Impulse/force in pounds for the time frame

[waynexk8]

Here is what Roger Enoka said to me, when I asked him the question.

http://www.amazon.com/dp/0736066799/?tag=pfamazon01-20

Roger Enoka wrote;
I do not think it is appropriate to perform such comparisons in terms of average or peak force as the force varies continuously during each action. The relation f = ma applies only at one instant in time. When force (f) varies, the force applied to the object must be expressed as either the work done (force x distance) or the impulse applied (force x time). Because of these requirements, I cannot answer your question without knowing how all the forces vary over time.

Sorry.

So can we work this out on work done (force x distance) or the impulse applied (force x time) ?

Roger Enoka wrote;
I'm sorry, but I do not understand the question. My guess is that you are asking about the relative muscle force when lifting and lowering a weight either quickly or slowly. Even if the amplitude of the average acceleration of the weight was the same for both phases of the lift, net muscle force would differ. To lift a weight, muscle force must be greater than the weight. In contrast, the weight can only be lowered when muscle force is less than the weight. As a result of these constraints, the muscle uses more energy to lift the weight than it does to lower the weight.

I hope this is helpful.
Wayne,

Fortunately, physics does work on the human body, we just need to formulate the question precisely. The problem with the rationale provided by Jeff is that there are too many assumptions, such as the use of presumed average values. As indicated by John Casler, a 250 lb maximum does not mean that this force is applied throughout the lift. Indeed, this represents the force at the weakest point in the lift. Physics does apply, but solutions require real data that varies over time.

Cheers.

Roger M. Enoka, Ph.D.
Professor and Chair
Department of Integrative Physiology
University of Colorado

Wayne

 

[sophiecentaur]

@ Wayne - don't forget to look at and respond to post 176.

Those replies are what I might have expected and I have said the same on many occasions. So who will you apply to next to get your non real answer to your non real question?

"So can we work this out on work done (force x distance) or the impulse applied (force x time) ?" Those two formulae relate to two different quantities. Which one did you want to work out and why? (and how many times a second?)

 

[waynexk8]

Wayne wrote:

Jeff could I have a look at the calculations for the offloading in the last 10% of the lift in a 1/1 you worked out please, could you as well keep it as layman’s terms as poss, if you can’t do that could you then in brackets put down what some of the equations/math/abbreviations are, what so and so actually mean. Hi Wayne,

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.

Jeff

Wayne again, but we had a problem here, as in real life, there was no 3 inch offloading, the weight did, and does NOT move out of your hands, however there was an answer, as it would move this far if a machine that could use the same force over the lift lifted and stopped. The reason that the physics did not work on the muscles and the force they produce, is that the physics LEFT OUT variables, as of the muscles have biomechanical advantages and disadvantages thought the range of motion, thus could not push up with full force all the time.

John wrote;
One thing for certain however, and that is, if the SLOWING is performed purposely, it WILL reduce muscle tension levels, and increase duration. That is not to say these are not valuable to some goals, but it is VERY difficult for many (if not most) people to maintain an awareness of how this affects the stimulus.

Also without discussing Power Expressions, Intensity Levels, and Work Output, we again cannot make any meaningful determinations as to what relationships the stimuli have.

Here is a video of a fellow Benching (FAST) over 400# 20 reps.

http://www.youtube.com/...rom=PL&index=20

It represents a significant level of Intensity, Power, and Work. It also due to the speed represents some EXTREMELY HIGH muscle tensions, and many exposures to them.

If this same fellow used the same load and increased the TUT x 2 or x 4 (twice or four times as long) by reducing the speed, the stimulus would have been FAR different, and the total stimuli except for fatigue, would have decreased.

Now it would be a difficult case to prove that the faster 20 RM effort would not offer superior results to a Slower 5 - 10 rep slower set with the same load.

If one finds a purposely slower rep speed "increases" hypertrophy it could be for several reasons, but seldom would they be related to "increased muscular tension force", but more an adaptation due to fatigue based stimuli.
Wayne

 

[waynexk8]

I was trying to get thought the old posts, have not read some of the new ones yet, but for now let’s jump to this, hopefully physics question. The concepts of work and energy are closely tied to the concept of force because an applied force can do work on an object and cause a change in energy. Energy is defined as the ability to do work, work is done on an object when an applied force moves it through a distance, mathematically, work is W = F • x, where F is the applied force and x is the distance moved, that is, displacement.

