Impulse/force in pounds for the time frame

[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.

 

[waynexk8]

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.

Well how as a physicist would you define/call it ?

I have said this many times, to move a car 1 mile in 30 minutes, with the force of a Man pushing, you will “have” to use a certain amount of force, right ? As if you knew the exact amount of force, and you used less force, you would not move the car 1 mile in 30 minutes, so you must be used a certain amount of force for 30 minutes. And if you moved the car 1 mile in 45 minutes, you “would” be using less overall or total force, yes ?

I am not being sarcastic here, but when I do ? It means I am asking you a question if you agree of not, and it’s very hard for me if you don’t answer, so please if you agree or don’t agree, please state why.

How can we add up total or overall speed. Right, let’s take an easy one, and measure speed by every .1 of a second. The object is moving at a constant speed of 100mps, so we could say JUST for arguments sake, the total or overall speed for 10 seconds = 1000. This is like they do all over the World in powerlifting, weightlifting, bodybuilding and athletics, if they have done a training day on squats and bench press; they say they have moved a certain amount of tonnage or poundage in a day. So if they did 10 x 10 squats and 10 x 10 bench press with 200 pounds, they will say they have moved 4000 poundage in one day, and if they want to get very technical, they will work or the exact amount of time they moved this poundage, and say they moved 4000 in 200 seconds, thus producing more power and force in less time than the week before hopefully.

But speed is a different thing, F = ma. If we work out the forces at work on some objects, it will be by multiplying the weight of the object by the acceleration of the object right.
The force at work on a cart pulled by a horse, the weight of the cart is 400 and an Acceleration of 20m/s to work out the Force pushing the cart is by multiplying the weight by the acceleration, 400 x 20 = 8000N in 1 second, if an acceleration of 40m/s its 400 x 40 = 16000N in 1 second. The more seconds you push the weight, the longer in seconds you will have to use this same force, as you cannot push the same weight for 10m with just using a 1 second push of 1600N can you ? Seems like some are forgetting to add in the force with respect to time, which equals more distance the weight moved, and longer the force is applied ?

Wayne

 

[Dale]

The more seconds you push with a fixed force, the larger the impulse of the push. Therefore a slow rep will have a greater impulse than a fast rep.

 

[douglis]

The more seconds you push with a fixed force, the larger the impulse of the push. Therefore a slow rep will have a greater impulse than a fast rep.

...consequently,when the same average force is applied for the same duration the impulse is always the same regardless if you lift fast or slow or if you do more or less reps.

That's the whole story that took so many pages and so many threads.

 

[sophiecentaur]

Well how as a physicist would you define/call it ?

I'd call it Impulse or Work Done or whatever else I actually meant. Your problem is that you haven't really got beyond the stage of deciding what you really want to know or what you mean. We have all told you answers that could apply for a range of things that you could mean. (Please please don't give us another story to illustrate what you mean - I think I shall run out screaming at another mention of Rep Rate)

I have said this many times, to move a car 1 mile in 30 minutes, with the force of a Man pushing, you will “have” to use a certain amount of force, right ? As if you knew the exact amount of force, and you used less force, you would not move the car 1 mile in 30 minutes, so you must be used a certain amount of force for 30 minutes. And if you moved the car 1 mile in 45 minutes, you “would” be using less overall or total force, yes ?

You start of talking about a force then you suddenly jump to "overall or total force". That's where the nonsense creeps in. Force is Force and it means just one thing. If the man pushed the car for 30 minutes, then I am assuming this would be at a steady speed on a flat road, just against friction? What you could say is that the WORK he did was Force times the distance the car traveled.
If he pushed with less force, then the car would not be going as fast and it would cover less distance. The WORK done would be the new force times the new distance.
You seem to have some objection to describing things that way. Why? It's the way that the rest of us talk because it fits in with the rest of Physics.
You would be doing less WORK but your description, involving the word 'force' makes no sense.

I am not being sarcastic here, but when I do ? It means I am asking you a question if you agree of not, and it’s very hard for me if you don’t answer, so please if you agree or don’t agree, please state why.

I have done that for you.

How can we add up total or overall speed. Right, let’s take an easy one, and measure speed by every .1 of a second. The object is moving at a constant speed of 100mps, so we could say JUST for arguments sake, the total or overall speed for 10 seconds = 1000.

Here you go again with 'total speed'. What you seem to mean here is that the DISTANCE is 1000 (100m/s for 10s). Speed times time = distance, doesn't it? So "total speed = distance" for you? Not good enough. You want to join the Physics club and get answers. Club rules apply, I'm afraid and you have to use the right terms or no one will understand what you're on about.

This is like they do all over the World in powerlifting, weightlifting, bodybuilding and athletics, if they have done a training day on squats and bench press; they say they have moved a certain amount of tonnage or poundage in a day. So if they did 10 x 10 squats and 10 x 10 bench press with 200 pounds, they will say they have moved 4000 poundage in one day, and if they want to get very technical, they will work or the exact amount of time they moved this poundage, and say they moved 4000 in 200 seconds, thus producing more power and force in less time than the week before hopefully.

No - they did more WORK and possibly in less time. This means more Power. The details of what forces were involved are not included in their statement.

