbearing activities. Maybe, in this patient, the high supination resistance was due to tonic peroneal muscle activity during bipedal standing that may be the cause of the peroneal tendinitis? If the clinician doesn't understand the biomechanical function of each individual structural component of the foot and lower extremity and understand what type of force has caused the patient's symptoms to that structural component (i.e. tension force, compression force or shearing force), then the supination resistance test will do little for the clinician in allowing them to design the proper orthosis for the patient and their symptoms.

    I do believe that the supination resistance test is a useful clinical test which, when combined with the patient's injury, skeletal structure, gait function, clinical and diagnostic tests, can help the clinician to best design the optimal treatment plan and foot orthosis for their patients. However, I believe that using the supination resistance test in isolation, without thinking about the other important factors that the clinician should consider when designing the proper foot orthosis for their patients, is an unwise method of prescribing custom foot orthoses.
  6. Good point, well made Kevin. I've actually started measuring pronation resistance too. Using a jig positioned under the lateral forefoot. I'm not sure if this is the best position, but if I'm treating peroneal (excessive supination moment) problems I tend to start with forefoot valgus posting, so I'm trying to get a feel for how much pronation force I might be adding with the wedging.

    Craig, I'm applying the supinatory load and taking it to the point of kinematic change at the rearfoot (I'm using the clamp from the STJ locator with a digital angle finder attached on the calcaneus) I then adjust the rearfoot angulation of the surface (increasing varus post) in intervals and measure decrease in supination resistance and repeat until supination resistance = zero. I know it's static, but a least we might be able to understand some of the kinetic effects of an element of the orthosis prescription. i.e. why did you choose a 5 degree post- because it reduced supination resistance by X Newtons in static stance, in this individual...
  7. DanthePod

    DanthePod Member

    I believe that evaluating supination resistance is a critical part in deriving an appropiate orthotic design selection. Though I agree with Kevin other observations and tests are essential in prescribing an orthotic design to achieve more predictable outcomes. :drinks
  8. What is critical about it? That is, precisely which orthoses design characteristics prescription variables are derived from this measure?
  9. Phil Wells

    Phil Wells Active Member

    I use the Supination resistance test to help optimize patient compliance.
    E.g. with patients with a low resistance to supination, I reduce MLA ORF to reduce the patient feeling as though 'they are being thrown onto the outside of my foot'.
    As Kevin said you need to know why the resistance is high/low but it does help to define the amount of strengthening or weakening required to get the best ORF.

  10. Phil and all, how do you reduce medial longitudinal arch reaction force with an orthosis? I don't understand the last sentence in Phil's post either, can someone explain: "it does help to define the amount of strengthening or weakening required to get the best ORF", please?

    What determines the reaction force which the orthosis exerts? Newton...come over here for a minute... we need your help. And bring Hooke, Cauchy and Young with you, they seem reasonably intelligent chaps...
  11. Phil Wells

    Phil Wells Active Member


    How about instead of reducing MLA ORF, we change it to minimising the ORF on the MLA.
    If I am putting something into the shoe that is aiming to have a clinical effect, I can optimise this by ensuring I am not applying unwanted forces e.g a supination force/moment where a pronation force/moment is needed to decrease load.
    Re determining the reaction force exerted by an orthosis, do we need to know unless we can measure it on a patient in the real world?
    Just playing devils advocate!

  12. Me too. So, if we put an orthosis in situ, will this increase or decrease the reaction force at the medial longitudinal arch in the majority of individuals? Since the majority of individuals probably don't have medial longitudinal arches that are in contact with the surface interface until we put an orthosis in situ, I'd suggest that, in many, the orthosis increases the reaction force at the medial longitudinal arch. Regardless, I am more interested in how orthosis characteristics influence such reaction forces and orthosis optimisation. How do we optimise orthosis reaction forces? How does the orthosis geometry and load/ deformation characteristics relate to the magnitude and distribution of the reaction forces at the medial longitudinal arch of the foot? i.e., how do the design features of the foot orthosis alter the magnitude, distribution and timing of the reaction forces at the foot-orthosis interface, when compared to the shoe alone? Is this "biomagic" or physics?
  13. RobinP

    RobinP Well-Known Member

  14. It is neither. It is neurophysiology and mechanics. Because without considering the neuromotor effects of how varying the location, magnitude and temporal patterns of ground reaction force on the plantar foot will affect the efferent activity from the central nervous system, we are only looking at half of the potential kinetic and kinematic effects that can occur from wearing foot orthoses inside shoes.
  15. Indeed Kevin, but let us start with the easy bit ;) the direct mechanical effects of foot orhoses... SO, I have two devices: identical in surface geometry at the foot-orthosis interface, and identical in materials used in their construction (same Young's modulus). Lets assume that orthosis 1 is constructed of a 2mm thick piece of the material, while orthosis 2 is constructed out of 6mm thick piece of the material. To keep it simple, lets assume that the material that the orthoses are constructed of is loaded within it's linear range and responds in accordance with Hooke's Law. At a given point under the medial longitudinal arch of the foot will orthosis 1 or orthosis 2 provide higher reaction forces under the foot at this point?