When I run 50m as fast as I can, then run 100m as fast as I can, I have used more overall, total force from my muscles, but let’s just say legs, running the 100m, do we agree there ? Now am I saying this right, or should I be asking which did the most work ?

As when energy is transferred either, to or away from an object by a force acting over a distance/displacement, work will be done on that object. The net work can also be expressed as the work done by the net force acting on an object,. W = Fd cos 0. The net work done on an object is the sum of the work done by each individual force acting on that object. The net work can also be expressed as the work done by the net force acting on an object. Can we work anything out on the work done by the constant and non constant force for the .5/.5 for 6reps moving the 80 pounds 12m and the 3/3 for 1 rep moving the 80 pounds 2m.


Wayne

 

[waynexk8]

I am not disagreeing with what the EMG people are saying. The problem is that you do not seem to understand what they are saying and how it applies to your question.

Ok, here is what I think. When you or the EMG takes the force readings, there are values of force above the weight of the weight being moved, for the accelerations, and values below for the decelerations, what the EMG and RMS does is adds all these up in the fast lifting and averages them out, and does the same on the slow, what D. and maybe you have done, is canceled the forces out of the values under the weight, and took them of the values above the weight ?

Why should you need to have RMS explained any more. It isn't an explanation that you need. What you need to do is to use the definition to work out some numbers yourself. Douglis has given you a perfect worked example.

First I need to know if you agree or why you think they used the RMS ? And the same for D. as he seemed, or did think the RMS was wrong, I need to know why he thought it was wrong, and now what he thinks as they say it’s the best way to average the forces out. I can’t just go away from this; I need to understand what he thought before and now, and why he was wrong, and why he tried to say RMS can not work.

What you need to do is to use the definition to work out some numbers yourself. Douglis has given you a perfect worked example.

Ok will have tried. What numbers has he given ? Or maybe I have not come to it yet.

=sophiecentaur;3783708]I cannot be bothered to answer that question because it is just a smoke screen to protect you from going to the trouble of working stuff out for yourself.

I know this is easy for you, but when I started and can now work out the power used in different reps with different weights speeds and times, I really enjoyed it, but I don’t know where to start on this. Please I need an example. But first I need to know if you agree or why you think they used the RMS ?

Wayne

 

[Dale]

Wayne, do these people know that their private correspondence with you is being posted on the Internet including their name? This seems like a very bad practice unless you have their explicit permission.

 

[douglis]

Wayne, do these people know that their private correspondence with you is being posted on the Internet including their name? This seems like a very bad practice unless you have their explicit permission.

Those discussion are from many years ago and have been posted in more than 10 different forums around the internet(without any permission of course).The same is also true for the discussions here!

Twelve pages and so many threads for a question that can be answered in just one line!
The average force is always the weight regardless the lifting speed.

 

[waynexk8]

What forces have you the slow got that can make up the propulsive forces in the studys, there are no extra force in the slow reps, are there ?

Back later to catch up on all posts.

Wayne

 

[jmmccain]

Wayne,

May I suggest an experiment? Do you have access to a bathroom scale and two video cameras?

Set up one camera to record the scale reading and perform your various fast and slow exercises. Afterwards, view the recording frame by frame and plot the scale reading for each frame (like douglis' graph, but it won't look the same). Don't forget to subtract your own weight from all the readings. This will rather conclusively demonstrate the difference in forces.

Add up all the readings for a repetition (again, adjusted for your weight). Then divide by the number of frames over which the readings were observed. This will be the mathematical average and should be very close to the weight you are lifting. How close depends upon how well the experiment is performed.

The second camera is used to record the movement of the weight with a tape measure, or other suitably visible scale, in the background. Again, go through frame by frame and plot the position of the weight. This should look something like douglis' graph. Next, plot the change in position from one frame to the next (velocity). Next, plot the change of this change (acceleration) from one frame to the next. This third plot should look just like the first plot of the scale readings.

F = ma

The first plot of the scale readings is force. The last plot is acceration. Since the mass doesn't change, the two remain proportional to each other and the plots should be very similar.

The scale and cameras won't lie. If you need help decipering the results and making the appropriate plots, well, I may be busy, but there are others around who will help.

 

[sophiecentaur]

Wayne,

May I suggest an experiment? Do you have access to a bathroom scale and two video cameras?