But speed is a different thing, F = ma. If we work out the forces at work on some objects, it will be by multiplying the weight of the object by the acceleration of the object right.
The force at work on a cart pulled by a horse, the weight of the cart is 400 and an Acceleration of 20m/s to work out the Force pushing the cart is by multiplying the weight by the acceleration, 400 x 20 = 8000N in 1 second, if an acceleration of 40m/s its 400 x 40 = 16000N in 1 second. The more seconds you push the weight, the longer in seconds you will have to use this same force, as you cannot push the same weight for 10m with just using a 1 second push of 1600N can you ? Seems like some are forgetting to add in the force with respect to time, which equals more distance the weight moved, and longer the force is applied ?

Wayne

Speed is a different thing, is it? But Maths applies to everything. The word 'total' means adding things together, whether it's cabbages or Joules. Some things just can't be 'added together' and adding speeds-at-different-times is as much forbidden as adding forces-at-different-times. This isn't poetry or stream of consciousness - it's strict and rigorous stuff.

Sorry, the last bit just reads like ramblings. I can't see what you're getting at except to tell me a story in non-technical terms. Why does a horse and cart have to be different from a man and a car? The man and car scenario says it all.

You are getting some damned good value out of all this you know.

 

[Dale]

...consequently,when the same average force is applied for the same duration the impulse is always the same regardless if you lift fast or slow or if you do more or less reps.

Agreed.

 

[Dale]

waynexk8, please try to learn the correct term for the different concepts in physics. If I came to your gym to ask for weightlifting advice you would be right to teach me the difference between a "curl" and a "squat". And if I called a "curl" a "total squat" you would be right to correct me and insist on using the correct language. And if I persisted in calling "curls" by the word "total squats" it would make things confusing, particularly if I began talking about how I am concerned that my knee surgery will interfere with my "total squats":

Me: "But my doctor says I won't be able to bend my knee as far"
You: "So what, curls don't use your knee"
Me: "Yes, but I am talking about total squats, and squats do use your knee"
You: "Yes, squats do use your knee, but you are talking about a different thing. What you call total squats are properly called curls. Curls don't use your knee"
Me: "But squats do use my knee and my doctor says I won't be able to bend my knee as far"
...

 

[waynexk8]

Not sure if I am doing the right thing here, going to answer some of the more recent posts, if you all think it’s a bad idea please say. I will get back to the other posts.

waynexk8, please try to learn the correct term for the different concepts in physics. If I came to your gym to ask for weightlifting advice you would be right to teach me the difference between a "curl" and a "squat". And if I called a "curl" a "total squat" you would be right to correct me and insist on using the correct language.

Very right of you, I do agree, but please I do find this hard at times, but if you tell me something I said wrong, I will try and remember, but if for some reason I get it wrong of forget, please tell me again and I will make a note. So sorry all, I am trying, but the most annoying is when people do not answer my writings, unless you have, and I have not got to them

And if I persisted in calling "curls" by the word "total squats" it would make things confusing, particularly if I began talking about how I am concerned that my knee surgery will interfere with my "total squats":

Yes see what you mean, could you point me out on the things I am getting wrong please.

Me: "But my doctor says I won't be able to bend my knee as far"
You: "So what, curls don't use your knee"
Me: "Yes, but I am talking about total squats, and squats do use your knee"
You: "Yes, squats do use your knee, but you are talking about a different thing. What you call total squats are properly called curls. Curls don't use your knee"
Me: "But squats do use my knee and my doctor says I won't be able to bend my knee as far"
...

Total get your point, and it’s a good one, and I do need to concentrate on this more.

Wayne

 

[sophiecentaur]

Perhaps I could suggest, Wayne, that you look up the definition of each of these 'physics' words, write that definition down for reference then only make comments using them as you've written them in your notes.

could you point me out on the things I am getting wrong please

Don't make me larf!

We spend most of our time telling you where you are using the wrong terms. If you are really so low in intelligence that you cannot recognise that then you will never get any of this. But look up the word 'intransigence'. It describes exactly what I see in your attitude to this problem.
You are very free with your apologies about this but there is no sign that you are actually doing anything about it. "Sorry" means you intend not to do something again. It's not just a get-out-of-gaol-free card.
We could be here till next Christmas and you will not get an answer in your own private made-up terms. Nor will you get it anywhere else.

Tell us, do you write all these put-downs in a book and then relate them to your weight lifting chums for a good giggle? You certainly know how to evoke them. You cause me to become very inventive!

 

[waynexk8]

Originally Posted by douglis

...consequently,when the same average force is applied for the same duration the impulse is always the same regardless if you lift fast or slow or if you do more or less reps.

Agreed.

But D. has not stated the average force, and some here are telling me, that there cannot be an overall or/and total force, so “please” where and how do you get this average force from ? I do not get where you get this from or how you work it out, or from which repetition ?

Here is how I sort of see it, let’s just say I used the following numbers of force in Newton’s, for 1 third of a second each. 7 then 13 then 91, average of the 3 numbers = 37, but the overall and or total of force used in the 1 second would be 111, so where do you get that the average is the same ?

PLEASE, you need to say which repetitions you are talking about. As if you measure the impulse from a weight lifted off the ground, lifted up 1m in .5 of a second, and then lowered down 1m in .5 of a second, and then you measured the impulse from a weight that started at the top, was lowered down 1m in .5 of a second, and then lifted up 1m in .5 of a second. Then there would be more force/impulse needed when lower and then lifting the weight.

As there will be huge deceleration/acceleration peak force from the transition from negative to positive, the MMMTs {momentary Maximum muscle Tensions} and the faster you lower the higher the force/impulse for a Milly second will be.

I would say 99% that you are taking the impulse/force from the weight lifted, and lowered "only" ?

