# Impulse/force in pounds for the time frame

by waynexk8
Tags: frame, impulse or force, pounds, time
P: 148
 Quote by waynexk8 That’s the way I envisage I do lift these weights on my fast repetitions, {will do some new videos if you like ???} I explode up and keep on pushing and pushing; I don’t or did not seem at all I was or had to decelerate, and until me and D. Found that study, we was trying to work this out with me decelerating for only 10 to 20% of the ROM. So could not I fully accelerate up, and then immediately reverse ??? As that’s what it seems I do. Wayne
The length of the acceleration phase is of little importance.As long as the terminal velocity is zero the acceleration is always balanced by the deceleration.Or...with your own words...the forces "make up".

However experimantally was found the acceleration phase lasts for about the half distance with the 80% of your maximum weight and much less with less weight when you try to lift as fast as possible.
In general,the faster the lifting the less the acceleration phase and the grerater the deceleration phase.
BTW at the study you're referring the lifting lasts ~1sec which is perfectly normal with such weight.Check it again.
P: 399
 Quote by douglis The length of the acceleration phase is of little importance.
It is of all the importance.

I just did the test Zula told of me. I attached a weight to the bar, and curled 80% up and down as fast as I could, the string and weight stayed basically the same, meaning high tension and little deceleration. I then accelerated up fast again, but stopped very suddenly, the string went slack and weight with was half a pound jumped up.

 Quote by douglis As long as the terminal velocity is zero the acceleration is always balanced by the deceleration.Or...with your own words...the forces "make up".
Do you mean as if with a lighter weight, and I lifted it so fast that is would leave my hand, are you calling the leaving of my hand, terminal velocity is zero ???

On paper it may look balanced/made up, however how do you know it is ??? As you will agree that a high force for a short time, can produced the force impulse on a low force for a longer time, so why do you think the forces balance out make up ???

If I was to press up into a very thick piece of clay at a high force velocity with an 80% dumb bell in my hand, my hand could go in 2 inch, then there will be an opposite reaction which it called the tension on the muscles, I do this 6 times in 6 seconds and make a 12 inch hole in the clay, very high tensions on the muscles. I then take the same dumb bell and press is up into a very thick piece of clay at a low force velocity with an 80% dumb bell in my hand, I do this one time, the dumbbell would not even go the first 2 inch, but still lots of tension on the muscles.

I think you coined the makeup.

 Quote by douglis However experimantally was found the acceleration phase lasts for about the half distance with the 80% of your maximum weight and much less with less weight when you try to lift as fast as possible. In general,the faster the lifting the less the acceleration phase and the grerater the deceleration phase.
Not the case. As I could accelerate the weight nearly right up if I wanted, as if I was throwing the weight, and if I did this with a light weight, there would be no deceleration until I left go, this is what I do with a weight that is more heavy, about 80%

 Quote by douglis BTW at the study you're referring the lifting lasts ~1sec which is perfectly normal with such weight.Check it again.
Sorry it “is” 1.5 seconds lifting 81%
There is something wrong with this study, or they were not testing for maximum force bench press, as I or most could lift 80% 20 inch in .5 seconds, so the question is why did they take 1.5 seconds, but as they did take 1.5 seconds, I can see why the large deceleration.

Page 455.

http://www.scribd.com/doc/2528195/El...he-bench-press

Wayne
P: 148
 Quote by waynexk8 Do you mean as if with a lighter weight, and I lifted it so fast that is would leave my hand, are you calling the leaving of my hand, terminal velocity is zero ???
Terminal velocity is the velocity at the end of ROM which is zero in the case of lifting(not throwing)

 On paper it may look balanced/made up, however how do you know it is ??? As you will agree that a high force for a short time, can produced the force impulse on a low force for a longer time, so why do you think the forces balance out make up ???
The change in momentum is zero so the average net force is zero...hence the forces "make up".Nothing more to talk about...I don't know what you can't understand.

If you're going to doubt it without any argument you'll just be ignored.

 If I was to press up into a very thick piece of clay at a high force velocity with an 80% dumb bell in my hand, my hand could go in 2 inch, then there will be an opposite reaction which it called the tension on the muscles, I do this 6 times in 6 seconds and make a 12 inch hole in the clay, very high tensions on the muscles. I then take the same dumb bell and press is up into a very thick piece of clay at a low force velocity with an 80% dumb bell in my hand, I do this one time, the dumbbell would not even go the first 2 inch, but still lots of tension on the muscles.
I also don't understand why you're repeating that irrelevant clay example.It just shows the higher peak force that causes the deformation of clay...nothing more.

