Supination resistance and orthoses prescription

More on energy store and return in the papers attached.

In our wedge example with the linear deformations, we should see greater kinetic energy transfer from the foot to the orthosis at the foot orthosis interface at the thick end of the wedge. Using the angular analysis it's the same across the wedge. What will change this is the Young's modulus of the material.

 

Mike, this seems to fit if you use a linear displacement model.

But in this thread, I'm really interested to hear Kevin's thoughts on how despite the load / deformations being lower at the thick end of the wedge compared to the thin end of the wedge if we use a linear displacement model, or the same across the entire wedge if we use an angular displacement model, this should still result in higher "contact pressure" at the thick end of the wedge, given Newtons' third law of motion. This thought experiment has been troubling me for some weeks now, and I'd really like to resolve it.

I suspect it comes down to the relative load/ deformation between the foot and the orthosis. If the foot is relatively more compliant than the orthosis at a given point on the interface between them, then the foot should deform more than the orthosis and more work is done by the orthosis on the foot than the foot does on the orthosis. If however, the orthosis is relatively more compliant than the foot at a point on their interface, then the orthosis should deform more than the foot and the foot will have done more work on the orthosis than the orthosis will have done on the foot. If both the orthosis and foot have the same stiffness at this point on their interface, then the work done by both on each other will be the same. Think too about how this influences energy transfer.

I suspect that the stiffness of the heel pad varies from medial to lateral, since the thickness of the fat pad probably varies too. As I've alluded to, Kevin's paper on anterior axial projection showed the medial tubercle of the calcaneus to project further inferiorly than the lateral tubercle during standing. As such, the fat pad should be relatively thinner under the medial tubercle. So, in the same way that we have a wedge of material in our orthosis under the heel, we also have a wedge of material in the form of the fat pad at the heel of the foot between the skin and the calcaneus. It is the interplay between these two "wedges" and their relative stiffnesses which determines the reaction forces and moreover, the work done and the energy transfer at the heel section of the foot-orthosis interface.

 

Simon:

I think I must not be understanding exactly what you are asking in regards to your original wedge question. Possibly if you included the part of the foot you are interested in on top of the wedge then I could help you more.

Eric said that when using the pressure mat that the center of pressure in the heel was biased medially toward the medial calcaneal tubercle which certainly makes sense in regard to the anterior axial study we did previously (Kirby KA, Loendorf AJ, Gregorio R: Anterior axial projection of the foot. JAPMA, 78: 159-170, 1988).

My original research interest in creating the anterior axial view back in 1984 during my Biomechanics Fellowship was that I wanted to see what the weightbearing surface contour of the calcaneus was. If you look through the Compendium and Root et al's Normal and Abnormal, all the illustrations of the plantar calcaneus from those books show that when the subtalar joint was neutral, the medial and lateral tubercles were resting equally on the ground, with the inference being, from these illustrations, that this particular position of the subtalar joint was inherently more stable since it was the one STJ rotational position where both the medial and lateral calcaneal tubercles would be "resting equally on the ground". The results from the study clearly showed that the lateral calcaneal tubercle was 21 degrees more "up in the air" relative to the medial calcaneal tubercle....probably bearing a fraction of the body weight that the medial tubercle bears.

Hopefully that helps provide a little better perspective on this interesting subject of the compressibility and morphology of the plantar heel.

 

Here's the problem Kevin:
Orthosis are supposed to alter the magnitude and distribution of reaction forces at the foots interface with the orthosis.

Newton's second law of motion tells us that force is equal to rate of change of momentum.

When we analyse the application of a series of compressive loads across the surface of a wedge of material and examine the deformation which those loads should produce per unit time from the time of load application to the time of equilibrium we see that the rate of change of momentum is either lower at the thick end of the wedge (if we analyse the linear deformations), or is exactly the same across the wedge if we view the deformations as being angular. Thus the angulation of the surface doesn't change the rate of change of momentum of the body loading the wedge anymore than would be expected by the body impacting on a flat sheet of the same material. So while the Young's modulus of the material may alter the magnitude of the forces, the surface geometry doesn't appear to make any difference.

