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waynexk8
waynexk8 is offline
#24
Jan14-12, 11:04 AM
P: 399
Quote Quote by douglis View Post
This can be understood even by common sense which is something you're not familiar with.
You throw an object when it's leaving your hands with velocity.You lift an object when it doesn't leave your hands so by definition at the end of lifting the velocity is zero and that's called terminal velocity.
Yes right.
As even with 80% moving it for 30 inch at 2m/s the bar will not leave your hands, so yes terminal velocity is zero.

Quote Quote by douglis View Post
Exactly!So you go from zero velocity to zero velocity
Here is where I cannot understand why you think this. As in our case, when you move a weight from a to b, you “must” start at zero, accelerate/decelerate go back to zero, for the transition, the accelerate/decelerate and so on. You can’t go from zero to zero. You “are” missing out the “actual” momentum/movement that we are debating/concerned about, the forces we are talking about are in-between both zero movements, and you talk about the zeros ???

Quote Quote by douglis View Post
....the change in momentum is zero.So the net impulse is zero and the net force zero too.
The change in momentum/movement can “not” be zero, as there “is” momentum/movement in-between both zeros, or from a to b.

What you say net force, as I said last time, you are on about the force moving the weight and the weight pushing back, the opposite reaction forces. We are “not” concerned with both forces cancelling each other out, we are concerned with the forces from the pushing force the muscles, which causes tension on the muscles, and the opposite reaction forces from the weight. So in our case they are both creating force from and on the muscles, thus we need to add these forces up, as of the tension on the muscles. You seem to be saying they cancel each other out, thus not force was applied from the muscles or from the weight and not tension on the muscles.

You hold a weight half way up, you then move the weight up 20 inch in the time you held it, in say 20 seconds, basically the same forces from your muscles, thus tension on your muscles. However when I am accelerating the weight up and down 20 times in the same time frame, I have {if we say for at least the acceleration is for 60% of the concentric} accelerated the weight 240 inch for just the concentric, how on Earth you can think you have used the same total or overall force in doing this is beyond me. As I asked before, if you claim to use the same total or overall forces, and they make up or balance out in the end, why does not the distance make up or balance out ??? Why do I use far far far more energy, why do I do more work, and work is the product of a force times the distance through which it acts, and it is called the work of the force. Also why do my EMG reading produce higher reading for the faster repetitions done in the same time frame.

Quote Quote by douglis View Post
The forces "make up".....the average applied force is equal with the weight regardless the length of the acceleration/deceleration phases and the distance covered.
You say this without any proof, evidence or an explanation, as we are on the physics forum, I think you need to at least try and prove this with proof, evidence with an explanation. I also repeat, why is the EMG higher ???

Quote Quote by douglis View Post
The ONLY forces available are the muscle force and the weight.There're no additional action-reaction forces to add as you fantasize.
Maybe you misunderstood, as the weight with the muscle force pushing up it, is the action-reaction forces !!!

Quote Quote by douglis View Post
Read carefully Phantomjay's last post.He's explaining perfectly that and your irrelevant clay example.
Will do that right now……….you mean this, I quote;
Phantomjay wrote;More force goes into the muscles only during the accelerating phase.
True.
Phantomjay wrote;For the slow case, you might have 110 pounds at the beginning during the accelerating phase...for the fast case, it might be 150 pounds during the accelerating phase, so more squooshh...and clay doesn't rebound.
Yes right, more squash, or should I say more compression on the clay for the fast. No, the clay does not rebound; it does not whatever you do to it, not sure why you say this.

Phantomjay wrote;The averrage force is still 100 pounds for both cases. But yes, at some point during the fast case, your muscles are receiving (and giving) more force during the accelerating phase, because it requires more force to accelearte to the higher speed in the same distance. Now I bet your really confused.
Yes this what it seems to say on paper, “however” I have contended from the beginning of this debate on other forums, “how” can you know, that as we all agree the fast repetition is using more and a LOT more force on the first say 60% {remember the fast does 6 repetitions in the same time frame as the slow moving the weight 6 times more the distance, using far more energy} than the slow, then when the fast is on its deceleration, it’s using less force than the slow, {and here is my main point} “how” do you know/think or work out, that the slow force makes up or balances out the forces here ??? as the slow force is “never” as high as the higher force of the fast acceleration forces, in my opinion the whole forces are not linear and cannot make up or be balanced out, as of my EMG reading, the more work/distance moved and more energy used.

It’s like a impulse; a small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important. So I say the small force would need far far far more longer time to make up or balance out the large force made by the faster accelerations of the faster repetitions.

That’s why you “always” fail far far far faster doing faster repetitions to slow, as the high acceleration forces, put far far far more tension on the muscles than the slow in the long run.

Quote Quote by douglis View Post
I'm out of this thread cause it's obvious you have some kind of obsession and I'm wasting my time.I hope Phantomjay will have the patience to explain it further to you.
I think/know we both have an obsession with this debate. I like to learn, and have learnt very much. I would have thought that as my other things as well as my EMG reading prove you very wrong, and “you” would like to know “WHY and HOW” they show you are wrong, do not you like the truth ??? Zula is constructing a machine to find out the forces that both velocities use.


Quote Quote by douglis View Post
page 459,figure 6a.
PM me for further explanation cause that's irrelevant with the thread.
It’s says nothing about the repetition taking 1 second on 6a, but does state very clearly on the page I gave in a graph that the whole concentric took 1.5 seconds, it’s as clear as day.

If I move a weight faster in the same time frame, and then I move the weight far far far faster in the same time frame, the acceleration will be higher “and” longer making for lager force impulse that in my opinion cannot be made up by the slower.


Yet again you miss out very many of the main issues of this debate, what about the weight and string test on the weight for deceleration ???


Wayne