N = newtons BW = body weight *data has been provided in the table*

1. compare the two maximal forces point 2 and four, between each condition with and without shoes. Briefly comment on any similarities and differences.

2. Compare the time taken from the point of heel-strike to the point of the heel-strike transient between the conditions of without shoes and with shoes. What does the time for the with shoes condition tell us in comparison to without shoes?

3. how is it possible that the force can be over 535N on the traces?

4. how can you apply these findings to athletic training or recreational runners?

5. Radin and others found the following heel strike transient rates: knee pain group = 68 BW/s and normal group = 48 BW/s. compare our findings with Radin what is the take-home message of this lab?

Without Shoes Condition Point N BW 1 240 45 With Shoes Condition N BW 250 .47 2 565 1.06 590 1.10 410 3 .77 400 1.75 2 Without Shoes Condition 4 575 1.07 With Shoes Condition 560 1.05 Measured distance (mm) x one second (s) Measured distance for one s (mm) Without Shoes Condition 1 = __.01 sec With Shoes Condition t = _.05 sec 5. Divide the heel-strike transient force in BW (point one) by the time you determined from number 4 to find the rate of the heel-strike transient force for each condition Heel-Strike Transient Force Time Without Shoes Condition 45/01= 45 BW/s Tith Shoes Condition 47/.05= 9.4 BW/S Without Shoes Condition Point N BW 1 240 45 With Shoes Condition N BW 250 .47 2 565 1.06 590 1.10 410 3 .77 400 1.75 2 Without Shoes Condition 4 575 1.07 With Shoes Condition 560 1.05 Measured distance (mm) x one second (s) Measured distance for one s (mm) Without Shoes Condition 1 = __.01 sec With Shoes Condition t = _.05 sec 5. Divide the heel-strike transient force in BW (point one) by the time you determined from number 4 to find the rate of the heel-strike transient force for each condition Heel-Strike Transient Force Time Without Shoes Condition 45/01= 45 BW/s Tith Shoes Condition 47/.05= 9.4 BW/S