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Kinetic effect of soft vs hard/rigid materials

Discussion in 'Biomechanics, Sports and Foot orthoses' started by JonathanH, Mar 29, 2013.

  1. JonathanH

    JonathanH Member


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    Hi Colleagues,

    Just another confusing thought element with general reading and would appreciate your perspectives, and advice....

    In the case of pain at plantar-central aspect of calcaneus due to compressive forces,

    1) what is the kinetic impact of placing a soft material (e.g poron) vs a hard material (high durometer EVA) on the plantar calcaneus?

    2) while the softer material may 'cushion' the heel, doesn't it prolong the timing of heel contact and hence the compressive forces on the heel?

    3) wouldn't a firm heel lift shorten the heel contact aspect of gait, whilst increasing the compressive force on the plantar heel?

    4) Where can I find out more information about the impact on forces/kinetics vs material characteristics?

    Thank you all and Happy Easter!

    Jonathan
     
  2. Andrew Ayres

    Andrew Ayres Active Member

    These are my thoughts:

    Force = Mass X Acceleration. If you have a soft material rate of loading will be slower therefore force will be decreased. If the material is elastic it will help accelerate the heel when it starts to lift (Hookes Law and Youngs Modulus). A bit like jumping on a trampoline compared to jumping on concrete

    Pressure = Force / Area. If a person is stood still their heel will push further into a soft material which will increase the weight bearing surface area therefore decrease the pressure.

    :confused:
     
  3. OK, a quick analogy: we have two identical cars both travelling at a constant 50 miles per hour, we hit the brakes in both cars at the same time but arrest the forward motion of one car over 50m and the other over 100m. For the car which we stopped over a shorter distance, we had to exert a higher braking force; for the car which we decelerated over a greater distance we exerted a lower braking force, but took longer to do it.

    So in our orthoses, stiffer designs will exert a "higher braking force" than more complaint designs. Our "braking force" with the more compliant device is lower but occurs over a greater time period. Impulse = force x time.

    Now, both cars had the same initial momentum and the same final momentum, so what does this tell you about the impulse applied to the cars?
     
  4. efuller

    efuller MVP

    Agree with what others have said about impulse. Although the question arises whether the injury is from high force or from impulse. My working theory, and I'd change my mind pretty quickly if better info came along, is that this injury is from a single high load rather than repetitive medium loads. The theory continues that reducing peak loads will lead to healing. So, I would go with the soft material. Again, I'd change my mind when presented with better data.

    I've heard that and again the question is what is better for healing. Is reduction of peak load better for healing or reduction of impulse? I'd bet a dollar on peak load.

    By heel contact do you mean the length of time between heel contact and forefoot contact? There are many factors that go into determining this time. One is touchdown angle (the sagittal plane angle of the plantar surface of the foot to the ground at the time of heel contact.) The velocity of ankle plantar flexion after heel contact. This is dependent on the moment from ground reaction force and the moment from the anterior tibial muscle (amongst others.)

    So enough of heel contact and on to compressive force on the heel. During heel contact, the force is dependent on the vertical velocity of the leg before contact and the material that the heel lands on. The velocity prior to impact is dependent on choice of running style. There are studies that show that people adjust their running style to the hardness of the surface that they are running on. So, it would be pretty hard to find a purely mechanical effect (as opposed to behavioral effect) of a heel lift.

    Eric
     
  5. You've also got to think about rate of loading and it's impact upon the viscoelastic tissues. So, with a reduced rate of loading the tissue will become less stiff, with a higher rate of loading the tissue will be stiffer. What is the impact of this on energy storage and ultimate tensile strength of the tissues? Certainly they will undergo less elongation when stiffer for a given load. So, it gets complicated: if the greater elongation which would occur with a more compliant tissue allows greater excursion of the centre of pressure, then the moment is potentially larger.... etc. you get my points Eric.

    P.S. isn't impulse thought to be of greater significance than peak load in the aetiology of ulceration in diabetics?

    Moreover, isn't the orthotic impulse the same with both soft and rigid orthoses if the angular velocity of a foot segment is the same at the initial time of contact and the same at some other point in time after contact? Change in momentum = impulse. What's different is the position the foot gets itself into and the stiffness and elongation of the tissues at any instant in time between initial contact and any point of "quasi-equilibrium" (by this I mean if we apply a quasi static analysis at any instant in time between initial contact and at any point in time thereafter).
     
  6. JonathanH

    JonathanH Member

    Thank you for the analogy

    So the impact/initial/peak force looks to be lowered as the time is increased with softer materials as to maintain momentum/conservation of energy and vice versa.....similar to the function of airbags? :D
     
  7. efuller

    efuller MVP

    Let's make sure we are talking about the same tissue. The original question was about plantar heel pain that was not at the medial calcaneal tubercle. I recall Kevin describing this as something like a heel contusion. I don't recall ever seeing this described in podiatry school. However, landing on heels at some level, should cause injury. So the injury would be under compression, rather than elongation. I need to think about viscoelasticity a bit.