  16. I do better with multiple choice rather than essay exams, Dr. Spooner.:pigs:;):eek:
  17. Not even willing to venture a guess, Kevin? ;) Lets, put it another way, if I load the point of interest on the two orthoses with the same force, the rate of change of velocity from the initial velocity to velocity = zero will stay the same per unit time in both orthoses for the point on the foot interfacing with the point on the orthoses. Force = rate of change of momentum (Newton's 2nd). Mass is constant. Viz. the reaction forces will be the same at any instant in time from impact to static equilibrium, regardless of the material thickness at a given point on the two orthoses. :eek: The only thing which will change this is by altering the Young's moduli that the two orthoses are made from or. Which leads to the conclusion, you've either got to change the material to change the magnitude of the reaction forces at a point on the foot-orthosis interface in order to alter the magnitude of reaction force at this point. Or maybe alter the surface geometry... but that's another story.

    Please, tell me why this is wrong. Anyone?
  18. efuller

    efuller MVP

    The quick answer that will be right some of the time is that the thicker one will exert more force. However, the answer is dependent upon the height of the arch of the foot in gait compared to the height of the arch of the orthotic. Say, the thin orthotic, after loading, is still higher than where the arch of the foot would have gotten then the thin orthotic could potentially apply as much force as the thick one. Then there is the pain avoidance response to an orthosis that is too high.... Arghh too many variables when you include a person with a brain standing/ walking on the orthotic.

  19. All other factors being equal, the stiffer orthosis will exert more reaction force on the plantar medial longitudinal arch of the foot than the more compliant orthosis. The stiffer orthosis exerts more plantar arch force since it will deform less than the more compliant orthosis and will thus maintain a higher arch height profile under a given loading force from the plantar arch of the foot.

    Of course there are many other variables to consider but this describes the basic mechanical characteristics of orthosis/foot arch load vs deformation.
  20. Excellent Prof. Kirby, go to the top of the class.

    Now consider a portion of the device which is in contact with the shoe and is subject to compressional loading. Again lets assume the material is homogenous and that we are loading it within it's linear region. This could be a section of the device through the heel-cup and rearfot post or in a shank dependent device it could be somewhere within the medial longitudinal arch. Lets say we zoom in on the section and it's basically an inclined plane as in the figure attached. We load each point on the surface of the inclined plane with 10N force. Which point will deform the most and which the least?

    Given your reasoning above, what should the distribution of reaction forces look like from the time of loading to the time of zero velocity?

    Attached Files:

  21. Phil Wells

    Phil Wells Active Member


    Can I just rewind slightly to your question different thickness of materials.
    I try to factor the torsional stability of the shell when optimising its function - how will its energy return properties be effected by loading/unloading?
    A less stiff material will deform more under the medial proximal aspect of the shell as the calcaneus plantarflexes (If the force applied is enough) which in the case of polyprop will result in a storage of energy (Mimicking the properties of the plantar aponeurosis in vivo). As the CoP moves anteriorly and the heel unloads , the energy is released back into the foot in the MLA (theoretically).The shell is now optimised to 'normal' PA function. The thicker material will not deform as much and therefore not store energy in the way the patient may need. This is more of a twist within the material rather than a simple deformation in the Z plane (sagital) an harder to model.

    I know it complicates things but dynamic 3D modelling may be the only way to explain how an orthosis applies ORF.

  22. I would think that the amount of deformation would be dependent on the elastic modulus of the EVA material relative to the magnitude of the deforming compression force.

    I think the best way to model this is to imagine 5 springs all lined up, like the wedge you illustrated, with each spring having the same spring stiffness but each having different lengths. Each spring will still deform depending on the spring constant of the spring, but each spring will have a maximum distance it can compress before the coils of the spring compress against each other, therefore preventing further spring compression by "bottoming out". Therefore, at low loads, each spring will compress the same amount since the spring deformation is determined by the spring constant. However, at higher loads, the shorter springs may "bottom out", since they have less room to compress before their spring coils collapse against each other when compared to the longer springs that may not bottom out before they have equilibrated under the deforming force with their spring coils not collapsed against each other.