Set up one camera to record the scale reading and perform your various fast and slow exercises. Afterwards, view the recording frame by frame and plot the scale reading for each frame (like douglis' graph, but it won't look the same). Don't forget to subtract your own weight from all the readings. This will rather conclusively demonstrate the difference in forces.

Add up all the readings for a repetition (again, adjusted for your weight). Then divide by the number of frames over which the readings were observed. This will be the mathematical average and should be very close to the weight you are lifting. How close depends upon how well the experiment is performed.

The second camera is used to record the movement of the weight with a tape measure, or other suitably visible scale, in the background. Again, go through frame by frame and plot the position of the weight. This should look something like douglis' graph. Next, plot the change in position from one frame to the next (velocity). Next, plot the change of this change (acceleration) from one frame to the next. This third plot should look just like the first plot of the scale readings.

F = ma

The first plot of the scale readings is force. The last plot is acceration. Since the mass doesn't change, the two remain proportional to each other and the plots should be very similar.

The scale and cameras won't lie. If you need help decipering the results and making the appropriate plots, well, I may be busy, but there are others around who will help.

I'd advise not getting too involved with this. All your suggestions have been made many times before, in this thread and earlier threads. Wayne does not believe in the accepted ideas of Physics. He has his own models and vocabulary of Physics.

Also, your thought experiment on the bathroom scales would not show the sampled forces accurately or frequently enough to convince Wayne. There would be errors which he would jump on and claim that the experiment showed him to be right. The scales and camera would lie in practice unless a much more sophisticated system were used.

In any case, what you say would only apply to free lifts and not to exercises on machines that introduce friction. That also confuses Wayne and strengthens him in his misconceptions.

 

[waynexk8]

I wish you and D. would do the same, and answer some of my questions ?

OK. You go from London to Bristol at an average speed of 62miles per hour. You could work that out easily enough. But how would you, Wayne, work out what your 'total speed' was? That's how daft your idea of total force is.

You got a point on that one, see what you mean, WILL think more on this one, BUT why then are you, and D. saying, the average force is the same ? I think that you have taken my maximum force from the acceleration, and then took away my minimum force from the deceleration, but “why” did/do you do this ? However ever, speed is a little different to force, take a look on this video, they measure the maximum force and minimum force, and then they calculate the average force, which will be the same, but we do NOT want this, this is what I keep ON telling D. over and over, we do not want the average force, we want all the three average force, the forces from the negative, the force when lowing the weight, the peak force, the huge peak force from the transition from negative to positive, and the positive force, the forces when lifting the weight.

NOW, if the weight was 80 pounds and my 1RM was a 100 pounds, if I start the lift from the top, lower in .5 of a second and lift in .5 of a second for the fast, and lower in 3 seconds and lift in 3 seconds.

1,
The fast,
The tension on my muscles to lower it would be just under 80, call it 79, then at the transition from eccentric to concentric, would have the maximum tensions on the muscle, say ? 140 ? Then the tensions on the muscles to accelerate the weight up would be close to 100. 79 + 140 + 100 = 319/3 = 106 over 1 second.

2,
The slow,
The tension on my muscles to lower it would be just under 80, call it 79, then at the transition from eccentric to concentric, would have the maximum tensions on the muscle, say ? 85 ? Then the tensions on the muscles to accelerate the weight up world be close to 80. 79 + 85 + 80 = 244/3 = 81 over 6 seconds.

3,
Divide the slow of 6 seconds by the fast of 1 second, 81/6 = 13 over 1 second.

Ok that’s wrong, but how would you do this please ? What if we added them up ? As in this debate/test/study, average means nothing, so it’s fast, 79 + 140 + 100 = 319 over 1 second. Slow 79 + 85 + 80 = 81 over 6 seconds, 13 over 1 second.

Take a look at this video, they say how they work out the average force on it, and they show the maximum force and minimum force, click on click to preview. What you and D. are doing, is taking to points, and taking the maximum and minimum forces and averaging them up, well of course the average force is going to be the same, but it means nothing here, you are just taking the maximum and minimum, and adding them together and averaging them out, that’s what we don’t want here, we NEED to add in the maximum acceleration force, the maximum peak force, {of the transition from negative to positive, this maximum peak force, can put up to 140 pounds of force on the muscles} and the maximum deceleration force, what you are doing is leaving out the most important, the maximum peak force from the transition, why are you leaving this force out, which put the huge tensions on the muscles, and the debate is, which puts the most over the same time frame, the most overall or total tension on the muscles.

http://webcache.googleusercontent.com/search?q=cache:Nze_pWKG-40J:www.fittech.com.au/products/ForcePlate.htm+force+plate+average+force&cd=2&hl=en&ct=clnk&gl=uk

Wayne

 

[waynexk8]

I have read the paper an I think I can see your problem. I find little to disagree with what is written.