Wayne

 

[sophiecentaur]

Originally Posted by douglis

...

But D. has not stated the average force, and some here are telling me, that there cannot be an overall or/and total force, so “please” where and how do you get this average force from ? I do not get where you get this from or how you work it out, or from which repetition ?

That statement just demonstrates that you do not know the meaning of the term 'Mean'. You think you can't have a Mean if you can't just add things up? Look up the true definition of the Mean and how it can be calculated. Don't apply your elementary School definition.

 

[waynexk8]

That statement just demonstrates that you do not know the meaning of the term 'Mean'. You think you can't have a Mean if you can't just add things up? Look up the true definition of the Mean and how it can be calculated. Don't apply your elementary School definition.

Well ok, this is more of the things I need to know, as yes I thought mean/average was adding up the numbers and dividing them, so how else can you get to or have mean/average ?

Wayne

 

[waynexk8]

I'd call it Impulse or Work Done or whatever else I actually meant. Your problem is that you haven't really got beyond the stage of deciding what you really want to know or what you mean. We have all told you answers that could apply for a range of things that you could mean. (Please please don't give us another story to illustrate what you mean - I think I shall run out screaming at another mention of Rep Rate)

So if its work done, do we all agree that the fast does 6 times more work in this example ?

You lift a 100 pounds 1 time in 1 second, you then lift the 100 pounds 10 times in 1 second, you have used more overall/total force lifting the weight the 10 times.

We all know it takes a force to move a weight, so if you use {let’s just go for a constant force now} a force, of 1N is equal to the amount of net/total/overall force required to accelerate a weight of 1kg to 1m in 1 second. So if I used less force say half a Newton for 1 second, the weight would not go so far. This is what I am trying to say, you need to use more Newton’s of force, to move the same weight further in the same time frame. But you and D. are saying you use the same Newton’s of force to move the weight half a meter in 1 second, and to move it 1 meter in 1 second ?

As I said before, if I “only” using 80% of my force for 6 seconds, I would be using 80% of my force for 6 seconds, agreed ? Let’s call that 80 force x 6 seconds. I then use 100% of my force for 6 seconds I would be using 100% of my force for 6 seconds, agreed ? Let’s call that 100 force x 6 seconds.

What I don’t get, is if you me or anyone else first used 80% of their force for 6 seconds, then used 100% of their force for 6 seconds, how can you say the person only using 80% of force to the person using 100% of force, has used the same force ? I mean you “are” using 20% less force, we all agree I “am” using 20% less force for the 6 seconds, but then you turn around and say we are both using the same force ? Is that not a total contradiction ? As we agree I am using 25% more force than you, but after we agree, you say that 80% force and 100% force are the same ?

Practical World experiment, I move my weight up and down 6 times in 6 seconds, it moves 12m in all, you move your weight up and down 1 time in 6 seconds, it moves 2m.

You start of talking about a force then you suddenly jump to "overall or total force". That's where the nonsense creeps in. Force is Force and it means just one thing. If the man pushed the car for 30 minutes, then I am assuming this would be at a steady speed on a flat road, just against friction? What you could say is that the WORK he did was Force times the distance the car traveled.

Ok, that sounds more like what I am thinking. That’s what I said a very long time ago to D. I said I moved the weight 6 times further in the same time frame.

If he pushed with less force, then the car would not be going as fast and it would cover less distance. The WORK done would be the new force times the new distance.

Yes I total agreed/agreed, this is what I say too. The WORK done would be the new force times the new distance, that sounds fantastic, thank you very much.

You seem to have some objection to describing things that way. Why? It's the way that the rest of us talk because it fits in with the rest of Physics.
You would be doing less WORK but your description, involving the word 'force' makes no sense.
I have done that for you.

So you call force over time/distance work done, that’s more like what I have been trying to say all along, so work will be product of a force times the distance through which it moves it by the time ? I did mention something like this, so that means I was right, and that’s why my EMG states what I said ?

Here you go again with 'total speed'. What you seem to mean here is that the DISTANCE is 1000 (100m/s for 10s). Speed times time = distance, doesn't it? So "total speed = distance" for you? Not good enough. You want to join the Physics club and get answers. Club rules apply, I'm afraid and you have to use the right terms or no one will understand what you're on about.No - they did more WORK and possibly in less time. This means more Power. The details of what forces were involved are not included in their statement.

Now you lose me again, as I thought you said; The WORK done would be the “new force” times the new distance. You did say “new force”

More power = more force used for the same time ?

Speed is a different thing, is it? But Maths applies to everything. The word 'total' means adding things together, whether it's cabbages or Joules. Some things just can't be 'added together' and adding speeds-at-different-times is as much forbidden as adding forces-at-different-times. This isn't poetry or stream of consciousness - it's strict and rigorous stuff.

Ok you say you can’t add force up, so why are you saying the average is the same, and why has D. been debating and saying the total/overall force is the same ?

Sorry, the last bit just reads like ramblings. I can't see what you're getting at except to tell me a story in non-technical terms. Why does a horse and cart have to be different from a man and a car? The man and car scenario says it all.

Man cart or horse, all the same thing, just changed it for variety.

But is the below not right ?