    Now, using the spring analogy, if we had an EVA material that was exceptionally soft, a relatively large compression loading force would deform the thicker portions of the EVA much more than the thinner portions of the EVA since the EVA is being "bottomed out" by the compression force and, as a result, the distance to "bottom out" in a thicker material is greater than the distance to "bottom out" in a thinner material.

    However, if we had a very high durometer EVA material (i.e. very stiff) and then a relatively small compression loading force would deform the thicker portions of the EVA about as much as it would as it did to the thinner portions of the EVA since neither of the portions of the EVA wedge is being "bottomed out" by the compression force.

    How am I doing Dr. Spooner? Do you think I will ever pass this class?;)

    By the way, I am greatly looking forward to Biomechanics Summer School with you, Craig and Trevor. Should be one of the best Podiatric Biomechanics Seminars of all time!:drinks
  23. Pretty good, so far Kevin.

    If we assume that the material is homogenous, does not bottom out and that each section is equivalent to a linear spring, then the resting lengths of the springs increases from sections 1-5. And if we assume linear displacement, then the longer the resting length of the spring, the greater the deformation for a given load. Thus, in the example we should see increasing deformation from sections 1-5. Given the analysis you made previously with higher forces being associated with less deformation per unit load, this should result in greater forces at the thin end of the wedge and decreasing from sections 1-5. this should result in a shift in centre of pressure toward the thin end of the wedge :eek:

    If we assume angular displacement of the orthosis at the foot-orthosis interface, then the displacement is the same across the surface of the wedge at each loading point and the reaction forces would be equal at each point. In this example, in which the material doesn't bottom out, the Young's modulus of the section may influence the magnitude of the reaction force but the geometry of the orthosis does not alter the reaction force distribution...:eek:

    What if we apply that same modelling approach to the heel cup and reafoot post section of an orthosis or the whole of a shank dependent device...food for thought?

    I should say that this example is a problem which Dave Smith has been greatly assisting and helping me try to work out. I've attached an image of a 3D model I made in my CAD system a few weeks ago in preparation for one of my presentations at Biomechanics Summer School.- Great minds...

    The bottoming out point is well made, this again should tend to increase reaction forces at the thin end of the wedge and shift the centre of pressure toward this end.

    So far we have three potential situations; none of them tending to shift the centre of pressure toward the thick end of the wedge.... it's about the foot's load/ deformation characteristics at each discreet contact point at the foot-orthosis interface- right?

    Looking forward to catching up with you and Pam in Manchester. :drinks

    Attached Files:

  24. Energy store and return is something I am very interested in. The key is in storing the energy and then returning it at the right place and time. The problem is that if we want to return the energy stored to drive the centre of mass forward we have to return the energy when the centre of mass is rising. In walking this is double limb support. So, take a right foot, we would really need to return the stored energy from the right orthosis from the time of left foot strike. At which time the heel of the right foot should have already left the floor. So we need to somehow store and transfer the energy from strike of the right foot's heel to the forefoot extension section of the right foot's orthosis. Running is easier since the centre of mass is rising during the whole of the second half of the contact period.

    It's probably a case of matching the natural frequency of the orthosis with the natural frequency of the limb. This is where the CNS mediated effects really start making it difficult, since if you alter the surface stiffness, you also alter the leg stiffness.

    As an aside, I also think that a potential reason for orthosis having detrimental effects could be due to the orthosis returning the energy out of phase with body. i.e. attempting to return too much energy at the wrong time, in the wrong place.
  25. Simon:

    I don't see how this could be the case. F = -kx. Therefore, spring length does not determine deformation, only the spring constant. Each section of the EVA block should deform the same amount under a given load, assuming the EVA is homogeneous.


    Check out this website and crunch the numbers. You will see that varying spring length does not affect deformation of a spring under a given load as long as the spring stiffness is constant.
  26. Kevin, let us go back to the wedge diagram that I attached earlier. Lets take section 2 and section 5 and assume that section 5 is exactly twice the height as section 2.

    Effectively we could say that section 2 is the equivalent of one spring and section 5 is two of these same springs joined together in series. If two similar springs are joined together in series and then compressed each must experience the same force compressing it. If each spring would normally compress 1mm for such a force the total compression of both springs would be 2mm. A long spring can be thought of as several shorter springs. For springs of the same stiffness acted on by equal loads, doubling the initial length of the spring doubles the compression produced.