    Yes, but is this the same injury as diabetic ulceration. That is believed to be due to ischemia, I think.

    Agreed the impulse is the same.

    Eric
     
  8. David Smith

    David Smith Well-Known Member

    The picture of local forces applied to the plantar surface of the heel at heel strike with varying shoe stiffness is a confounding one. If your foot strikes a rigid surface then your leg stiffness will be reduced but if you foot strikes a soft compliant surface then you leg stiffness tends to increase, therefore plantar forces are attenuated by the CNS decision on the appropriate leg stiffness. This is obvious if you try walking or running barefoot on cobbles or stony ground.
    So you can have the situation where there is a soft insert in the shoe intended to cushion the heel and the CNS detects this compliant surface and adjusts the leg stiffness accordingly but then the soft material bottoms out and suddenly you are on a hard surface with a stiff leg and so the peak force ends up much higher than if you had stuck to a rigid shoe, insole or orthosis. Prof R Aboud done much research into this aspect of shoe wear at Dundee and Edinburgh Uni's.https://medicine.dundee.ac.uk/staff-member/professor-rami-j-abboud

    regards Dave
     
  9. Andrew Ayres

    Andrew Ayres Active Member


    That makes sense. Does that also mean that the GRF will be the same in both situations?
     
  10. no...
     
  11. Andrew Ayres

    Andrew Ayres Active Member

    Oh. Well that ****** on my fire, I read part of your post wrong which through me. This is whats going throughmy head am I right?

    Impluse = force x time interval.

    2 feet exactly the same with the same velocity make contact with different orthotics. The softer orthotic will apply a smaller force than the stiffer orthotic. The softer orthotic will apply its smaller force for longer as it will take longer to compress and change the momentum of the foot.
    So if you measured the velocity of the foot segments anytime between contact and the end of momentum change the velocities will be different.
     
  12. Jonathan:

    Thanks for starting such an interesting thread. Simon, Eric and Dave have provided very good answers. Let me see if I can explain things in a slightly different way.

    Any time you analyze a dynamic situation, such as a calcaneus hitting a hard concrete floor or hitting a soft padded mat on the same concrete floor, you are dealing with an aspect of physics that has been studied for many years due to its utility in designing car bumpers, vibration-reducing materials and running shoe midsoles. That aspect of physics deals with collisions.

    The first concept to understand is impulse.

    http://hyperphysics.phy-astr.gsu.edu/‌hbase/impulse.html#c4

    Impulse is the mathematical product of the average force and the time it is exerted:

    Therefore, a force that is larger and/or is applied over a greater period of time will cause a greater impulse.

    Impulse can also be described as the change in momentum of an object. In other words, since:

    Momentum = Force x velocity;

    Impulse = Force x (change in velocity of object).

    Therefore, if a heavy object is slowed down in a short period of time, then a very large impulse would need to be exerted on that object to decelerate it. However, if the light object is slowed down in the same period of time, then a smaller impulse would been to be exerted on that lighter object to decelerate it.

    In order to reduce any impact force, such as a calcaneus hitting an object (whether that object is an orthosis, a shoe midsole or a concrete floor), we can use the concept of impulse to understand how to minimize the compression forces on the plantar calcaneus during that impact.

    One way to reduce the impact forces on the plantar calcaneus is to increase the time over which the calcaneus is decelerated to a zero velocity. This could be done by any of the following methods to reduce plantar calcaneus impact forces:

    1) by adding a 3 mm piece of neoprene on top of a foot orthosis that decelerates the calcaneus to zero translational velocity in 10 msecs,

    2) by having the foot inside a shoe with a 15 mm thick midsole made of EVA that decelerates the calcaneus in 50 msecs, or

    3) by having the calcaneus land on a mini-trampoline that decelerates the calcaneus over a 45 mm distance in 150 msecs. [Times of collision are approximate and used for illustration purposes here.]

    As long as we are "extending the time of collision" of the plantar calcaneus from its initial translational velocity to its zero translational velocity condition, then, by definition, the average compression forces acting on the plantar calcaneus during that collision will be reduced in magnitude.

    The second consideration for peak compression forces acting on the plantar aspect of the calcaneus (at the medial calcaneal tubercle) is how plantar pressures are distributed on the calcaneus by the material that contacts the plantar skin of the calcaneus. A material that conforms to the plantar calcaneus will increase the surface area of contact between the plantar calcaneus and will, all other factors being equal, reduce the magnitude of pressures acting on the plantar calcaneus.

    Remember that pressure is force over surface area (P = F/A). Therefore, for a given impact force between the plantar aspect of the calcaneus and ground, if the material interfaced between the plantar calcaneus and the ground (e.g. foot orthosis, silicone heel pad, shoe midsole) is congruous to the plantar aspect of the calcaneus, then the surface area over which the impact force is acting on the plantar calcaneus will increased and the plantar pressures will, therefore, decrease in magnitude on the plantar calcaneus.

    Now, since pressures and stresses have the same units of measurement (i.e both are measured in Pascals) and are the physical factors that cause tissue damage, then all of these mechanical principles must be considered when asking how cushioning materials effect the plantar forces and pressures acting on the plantar calcaneus during weightbearing activities.