 Not the case. As I could accelerate the weight nearly right up if I wanted, as if I was throwing the weight, and if I did this with a light weight, there would be no deceleration until I left go, this is what I do with a weight that is more heavy, about 80%
The length of the deceleration phase is defined by the laws of physics...not by you.

 Sorry it “is” 1.5 seconds lifting 81% There is something wrong with this study, or they were not testing for maximum force bench press, as I or most could lift 80% 20 inch in .5 seconds, so the question is why did they take 1.5 seconds, but as they did take 1.5 seconds, I can see why the large deceleration. Page 455. http://www.scribd.com/doc/2528195/El...he-bench-press Wayne
Sorry it "is" 1 sec lifting.Check page 459 figure 6a....although it's of little impotance.
P: 399
 Quote by douglis Terminal velocity is the velocity at the end of ROM which is zero in the case of lifting(not throwing)
Please explain how “my” lifting it different to throwing ???

Yes, but please let us know “your” actual meaning of terminal velocity here, also why you include it, if what you think it is, is zero. As in weightlifting and physics it’s different.

Lifting and throwing a weight up in say the bench press, in my form of explosive lifting with 80% is basically the same. As I try to get the weight from a to b as fast as possible, that the same as if I was trying to throw the weight on the bench press.

 Quote by douglis The change in momentum is zero so the average net force is zero...hence the forces "make up".Nothing more to talk about...I don't know what you can't understand.
The change in concentric momentum/movement is from zero velocity, to acceleration, {80% moved 20 inch in .5 of a second} to a deceleration, then zero for the transition, then reverse down for the eccentric.

We are not looking for the net force the way you are saying it, and never have been, why do you mention this ??? You are saying if both forces are equal they cancel each other out the net force is zero. The force we are talking about are the forces from the muscle for holding the weight, for moving the weight, for accelerating the weight and for decelerating the weight, then add in the opposite reaction force, as in the more force you produce the more the clay would get squashed, or the more tension on the muscles. Like my EMG reading proved.

 Quote by douglis If you're going to doubt it without any argument you'll just be ignored.
Doubt what, I don’t understand what you think I will not doubt ??? What would be nice if you understood my far higher reading in a practical World experiment, my EMG reading ???

 Quote by douglis I also don't understand why you're repeating that irrelevant clay example.It just shows the higher peak force that causes the deformation of clay...nothing more.
You are quite wrong there, the deformation of clay in both lifts, will show all forces, or if you would like to state how/why you think that the force of you moving very slowly will not be shown in the clay ??? As the law states that for every reaction there is an opposite reaction. Hold the weight with the clay, then move the weight very slow up and down with the clay, you will see the clay is deformed more when you move the weight. It has to show all, please explain why you think it will not ???

 Quote by douglis The length of the deceleration phase is defined by the laws of physics...not by you.
I am, and we are all physics. I can, as I did earlier, choose the length of the acceleration and deceleration. Simple test, punch with as much force and velocity upwards as you can, and slow and stop as close to arms length as you can, now with as much force and velocity upwards as you can, but try and stop half way up, you have just changed both acceleration and deceleration.

The strength of the person will work out the acceleration and deceleration. As I said, I lift explosively and it’s basically the same as if I was throwing explosively, I would try to do both as fast as possible.

 Quote by douglis Sorry it "is" 1 sec lifting.Check page 459 figure 6a....although it's of little impotance.
Sorry, on the graph it clearly sates 1.5, also on your page, where are you looking ???

It’s of the greatest impotence, why do you think not ??? If I move the weight 1m in .5/.5 of a second 6 times to your 1m in 3/3 seconds, I have created 6 large force accelerations and peak forces. I will tell you why more, but no time now.

Why do you think your lower forces can make up or balance out my higher force ??? You still have not explained this, how can say 80 force make up or balance out a 100 force, if both forces are used for close to the same time frame ???

Did you try the weight on a barbell with string, to show very little deceleration ???

Wayne
P: 148
 Quote by waynexk8 Please explain how “my” lifting it different to throwing ??? Yes, but please let us know “your” actual meaning of terminal velocity here, also why you include it, if what you think it is, is zero. As in weightlifting and physics it’s different.
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.