    In our example the compression under loading at each point would be given by:

    Compression = Young's modulus X (original length) X load / cross-sectional area

    Young's modulus is the same at each point, so too are the loads and cross-sectional areas. What changes is the original length which increases from 1-5. Try putting in some values, Kevin. If I get time today I'll do a physical demonstration of this and take some photographs or a video. I've already done this as an FEA for Manchester. Dave Smith did this with some elastic bands and a mallet and a cardboard box and a set of bathroom scales in this thread: http://www.podiatry-arena.com/podiatry-forum/showthread.php?t=62683&page=4 post number 97

    See also: http://scienceworld.wolfram.com/physics/SpringsTwoSpringsinSeries.html
  27. Then, from your logic above, and since F = -kx, when you add two springs together in series, each with a spring constant of k, the new spring constant of the two springs combined in series must now be 0.5k in order to satisfy your argument above. Is this correct?
  28. Kevin, seems to be right.
  29. Just found confirmation of this concept:


    Makes more sense now. I didn't realize the spring constant would change when the springs were added together in series. See what happens when you have never had a single engineering class??
  30. Yes, Simon, you were right and I was wrong.:drinks
  31. I don't really care about being right or wrong. I'm interested in getting to the bottom of how orthoses work and having you help with this problem will only have a positive influence, I'm sure. So, given the analysis we have been looking at Kevin, would you agree that an homogenous wedge of material should tend to shift the centre of pressure toward the thin end of the wedge if we consider linear deformation and/ or should make no difference to the centre of pressure position if we consider angular deformation?

    If so, then how do we explain results like these: http://www.japmaonline.org/cgi/content/abstract/94/1/1

    I suspect that the answer lies within your anterior axial projection paper.
  32. I would suspect that any shift in center of pressure towards the thin end of the wedge by differential compression of the wedge would be insignificant relative to the shift in center of pressure toward the thick end of the wedge caused by the thicker (i.e. more prominent) aspect of the wedge creating more reaction force on the plantar foot.
  33. How does it do that, when we've just established that the wedge will tend to have higher reaction forces at the thin end and shift centre of pressure towards it's thin end?

    Newton's Third.
  34. Hello Gents - I asked this on another thread - Simon replied not really but he would get back to it, thought I might post it up again see what the thoughts are.


    Attached Files:

  35. Mike, I didn't understand your point nor your diagram last time. I still don't understand it.
  36. Simon:

    We have not established that "the wedge" will tend to have higher reaction forces at the thin end of the wedge. All we have established is that the wedge will compress slightly more on its thicker side than on its thinner side. Unless the wedge is made of extremely soft material, once the material of the wedge (e.g. high durometer EVA or Plastazote #3 or polypropylene) has compressed to an equilibrium condition under loading from the foot, then the final shape is still a wedge that will, all other factors being equal, show higher contact pressures from the foot at the thicker end than at the thinner end.
  37. But I thought, like you that:
    So, given that the thick end of the wedge is more compliant and deforms more for a given load than the thin end, shouldn't the higher reaction forces be at the thin end?

    As I said earlier, my statement above only relates to an analysis of linear deformations, If we look at the problem in terms of angular deformation then the angular deformation is the same across the whole of the wedge. Like you said, "the final shape is still a wedge". Given this, the force distribution should be the same across the whole of the wedge from time of contact to time of static equilibrium, since the rate of change of momentum will be equal at all points from initial velocity to zero velocity.

    At the time of static equilibrium, the force acting at each point of the wedge, 1-5 has to be 10N, because that was the load applied. Pressure = force / area. Since the contact area at each section of the wedge is the same and since the loads are the same, the contact pressure has to be the same too. If we took a flat piece of the material that the wedge is made from and loaded it in the same way as the wedge, the results would be the same as they are for the wedge.
  38. Fair enough-

    If we model a device as 2 wedges 1 from heel to the high point of the MLA and the other Top of the MLA to the distal edge - the picture and your point is - this should result in greater forces at the thin end of the wedge and decreasing from sections 1-5. this should result in a shift in centre of pressure toward the thin end of the wedge By shifting the COP towards the thin end of the wedge we will have accelerated the COP from the thick edge towards the thin edge - right ?

    So if we model the device as 2 wedges, as the foots COP moves from heel strike to toe off / or proximally to distially - the wedge effect will on the 1st wedge decelerate the speed of COP ( by attempting to move the COP towards the thin edge of the wedge ), after the foot has over come the high point the 2nd wedge will accelerate (or shift the COP) towards the thin edge of the wedge.

    does than make any sense ?

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