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Homework Help: Eccentric to the concentric force calculation

  1. Apr 17, 2010 #1
    Hi all,

    I was trying to work out the force needed to lift a 100kg weight. However the lifter first is lowering this weight at .5 of a second, in 1m, and at the last instant has the accelerate this weight up 1m at .5 of a second.

    Someone worked out for me the lifting and lowering of a 100kg for 1m each way at .5/.5 and 2/4. However, as the force need is going to be higher in the transition from the eccentric to the concentric, in each repetition of the lifting and lowering in weightlifting, I was hoping to work this out, as it’s a bit over my head. It’s nothing to do with homework by the way, I have posted similar questions before, it’s just for a debate we are having. Thx in advance for your time if anyone is able to work it out, here are the calculations a friend worked out, hope they are right not sure.


    Let's assume we're comparing 2 sets using the same weight, but different rep speeds. set 1 uses 0.5/0.5 and set 2 uses 2/4. For simplicity's sake we'll assume the weight accelerates 100% of the way up and down for both sets.

    Known: Mass(m) =100kg (220lbs) Acceleration(a)=?? Distance(d)=1m

    First let's solve for the acceleration required to move the weight 1m in the time frames of the sets.

    Calculate a, to travel 1m in 0.5s (Raising and Lowering of set 1):
    d=1/2at^2
    1m=1/2*a*(0.5s)^2
    a=2(1m)/(.25s^2)
    a=8 m/s^2

    Calculate a, to travel 1m in 2s (Raising set 2):
    d=1/2at^2
    1m=1/2*a*(2s)^2
    a=2(1m)/(4s^2)
    a=0.5 m/s^2

    Calculate a, to travel 1m in 4s (Lowering set 2):
    d=1/2at^2
    1m=1/2*a*(4s)^2
    a=2(1m)/(16s^2)
    a=.125 m/s^2


    Now lets solve for the forces required to accelerate the weight

    Calculate the force required to raise 100kg, 1m, in 0.5s:
    Sum of forces=ma
    F1-mg=ma
    F1-(100kg)(9.81m/s^2)=(100kg)(8m/s^2)
    F1=(981+800) kg m/s^2
    F1=1781 kg m/s^2 required to raise the weight 1m in 0.5s

    Now compare it to the force required to raise 100kg, 1m, in 2s
    Sum of forces=ma
    F2-mg=ma
    F2-(100kg)(9.81m/s^2)=(100kg)(0.5m/s^2)
    F2=(981+50) kg m/s^2
    F2=1031 kg m/s^2 required to raise the weight 1m in 2s

    ***As predicted, it takes more force to raise the weight 1m in 0.5s than it does to raise it in 2s. It takes F1/F2=1.73 times as much force to do so. You're right about this but nobody is disagreeing with you here

    However, now let's look at what happens on the way down.

    Calculate the force required to lower 100kg, 1m, in 0.5s:
    Sum of forces=ma
    mg-F3=ma
    (100kg)(9.81m/s^2)-F3=(100kg)(8m/s^2)
    F3=(981-800) kg m/s^2
    F3=181 kg m/s^2 is required to lower the weight 1m in 0.5s

    Now compare it to the force required to lower 100kg, 1m, in 4s
    Sum of forces=ma
    mg-F4=ma
    (100kg)(9.81m/s^2)-F4=(100kg)(.125m/s^2)
    F4=(981-12.5) kg m/s^2
    F3=968.5 kg m/s^2 is required to lower the weight 1m in 4s

    ***Contrary to your belief, it takes MORE force to lower the weight 1m in 4s than it does to lower it in 0.5s. In fact, it take A LOT more. It takes F5/F3=5.35 times as much force to do so! That's 5.35 times more force to go slow than fast!!!!