 The change in concentric momentum/movement is from zero velocity, to acceleration, {80% moved 20 inch in .5 of a second} to a deceleration, then zero for the transition, then reverse down for the eccentric.
Exactly!So you go from zero velocity to zero velocity....the change in momentum is zero.So the net impulse is zero and the net force zero too.
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.

The ONLY forces available are the muscle force and the weight.There're no additional action-reaction forces to add as you fantasize.

Read carefully Phantomjay's last post.He's explaining perfectly that and your irrelevant clay example.

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.

 Sorry, on the graph it clearly sates 1.5, also on your page, where are you looking ??? Wayne
page 459,figure 6a.
PM me for further explanation cause that's irrelevant with the thread.
P: 399
 Quote by douglis 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 by douglis 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 by douglis ....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 by douglis 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 by douglis 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 by douglis 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 by douglis 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 by douglis 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
 P: 148 Wayne...let's start it all over again. Do you understand that the average force that's applied by the muscles is always equal with the weight regardless the repetition speed?Yes or no?
P: 399
 Phantomjay wrote;More force goes into the muscles only during the accelerating phase.[/i]
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 by PhanthomJay Your body is not telling you to decelerate it; the laws of physics demand it if you are to slow it.
Yes very true.

 Quote by PhanthomJay Your body tells you that your muscles are getting strained with too much force, and it will tell you that even if you were to hold the weight still and motionless for a long period of time, without speed , acceleration, or deceleration,
Again true.

 Quote by PhanthomJay only a force of 100 pounds on your muscles.Sorry, i misspoke, the average force (for what it's worth) is always just 100 pounds.
I do not think average means much.

The average net force is equal to the weight, regardless to duration and time ??? So if I move a weight for an hour, for a 1000m and then for 1 sec, 1mm. And if the net force is the same, the moving of the weight for an 1 hour will put a very lot of force out from and onto the muscles, even thou the average is the same.

 Quote by PhanthomJay Even though for most of the time the applied force is 150 pounds while accelerating, there is very little force that you exert when decelerating (the weight does the workfor you),
I contend this, the weight does “no” work for me, I am moving the weight, nothing else is, what else could be moving the weight, I do not understand what you mean or think ??? If I stopped at anytime in the lift, the weight “would” stop moving, it would “not” travel or move on its own, how could it ??? Someone worked out below, that if you lift 200 pounds 15 inch at 1m/s that the weight itself would {but it does not, please read on as too why} move 20% or 3 inch if you immediately stopped the pushing force at the top. However this does “not” happen, because of the biomechanical advantages and disadvantages thought the ROM, {Range Of Motion} so this means that there in “not” very little force, but a “very” lot of force being used thought the ROM. Even if a machine was pushing it, the weight cannot move on its own ???

OK, let's see if I can reproduce this. Let's assume we're doing a bench press with 200 pounds, and our 1RM is 250 pounds (80% 1RM). Furthermore, the ROM is 15 inches.

The first thing we must do is convert pounds into the English unit of mass - the "slug". That is - 200 pounds/32 ft/sec^2=6.25 slugs.

Now, from F(net)=ma, we have 250-200=6.25 x a, or a=8ft/sec^2. This is the acceleration of the bar during the concentric.

Now, let's assume that we will push with our maximal force (250 pounds) up to the 12" point. We next need to find the velocity of the bar at this point from the equation v^2=2ad. Plugging in "a" from above and "d"=1ft, v=4 ft/sec. Note that this is not the top of the lift, but 3" from the top.

Next we want to find the time it takes to get to the 12" point, from the equation d=1/2at^2. Plugging in our values of "a" and "d", t=.7 seconds (again not the top).

Now, we assume that we stop pushing the bar at the 12" point, and let gravity slow it down, so that it comes to rest at the top. From v^2=2ad, we plug in our value of "v", but here we use a=32 ft/sec^2 (acceleration of gravity). From this we find that d=3", so that the bar comes to a perfect halt right at the top of the ROM - the 15" point.

Finally, we want to find out how long it takes gravity to stop the bar, from d=1/2at^2. Plugging in d=3" and a=32 ft/sec^2, t is found to be about .1 seconds.