But speed is a different thing, F = ma. If we work out the forces at work on some objects, it will be by multiplying the weight of the object by the acceleration of the object right. The force at work on a cart pulled by a horse, the weight of the cart is 400 and an Acceleration of 20m/s to work out the Force pushing the cart is by multiplying the weight by the acceleration, 400 x 20 = 8000N in 1 second, if an acceleration of 40m/s its 400 x 40 = 16000N in 1 second. The more seconds you push the weight, the longer in seconds you will have to use this same force, as you cannot push the same weight for 10m with just using a 1 second push of 1600N can you ? Seems like some are forgetting to add in the force with respect to time, which equals more distance the weight moved, and longer the force is applied ?

You are getting some damned good value out of all this you know.

Yes I do honestly thank you for your time and help, please if you have looked at some of my videos you know I do mean that.

I am not sure if anyone has answered this, thus I will say it again. If “I” lift a weight up and down slow, and it takes me 1 minutes to hit momentary muscular failure, this is I cannot lift the weight again. I then lift the weight up and down very far, and it takes me 30 seconds to hit momentary muscular failure, I “HAVE” used more of my temporary force per unit of time, I have used my force up faster, yes ?

That’s why the EMG stats more muscular activity in the same time frame for the faster reps.

Wayne

 

[sophiecentaur]

Sorry, that's just wall to wall 'instances' again and you have not condensed your question into one that is answerable (or even understandable). I CAN say that, if you do something six times more often then it's six times as much work (ON the weights), irrespective of how long you took over it. FACT
However, the work done by your muscle fibres amongst themselves does not relate to what I just wrote. That also seems to be a problem for you.

Yes I total agreed/agreed, this is what I say too. The WORK done would be the new force times the new distance, that sounds fantastic, thank you very much.

I referred to your situation where a car is pushed with two different forces. Less work is done and the car doesn't get as far because it will be going at a slower speed. READ WHAT I WROTE - it's in English.

More power = more force used for the same time

No, not at all. READ THE DEFINITIONS. More Power does not have to mean that. It would be more force over the same Distance in the same time.

Did you hope that the man and the car were sorted out then? They, apparently were not.

Ok you say you can’t add force up, so why are you saying the average is the same, and why has D. been debating and saying the total/overall force is the same ?

He was just indulging you by going along with a nonsense term and has managed to shoot himself in the foot. I knew it was a bad idea haha.
[Ok you say you can’t add force up, so why are you saying the average is the same, and why has D. been debating and saying the total/overall force is the same ?
[/QUOTE]
Because you don't just add em up and divide by 'some number'. Look it up: Definition of the MEAN in wikipedia.

Seriously, if you can't get any of this you have no hope of understanding any answer I could give you. You may just have to accept it.

It would be a good idea if you stopped trying to claim you are right by calling on the EMG results as proof. I understand what your machine is saying, to some extent but you, clearly, do not. The makers are providing you with some numbers which represent how hard you are 'working' your muscles. They do not claim any more. Hope the money they are making is sufficient to make up for all the daft questions, from other people, as well as yours they must be getting.

 

[douglis]

As I said before, if I “only” using 80% of my force for 6 seconds, I would be using 80% of my force for 6 seconds, agreed ? Let’s call that 80 force x 6 seconds. I then use 100% of my force for 6 seconds I would be using 100% of my force for 6 seconds, agreed ? Let’s call that 100 force x 6 seconds.


Wayne

Unbelievable!You still can not get this!

YOU DON'T USE 100% FORCE FOR THE WHOLE 6 SECONDS.
You use more force than the weight when you accelerate and less force than the weight when you decelerate.For these 6 seconds...either you lift fast or slow...the average force per second is always the weight.
In any case you apply 80% of your force for 6 seconds.The impulse(let's leave that stupid "total/overall force" term) is always the same.

 

[Dale]

Very right of you, I do agree, but please I do find this hard at times, but if you tell me something I said wrong, I will try and remember, but if for some reason I get it wrong of forget, please tell me again and I will make a note. So sorry all, I am trying, but the most annoying is when people do not answer my writings, unless you have, and I have not got to them

Yes see what you mean, could you point me out on the things I am getting wrong please.

Total get your point, and it’s a good one, and I do need to concentrate on this more.

Hi waynexk8, the first thing that you need to correct is that there is no such thing as total force. There is impulse which is defined as:
\mathbf{I}=\int_{t_i}^{t_f} \mathbf{f}(t) \, dt
Impulse has units of momentum, not units of force.

There is also average force which is defined as:
\overline{\mathbf{f}}=\frac{\int_{t_i}^{t_f} \mathbf{f}(t) \, dt}{t_f-t_i}
Average force does have units of force.

Second, you are under the mistaken impression that the impulse is higher if you do a rep faster. This is incorrect. The impulse for one rep depends only on the weight and the amount of time, so one slow rep (10 s) might have the same impulse as 5 fast reps (2 s each). Similarly, you seem to be under the impression that the average force over a rep is higher if you go faster. This is also incorrect. The average force for one rep depends only on the weight. Notice that there is a close relationship between impulse and average force.

Personally, I think that the question you should be asking is the following:
"What different physical quantities are in fact higher in a fast rep than in a slow rep?"

 

[sophiecentaur]

You might just as well have written that post in Russian, you realize!

 

[waynexk8]

Let’s say {and this is the debate} we had a machine lifting up a weight, and you could set this machine to any lifting force. First we set it at 100f {lets just call it force} up 1m and down 1m for 6 seconds, then we set it at 50f up 1m and down 1m for 6 seconds. The 100f as it had more force would move faster, so it would move the weight up and down more times.

At the end of the test, you would not say that the machine used the same force for 6 seconds, as it was set for 100f and 50f. The 100f would as you said do also more work, as you say, work is the product of a force times the distance through which it acts, and called the work of the force.

