- #1
Simon Spooner
- 2
- 0
Please be gentle, I'm not a physicist, I'm a podiatrist trying to find the answer to a problem that has been confusing me! I hope you can help.
In podiatry we often use foot orthoses in an attempt to alter the magnitude, location and timing of reaction forces at the foot's interface. A simple 2D cross-section model of the heel section of an orthotic could be a wedge of homogenous material. If we assumed that under loading conditions the wedge of material functions in a linear fashion in accordance with Hookes Law, then effectively we could model the wedge as a number of springs increasing in resting length from the thin end of the wedge to the thick end. As the resting length of each of these "spring" columns of material increases so the spring constant (K) should decrease.
If the heel of the foot is modeled as an homogenous mass and impacts vertically downward onto the wedge then the foot should come into contact with the highest side of the wedge first. Assuming that the foot does not rotate but continues to load onto the wedge until static equilibrium is achieved, there should be a longer period of compression between the foot and the wedge at this point of initial contact than lower areas of the wedge-foot interface, which came into contact later as the foot and wedge deformed under loading. Yet, there should also be greater vertical linear displacement per unit load at the thick end of the wedge than at the thin end- right?
So, while the thick end of the wedge may have deformed more than the thin end by the time static equilibrium exists (velocity = zero) between the foot and the wedge, but the thin end of the wedge will push back with a higher reaction force per unit deformation. So would the centre of pressure always be under the point of initial contact at the time of static equilibrium between the foot and the wedge?
I hope that makes sense. My maths isn't so hot, so explanations without too much algebra would be really helpful. Many thanks for any help you can provide.
In podiatry we often use foot orthoses in an attempt to alter the magnitude, location and timing of reaction forces at the foot's interface. A simple 2D cross-section model of the heel section of an orthotic could be a wedge of homogenous material. If we assumed that under loading conditions the wedge of material functions in a linear fashion in accordance with Hookes Law, then effectively we could model the wedge as a number of springs increasing in resting length from the thin end of the wedge to the thick end. As the resting length of each of these "spring" columns of material increases so the spring constant (K) should decrease.
If the heel of the foot is modeled as an homogenous mass and impacts vertically downward onto the wedge then the foot should come into contact with the highest side of the wedge first. Assuming that the foot does not rotate but continues to load onto the wedge until static equilibrium is achieved, there should be a longer period of compression between the foot and the wedge at this point of initial contact than lower areas of the wedge-foot interface, which came into contact later as the foot and wedge deformed under loading. Yet, there should also be greater vertical linear displacement per unit load at the thick end of the wedge than at the thin end- right?
So, while the thick end of the wedge may have deformed more than the thin end by the time static equilibrium exists (velocity = zero) between the foot and the wedge, but the thin end of the wedge will push back with a higher reaction force per unit deformation. So would the centre of pressure always be under the point of initial contact at the time of static equilibrium between the foot and the wedge?
I hope that makes sense. My maths isn't so hot, so explanations without too much algebra would be really helpful. Many thanks for any help you can provide.