    Wayne
     
  2. jcsd
  3. Apr 17, 2010 #2
    First, you can't assume that you can accelerate 100% of the way up or down. There must be a deceleration portion to each move. To minimize your accelerations, assume that accel time and decel time are equal (accel and decel distances are also equal). For the 0.5 second raise (or lower), assume the accel time to be 0.25 sec, and the accel distance to be 0.5 m. Recalculate all acceleration rates based on these numbers (deceleration rates will be identical in magnitude, opposite in sign).

    Second, once you pick a direction as positive, you must to stick with it. Assume up is positive for force, displacement, velocity and acceleration. Gravity will thus be negative.

    Your final statement is incorrect (regardless of typos and other errors); if you accelerate 100 kg down at 8 m/s^2, you will still have to resist gravity (push up) with 181 kg*m/s^2 (newtons). If you accelerate 100 kg down at 0.125 m/s^2, again you have to push up (against gravity) with 969 N. The "slower" you accel down during your move, the harder you have to push UP (not down). This is all non-sense though since your initial assumption was wrong. You need to recalculate your accelerations, before you can re-visit the resulting force requirements.
     
  4. Apr 18, 2010 #3
    Are we talking about weight lifting here?If so I don't think we can consider that we accelerate for all the way up or down.
    In fact the weight always starts and ends at rest.That means the average acceleration is always zero and the force is always equal with the weight(F=mg) both in the way up and down regardless the speed.
     
  5. Apr 18, 2010 #4
    Hi all and thx.

    Wow this is getting deep and complicated, but that’s all the better.

    First, I would like to say {hope you’re interested, it was not written by me} that if you did not know, the eccentric action, in normal training is always "underloaded" to its capacity. In eccentric actions, the muscle fibers activation distribution is far less, due to the greater efficiency of the act of braking compared to the muscle fibers act of shortening. This difference produces a lower Motor Signal, and fewer MFs are used to control the load/force.

    The strength of the eccentric action lies in its contribution to the concentric/eccentric - eccentric/concentric cycle, and how it contributes force creation and MMMT (Momentary Maximum Muscle Tensions) that are then applied to the elastic components and harvested during the transition to concentric action. This means "if" we are looking to load the eccentric with loads that move closer to the capabilities of the action, and also those which can create the right speed of the action to cause a more effective SSC (Stretch Shortening Cycle) then "allowing" a controlled acceleration to achieve that force in the desired ROM of the eccentric transition to concentric action will provide significant higher force needed by the muscles in eccentric transition to concentric action
    .
    Some would suggest a "slower" lowering of the eccentric load, to somehow capitalize on the aspect of fewer muscle fibers handling the load, but in fact, that then has a tendency to "reduce" the tension, by reducing the "load" seen by the muscle, so it should be best to try for controlled acceleration, say ¾ the way down the eccentric, then to blast up on the concentric, for the highest force to be on the muscles, and you can also get more repetitions done in the same time frame as a slow repetition, thus creating more tensions on the muscles, that’s why I want to find this out. As when a given load is lowered faster, I would have thought the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.

    Just now going to answer the questions from the last post.

    Wayne
     
  6. Apr 18, 2010 #5
    Hi tvavanasd and aris1.

    When we was first talking about this, for the faster rep we were saying that you accelerate 80% of the concentric, and use 20% for the deceleration and transition to eccentric, and the same for the eccentric to concentric transition, and maybe the same for the slower rep ??? But the person working these numbers out did not take that into account, as to make things a little easier. I do not know how to work these into to the calculations, wish I did as I would, hope one of you would be so kind as to do, as the transition from negative to positive in the faster rep is what I am most interested in now.

    Yes up would be positive going against gravity. But you got me where you then said gravity will thus be negative ??? Would not I be resisting gravity more on the way up ???


    Right get that. However, please remember I did not work the calculations out. And when you say The "slower" you accel down during your move, the harder you have to push UP, I was wondering about the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load, as if something is falling slightly faster would not it appear heaver to the muscles ???. Here Per Aagaard says it far better than me, however here he is taking about accelerating up.