Torque. When a force of 1 Newton is applied at a distance of 1 meter from a pivot at right angle to the radius then a torque of 1 Newton Meter is present.

So if I applied 2N of force at right angles 1 m from the pivot point, for 2 seconds to turn a very tight nut, to applying 1N of force at right angles 1 m from the pivot point, for 2 seconds to turn a very tight nut. You would not say that both times I applied the different forces I actually used the same force in the same time ?

As I said before, I push up and lower the weight under control, with as much force {I call this strength} as I can for 6 seconds, I then push up and lower the weight under control, with 20% less force for 6 seconds. I don't get why you say I have used the same force both times ? Also if I lifted until I could not lift the weight again, I WOULD fail far faster when lifting faster, proving I have lifted with more force what I said I was using as much force as I could. Thus why the EMG machine stated this as well


Wayne

 

[waynexk8]

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,

Interesting, so Impulse is something that changes the momentum/movement of an object; the integral, part of the whole of a force with respect to time, but if it’s a part of force, is not it a force ?

This is what I keep saying; a small force applied for a long time can produce the same momentum/movement change as a large force applied briefly, because it’s the product of the force and time for which it is applied that is important.

But this is my point, the small force is not applied longer then the large force, it’s applied for the same time. So the small force applied for the same time as my large force for a brief time, can’t not as we know cannot produce the same momentum/movement

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.

No, as both are applied for the same time, the fast force {which is let’s just call/say force now} is 100% force from the Man, or a 100 pounds of force moving the 80 pounds up 1m and down 1m for 6 times = 12m in 6 seconds. The slow force is 80% force from the Man, or 80 pounds of force moving the 100 pounds up 1m and down 1m 1 times = 2m in 6 seconds.

The fast uses 100 pounds of force for 6 seconds, and the slow uses 80 pounds of force for 6seconds. A 100 pounds of force = more than 80 pounds of force.

Wayne

 

[waynexk8]

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.

Before I bought the EMG, you stated that a EMG is the only way to sort this out, now that three experts on EMG state RMS is the best way to do the EMG, why do you now think it’s wrong ?

The physics are trying me there cannot, or physics does not have an equation for total/overall force, so how/why have you ? How can it be equal if I use more energy, work, and distance ?

How can it be equal If say I am using all my force, a 100% force a 100 pounds of force for 6 seconds, I then say out loud, I am now going to use less force than before, I am now going to use 80% force, 80 pounds of force for 6 seconds, and let’s say I am a machine, how is 80 as much as 100 ?

Then if you want more proof, if I lifted the first time using a 100% force, a 100 pounds, I “would” fail at lifting the 80 pounds that we are lifting, say 50% faster than lifting with the 80% of 80 pounds of force, so as I fail 50% faster, that proves I have been using more force, cos all my temporary force is used up faster.

Sorry to repeat myself.

Wayne

 

[Dale]

First we set it at 100f {lets just call it force} up 1m and down 1m for 6 seconds, then we set it at 50f up 1m and down 1m for 6 seconds.

The problem is that this is impossible as you have described it. You cannot move a weight up and down using a constant force. Your force has to be higher than the weight to accelerate it up and lower than the weight to accelerate it down.

If your lifting force is higher then your lowering force is lower, so the average force over one rep is always equal and opposite the weight.

This is an important point, so let's stop here before going on with your questions about impulse. Does what I said here make sense?

 

[douglis]

If your lifting force is higher then your lowering force is lower, so the average force over one rep is always equal and opposite the weight.

DaleSpam...it doesn't have to be for over the whole rep.
Even if you examine separately the lifting and the lowering phase the average force is always the weight.In both phases the weight starts and ends at rest so the average acceleretion is always zero.

 

[douglis]

Before I bought the EMG, you stated that a EMG is the only way to sort this out, now that three experts on EMG state RMS is the best way to do the EMG, why do you now think it’s wrong ?

The physics are trying me there cannot, or physics does not have an equation for total/overall force, so how/why have you ? How can it be equal if I use more energy, work, and distance ?

How can it be equal If say I am using all my force, a 100% force a 100 pounds of force for 6 seconds, I then say out loud, I am now going to use less force than before, I am now going to use 80% force, 80 pounds of force for 6 seconds, and let’s say I am a machine, how is 80 as much as 100 ?

Then if you want more proof, if I lifted the first time using a 100% force, a 100 pounds, I “would” fail at lifting the 80 pounds that we are lifting, say 50% faster than lifting with the 80% of 80 pounds of force, so as I fail 50% faster, that proves I have been using more force, cos all my temporary force is used up faster.

Sorry to repeat myself.

Wayne

No Wayne...we will not forgive you for repeating yourself.You ask again the same nonsense ignoring all the answers.

For some strange reason you're unable to understand that it's impossible to use the 100% of your force for the whole set.Regardless if you lift fast or slow you use the same average force for the same duration.
The impulse(what you stupidly call "total/overall force") is always the same.
I don't care if you don't want to accept the truth or you just don't have the intelligence to understand it.
It's a fact...accept it.

 

[Dale]

DaleSpam...it doesn't have to be for over the whole rep.
Even if you examine separately the lifting and the lowering phase the average force is always the weight.In both phases the weight starts and ends at rest so the average acceleretion is always zero.

This is true, but Wayne is interested in doing reps, and over a rep we can also make statements about the work done. So I just mention reps for simplicity.