Therefore, the total time for the concentric in this case would be .7 seconds (for the acceleration phase, or "onloading") plus .1 seconds (for the decceleration or "offloading" phase), for a total of .8 seconds. This is a bit faster than 1/1, but the best I could do this late at night. If your ROM were a bit longer then 15", then the speed would be closer to 1/1.

So, we see that in this case the offloading relative to the ROM is 3" out of a total of 15", or 20%. The offloading relative to the time is .1 sec out of .8 sec, or about 12%. Note that this is worst case...that is the first rep. As the set progresses, the offloading will reduce with each rep, due to fatigue, and towards the end of the set will be negligible. Therefore, you could state that the "average" offloading of the entire set relative to the ROM would be about 10%.

Note that this also assumes we can push with maximal force all the way to the 12" point. If our strength curve is such that our force output diminishes towards the top, then the offloading will be less than given above.

 Quote by PhanthomJay so the average force averages out, if you know what I mean (in eight equal intervals on the upstroke, force you apply might be 125, 150,150,150,150, 25, 25, 25; average = 800/8 = 100 pounds). You can be moving at constant speed slowly (say 1m/s) or at constant speed quickly (say 3m/s),and the force you exert is still the same 100 pounds for both cases.
As I said above, I can’t see how the weight would move on its own ??? it would be more like 125, 150, 150, 150, 150, 125, 125, 80, 20, zero movement for the transition.

“If” the forces averaged out, “why/how” is the EMG readings higher, why/how is power higher why/how do you use more energy, why/how does the fast move the weight 6 times further, why/how do you fail on the fast far faster ???

Play from 5.00
Note that the differences.
Slow,
Power 649
Force 546
Velocity 161.

Fast,
Power 829
Force 579
Velocity 192.
Imagine if the person had done 6 repetitions fast and one slow, the force on the fast would/could be far far far higher, as on the video, the speeds were basically quite close.

 Quote by PhanthomJay Your muscles are tiring more easily in the high powered fast case, because internal energy is being lost from the complex muscular activity..
Yes, that’s because I have to be using more force ??? If not how/why else would I need to use more energy ???

 Quote by PhanthomJay if your muscles were robotic, it wouldn't matter..or a machine getting powered electrically...but humans are not machines, they need chemical/biological/food energy to operate.
The machine too would need more energy to use more force, its impossible otherwise to use more force without more energy, is it not ???

 Quote by PhanthomJay Don't confuse the two.You are not correct.
So why is more energy used when force goes up then, as its “not” a coincident that the “exact” moment you use more force you use more energy, is it ???

 Quote by PhanthomJay However, the power required is greater when you are moving faster with the same applied force.
You would have to use more force to move faster. Where Fnet is the total external force.

Wayne
P: 399
 Quote by douglis Wayne...let's start it all over again. Do you understand that the average force that's applied by the muscles is always equal with the weight regardless the repetition speed?Yes or no?
I understand to well what you are thinking/saying. Your saying that, {these numbers are just for me showing you this} if I moved a weight 1000mm to you moving it 166mm in the same time frame, that all the higher force I use acceleration the weight for say the first 60% you make up or balance out this force when I am on the deceleration. Or if I use 100N 100N 100N 100N 100N 100N for the first 60% then use 50N 50N 50N 50N = 800N. You use 80N 80N 80N 80N 80N 80N 80N 80N 80N 80N = 800N.

1,
This seems to work on paper, but not in practice.

2,

However, the following states the force do not balance out or make up.

3,
However not on EMG reading, which test the electrical activity of muscles.

4,
Not on distance as the weight is moved six times further in the same time frame.

5,
I accelerate 600mm to you moving at a constant velocity for only 166mm.

6,
You always fail far faster in the faster reps.

7,
Why do you use more energy the moment you use more force ??? If as you claim your forces make up or balance out, should not the energies ???

8,
I know we went over this once, but there is still things bugging me, could we go over here on the physics forum, as other members can comment.

As you know, if I immediately stopped at any part of the acceleration, the weight would stop immediately, and would not carry on at all, so “HOW” do I as you claim I use less force than the weight ??? As if you less force, does that “NOT” mean that I would be then reversing the direction, not still going forward but backwards ??? As deceleration is negative acceleration or decreasing velocity over time.

Why can not the fast repetition, coming off the eccentric portion, hitting the peak forces on the transition from negative to positive. Could it not be like this, maximum force 100 pounds, weight used 80%

130, 100, 100, 100, 100, 100, 95, 90, 85, 80, zero, for the transition from positive to negative.