Given Newton's third law, the reaction forces at the orthosis have to be exactly equal and opposite to those at the foot's interface. So how does the wedge increase the reaction force acting on the foot beneath the thick end of the wedge, when compared to the reaction forces at the thin end?

 

The same way a glass marble would increase the compression force on the area of the foot that the marble was stepped on by a barefooted person.....by changing the surface geometry of the supporting surface of the foot.

 

That's not about the geometry, it's about the Young's Modulus of the glass relative to the foot. What if the marble had the same or lower stiffness than the foot? Indeed, if we measured the force distribution across the whole of the interface between the foot and the marble, would it vary? Or would the force be the same wherever the foot contacts the marble? Moreover, how does the wedge in the example increase the compression force acting on the foot more at the thick end than at the thin end? That is the key to this and so far this has not been resolved.

Put another way, lets say we stand with our heel on a wedged surface such that the thick end of the wedge is under the medial side of the heel. How does the wedge create more compression under the medial heel than under the lateral heel? To do this, the force has to be higher there- how does the wedge create higher forces under the medial heel than the lateral heel?

 

Kevin,

I've attached an FEA plot showing deformations of an orthosis under uniform vertical loading. I'd like to draw your attention to the heel cup area of the device which is subject to compressional loading. Why is there no variation in the deformations in this area? I can pull up models with rearfoot posts of various angles (I will at Biomechanics Summer School), the results all look the same in this portion of the orthoses models- why?

If the load-deformation characteristics determine the reaction forces, as you've previously stated, there has to be variation in the load-deformation characteristics across the heel cup section of the orthosis under loading if the rearfoot posting angle and geometry of the heel-cup are to play a role in shifting the position of the centre of pressure and net ground reaction force vector- right? Otherwise, the force distribution should be pretty much the same as it would be on a flat material of the same Young's modulus. If we wish to shift the centre of pressure medially then there has to be less deformation per unit load on the medial side of the heel-cup, given your previous assertion. My reasoning on this leads to the conclusion that there probably isn't much of a change in load/ deformation characteristics due to geometrical differences at the heel cup section of orthoses. Therefore, there isn't much change in the force distribution under this section of the orthoses either as a result of variations in their geometry. Rather the shift in centre of pressure and the line of action of the net ground reaction force vector is accomplished via the increased contact area in the medial longitudinal arch section of the device at the foot-orthosis interface and the variation in load-deformation characteristics in this portion of the device.

Furthermore, my thoughts presently are that rather than varying the posting angle of a device, in many respects we should stand a better chance of shifting the centre of pressure beneath the heel by varying the Young's moduli of the materials that the rearfoot post is constructed from. i.e. to shift the centre of pressure medially at the heel cup, it should be more effective to have a material with a higher Young's modulus on the medial side compared to the lateral side of the rearfoot post rather than increasing the angulation. Viz. put a higher Young's modulus on the medial side of the STJ axis than on the lateral side at the heel cup to help in the treatment of pathology related to excessive pronation moment, reverse the materials in terms of the Young's moduli beneath the heel cup for treatment of pathology related to excessive supination moment.

Why do we have angled rearfoot posts? - Historical hangover from Rootian philosophy?

Why do running shoe manufacturers tend to use dual density midsoles to "control pronation"? They shift the position of the net ground reaction force vector more effectively than does varying surface angulation?

I suspect that this is where friction comes into play...

 

Because this portion of the orthosis has a rearfoot post.

No. Three dimensional alterations in the geometry of the heel cup portion of the orthosis plate may alter the position of the center of pressure (CoP) without significant load-deformations of the heel cup of the orthosis itself. The medial heel skive in a polypropylene orthosis with a flat rearfoot polyproplyene post would be a good example of this.

I'm not following you here.

I prefer to have the lab add in a medial heel skive and use a homogenous, durable rearfoot post material made of polypropylene. Less costly, less time consuming and more durable. However, there is no reason that you can't knock yourself out and create a dual density rearfoot post. Seems like overkill to me and it probably wouldn't make any better of an orthosis by going to the trouble to do so.