They are not considering the whole cycle (lift/lower) when they refer to mean force.

Right could we get to this point, as D. would not explain, could you explain on what you think they are leaving out ? as I do not get it.

They are not only discussing free lifts but machines that present RESISISTIVE loads. Everything changes in that case because you are not just changing gravitational potential energy in that case but work is being done in overcoming friction.

Please forget the machines, this at this moment is just about free weights.

If you had read what they say then you would not think they are disagreeing with established physics at all. You did not understand what you were arguing about because you insisted on giving details instead of condensing your questions into something meaningful.

I did not think they are disagreeing, or never said they were disagreeing with established physics.

What I said, was what forces do you think you have that can make up of balance out the higher propulsive forces of the fast in the studies ?

Let’s take the mean propulsive forces, slow 6.2mean in 10.9 seconds. Fast 45.3mean 2.8 seconds, now let’s divided the mean slow of 10.9 seconds by the fast 2.8 = 3.8, so now let’s divide the slow mean by 3.8 = 1.6.

Fast mean for 2.8 seconds = 45.3.

Slow mean for 2.8 seconds = 1.6.

Please what forces have I left out that the slow has to make up or balance out these ?

http://www.jssm.org/vol7/n2/16/v7n2-16pdf.pdf

Wayne

 

[waynexk8]

Will get back to jmmccain and Dalespam tomorrow, funny I have two very good high definition cameras.

Wayne

 

[waynexk8]

No time just now, to read the posts and answer.

Just wanted the state why average force means nothing in this debate

I lift 80 pounds at “any” speed, let’s go for 1/1, {1 second positive and 1 second negative} you all are saying that the average force for 1 repetition, and the average force for a 100 repetitions are the exact same ? However, we all know that you will exert far more total or overall force doing the 100 repetitions. You will exert a force the same as you do for 1 repetition {forgetting fatigue here} = 2 seconds, for 200 seconds, so its overall or total force output x 200 seconds.

So the question is, why does everyone keep commenting on the average force is the same, when it has no reverence in this debate ? What we are looking for, is in which repetition speed, in the same time frame, puts the most overall or/and total tensions on the muscles ?

As I have said, as you fail 50% faster using the .5/.5 to the 3/3 using 80% this does mean thet you have put more tension on the muscles faster, right ? And to put more tension on the muscles, and faster, you will have to use more force, if not please say why you think not ? PLEASE will someone try to answer this.

Are there anyone out there that thinks if you fail to lift a weight up after say 10 seconds, as you are lifting it faster, that it is not because you have put more tension on the muscles than the slow lifting, if so say why please.

Wayne

 

[sophiecentaur]

You were the one who asked what average force was. You also asked about Total Force.
Do keep up!

 

[sophiecentaur]

Wayne
Can you quote me a single statement, from anyone else, on this forum or in the communications with others that you have shown us, that indicates what you are talking about makes any sense in terms of the Physics? Everything I have read looks like non-committal and polite put-downs. When will you get the message?
Have you not realized that no one wants to discuss your endless, blow-by-blow descriptions of a lifting session? If you cannot be bothered to condense your questions into a digestible form then I, for one, can't be bothered to read them. I (/we all) have given you all the facts about the very straightforward Physics that relates to the process.
You haven't been prepared to accept it because it "doesn't feel right to you", somehow and you haven't been prepared even to use 'our' vocabulary. Can you be surprised that you aren't getting satisfaction?

 

[waynexk8]

Hi waynekx8, I didn't notice that you were back.

It looks like you are still confusing "work done" and "energy expended" as we discussed last year: https://www.physicsforums.com/showthread.php?p=3190515#post3190515

Hi again Dalespam,

Actually I am not confusing work done and energy expended, "unless" work done means the overall or total force used ? And we know for a fact that the fast does more work and uses more energy.

Average means nothing in this debate. What I am looking for is the higher overall or total forces used, thus more overall or total force on the muscles, just take this example.

Lift a weight up and down for 10 seconds, lift a weight up and down for 6o seconds, the average force is the same, but lifting the weight for 60 seconds WILL and DOES put more total or overall tension on the muscles, this means the total or overall force was more.