    Per Aagaard Professor, PhD
    Institute of Sports Science and Clinical Biomechanics
    University of Southern Denmark

    When a given load is lifted very fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.
    Momentum, or as I like to say, off loading may be an issue, if the weight is to light, and you are not accelerating the weight enough.

    Acceleration is not momentum, its the opposite of momentum, acceleration requires more force/strength, where momentum requires less force/strength.

    For instance, a 120 kg squat can easily produce peak vertical ground reaction forces (beyond the body mass of the lifter) of 160-220 kg's when executed in a very fast manner! Same goes for all other resisted movements with unrestricted acceleration (i.e. isokinetic dynamometers (and in part also hydraulic loading devices) do not have this effect).

    This means that higher forces will be exerted by MORE muscles fiber when a given load is moved at maximal high acceleration and speed - i.e. contractile stress (F/CSA) will be greater for the activated muscle fibers than when the load is lifted slowly....
    best wishes
    Per


    I honestly wish I could do this, but it’s a bit over my head, but I am trying to learn. As I said I was hoping someone would do this for me, and add in the forces needed from the faster rep for the transition from negative to positive, As you know far better than me, its takes far more force to stop a 100kg accelerating, or should we say decelerating down 1m at .5 of a second, then relift it back up 1m in .5 of a seconds, than to just lift it up from rest.

    Again thx for your time, this is getting very interesting. I did find a site that works these things out, but it does not take gravity into account, but I see no point in that.

    Wayne
     
  7. Apr 18, 2010 #6
    SORRY, I should have put this in sooner; I actually made a video of the sort of thing I am actually talking about. However, as you will see on the slower rep I did the negative a little faster than the positives. I think if I had done the negatives a little slower, I would have done that last rep or maybe one more, making time to failer later.

    But in all I failed roughly 55% faster on the faster rep.

    http://www.youtube.com/user/waynerock999

    One more doing lateral raise.

    http://www.youtube.com/user/waynerock999#p/u/12/ywqBFfk5fHo

    Wayne
     
  8. Apr 18, 2010 #7

    sophiecentaur

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    If you want to work out the maximum force and you have a video, look at the distance traveled frame by frame. From that, you can work out the velocity between frames - the difference in velocity between adjacent pairs of frames will give the acceleration.
    Find the maximum over the cycle in the video and then add g to it. Then multiply be the mass of the weight you lifted.
    Force = Mass X Acceleration

    Alternatively use an accelerometer to measure and record the acceleration? They are available to hook up to a PC and will measure a whole range of acceleration values - conveniently, too.
     
  9. Apr 18, 2010 #8
    That sounds very interesting, and would love to be able to do that, is there any online sites telling you how to do this ??? I will have a look around.

    Have to work out how to do the first part first.

    Again, this sounds very interesting, where do I buy one of these please ???

    If anybody has time to work the calculations for me please, that’s if it’s not too difficult and can actually be done ???

    Wayne
     
  10. Apr 18, 2010 #9

    sophiecentaur

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    Even the humble ipod touch has accelerometers in it and there are some (free) apps that can access the data. I'm sure you could find a School Lab data-logging peripheral that would give you what you want. Try an educational supplier in your country.
     
  11. Apr 19, 2010 #10
    This is all very interesting, will have to look into this.

    What I was hoping for now, that’s if it can be worked out, was the force need to stop a 100kg after accelerating down 1m at .5 of a second, and to immediately relift it back up 1m at .5 of a second.

    In other words, I am lowering the 100kg at .5 of a second then at the last say 20%; I then have to stop it and immediately accelerate it back up at .5 of a second for 1m. I need the forces needed for that transition from negative to positive, as I have the force just to lift a 100kg from a standing start, but the forces will but much greater if the weight is all ready dropping that fast, these will be the peak forces of a milly second.