We don't need to with soft soled shoes but they do work well. Rich Blake uses all flat rearfoot posts on his patient's orthoses, the last time I talked to him.

Like many of the other things we do in podiatry, what we do with orthoses is very much based on the previous works and ideas of others, with no research basis to support those methods.

Brooks once made a shoe, the Vantage, with a 4 degree varus rearfoot wedge.

http://www.runshoehistory.com/2010/11/birth-of-stability.html

However, I don't think you will see a varus heel midsole wedge occurring anytime soon again due to the potential legal ramifications of using a "prescription shoe modification" in over-the-counter running shoes. I call the dual density midsole running shoe, a shoe design feature which has been around now for over 30 years, a shoe with a "dynamic rearfoot varus wedge". This way shoe manufacturers can get the dynamic effects of a varus rearfoot wedge into their running shoes without making a shoe sole dorsal surface that is not parallel to the ground, which may create shoe liability issues if, for example, an inversion ankle sprain occurs in the shoe.

 

How does the geometry of the heel cup alter the centre of pressure?

Lets go back to fundamentals: how does a material generate a reaction force? When I stand on an orthosis, the orthosis pushes back with an equal and opposite force. This reaction force is developed by the stress and the strain in the material- right? So how does the geometry of the heel cup change the stress and strain developed?

 

One way an object can generate a force on another object is by making physical contact with the other object.

If the geometry and elastic modulus of the heel cup of the orthosis is such that it creates more force medially when it comes into physical contact with the plantar heel, then greater medial plantar tissue strain and stress develops. Differing geometrical shapes and elastic modulus of the heel cup will change the contact forces on the heel cup of the orthosis due to differential changes in plantar heel soft tissue and osseous stress and strain which, in turn, are due to the three dimensional geometrical shapes and elastic modulus of the plantar heel soft tissue and osseous structures. These physical contact interactions create compression and shearing forces both on the orthosis and on the tissues of the foot.

 

Kevin, ultimately reaction forces at the foot-orthosis interface have to be equal and opposite- Newton's third law. Reaction forces in the orthosis are generated by the stresses and strains within it's materials and in the foot by the stresses and strains within the tissues. The concepts of stress and strain were introduced by Cauchy to overcome the problems arising from comparing structures of variable size and geometry. Thus, by their very nature stress and strain within the orthosis and foot are independent of their respective geometry.

If, as we did previously, we apply Hooke's principles which are maybe more applicable to the gross responses of structures such as foot orthoses and feet, then greater deformation per unit load will occur at the thicker end of the wedge, or in areas of the heel cup where the material(s) are thicker. We showed this previously. Similarly, greater deformation per unit load might occur in the fat pad of the heel in areas where it is thicker when compared to areas where it is thinner (this is why I have been mentioning your anterior axial projection paper)- agreed?

While you keep saying that if the geometry and elastic modulus of the orthosis creates more force medially then it will tend to shift the centre of pressure medially (no argument from me- unless the centre of pressure actually shifts medially due to the foot pronating more, which is another possible explanation for medial shift in centre of pressure), you still haven't told me how contemporary orthoses designs, nor our simplified wedge model, manage to achieve this medial shift in centre of pressure. Indeed, our analysis thus far seems to show the opposite. If we assumed the foot was one homogenous block of material applying uniform loading to the wedge, the varus wedge should tend to shift the centre of pressure at the foot-wedge interface toward the thin end of the wedge, i.e. more laterally, when compared to this "homogenous foot" loading a flat surface since the wedge will deform less per unit load at its thinner (lateral) end. This is what our analysis has thus far demonstrated. This is the opposite of the design intent. And in this scenario we should be better off either using a valgus wedge to shift the centre of pressure medially or a flat sheet made up of materials which increase in Young's moduli from lateral to medial- agreed?