These next numbers mean nothing, and are just my example.
.5m/s, 2m/s, 2m/s, 2m/s, 2m/s, 2 m/s, 1.6m/s, 1.2m/s, .8m/s, .6m/s, zero, for the transition from positive to negative.

I can’t seem to find any studies of moving 80% at 2m/s for say 20inch, or there about, will look more under weightlifting.

9,
What I have said all along, is my higher force impulses, cannot be made up or balanced out by your lower force impulses, and you have no way to prove it, however I do as in all the examples above.

10,
Just thought of something, eccentric muscular action for the deceleration, this will take force from the muscles, muscles are able to withstand greater eccentric than concentric loading due to increased fast-twitch muscle fibre and motor unit recruitment. Will have to think about this one.

Wayne
P: 399
 douglis wrote; I'm not saying that your EMG is wrong.What I say is what I know from physics.The RMS {Root Mean Square}is NOT the technical average.It's the ~70% of the peak values.When we say the average force is always equal with the weight we're talking about the technical average which is shown only from the EMG normalized force-tension graph.

Quite wrong I think, please read this. Seems like it’s the average.

http://www.raeng.org.uk/education/di...ring/8_RMS.pdf

1, Fast 409, Slow 349,
2, Fast 437, Slow 346,
3, Fast 0.1, Slow 0.3,
4, Fast 0.6, Slow 0.7,
5, Fast 1104, Slow 1114,
6, Fast 146.0, Slow 193.4,
7, Fast 175.0, Slow 173.0,

1/ WRK This is the work average for the session measured in [µV]
AVG microvolts. The average readings will vary from one patient to
another.

2/ RST This is the rest average for the session measured in µV - Below
AVG 4 µV a muscle is beginning to rest.

3/ AVG This is the average onset of muscle contraction measured in
ONST seconds, readings below 1 second can be considered normal.

4/ AVG This is the average muscle release measured in seconds,
RLSE readings below 1 second can be considered normal.

5/ W/R This is the average peak value measured in µV - The value will
PEAK vary from one patient to another.

6/ WRK This is the average muscle deviation when contracting the
AVDV muscle. Readings of below 20% of WRKAVG can be
considered adequate, below 12% can be considered good.

7/ RST This is the average muscle deviation when the muscle is at rest.
AVDV Below 4 µV a muscle is beginning to rest.

Wayne
P: 399
Hi D.

Hope we both are both are seeing understanding force/strength in the same way ??? As I do, as do explain things to you, but at times you never acknowledge that you understand or agree. We all agree that when using 80% for the fast at .5/.5 for 15 repetitions = 15 seconds, and the slow at 3/3, that you will be able to go for longer using the 3/3, as you fail faster doing the .5/.5 yes ???

So this means I “HAVE” used up my force and faster, yes ??? Or as I call it strength faster, so if we are Clones, and have a 1000N of force to use up, at using 80% as explosively/dynamically, I “have” used up “all” my force, “but” as you are “STILL” exercising, repping the weight, you “MUST” still have force/strength left, right ???

 doulas wrote; Hi Wayne....I believe you got all the answers you wanted on the physics forum.It's clearly stated by the physicists that you are not correct.I believe that you just don't want to accept the truth and that's why nobody(including me) is bothered to answer.

I have not had many answer yet, waiting for the next reply, but as I know these people are busy.

Zula is constructing a force machine.

If I fail faster, as told you above, if I also use more energy, more the weight more distance, I fail understand how or why you think you are right ???

Please read chapter 4, just the first 2 pages, it says like I have been trying to say all along, these forces that I talk about cannot be easily equated with physics.

One more,
If I try and lift as much as I can fast, to you lifting slow, I will lift more, thus I create more force and for longer.

http://www.findphysio.com/E-books/Bi...20Exercise.pdf

 doulas wrote; Just one thing is left to clarify. Check carefully the force-time graph(figure 6a) at page 459.It's clearly shown that at the ascent phase lasts from 1.2sec to 2.2sec.The duration of the ascent is ~1sec which is perfectly normal for the 6-7RM(81% of 1RM)http://www.exrx.net/Calculators/OneRepMax.html. Also you can clearly see that the acceleration phase is only for the first half(maybe even less) of the ascent.It's also shown perfectly how the forces "make up".For the first half the force is above the bar weight and for the second is below the bar weight.