    Fast rep, moving up from a standing start, moving 1m in .5 of a second.
    Concentric force =1781 kg m/s^2

    Wayne
     
  12. Apr 19, 2010 #11
    So you want to find the peak value of force if,for example,let the weight fall free and then at the last inches you try to stop it immediately?
    Maybe you can calculate the speed at this point either by the equation v=gt (if you count the time) or by the equation v=(2gh)^.5 and for h you can put .8m in you example.
    Then from the speed you can find the acceleration and then the force.

    Also I believe that's the force required to throw the weight in the air with speed 2m/sec(1m in .5sec) and not the force to lift the weight which I think is what you're looking for.
    If you just lift the weight means the ending speed is zero and in that case the average force is always 981N(100kgr) either you lift it in .5sec or 2sec.
     
    Last edited: Apr 19, 2010
  13. Apr 19, 2010 #12
    Well not exactly free falling, but coming down as say a speed of .5 of a second per 1m, and at the last 20% I use a force to stop the weight and accelerate up back up 1m in .5 of a second. But I am unsure how to do this. When you say things like the above, you lose me, even that I would love to be able to do the calculations I do not know how, and was hoping someone would do them for me, then I can learn from the other equations I put in and from these.


    Here is what the person who worked then out said they were, they are the forces required to accelerate the weight, one at .5 of a second per 1m, and one at 2 seconds per meter.

    Here is my video, if this helps;

    http://www.youtube.com/user/waynerock999

    The forces required to accelerate the weight of 100kg, one at .5 of a second per 1m, and one at 2 seconds per meter, and now the peak force from the transition from negative to positive.

    Thx for your help. Seems this is harder to work out than I thought, was thinking lots would be able to do this easy, hmm.

    Wayne
     
  14. Apr 22, 2010 #13
    Any one on this ine please ???

    I would have thought if you say lowered 100kg slowely as in 8 seconds for 1m, and then lowred the same weight for 2 seconds, that as of the acceleration component, That when a given load is lifted slightely faster, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by will exceed the nominal weight of the load ???

    But what I was really trying to work out was the below, if you have time please could you help.

    I was trying to work out the force needed to lift a 100kg weight. However the lifter first is lowering this weight at .5 of a second, in 1m, and at the last instant, say the last 20% he has the accelerate this weight up 1m at .5 of a second.

    Someone worked out for me the lifting and lowering of a 100kg for 1m each way at .5/.5 and 2/4. However, as the force need is going to be higher in the transition from the eccentric to the concentric, in each repetition of the lifting and lowering in weightlifting, I was hoping to work this out, as it’s a bit over my head.

    Wayne
     
  15. Apr 24, 2010 #14

    sophiecentaur

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    The force needed to 'lift' a 100kg Mass could be as little as gX100+ whatever is needed to accelerate it.
    This acceleration will, I'm sure, be different from person to person - depending on their optimum lifting speed. You just have to measure their max acceleration. If you were to make a machine to lift something, the force could be only a tiny bit over the straight weight of the object -i.e. you could accelerate it as slowly as you like. But that would Kill your muscles, wouldn't it?
    I think you are, in fact, asking an unanswerable question, although it may appear to be quite straightforward. You might actually have to measure the force. There will be force meters that will register a maximum force, which would be what you need.

    You're correct about it being over your head. That's what weightlifters aim at isn't it - at least for part of the time? :wink:
     
  16. Apr 24, 2010 #15

    sophiecentaur

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    The force needed to 'lift' a 100kg Mass would be gX100+ whatever is needed to accelerate it.
    This acceleration will, I'm sure, be different from person to person - depending on their optimum lifting speed. You just have to measure their max acceleration. If you were to make a machine to lift something, the force could be only a tiny bit over the straight weight of the object -i.e. you could accelerate it as slowly as you like. But that would Kill your muscles, wouldn't it?
    I think you are asking an unanswerable question, although it may appear to be quite straightforward. You might actually have to measure the force. I believe you can buy force meters that register the maximum force.

    You're correct about it being over your head. That's what weightlifters aim at isn't it?
     
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