I spent an hour with a retired professor of engineering today and discussed this problem with him. He basically agreed with the analysis above. Me and the retired Prof. also both agreed that it was the load/ deformation of the foot which was missing from this analysis; that the foot wasn't an homogenous mass. So we need to show that the load/ deformation characteristics of the plantar heel vary from medial to lateral too, in order to work out how varus rearfoot wedges might shift the centre of pressure medially in-vivo. It seems to me, that given the relative inferior protuberance of the medial tubercle of the calcaneus during standing x-ray as observed in the anterior axial projection paper, the fat pad of the heel may be thinner beneath this bony area, compared to the lateral heel. Or the loading on this area of the fat pad is higher and it deforms more due to this factor on weightbearing. Either way, this area of the heel probably exhibits higher load/ deformation characteristics than the lateral side of the heel. It's probably this factor which overcomes the tendency of the varus orthosis wedge to shift the centre of pressure beneath the heel in a lateral direction and to provide a net medial shift in centre of pressure at the foot-orthosis interface with varus wedging. And if we were very clever we could work out the load/ deformation across the heel of the foot and work out the required load/ deformation characteristics across the heel cup of the device to shift the centre of pressure where we wanted it. However, all of this still tells us that in order to maximise the medial shift in centre of pressure under the heel of the foot, the area of the orthosis which this medial area of the heel interfaces with should also have higher/ load deformation characteristics than the lateral portion of this interface area. Our current orthosis design features which attempt to shift the centre of pressure more medially, i.e. adding a medial heel skive or higher varus wedge in order to shift the centre of pressure medially, should tend to do the opposite of this, since these designs will result in relatively lower load/ deformation characteristics were the material is thicker in the device, i.e. beneath the medial heel, when compared to flat sheets of the same materials.

 

Simon:

I can see we are not making any progress here...we'll need to discuss these concepts further in Manchester.

 

Ok, I can see that too. Thanks Kevin.

So, can anyone else tell me how the geometry and elastic modulus of the heel-cup section of an orthosis creates more force medially and tends to shift the centre of pressure medially? Or how a simple wedge of homogenous material positioned beneath the heel increases the reaction forces medially at the foot's interface?

 

Can you report back to us where you get to, somewhat confused at this stage.

 

It's very simple Mike, the orthosis has to be stiffer than the heel pad of the foot to move the centre of pressure medially in the presence of a varus wedge. If the foot's heel pad is stiffer than the orthosis the centre of pressure will move laterally. It's not really that simple, but hey, life never is.

 

I think it can be simplified even more than this. All that needs to occur to shift the center of pressure medially on the plantar heel is an increase in contact forces medially on the plantar heel, irregardless of the stiffness of the heel fat pad. The calcaneus lies deep to the calcaneal fat pad and it is the geometry of the plantar calcaneus that also largely determines how forces are transferred to the heel cup of the orthosis.

 

Same question then: how does the orthosis increase the contact forces medially on the plantar heel?

 

Now that I have been answering questions for the last week, I think it is time now for me to be asking the questions.

How does the surface of your thumb increase the contact forces on the medial aspect of the plantar heel of a patient when you push against the medial aspect of the plantar heel of the patient with your thumb?

 

I don't know, Kevin. How does it do that?

Anyway, the muscles in your arm and hand actively generate force to push the thumb into a discreet area of the heel of the foot and there is a transfer of momentum from the hand to the foot. Orthosis are passive and inert as such they cannot actively generate forces as the muscles in your arm and hand do. Rather, they can only react to the forces applied to them and provide reaction forces via the stress and strain within their structures through the atomic and molecular bonds within them. So, how does the orthosis increase the contact forces medially on the plantar heel?

To do this the rate of change of momentum between the foot and the orthosis has to be higher under the medial heel than the lateral heel- agreed? This has to be so, in order for there to be higher forces under the medial heel since this is Newton's second law. So, what causes a higher rate of change of momentum in the medial side of the heel cup of the orthosis compared to the lateral heel cup during loading? Is it the geometry of the orthosis? If so how can this be, given that we have shown that longer columns of a material are less stiff than shorter ones of the same material? So, the foot's momentum is not likely to be transferred to the orthosis over a shorter distance or time period on the medial side of the heel due to load/ deformation characteristics of the orthosis, which themselves are determined in part by the geometry of the device.

