burnthesheep wrote:
You could model it in mechanical vibrations as a "spring-mass-dashpot" system.
Imagine your leg to be like the spring/shock absorber system on a car and your body is the mass. That system has a mass, a spring constant for each connected component, and dampening for each component.
Imagine also the car system going over a speed bump and exerting a force into the system.
That's about the easiest to understand I can convey it in relating it to running. The spring constant "K" of the asphalt or concrete or steel plate is going to be extremely similar since it is effectively a rigid body in the system. The shoe, sock, the skin and fat and muscles in your foot, the muscles/tendons and other things up through your leg........they all will have an effective dampening and spring constant.
Those are not rigid. They measurably give.
To have a spring constant to begin with, the item has to deform. You apply force to a spring, it compresses or extends. Same for a shoe, the fat/skin/muscles in your foot, and the muscles and tendons in your leg.
Asphalt and concreted do have a spring constant "K", but it is essentially infinite in both cases. Large forces and minor deformation: F=kx, so k=F/x.......F super large and x is infinitesimally small......so assume infinity.
In the vibrations formulas, the springs in series (your shoes, the surface, your legs/foot/sock) are: 1/keq = 1/k + 1/k.......
If the second "k" is effectively infinite (concrete or aspahalt), the equivalent spring constant is equal to just the spring constant of the rest of the system. One divided by infinity is effectively zero. Complete the equation and you're left with the original spring constant of the rest of the leg/foot/sock/shoe spring constants.
This is why it is nonsense about the asphalt versus concrete.
Thanks guys. I appreciate it. I never thought I would say this but boy I wish I had taken physics in high school. :)