    Hope this helps.:drinks
     
  13. efuller

    efuller MVP

    To complicate things further, part of the force goes deceleration and part goes to resisting the force of gravity acting on the body. So, when the heel has a downward velocity just prior to contact, this mass x velocity is stopped by an impulse. At the same time gravity is pulling the whole body downward and the ground is pushing the body upward an equal amount (when the acceleration has stopped.) So in static stance there is force on the heel even though there is no acceleration. The heel can't tell the difference between the acceleration part of the force and opposition of gravity part of the force.

    If I recall the Bobbert paper correctly, they calculated a landing force peak, in running, that correlated to the mass of the lower leg. The mass of the body above the lower leg was slowed by knee flexion and didn't contribute much to the initial force peak. If I recall correctly....

    Eric
     
  14. David Smith

    David Smith Well-Known Member

    I know your trying to express the concept of heel strike force in an understandable way but it should be iterated that any force must by definition be the result of acceleration since f=ma. When there are two equal and opposing forces (as there always are) then the two equal acceleration cancel out so there is no displacement. There is one exception which is when the opposing force is inertial force sometimes described as a virtual or fictitious force because it only appears as the result of an accelerated mass, when there is an inertial force then the displacement is in the opposite direction to the inertial force and acceleration is at the rate of iF/m.

    The initial heel strike peak in the force time curve (FTC) of GRF, often know as the claw back peak, can be imagined to be some function of the foot and shank mass, i.e. lower leg below knee, if one assumes no moments are transmitted across the knee at the time of claw back. Claw back if that part of gait where just before initial contact the foot drops back in the opposite direction to the leg swing direction and is like a hammer strike. This is the small spiked peak seen about half way up the initial FTC curve.

    The next part of the FTC is a function of CoM velocity and how fast it is braked, i.e. acceleration in the opposite direction to the CoM velocity. This is seen as the first major peak of the 'M' shaped FTC curve (in normal gait) and this curve will be attenuated by the relative stiffness of the knee joint and the hip joint. If the hip and knee joints are compliant then the FTC will be very attenuated, i.e. low peak and extended time, if the knee and hip are stiff then the FTC has a higher peak for a shorter time, of course there are all permutations in between.

    Now if you add a soft landing by the use of a compliant shoe sole or insert the this will attenuate the FTC even further but the CNS may then decide to stiffen the knee and hip joints so maintaining the same FTC as without the cushioning, however since we assume no transmission of moments across the knee at initial heel strike claw back then, at this point, there is no stiffening relative to muscular action mediated by the CNS and so the cushioning under the heel will tend to attenuate the claw back peak.

    hope that is clear enough

    Regards Dave
     
  15. efuller

    efuller MVP

    Dave, I liked the rest of your post, but have quibble about on part.
    Take static stance. There is no acceleration. The force of gravity is pulling the body downward and ground reaction force is pushing the body upward. Therefore, you can have force without acceleration. You have to take this into account when your measuring device is measuring force as we were discussing heel strike. You just can't say that there is no force because the accelerations cancel. The needle on the scale has moved.

    In terms of Newton's third law; for every action there is an equal and opposite reaction, you have to clearly describe where the forces are applied. I think you get this Dave, but for the others. Say that person who was standing in static stance decides to jump. They bend their knees and accelerate downward and ground reaction force will decrease. Then they extend their knees and plantar flex their ankles. As they do this ground reaction force will be greater than body weight and the fact that ground reaction force is greater than the force of gravity will cause the person to accelerate upward. So, gravity does not equal ground reaction force in this instant. However, the force from the foot applied to the ground is equal the force from the ground applied to the foot.
     
  16. Eric:

    I think we must be very clear of our terminology here. Gravity is a constant acceleration of the masses of our bodies here on Earth. Therefore, there is acceleration in static stance.

    I think Einstein's thought experiment for general relativity of the man in the elevator being accelerated upward in space is very instructive in these types of discussions.

    http://www.thebigview.com/spacetime/spacetime.html
     
  17. efuller

    efuller MVP

    Ok, I can see how you can say the accelerations cancel. But how does that help us explain the reading on the scale or force platform? When we stand on a scale the scale thinks that there is a force applied to it. If we put some force sensors in our shoes we will get a different force measurement when we land from a jump on earth as compared to being in space and "jumping" off one wall and "landing" on another wall. In space, once we have decelerated the force sensors will read zero and on earth they will read body weight.

    Eric
     
  18. Actually a new thread would probably be better for this discussion. If you want to start one Eric, then you, Dave and I can probably discuss this problem to greater benefit of everyone following along.
     
  19. That is because, on Earth, our mass is being continuously accelerated by gravity. Without acceleration, there is no force, regardless of the mass. F = ma.

    In space, with negligible gravitational acceleration, the force sensor will read zero, since acceleration is close to zero.

    I know you know this, Eric. However, I'm not quite sure where we are in disagreement with each other or where you are in disagreement with Dave. confused:
     
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