What we WANT to know is the same for lifting the weight up and down 6 times in 6 seconds, and lifting the weight up and down 1 time in 6 seconds.

Now I say without a shadow of a doubt, it’s the fast, and here is why. If you took both repetitions speeds to momentary muscular failure, meaning if you lifted the weights at the two given speeds until you could not move the weight, you would fail about 50% faster lifting the weight the fast way. This can ONLY mean one thing, you put more tension on the muscles faster, thus the muscles failed faster, thus more force must have been used in not only the same time frame, but in less time frame, you also use more energy in the fast, why ? It’s because your putting out more force and tension on the muscles, if not, why please ? As no one here is giving me a direct answer, they seem to not want to, or can't ?

Wayne

 

[sophiecentaur]

Hi again Dalespam,

Actually I am not confusing work done and energy expended, "unless" work done means the overall or total force used ? And we know for a fact that the fast does more work and uses more energy.

Wayne

That statement is meaningless. There is no "debate" possible on that basis. Why not do us all a favour and use PF language?

Total force is as daft as total speed. Come to terms with that.

 

[Dale]

Actually I am not confusing work done and energy expended ... And we know for a fact that the fast does more work and uses more energy.

Yes, you are still confusing them. In fact, this quote proves that you are confusing them since you finish with the confused statement that the fast does more work and uses more energy.

The correct statement is that the fast does the same amount of work and uses more energy (i.e. the human machine is less efficient at doing fast work). If you were not still confusing the concepts then you would not still be making the same incorrect statements.

What I am looking for is the higher overall or total forces used,

Let f(t) be the force exerted by the human on the weight at time t. Please define overall or total force in terms of f(t).

For example, average force from time $t_i$ to time $t_f$ is:
$$\overline{\mathbf{f}}=\frac{\int_{t_i}^{t_f} \mathbf{f}(t) \, dt}{t_f-t_i}$$

Please provide a similar rigorous definition for total or overall force.

 

[sophiecentaur]

@Dalespam
I see you are being drawn into this. Beware - madness lies there.

 

[Dale]

@Dalespam
I see you are being drawn into this. Beware - madness lies there.

Thanks for the warning sophiecentaur, but I have worked with wayne before, same topic different thread. I will only stick around while it is fun.

 

[sophiecentaur]

It's quite incredible how he bounces back with the same question, which, in itself, has no meaning.

 

[Dale]

I was trying to think of weight lifting analogy to describe his behavior in terms that he could understand, but my weight lifting vocabulary is fairly limited.

Basically, I want to get across the point about how stupid I would be if I were to go to him and a bunch of other weight lifting experts and ask for their advice on a weightlifting problem and when all of them tell me the same thing I then ignore their expertise and continue doing the opposite. Particularly if they are united in their opinion and persist in their opinion after detailed questioning.

They are experts in weight lifting, I know that I am ignorant on the subject, and since they all agree with each other, and my opinion disagrees with them, the only reasonable thing to do is to recognize that my uninformed opinion is almost certainly wrong and try to understand their advice.

Also, it wouldn't be helpful for me to call a weight machine a dumb bell, nor would it be helpful for me to talk about things they have never heard of like the supercalifragilistic rep without describing in detail what I mean by that. Furthermore, if they corrected me on my incorrect use of the term weight machine and dumb bell and I continued to use it incorrectly (for years) they would reasonably become frustrated.

 

[sophiecentaur]

Precisely. For "force" substitute "bannana" and his physics would make the same amount of sense.

 

[waynexk8]

Wayne,

May I suggest an experiment? Do you have access to a bathroom scale and two video cameras?

Hi jmmccain,

Yes,

Set up one camera to record the scale reading and perform your various fast and slow exercises. Afterwards, view the recording frame by frame and plot the scale reading for each frame (like douglis' graph, but it won't look the same). Don't forget to subtract your own weight from all the readings. This will rather conclusively demonstrate the difference in forces.

How do I do this please ?

Add up all the readings for a repetition (again, adjusted for your weight). Then divide by the number of frames over which the readings were observed. This will be the mathematical average and should be very close to the weight you are lifting. How close depends upon how well the experiment is performed.

However, as I said, there is a problem using average force, as if you do 1 repetition at any speed, say 1/1, that’s 1 second up and 1 second down, you will get the same average force if you do 1 repetition or a 100 repetitions. So average will not tell us anything, or will it ?