 

[waynexk8]

Unbelievable!You still can not get this!

I think you to not still get this, you need to try and counter real Would tests that show you wrong, not just try and say theory. {why do I have to repeat ?}

Practical proof,
1,
The EMG, that reads the faster in the same time frame, with a higher reading.

2,
The fast uses more energy.

3,
The fast moves the weight 6 times as far.

4,
If you lift the weight for as long as you can, in the slow and fast cadence, when using 80% of your 1RM, you will fail roughly 50% faster in the fast, “proving” you have used more of your temporary force up faster, using more force per unit of time. As when I can’t lift the weight, as all my temporary force has gone, you have still got force left, thus you have not used it up so fast.

Or to say it answer way, I have lifted the weight say up and down 20 times in 20 seconds = 40m, but in this same time frame, you have only moved the weight up and down 5 times = 10m, and as you know, it takes less force to move a weight less distance in the same time frame.

http://www.youtube.com/watch?v=sbRV...DvjVQa1PpcFP45dm44zL2NgfO5GmExvyhvSGywVHw7f8=

YOU DON'T USE 100% FORCE FOR THE WHOLE 6 SECONDS.

I try to use 100%, you also have to accelerate and decelerate, the difference is, in 6 seconds I am accelerating say 80% of the positive, and so with 6 repetitions at 1m each way, I have accelerated for 4.8m, to your basically constant speed of nearly 1m. In time I accelerate for 2.4 seconds, you move at a constant speed for nearly 3 seconds. It takes far more force to accelerate something than to move it at a constant speed.Maybe you will understand it this way better
Lets put it another way. A very big enclosed piston is going up and down, when it reaches to top it hits the cylinder casing and reverses back down, when it hits the bottom hits the cylinder casing and reverses back down.

This piston has a 100 pounds of force, and its moving 80 pounds, the piston lift the weight, hits the top {same as my hands are griped tightly around the weight bar} then immediately pulls down in the opposite direction, it does this 6 times moving the weight 12m in 6 seconds. Then the piston moves the weight up once and down all in 6 seconds, moving the weight 2m in all.

LOOK the force to hold the weight half way up, is basically the same as you moving it very slow, then tell me “how” you make any of these forces up. Just PLEASE for ones tell me how you make these forces up ?

Fast = 45.3N in 2.8 seconds.

Slow = 1.6N in 2.8 seconds.

I do not understand how you think you make up roughly 44N ? In the same time frame.]

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

You use more force than the weight when you accelerate and less force than the weight when you decelerate. For these 6 seconds...either you lift fast or slow...the average force per second is always the weight.
In any case you apply 80% of your force for 6 seconds.The impulse(let's leave that stupid "total/overall force" term) is always the same.

I explained why average means nothing, and if total/overall means nothing, just try and show me how you would out the average ?

I am fed up with repeating, as you do not answer questions, thus we have to go back to them.

Wayne

 

[waynexk8]

Sorry, that's just wall to wall 'instances' again and you have not condensed your question into one that is answerable (or even understandable). I CAN say that, if you do something six times more often then it's six times as much work (ON the weights), irrespective of how long you took over it. FACT

Ok, so I did 6 times more work, but if that’s in the same time frame, or less time, your saying time means nothing ?

What if I said it like this, imagine force as the fuel in your car tank, the fast and the slow has 10 Gallons each, they both move for 6 seconds, the fast moves 6 times further than the slow, let’s say the fast used up 6 gallons and the slow 1 gallon in the same time frame.

As the muscle use force or/and strength to move a weight, and too accelerate a weight further in the same time frame as a person moving it at a constant speed, that needs more Newton’s. The force at work on a cart pulled by a horse, the weight of the cart is 400 and an Acceleration of 20m/s to work out the Force pushing the cart is by multiplying the weight by the acceleration, 400 x 20 = 8000N in 1 second, if an acceleration of 40m/s its 400 x 40 = 16000N in 1 second. The more seconds you push the weight, the longer in seconds you will have to use this same force, as you cannot push the same weight for 10m with just using a 1 second push of 1600N can you ? Seems like some are forgetting to add in the force with respect to time, which equals more distance the weight moved, and longer the force is applied ?

So faster push = 1600N.

Slower push = 800N.

Both in the same time frame.

I don’t get what’s wrong with that please ? Is it because I have to decelerate. SORRY, just seen what you wrote below.

=sophiecentaur;3803479However, the work done by your muscle fibres amongst themselves does not relate to what I just wrote. That also seems to be a problem for you.

I know quite a bit on muscles fibers, best not get into this, as we can easy say this is a machine pushing up and down.

I referred to your situation where a car is pushed with two different forces. Less work is done and the car doesn't get as far because it will be going at a slower speed. READ WHAT I WROTE - it's in English.

I missed this, will go back and check, thank you.

No, not at all. READ THE DEFINITIONS. More Power does not have to mean that. It would be more force over the same Distance in the same time.

But does not it mean more force over more distance in the same time frame too ?

Did you hope that the man and the car were sorted out then? They, apparently were not. He was just indulging you by going along with a nonsense term and has managed to shoot himself in the foot. I knew it was a bad idea haha.

He has done that many times.

I lift my 1RM, my maximum weight I can lift for 1 repetition, to do this, I have to move as relatively as fast as I can, if NOT the weight will not go up. Let’s call that a 100 pounds, lifted up and lowered down in 6 seconds using my maximum force.

