What would be the percentage/degree of absorbancy per impact?

Like walking, if the weight is 60pounds x 3= 180 pounds but lets say the walking is on a concrete surface
What amount of absorbancy is needed to absorb the impact in both directions?
And is it correct that concrete inflicts more of an impact back then that initially given and more than dirt would?

I'm not sure what you're asking here, but you can see the difference when working out the force applies to the foot by the ground:

Just use F=ma, you can see that when you hit concrete, because it is more solid, a is larger and therefore so is the force, You can feel that it's harder when you walk on it because of this.

For dirt, it slows your foot at a lower rate, because it's soft, and so the force is less.

Please note that while the force applied is different, the dirt applies the force for a longer time, so that the Impulse (I=Ft) is the same in both cases.

Does this help or were you asking something else? Please clarify.

Like a foot in a shoe,
the shoe absorbs so much shock from the initial impact from within and outside of the impact?
If the impact of walking is 3 x's ones weight, say a child at 60 pounds x 3= 180
If this is the impact going out, Is it the same or more coming back?

How do I determine how much padding is needed per weight to Reduce the shock felt in either direction?
Is there a method that you can put in a word formula and not abbreviations, to help me better understand?

Does this make more sense? Sorry for the confusion. Thank you for the clarity on concrete and dirt, that does help me with that part of my question.

The impact your foot feels can be calculated from F=ma, that is: the force on your heel equals your mass times the acceleration the heel experiences as you walk. One way to calcuate the acceleraton is using the formula a = v^2/2d, where v is the velocity your foot is moving vertically at the moment of impact with the floor, and d is the distance the feel continues to move vertically after the shoe touches the floor due to the deflection of the heel and shoe padding. Putting it together :

F = mv^2/2d

Don't know how you'd calculate v, which remember is the vertical velocity of the person's foot when the heel strikes the ground. But as an example - suppose that the child runs at 5 MPH (which is 7.3 ft/sec), and that the vertical speed of the child's foot when it hits the ground is 20% of that - I have no idea whether this is correct or not, but let's go with it. Further, assume that the foot depresses into the shoe padding 1/8 inch, and that the child's heel also deforms 1/8 inch upon impact. That's a total distance of 1/4 inch. So the vertical velocity of the foot must be stopped in 1/4 inch. Do the math: a = (0.73 ft/s)^2/(2*.25*1/12 ft) = 51.2 ft/s^2, or about 1.6 times gravity. So in this example the 60 pound child feels a force of 1.6 * 60 = 96 pounds on his heel when his foot strikes the ground. That shock is felt stating from the moment the paddings starts to deform, through the point when it is fully deformed (absorbed the full blow of the impact), and will continue in a diminished capacity as the padding rebounds until it is holding the child's weight alone with no more dynamic forces. If the padding was perfectly elastic, then it would shove the child back into the air due to its "springiness."

Sparklejunk, you should reply in a thread where you got an answer if you have more questions. I will move this topic and answers to your previous question, as it is a continuation.