Not sure what you want me to look at here ??

 doulas wrote; This EMG graph is the perfect answer for everything you ask.Nothing more to add. http://www.scribd.com/doc/2528195/El...he-bench-press
I “have” told you and all that look at this can quite easy see, on page 455, the this bench press took 1.5 seconds this kind on very slow bench press world have a acceleration and deceleration phase like that, as of the sticking point in the bench press. As if the press was done that slow. My bench press, as I keep telling you, and you don’t want to listen, lasts as you know, for three times less than this, as mine goes three times faster, its move the whole concentric repetition in .5 of a second.

Wayne
 Sci Advisor Thanks PF Gold P: 12,262 This looks like the same question, yet again Wayne. Will you keep asking it until you get the answer you want?
P: 399
 Quote by sophiecentaur This looks like the same question, yet again Wayne. Will you keep asking it until you get the answer you want?

Hi sophiecentaur,

I am honesty not like that, as I recently bought an EMG machine, {the results are on this thread} it reads the electrical muscle activity, the more average or total force/strength you use it reads out, and I did many tests on it, and everyone came out stating that you use more force when moving faster.

Also we all agreed that you use more energy when moving a weight faster, so as soon as the force and average force goes up, the energy does also, this can’t just be a coincidence ??? Then you move the weight 1000mm moving fast, and only a 166mm moving slow, that means “if” I accelerated for only 60% I would have accelerated the weight 600mm to the slow only moving the weight at a constant velocity for 166mm.

You also fail far far far faster on the faster reps, you hit muscular failure faster.

We as we all agree that when using 80% {of your 1RM repetition maximum} for the fast at .5/.5 {.5 of a second concentric and .5 of a second eccentric} for 15 repetitions = 15 seconds, and the slow at 3/3, that you will be able to go for longer using the 3/3, as you fail faster doing the .5/.5 yes ??? And you fail roughly 50% faster on the fast repetitions.

So this means I “HAVE” used up my force and faster, yes ??? Or as I call it strength faster, so if we are Clones, and have a 1000N of force to use up, at using 80% as explosively/dynamically, I “have” used up “all” my force, “but” as you are “STILL” exercising, repping the weight, you “MUST” still have force/strength left, right ???

So, as well as 99% of the Worlds bodybuilders, strongmen, weightlifters, and all sportspeople of every sport train in the faster repping way, and the evidence above, I just don’t see how some here think other ??? It’s basically physic equations have not got all the variables, thus they are adding it up wrong. Bottom line is, the slow constant medium forces cannot balance out or make up the high peak forces of the fast on their accelerations, in the same time frame, it’s like impulse, impulse is defined as the integral of a force with respect to time. When a force is applied to a rigid body it changes the momentum of that body. 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.

We are now constructing a force plate.

Thank you for your help on the last thread, and hopefully on this one, as maybe you can spot where the equations are going wrong.

Wayne
 Sci Advisor Thanks PF Gold P: 12,262 The equations are 'gong wrong' because you can't expect them to apply in a situation like this (as in all of your questions). You confuse the terms you use and seem to think that 'work done on' is necessarily related to 'energy transferred from muscles'. I really don't know why you want to include Physics in your training if you won't learn Physics to an appropriate level. If you continue to talk in the wrong language, how can you expect to have a proper conversation . Quote "I have used up all my force" - this is a meaningless statement in terms of Physics. Your posts are full of such statements but you refuse to modify them or learn the appropriate way of putting things. If someone can only play the Piano with one finger then it's no surprise that a Piano Concerto gives them a problem. Like I have often said, you can't bend Physics to your will. You just have to follow where it takes you. You don't seem to want to, though.
 P: 253 Would anyone else agree that if you apply only an upward force(against gravity) during the upward rep, then yes, it is exactly equal to gravity once you come to a stop. Would you also agree that if however, you apply a sharp upward force and then ALSO a force WITH gravity, the net impulse applied to the weight is still 0, consisting of +your force against gravity, and -gravity and -your force with gravity, thus, the overall force you are applying is greater than gravity. I saw wayne mentioned the experiment we are throwing together, and I was hoping this made sense to you sophie and douglis. It seems that is where this conversation is hung, in the assumption that you are only working against gravity. With a fast enough rep it is possible that is not the case. I am getting together a cheap accelerometer with which we will be testing this theory out.
P: 148
 Quote by Zula110100100 Would you also agree that if however, you apply a sharp upward force and then ALSO a force WITH gravity, the net impulse applied to the weight is still 0, consisting of +your force against gravity, and -gravity and -your force with gravity, thus, the overall force you are applying is greater than gravity.
Zula...we have discussed this case before in another of dozens Wayne's same threads.
You're referring to the case where you also use the antagonists muscles in order to stop the weight and you don't rely only on gravity for the deceleration.
Of course you're right...in that case the overall force is greater than gravity and probably that's what he's doing in his experiments.