1 repetition = 100 force up and 60 force down, 100 + 60 = 160/1 = 160, average force = 160.
100 repetitions, 100 x 100 = 10000, 100 x 60 = 6000, 10000 + 6000 = 16000/100 = 160.

Some the average force is the same, BUT we all know that the doing the 100 repetitions is going to use more muscle force and put more tension on the muscles

I made a thread on this and told D. this average means nothing in this debate, but he and some others still insist it does, but can’t not say ?

The second camera is used to record the movement of the weight with a tape measure, or other suitably visible scale, in the background. Again, go through frame by frame and plot the position of the weight. This should look something like douglis' graph. Next, plot the change in position from one frame to the next (velocity). Next, plot the change of this change (acceleration) from one frame to the next. This third plot should look just like the first plot of the scale readings.

F = ma

The first plot of the scale readings is force. The last plot is acceration. Since the mass doesn't change, the two remain proportional to each other and the plots should be very similar.

The scale and cameras won't lie. If you need help decipering the results and making the appropriate plots, well, I may be busy, but there are others around who will help.

This seems very interesting, and have heard of this before, but not sure this will give use the right results. PLEASE what results will this tell us, I don't understand what this will show ?

Wayne

 

[sophiecentaur]

Do you guys know about sampling theory? To do this investigation to any satisfactory degree (to avoid misleading results) you will need a slo-mo camera, at the very least, and a set of scales with ms response times. I don't think the average bathroom scales has been designed that way. Also, you would need many samples and some good data analysis.
If it were really as trivial as you suggest, it would be an AS level Physics practical.
The last thing we would want would be a naff experiment that could yield results either way. If they went the wrong way, we could never ever convince Wayne that he's barking up the wrong tree.

And what about the 'total speed' question, Wayne? If you can't justify total speed then you can't justify total force. I await your specific and detailed response.

 

[waynexk8]

I'd advise not getting too involved with this. All your suggestions have been made many times before, in this thread and earlier threads. Wayne does not believe in the accepted ideas of Physics. He has his own models and vocabulary of Physics.

I have heard of this before, but never knew how to do it, and not sure what the results will be ?

Wayne does believe in accepted ideas of Physics, I don’t understand why you say that, what I am saying, and have proven, that the avenger force can show us “nothing” unless you think is can, if so, please say what ?

Also, your thought experiment on the bathroom scales would not show the sampled forces accurately or frequently enough to convince Wayne. There would be errors which he would jump on and claim that the experiment showed him to be right. The scales and camera would lie in practice unless a much more sophisticated system were used.

I would not jump on small errors ?

You a D. have dismissed a real World EMG test, but you cannot say why, I think because it shows what I say is right, and what you cannot work out with physics, in that there HAS and IS more force in the faster repetitions in the same time frame. Or do you and D. think that I fail faster when doing the faster repetitions because there is less force output by the muscles, and less tension on the muscles ? If so please say and explain why you think that or other, the point is you do fail 50% faster, thus there can ONLY be ONE reason for this, you produce for force in the same time frame, putting more tension on the muscles, making the muscles fail faster, if you do not think this, please say why ?

In any case, what you say would only apply to free lifts and not to exercises on machines that introduce friction. That also confuses Wayne and strengthens him in his misconceptions.

This does, or cannot confuse me; I train on free weights, Nautilus machines, and round circular pulleys, and have made many of my own machines and know the effects of all of them on the muscles. This debate is basically on free weights, with a machine doing the exercise or a Human.

Wayne

 

[sophiecentaur]

You so much do not believe in the ideas of conventional Physics, Wayne, that you even refuse to use the correct terminology.
You seem to ignore the fact that everyone agrees that the Maximum force for fast lifts must be higher because of the acceleration - that's proper Physics.
Just imagine - to change tack- that you had replaced your arms with a strong pair of springs of the appropriate stiffness. If you lifted the weights to your normal lift height and then let go, the weights would go down and up and down and up for a long time until friction became apparent. No work done at all if you recover the weights when they are at the top of their bounce. That's Physics. It's true and it just doesn't represent a good model of your muscles.
If you used a stiffer spring, the oscillations would be at a faster rate or, for a less stiff spring, the oscillations would be at a slower rate. No difference, in any of the cases, with the energy involved (total = zero).