I hope Kevin that you can understand my desire and tenacity to get to the truth here and that anyone following can understand why this is so counter-intuitive. I do appreciate your help, Kevin.

 

Simon:

I simply don't understand what you are having difficulty with here....maybe a mental block or something on one or both of our parts? Maybe someone else following along can help us here?

If it were me trying to understand this problem, I would try to use equilibrium conditions, velocity = 0, rather than a velocity-dependent parameter such as momentum to understand what is mechanically happening here in your examples. I believe using free body diagrams and static equiilibrium modelling would greatly clarify your thought process and allow you to then make progress toward the more dynamic process that actually occurs during gait that could then be used when you start analyzing momentum.

For now however, I simply don't have a lot of time right to offer anything else for the discussion. Good luck and looking forward to seeing you again in a few months in Manchester.

 

I don't know the answer either Kevin, I am quite happy to admit that. I suspect that I shall need to do some study of contact mechanics and Hertzian models. http://en.wikipedia.org/wiki/Contact_mechanics to overcome my "mental block".

Unless you can spare me the time to draw out the required free-body diagrams?

What I'm having difficulty with here is how a varus heel wedge or varus heel cup of a foot orthosis shifts the centre of pressure medially beneath the heel of the foot. If there is a simple answer, I'd be more than happy for anyone to tell me. Or, even a complicated answer... any answer to this question would be a start because the only answers I've got at the moment seem to be "it just does" and "biomagic". Not denying that it might do that, just be nice to know how.

 

All,

Can anyone help to solve this little problem. In the diagram attached let the angulation of surface 1 be 10 degree from the horizontal and let the angulation of surface 2 be 40 degree from the horizontal. Let CM be the centre of mass of the circle. Can anyone tell me how to calculate the contact forces at A and B, please?

Many thanks.

 

Simon:

You have modeled an unstable situation where an equilibrium situation won't exist unless there is a tension force between the ball and the wedge that holds it in place.

 

Simon:

It is quite simple, now that you have framed your question this way. The varus heel wedge, because the medial side of the wedge is thicker than the lateral side of the wedge will, with weightbearing by the heel onto the wedge, will caused increased compression force, increased compression stress and increased compression strain on the medial aspect of the plantar heel. In much the same way if you were to step onto a surface irregularity, such as stepping onto a marble, the compression force on your foot at the area of the marble would increase since your foot hits the marble first probably before the other areas of the foot make contact with the ground. Surface geometry largely determines plantar pressure.

 

How?? With the greatest respect, saying what you've said above isn't answering the question, Kevin. The thicker side of the wedge will compress more under a given load etc. etc. Remember...

Show me the maths to solve the problem, please..... how is the increased compression force generated on the medial side, given that the wedge will have lower load/ deformation characteristics on the medial side?

Newton's third law tells us that the wedge will push back with the exact same forces with which it is loaded across it's surface.... so why does the foot load the medial side of the wedge with more force than that which it loads the lateral side?

Particularly, since the foot usually strikes the ground posterior laterally beneath the heel.

 

As it's Easter, lets take this egg in it's homogenous foam egg cup and drop it to the ground so that it hits the ground at the same time and evenly across the egg cup's surface 1-2, why is the rate of change of velocity higher at B than A? In other words, why would the impact force be higher at B than A? Why would this egg be more likely to break at point B rather than point A, provided that the foam doesn't bottom out?

Isn't this counter-intuitive? Hope y'all can understand this problem now because we've been saying that our orthoses (egg cups) work by increasing the force at point B of our eggs (feet) so that it is greater than the force at point A under our eggs (feet)- is that right? Do you trust your instincts or that which you have been led to believe..?

Now, when I went to school I was a high jumper (didn't do distance running, as I got bored running round and round and round... Plus you could smoke in between jumps in high jump, difficult to light up when you are running round and round and round), and when I jumped... it was a lot nicer to land on a thick mattress than a thin one.... why? I watched as a boy missed the mattress completely and landed on the concrete floor once, Fosbury flop (true story)... he screamed... why?