However....Wayne denies that case.He insists(without any reasonable argument) that without any "help" from the antagonist muscles(muscles apply only upward force) the overall force is greater with fast lifting ignoring the fact that the net impulse is always zero.
Please,if you have the patience(cause I don't) explain to him that he's totally wrong.
P: 399
Hi all, and thank you for the replies, “no” time to read any of the posts, but this one came up on my mail.

 ---Quote (Originally by Zula110100100)--- Would you also agree that if however, you apply a sharp upward force and then ALSO a force WITH gravity, the net impulse applied to the weight is still 0, consisting of +your force against gravity, and -gravity and -your force with gravity, thus, the overall force you are applying is greater than gravity. ---End Quote--- Zula...we have discussed this case before in another of dozens Wayne's same threads. You're referring to the case where you also use the antagonists muscles in order to stop th weight and you don't rely only on gravity for the deceleration. Of course you're right...in that case the overall force is greater than gravity and probably that's what he's doing in his experiments. However....Wayne denies that case.
NO I do not, why would I ??? I have been training for about 40 years, and it’s well known that you have to use antagonist’s muscles in order to slow, stop and very slightly reverse the weight

 He insists(without any reasonable argument) that without any "help" from the antagonist muscles(muscles apply only upward force)
I never said this ???

 the overall force is greater with fast lifting ignoring the fact that the net impulse is zero. Please,if you have the patience(cause I don't) explain to him that he's totally wrong. He insistswithout any reasonable argument.
I have more than a reasonable debate, see above posts and below. Please remember, I am doing 6 more repetitions in the same time frame, and moving the weight 6 times as far.

I “proved” in my opinion I was right in my last post. Or could anyone here please try to explain that when you fail {and you do this is fact} in the faster repetitions, too which that means you have no force/strength left, but the person doing the slower repetitions does have force/strength left, how can I have not used all my force/strength up faster when I have used my force/strength up 50% faster ??? Proof, evidence and facts, I have used my force/strength up first, thus if we both had 1000N of force/strength, I have used it up faster, to say other would be a full English and physics contradiction, it would be like say the Man who ran the 100m the slowest was the fastest.

Meaning the physics calculations need to add in the other branch of physics, Kinology kinology/physics which treats of the laws of motion, of moving bodies to the calculations. And Biomechanics the study of the structure and function of biological systems. These with the fact that you use more energy the instant you use more force/strength, also that you move the weight further in the same time frame, and that you fail faster.

I repeat, the physics calculation cannot be right, as the situation cannot exist, as the faster repetitions do indeed finish way before the slow, this is why my EMG readings in the same time frame, put the fast repetitions higher, because on the tests I did not go to failure, as we needed the results of the same time frame. Please take a look at the videos, the weight here was too light.

Wayne
P: 148
 Quote by waynexk8 NO I do not, why would I ??? I have been training for about 40 years, and it’s well known that you have to use antagonist’s muscles in order to slow, stop and very slightly reverse the weight
The case,for example,that you do a bicep curl and you decelerate with your triceps was not a part of the discussion.

 I “proved” in my opinion I was right in my last post. Or could anyone here please try to explain that when you fail {and you do this is fact} in the faster repetitions,..... Proof, evidence and facts, I have used my force/strength up first, thus if we both had 1000N of force/strength, I have used it up faster, Wayne
I don't know what you think you prove with this example.
The fact that you fail faster with fast lifting just proves that you use energy faster(greater rate of energy expenditure) and not that you "use force/strength faster".
Greater rate of energy expenditure doesn't equate greater overall force.
Practically means that the fluctuations of force in fast lifting require more energy than the equal constant force of slow lifting.

 Related Discussions General Physics 69 Introductory Physics Homework 2 Introductory Physics Homework 16 Introductory Physics Homework 2 Introductory Physics Homework 2