I then don’t use my maximum force, as I lift slower, so this time I only lift 90 pounds up and down in 6 seconds.

I used more force per unit of time lifting the weight the fastest with more force, right ?

Wayne wrote=sophiecentaur;3803479Ok you say you can’t add force up, so why are you saying the average is the same, and why has D. been debating and saying the total/overall force is the same ?
Because you don't just add em up and divide by 'some number'. Look it up: Definition of the MEAN in wikipedia.

I will have to go over that tomorrow.

Seriously, if you can't get any of this you have no hope of understanding any answer I could give you. You may just have to accept it.

It would be a good idea if you stopped trying to claim you are right by calling on the EMG results as proof. I understand what your machine is saying, to some extent but you, clearly, do not.

Could you tell me what you think it means then ?

As I have known and studied EMG tests for many years, I know all too well what they mean.

When I move faster I put out more force, be this moving up and down at .5/.5 to 3/3 or 5/5 to 10/10. So as I am using more force to move faster, this puts MORE tension on the muscles, and the only way the slower moving can make up these higher forces for the moving faster, is to be used for a longer time than the fast, but they are both used the same time in this debate.

It’s like I put out a 100 force, and the slow puts out 80 force, 100 force can NOT be made up by the 100 force in the same time frame, even if for some of the repetition I am on the deceleration, as I HAVE put out a 100 force thus 100 tension on the muscles, in the whole time frame, the slow has NEVER put out a 100 force, it’s just put out 80 force, 80 is not as high as 100. Thus as I do more and more repetitions in the same time frame, the total/overall reading on the EMG MUST be higher with the fast.

The makers are providing you with some numbers which represent how hard you are 'working' your muscles. They do not claim any more. Hope the money they are making is sufficient to make up for all the daft questions, from other people, as well as yours they must be getting.

Right, and that’s the debate.

I don’t like to repeat myself, But as you fail roughly 50% faster when both using 80% or your 1RM, that means on the fast you have worked the muscles harder, using more force in the same time frame, unless anybody thinks you fail faster because its easier ?

Wayne

 

[douglis]

The same as always!You ignore everything that has been answered.
Let's redefine the question.You asked which lifting speed has greater effect of force over time(impulse) which in Wayne's world is defined as "total/overall force".

Everyone explained to you that the impulse is identical regardless the lifting speed but you deny that fact based on some "practical proofs".Let's see them once again.

Practical proof,
1,
The EMG, that reads the faster in the same time frame, with a higher reading.

The EMG reads the Root Mean Square of the values.
Does the higher RMS somehow change the fact that the impulse is the same?NO!

2,
The fast uses more energy.

Does the higher rate of energy expenditure somehow change the fact that the impulse is the same?NO!

3,
The fast moves the weight 6 times as far.

Does the greater distance somehow change the fact that the impulse is the same?NO!

4,
If you lift the weight for as long as you can, in the slow and fast cadence, when using 80% of your 1RM, you will fail roughly 50% faster in the fast, “proving” you have used more of your temporary force up faster, using more force per unit of time.
Wayne

Does the higher rate to fatigue somehow change the fact that the impulse is the same?NO!

So Wayne...quit these nonsense and see what everybody's telling you.Your above pseudoarguments are just a sad attempt to ignore the facts and keep leaving in your own world.

 

[Dale]

Wayne, please focus. My post 231 is very brief. Please go over it and let me know if it makes sense to you. You need to understand how the average force is the same before we can continue. This is a simple fact of physics that you cannot avoid and need to confront.

If I went to you for help on my biceps but insisted on doing squats all day and kept on giving you all sorts of examples about why curls don't make sense to me then I would not be able to progress, would I? It is time for you to stop squatting and do a few mental curls instead. Please go back to 231 and let's confront the issue directly.

Do you understand that it is physically impossible to move a weight up and down using a constant force?

 

[sophiecentaur]

Let’s say {and this is the debate} we had a machine lifting up a weight, and you could set this machine to any lifting force. First we set it at 100f {lets just call it force} up 1m and down 1m for 6 seconds, then we set it at 50f up 1m and down 1m for 6 seconds. The 100f as it had more force would move faster, so it would move the weight up and down more times.Wayne

This demonstrates well how you have got this wrong from beginning to end. Your muscles are not a machine. I can design a machine that uses NO energy whilst doing 'reps' at any rate and for any length of time (except to overcome some friction - and we could reduce this to an arbitrarily small amount). The way the forces vary as the weights accelerate will depend on the rate but the mean velocity is zero for each cycle, and so is the energy transfer.

btw, when you describe your machine as having a certain "lifting force" you do not specify for how long or over what distance this force is applied - nor what happens on the way down; the model is incomplete.

All this machine needs to consists of is a wheel, with the weights hung on the periphery. You could modify it, if you like, so that the weights are on a crank and push rod so that they are constrained to go just up and down. Over one cycle of operation, ZERO work has been done on the weights and Zero energy is expended by the energy source. All that is necessary is to start the machine up and bring it to the desired speed. This (kinetic) energy could all be reclaimed at the end of the exercise.
Perhaps this will help you to see that there is absolutely no parallel between what your muscles are doing and a simple mechanical model. Anything you may 'think' that your muscles are doing is entirely subjective. However many times you repeat a story about reps and rates and what you reckon you can tell us about what your muscles feel like, there is no direct correspondence between your body and simple mechanics. I believe that is what you have been trying to show all along.