I'm still waiting for your detailed reply about Total Speed. :zzz:

 

[waynexk8]

Hi sophiecentaur, you should get this debate and what I am getting at if you read 8, you too D.

You so much do not believe in the ideas of conventional Physics, Wayne, that you even refuse to use the correct terminology.

Please could you try and answer my questions, I am not interested in a mocking match, and you not answering seems to show you are unsure.

1,
You a D. have dismissed a real World EMG test, but you cannot say why ? You seem to be against RMS way, but cannot say why ??

2,
Do you agree that a muscle that if I do 10 repetitions at 1/1 = 20 seconds, and 1 repetition at 1/1 = 2 seconds. That the 10 repetitions will use more overall muscle force, more total muscle force, longer muscle force ? If and when you and D. understand this question, you will then understand my question, as from what you and D. say, I am sure you don’t understand. As I fail faster in the faster repetitions, does that not tell you anything ?

3,
I 100 physics had a debate, and did extensive tests, and all agreed I was right, would you be able to say I was right, or are you too deep in ?

4,
Do you see where/why average force means nothing in this debate ? If you do 1 repetition at any speed, say 1/1, that’s 1 second up and 1 second down, you will get the same average force if you do 1 repetition or a 100 repetitions. So average will not tell us anything, or will it ?

1 repetition = 100 force up and 60 force down, 100 + 60 = 160/1 = 160, average force = 160.
100 repetitions, 100 x 100 = 10000, 100 x 60 = 6000, 10000 + 6000 = 16000/100 = 160.

Some the average force is the same, BUT we all know that the doing the 100 repetitions is going to use more muscle force and put more tension on the muscles

Do you see what I mean ?

5,
Do you and D. think that I fail faster when doing the faster repetitions because there is less force output by the muscles, and less tension on the muscles ? If so please say and explain why you think that or other, the point is you do fail 50% faster, what both use 80% on the repetitions, thus there can ONLY be ONE reason for this, you produce for force in the same time frame doing the faster repetitions, putting more tension on the muscles, making the muscles fail faster, if you do not think this, please say why ?

6,
You use more energy in the faster, why ? As immediately you move faster, using more accelerations using more force, you use more energy, are you saying you use more energy because you don’t use more force ?

7,
You move the weight 6 times further in the same time frame, you have to use more force to move a weight further in the same time frame, if not, how do you move the weight further if by not using more force/accelerations ?

You seem to ignore the fact that everyone agrees that the Maximum force for fast lifts must be higher because of the acceleration - that's proper Physics.

I am not ignorant to that, I know it, and I know everyone else knows it, this is not the debate.

Let me try and explain again.

8,
We are the exact same strength, we are moving 80% of our 1RM, {Repetition Maximum} We are both going to move this weight until momentary muscular failure, meaning we are going to move the weight until we cannot lift it again.

I use my 100% maximum force ALL the time, which is a 100 pounds, you on the other hand only use 80% of your force all the time.

I hit momentary muscular failure about 50% faster than you do, because I am using 100%, but you are not using 100% you are using 80% yes 80%, and using 80% consistently for the exact same time as me moving the weights, {lets cal that 30 seconds} HOW can you think or claim that you ONLY using 80% consistently for the same time frame as I using 100% consistently, will use the same overall or total force output ? Or will make the same momentum/movement change ?

1,
I use 100% force for 30 seconds.

2,
You use 80% force for 30 seconds.

3,
How can using 80% of force for 30 seconds be the same as using 100 force for 30 seconds ? Please state why.

Just imagine - to change tack- that you had replaced your arms with a strong pair of springs of the appropriate stiffness. If you lifted the weights to your normal lift height and then let go, the weights would go down and up and down and up for a long time until friction became apparent. No work done at all if you recover the weights when they are at the top of their bounce. That's Physics. It's true and it just doesn't represent a good model of your muscles.
If you used a stiffer spring, the oscillations would be at a faster rate or, for a less stiff spring, the oscillations would be at a slower rate. No difference, in any of the cases, with the energy involved (total = zero).

Not sure what you mean there ? Or are getting at, as I HAVE to use force and energy on the way down, or lowering the weight as well as lifting.

I'm still waiting for your detailed reply about Total Speed. :zzz:

I answered that, is a different quantity, but if you want I will come up with a better one, please try and answer 8, and the rest, please you too D. and anyone else here.

Wayne

 

[sophiecentaur]

Nothing more from me until you answer my 'simple' question.

