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Fast And Slow Muscle Fiber Conundrum

  1. Nov 18, 2008 #1
    Question 1,

    WHICH SET WILL PRODUCE MORE, AND MORE OVERALL FORCE FROM THE SLOW MUSCLE FIBERS, THUS IN TURN MORE TENSION TO THEM.

    A slow set of slow reps 3/3, using 75% of your 1RM, let us call this 100pounds, 6 repetitions = 36 seconds.

    You fail faster in the faster rep with the same weight.

    A fast set of fast reps, .5/.5, using 75% of your 1RM; let us call this 100 pounds, 25 repetitions = 15 seconds.

    Overall poundage moved.

    Slow set 600 pounds,

    Fast set 2500 pounds.

    Question 2,

    How much force would the slow muscle fibers be producing in the slow set and the fast set ??? Low, medium, high or all.

    I myself would say the slow muscle fibers generate far more force in the faster rep.

    3/3 and 1/1 means three seconds up for the positive and three seconds down for the negative in each repetition of the set. 75% means using 75% of a person’s 1 repetition maximum attempt in a lift.

    Thank you for your time.

    Wayne
     
  2. jcsd
  3. Nov 20, 2008 #2
    I am quite surprised at no comments.

    What if I changed the question slightly, too, which set will produce more overall tension to the slow muscle fibers, also fatigue or work the slow muscle fibers harder. And make them try and produce more force output.

    The power/strength overall output in the faster/slow reps and sets.

    Lets calculate how much power/strength I would be used on both rep speeds. Distance weight moved 1.85 M. Since we are using the metric system, we will first need to convert the mass of the barbell into kilograms (200 lb ÷ 2.2 = 91 kg). Secondly, to determine the force we will need to figure out what the weight of the barbell is (W = mg = 91 kg x 9.81 m/s² = 892 kg.m/s² or 892 N). Now, if work is equal to Force x distance then, U = 892 N x 1.85 m = 1650 Nm.

    We can calculate that lifting a 200 lb barbell overhead a distance of 1.85 m required 1650 J of work. You will notice that the time it took to lift the barbell was not taken into account.
    Let us only add up the positive part of the lift.


    The concept of power however, takes time into consideration. If for example, it took .5 seconds to complete the lift, then the power generated is 1650 J divided .5 s = 3300 J/s.


    If it took 2 seconds to complete the lift, then the power generated is 1650 J divided 2 s = 412.5 J/s.



    Slow set,

    412.5 x 6 = 2475Joules.



    Fast set,

    3300 x 25 = 99000Joules

    Thank you for your time.

    Wayne
     
  4. Nov 23, 2008 #3
    I think I have solved the problem.

    On asking your opinions on which set will produce more overall tension, stimuli, stress to the slow muscle fibers ONLY, also fatigue or work the slow muscle fibers harder.

    And also make them try and produce more overall force output, in free weights and their dynamics, not Isokinetic/Isometric, so please no studies from Isokinetic Machines that have no relevance to this debate, if you think they have, please say why when you use them.

    And when moving free weights, as long as you are not breaking terminal velocity, a faster movement will result in more force being used. A force is required first to maintain static equilibrium and second to generate acceleration.

    Additional force results in acceleration of a mass, since the slow movement reduces acceleration; the force used to accelerate the object is near zero above the load its self.

    A slow set of slow reps 3/3, using 75% of your 1RM, let us call this 100pounds, 6 repetitions = 36 seconds.

    You fail faster in the faster rep with the same weight.

    A fast set of fast reps, .5/.5, using 75% of your 1RM; let us call this 100 pounds, 25 repetitions = 25 seconds.

    Overall poundage moved.

    Slow set 600 pounds,

    Fast set 2500 pounds.

    Lets calculate how much power/strength I would be used on both rep speeds. Distance weight moved 1.85 M. Since we are using the metric system, we will first need to convert the mass of the barbell into kilograms (200 lb divided 2.2 = 91 kg).

    Secondly, to determine the force we will need to figure out what the weight of the barbell is (W = mg = 91 kg x 9.81 m/s = 892 kg.m/s or 892 N). Now, if work is equal to Force x distance then, U = 892 N x 1.85 m = 1650 Nm.

    We can calculate that lifting a 200 lb barbell overhead a distance of 1.85 m required 1650 J of work. You will notice that the time it took to lift the barbell was not taken into account.

    Let us only add up the positive part of the lift.

    The concept of power however, takes time into consideration. If for example, it took .5 seconds to complete the lift, then the power generated is 1650 J divided .5 s = 3300 J/s.

    If it took 2 seconds to complete the lift, then the power generated is 1650 J divided 2 s = 825 J/s.

    Slow set,
    825 x 6 = 4950Joules.

    Fast set,
    3300 x 25 = 82500Joules

    Wayne
     
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