Here's an experiment to try at home: buy a dozen or so eggs. Buy some low density 1mm eva sheet. Set the eggs so they can drop from the same equivalent height onto the surface. Drop each of the eggs onto the surface, but increase the thickness of the eva on the surface 1mm at a time with each egg. I'm guessing you'll reach a certain thickness of eva and the eggs will stop breaking... but what our current understanding of orthosis tells us is that the thicker the eva, the more likely the eggs should be to break because the impact forces should be higher in association with thicker eva... apparently.

Now do the same experiment, but replace the sheets of eva with sheets of concrete, increase 1mm at a time..... see what happens. I'm guessing the eggs will keep braking. Why the difference? The surface geometry is the same... try it again, this time angle the surface....

 

One more... Looking at the diagram assuming we have a 45 degree wedge of homogenous material, we apply a force of N on the left hand side, what forces would we need to apply at each point from left to right to turn this piece of material into a flat horizontal surface, in terms of N?

 

More random thoughts... If I took a balloon and filled it with water and placed it into the heel cup of an orthosis or an asymmetrical egg cup, would the pressure be the same across the contact surface between the balloon and the egg-cup (orthotic)? My guess is that this would be the case, regardless of the shape of the egg cup. Why would the heel pad of the foot react differently to this situation than the balloon?

This on the initiation of foot loading in rigid and soft devices: http://www.sciencedirect.com/scienc...c98030a06a5e4c0481190db4fba6b675&searchtype=a

This on GRF vectors with and without orthotics:
http://www.staffs.ac.uk/isb-fw/Abstracts/RAZEGHI.pdf

 

Anyone got a copy of this they could mail me, please?
http://www.japmaonline.org/cgi/content/citation/86/7/307

 

A simplified solution might lie within this paper Kevin. This one looked at a sagittal plane model of the foot with load under the forefoot and rearfoot. Maybe a similar approach could be adopted for a frontal plane analysis of discreet areas of the heel within an orthotic heel cup?

 

You will have mail in a minute or two

 

I'll take a stab at the question. I wasn't able to keep up on the wedge discussion. Something did seem quite right about the initial assumptions, but I couldn't quite put my finger on it.

So, you take a heel and lower it onto the ground and then on to a varus wedge. When the heel is lowered on to the wedge (10% body weight), the initial contact point will be more medial. Then add more load 25% body weight. If the heel keeps the same orientation to the ground in the frontal plane, there will be more compression of the soft tissue of the heel, and the wedge, medially, because as you add more load there will be more compression at the points where there was contact earlier. So now you "lower" the heel further till you get full body weight. Again if the heel maintains the same orientation there will have been force in the area that contacted first assuming that plantar heel soft tissue is uniform. Or, if you use the anterior axial x - ray data there is less tissue to compress under the medial calcaneus when compared to the lateral.

I did some preliminary tests with people standing on an EMED with and without a varus wedge. The data was trending toward being more medial with the varus wedge. In looking at a lot of roll over processes with the EMED there was present most of the time, higher pressures under the medial heel when compared to the lateral heel. I'm pretty confident in saying that a wedge will increase the force on the high side of the wedge. We have to match the explanation to the observation.

Eric

 

Thanks Eric, there are a number of assumptions being made, one is that we will contact the high side of the wedge before we contact the middle or low side. Given that we tend to strike the ground posterior-laterally, I don't think this can be taken as given- what do dynamic in-shoe pressure studies show? Regardless, lets assume we do indeed contact on the high side of the wedge first. We have already established that the spring stiffness (K) of the wedge will be relatively lower as the thickness of the material increases. Thus, for a given deformation the reaction force will be lower at the thick end of the wedge than at a thinner portion of the wedge. So once we start to make contact lets say from 25% through to 100+% bodyweight, despite there being relatively more compression on the medial side of the wedge, the reaction force may not be any higher here than in areas of the wedge which have undergone less compression- Hookes Law. So at a given time, despite the compression being greater medially, the reaction forces might be higher laterally. Indeed the impulses of the forces could be the same across the wedge from the time of initial contact to static equilibrium- right?