 

[douglis]

I use my 100% maximum force ALL the time, which is a 100 pounds, you on the other hand only use 80% of your force all the time.

I hit momentary muscular failure about 50% faster than you do, because I am using 100%, but you are not using 100% you are using 80% yes 80%, and using 80% consistently for the exact same time as me moving the weights, {lets cal that 30 seconds} HOW can you think or claim that you ONLY using 80% consistently for the same time frame as I using 100% consistently, will use the same overall or total force output ? Or will make the same momentum/movement change ?

1,
I use 100% force for 30 seconds.

2,
You use 80% force for 30 seconds.

3,
How can using 80% of force for 30 seconds be the same as using 100 force for 30 seconds ? Please state why.[/b]



Wayne

Everyday you prove that you don't have a clue what everybody is trying to explain to you.

YOU DON'T USE 100% FORCE FOR 30 SECONDS.
You use more force than the weight when you accelerate and less force than the weight when you decelerate.For these 30 seconds...either you lift fast or slow...the average force per second is the weight.

Stop saying nonsense like "average force means nothing in this debate".It's the same average force per second and this answers EVERYTHING everything you asked.
The effect of force over time(which is what you described as "total/overall force") is identical.

 

[douglis]

Let f(t) be the force exerted by the human on the weight at time t. Please define overall or total force in terms of f(t).

For example, average force from time $t_i$ to time $t_f$ is:
$$\overline{\mathbf{f}}=\frac{\int_{t_i}^{t_f} \mathbf{f}(t) \, dt}{t_f-t_i}$$

Please provide a similar rigorous definition for total or overall force.

Hi DaleSpam,
he described the "total/overall force" as the effect of force over time so the ∫f(t)dt is exactly what he means.
...and of course is the same regardless the lifting speed since the average force is the weight in any case.

 

[sophiecentaur]

Let's be accurate about this:
∫f(t)dt (a definite integral) is the impulse (change of momentum), which varies, depending which part of the lift cycle your integration extends. There is greatest impulse during sections of the fastest repetitions because the velocity change is greatest.

I wouldn't accept some cranky alternative language from douglis or dalespam so why should I accept it from Wayne?

Some of the published stuff uses terms similar to what he uses but makes it clear that it is only over the lift part. I think Wayne is over / mis interpreting a lot of it because his terminology is so inaccurate that I don't think he can be understanding the true messages.

Funny thing is that I find little to argue with in any of the references I've looked at. Yet he seems to find a great deal of support within the same texts. They obviously steer clear of making non-physics gaffs because they are not interested in Waynes "physics approach". They clearly know the business better than he.

I can see that he reckons he works harder doing faster lifts but I can't see how he can assume that he knows what his muscles are doing whilst he's lifting. That is what the EMG tries to do - but his interpretation of those seems a bit fanciful, to be honest (and polite).

I think his posts must qualify for a record for the consistently long yet vague posts I have read on PF.

 

[Dale]

he described the "total/overall force" as the effect of force over time so the ∫f(t)dt is exactly what he means.

Ah, ok. That is a defined term in physics, called impulse:
http://en.wikipedia.org/wiki/Impulse_(physics )

Edit: I see sophiecentaur already mentioned it.

I just want to point out that impulse does not have units of force, it has units of momentum. Assuming that we are talking about the impulse over one full rep then the change in momentum is 0 so the impulse on the weight is 0.

The only two forces acting on the weight are gravity and the human, so the impulse provided by the human is equal and opposite to the impulse provided by gravity. So, the impulse is equal to the weight times the duration of the rep. Therefore the human provides a larger impulse on a slow rep than on a fast rep.

waynekx8, if by "total force" you mean "impulse" then it is greater for a slow rep than for a fast rep. If you mean some other quantity then please define it explicitly.

 

[douglis]

The only two forces acting on the weight are gravity and the human, so the impulse provided by the human is equal and opposite to the impulse provided by gravity. So, the impulse is equal to the weight times the duration of the rep. Therefore the human provides a larger impulse on a slow rep than on a fast rep.

waynekx8, if by "total force" you mean "impulse" then it is greater for a slow rep than for a fast rep. If you mean some other quantity then please define it explicitly.

That's the whole point of the discussion and it's been explained to Wayne many times.
Regardless if you lift the weight fast or slow for 30 seconds...the "total/overall" force is always the same and equal with gravity's impulse for that duration.