S.Litsos, E.Verzili Laboratorium 1 - Frorce Platform - Biomechanics

7/29/2019 S.Litsos, E.Verzili Laboratorium 1 - Frorce Platform - Biomechanics

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FINAL REPORT FOR FORCE-PLATE LAB TEST GROUP 4 -15 SEPTEMBER2011

Students:

Stavros LitsosEmiliano Verzilli

ANSWERS TO THE QUESTIONS

1. Calculate the average weight of the quiet standing trial.

1. The average value of the quiet standing trial is the sum of all measurementsdivided by the number of intervals and is 575,22 (N). This is also the averagevalue of the weight.

2. Create one figure showing 3 graphs: vertical GRF during quietstance, during walking and during running. Discuss interestingfeatures about these graphs.

2. See "3 graph=stance, running, walking" sheet.

By looking at the 3 graphs we notice immediately that the slope representingthe stepping moment is more rounded in the quiet standing trial than in thenext two graphs; the reason for this is that the time the force is applied to the

plate is longer in the quiet standing trial than in the following two tests.Another noticeable aspect of these graphs is that the slope becomesprogressively steeper from the standing trial graph to the running graph; thereason for this is that the magnitude of the force applied to the plateincreases.Finally, we can point out that the running graph has one peak while thewalking one has two. This is due to the fact that the walking graph shows boththe heel and the toe touch. On the other hand, the running performance isrepresented by one peak due to the velocity of the performance. What itseems like a second peak in our graph is only due to the instability related to

the shoes used by our performer.

3) Create two graphs of Ground Reaction Force (GRF) vs Time (one foreach jump Draw and label lines on the graphs to represent thefollowing:

1. Body Weight line2. Takeoff (TO)3. Landing (L)4. Flight Time (FT)


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3. See "countermove" sheet and "squat" sheet for answers to questions 3.1-3.4)

3.5) Impulse up to the point of takeoff?

Countermove jumpANSWER -168,37 (Ns) WITH TAKE OFF TIME AT 3,563 s found graphically(THE SUM OF ALL IMPULSES FOR INTERVALS UP TO TAKE OFF)

SquatANSWER -424,31 (Ns) WITH TAKE OFF TIME AT 3,66 s found graphically(THE SUM OF ALL IMPULSES FOR INTERVALS UP TO TAKE OFF)

3.6) Take off velocity ?

Countermove jumpANSWER 2.87 m/s

(WE USE THE IMPULSE LAW F*t=m*v; first we calculate the mass m from theaverage weight w= 575,22 (N) and the acceleration of gravity and we obtainm= 58,63 kg; then we know that F*t at take off is -168,37 Ns, therefore v=F*t/m = 2.87 m/s)

Squat jump

ANSWER 7.23 m/s

(WE USE THE IMPULSE LAW F*t=m*v; first we calculate the mass m from theaverage weight w= 575,22 (N) and the acceleration of gravity and we obtainm= 58,63 kg; then we know that F*t at take off is -424,31 Ns, therefore v=F*t/m = 7,23 m/s)

3.7) Flight time?

Countermove jump

ANSWER 0,58 s

(WE USE ONE OF EQUATION OF CONSTANT ACCELERATION ;the jumper is only affected by gravity when airborne; his take off velocity isknown and set at 2,87 m/s; his velocity at the apex of his flight is zero;therefore we can use the equation 0=v+gt, 0=2,87m/s - 9,81m/s2 * t, t= 0,29s; since the projection height and landing height are equal the time it takes toreach the apex is one half of the total flight time and FT=2t=0,58 s).

Squat jumpANSWER 1.52s

(WE USE ONE OF EQUATION OF CONSTANT ACCELERATION ;


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the jumper is only affected by gravity when airborne; his take off velocity isknown and set at 7,5 m/s; his velocity at the apex of his flight is zero;therefore we can use the equation 0=v+gt, 0=7,2m/s - 9,81m/s2 * t, t= 0,73s; since the projection height and landing height are equal the time it takes toreach the apex is one half of the total flight time and FT=2t=1.46s).

3.8) Flight height?

Countermove jumpANSWER 0,41 m(same reasoning as before but this time we need to find the maximum height;therefore we use the equation 0=v12+2gh with v1=2,87m/s and g=-9,81m/s2;we find that h= 0.41 m)

Squat jump

ANSWER 2,66 m(same reasoning as before but this time we need to find the maximum height;therefore we use the equation 0=v12+2gh with v1=7,23m/s and g=-9,81m/s2;we find that h= 2.66 m)

Answer in a few words the following questions:

1. What does a longer flight time imply about jump height?

1) the jump height is higher

2. Compare the calculated and measured (from the graph) flight time.What might be the reason for differences?Countermove: FT from graph is 0,62 s; calculated FT is 0,58 s.Squat: FT from graph is 0,59 s; calculated FT is 1,46s.The reason for differences might be approximation related to measuringdirectly from the graph.

3. Compare the force-time curves for the two vertical jumps. What

might be the reason for the differences in jump height?

The following explanation is based on both mechanical and physiologicalreasons.

In the countermoving jump the total mechanical work performed is greaterthan in the squat jump.How high a person can reach after a jump, depends on the Kinetic Energy attake-off which is equal to the mechanical work carried out during the jump(see principle of work and energy).


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The mechanical work (F.s) in a countermoving jump is greater because F (themuscular force) is greater and this is due to two important physiologicalreasons:

- the ability to store elastic energy in the eccentric phase increases;

- stretch reflexes leading to the activation of higher-lying motor units areactivated.

See Hills curve.

As a result the Kinetic Energy Ek=1/2mv2 at take-off increases; because themass stays the same it is v2 that increases, in other words the take-off

velocity increases.Because the jump height is given by the equation h=v2/2g, we see that as theKinetic Energy increases so does the jump height.

(NB: the above equation for the jump height comes from the principle ofconservation of mechanical energy applied to the flight; we have that theEnergy at take off is equal to the Energy at the peak;that is Ek + Ep at take off = Ek+Ep at peak; that is 1/2 mv2 + mgh at take off= 1/2mv2 + mgh at peak;then we take away m from both sides of the equation and set v at peak to 0 sothat we only have1/2 v2 + gh at take off = gh at peak,gh at peak - gh at take off= 1/2 v2 ,h at peak - h at take off=v2/2g,

h=v2/2g)

NB: we are actually getting higher values for FT, flight height and TakeOff Velocity for squat jump rather than countermove. The reason tothis unusual results ia at present uknown to us.Besides, we want to point out that there was an error in the execution

of the countermove jump because the performer swung his arms backrather than keeping them on the hips.


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S.Litsos, E.Verzili Laboratorium 1 - Frorce Platform - Biomechanics