With regard to the compression of the heel pad, it depends on the relative stiffness of the heel pad to the stiffness of the heel cup of the orthosis. But the same concept as above probably applies. That is thinner areas of the fat pad may have higher K values than thiner areas. i am yet to find a study which has measured the load/ deformation characteristics of the heel pad in multiple areas of the heel pad.

There must be a relationship between the angle of attack of the foot; the load/ deformation characteristics across the heel pad; the location of initial contact on the heel; the location of initial contact on the wedged surface; the angulation of the wedge and the load-deformation characteristics of the wedge; and the co-efficient of friction at the interface which determines the position of the centre of pressure from impact to static equilibrium?

BTW. What happens if the heel changes orientation? I'm guessing, similar to the observations here: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2699559/
That is, that as the foot pronates in relation to a valgus (lateral wedged) surface the centre of pressure moves medially and in response to a varus (medially) wedged surface when the foot supinates the centre of pressure moves laterally. Centre of pressure position is a rather blunt instrument when viewed in isolation.

You are absolutely right though, we do need to match the explanation to the observation. Which studies show that initial contact between the foot and the orthosis is more medial when the devices are posted in varus; that centre of pressure is more medial with rearfoot varus posting (and moreover, show that this isn't due to the foot pronating more)?

 

Simon:

How much does a polypropylene varus heel wedge deform medially compared to laterally in millimeters? I would estimate less than a fraction of a millimeter even at 200 pounds of force.

So, if this is the case, why are you so concerned about the "spring stiffness of the high side of the wedge" when the deformations are so insignificant compared to the geometry of the varus wedge in a relatively non-compressible material such as a polypropylene orthosis with a polypropylene rearfoot post [which is the standard orthosis I make for my patients]?

 

Michael could I trouble you for that paper as well?

 

I'm interested in this problem for a number of reasons, not least because not all orthoses are constructed of polypropylene with polypropylene rearfoot posts. Certainly here in the UK, many devices are constructed of more compliant materials. Anyway, it's all relative even with the polypropylene. The reaction force is generated by the stress and strains within the material, will the stress and strain be equal through the polypropylene rearfoot post?

Kevin, if we add your polypropylene device to a running shoe with an eva midsole, what influence does the midsole have on the K values beneath each point of the heel cup?

 

Done :drinks

 

Simon not sure if this thinking is correct.

If we have a wedge broken up into 5 point like you have ( say a skive ) with a heel over it.

At each point we will have the reaction force and as you have described the thicker less reaction force - right ?

And to every force the will be an equal and opposite force - so the force from the heel on the wedge should be at it´s highest at the thicker part of the wedge - right ?

So the net force acting about the wedge ie heel and reaction force from the wedge be the same at all 5 points ?

So this then shows the importance of friction on more compliant devices

and also the importance of geometry with the more compliant section coming in contact etc as Eric discussed.

I must admit I am pretty lost on this thread

 

I might as well do something useful while I'm watching the heavyweights slug this one out...

According to wikiepedia the Young's mod for polypropylene is about 250000psi. If we assume a starting thickness of 1/4" and a cross-sectional area of 2 sq ins, then the deformation from a force of 200lbs will be (200 x 0.25)/(2 x 250000) = one ten-thousandth of an inch.

EVA is pretty variable but for ease of calculation assume a high density EVA has a YM of about 25000, so you'd get ten times more compression i.e. one thousandth of an inch.

The difference in deformation doesn't, I think, have any effect on the reaction force - this is going to be equal to the perpendicular component of the force landing on it. Although, as someone has already commented, there is a danger in trying to apply static mechanics to a dynamic problem.

 

How does the orthotic generate reaction forces? Does the load/deformation of the orthosis play any role in the development of the reaction forces? Will reaction forces be higher or lower when load/deformation is higher, or doesn't it make any difference? What factors will alter the load/deformation characteristics of the orthosis?