 

[waynexk8]

Hi waynexk8, the first thing that you need to correct is that there is no such thing as total force. There is impulse which is defined as:
\mathbf{I}=\int_{t_i}^{t_f} \mathbf{f}(t) \, dt
Impulse has units of momentum, not units of force.

Hmm, ok, seems tricky, maybe I am saying it wrong then, as I think of me using force or strength more like this. If I can only lift 100 pounds 1 time, I can only lift 80 pounds several times, let’s say 10 times up and down at 1/1, this takes me 20 seconds, here is how I think this, say like a car, I have 20 gallons of fuel {or as I think force/strength} only, then I run out of fuel/force, BUT if I lifted slower, I would not used up my fuel/force so fast, or so fast per unit of time. So I have to use more fuel/force to lift the weight more times in the same time frame.

There is also average force which is defined as:
\overline{\mathbf{f}}=\frac{\int_{t_i}^{t_f} \mathbf{f}(t) \, dt}{t_f-t_i}
Average force does have units of force.

Thing is, I don’t understand that, if you have time please, is there any way you could explain this or explain it in layman’s terms, otherwise it had for me, as its just numbers.

Second, you are under the mistaken impression that the impulse is higher if you do a rep faster. This is incorrect. The impulse for one rep depends only on the weight and the amount of time, so one slow rep (10 s) might have the same impulse as 5 fast reps (2 s each).

Could you clear something up before I comment on that please ? As if I lift a weight up and then down, and then just lower the weight and then lift it up, it’s going to take far more force to stop a weight that is being lowered say .5 of a second for 1m, and then immediately lift it back up, we call this the transition from negative to positive, where the peak forces and tensions are on the muscles, the MMMTs. {Momentary Maximum Muscles Tensions}

As if you lower the weight and then lift, on using 100 pounds, you might have 150 pounds of force and tension on the muscles for .1 or .2 of a second, when just lifting the weight and lowering it, you will have nowhere near that force and tension on the muscles.

So please which repetition has the same impulse as the slow one ?

Similarly, you seem to be under the impression that the average force over a rep is higher if you go faster. This is also incorrect. The average force for one rep depends only on the weight. Notice that there is a close relationship between impulse and average force.

I explain before, that average means nothing in this debate. As if I do 1 repetition in 1 second, and a 100 repetitions in a 100 seconds, they say the average is the same, but we all know the force, work, energy, velocity and accelerations were far harder doing the 100, AND far harder on the muscles.

Personally, I think that the question you should be asking is the following:
"What different physical quantities are in fact higher in a fast rep than in a slow rep?"

Yes you could be right there, or to put it very blunt, which repetition cadence is harder and puts more tension on the muscles. As when using 80% you fail roughly 50% faster, I think we all know it’s the fast.

Is not power the mathematical product of force and velocity ? So more power more force and velocity, more force and velocity more power. I said this before, and someone said that more power does not equal more force and velocity, but could not prove what they thought. So work is the of a force over a distance, lifting a weight up and down is an example of work. The force is equal to the weight of the object, and the distance is equal to the height lifted {W= Fxd}
So energy capacity for doing work, so mechanical work is when an object is standing still and we force it to move.

Let’s take these scenarios.

1,
I lift a 100 pounds up and down 20 times in 20 seconds, then lift 50 pounds up and down 20 times in 20 seconds.

2,
I lift 100 pounds up and down 10 times in 40 seconds.

Which is harder ? 1, would need more work, force, energy, velocity and acceleration used.

So maximal muscle power production is the dominant factor in movements which aim to produce maximal velocity at the point of release, takeoff, or impact. Therefore the ability to perform a large amount of mechanical work in a short period of time, or the ability to produce high force output at fast movement velocities, is critical to lifting a heaver weight more times, so if the slow and the fast have to pick a weight to do any number of repetitions, the fast with be able to use the higher poundage.

Power can be defined as the force applied multiplied by the velocity of movement (Knuttgen
and Kraemer, 1987). As the work done is equal to the force times the distance moved (Garhammer,
1993) and velocity is the distance moved divided by the time taken, power can also be expressed as
work done per unit time (i.e., the rate of doing work) (Garhammer, 1993).

work = force x distance
velocity = distance/time
power = force x velocity
therefore:
power = force x distance/time = work/time.

And that’s what I have been saying all along ?

Wayne

 

[Dale]

Wayne, for the 3rd time now, please answer this question:

Do you understand that it is physically impossible to move a weight up and down using a constant force? I need to know if that makes sense to you before we can proceed.

 

[waynexk8]

Wayne, for the 3rd time now, please answer this question:

Do you understand that it is physically impossible to move a weight up and down using a constant force? I need to know if that makes sense to you before we can proceed.

That’s an odd question.

Yes, I know that. So if you are lifting a weight up 500mm in 3 seconds, and going to reverse the direction, as in weightlifting, lifting a weight up and down. First you have to accelerate, then “try” to move it with a constant force, and then decelerate the weight.

Now with the human body it’s far harder than with a machine. As take the bench press, there are several biomechanical advantages and disadvantages thought the ROM {Range Of Motion} of the exercise, there is point in the movement, where the leverage of certain muscles acting about the shoulder joint and chest, are at a mechanical disadvantage, and thus creating a disadvantage at about half way to three quarters the way up the lift, this point will decrease the force potential and thus deceleration to the bar.


Note on a 1RM, the acceleration was for roughly 65%, so on using 80% like we are; the acceleration would be at least ? 80%

https://docs.google.com/viewer?a=v&...BmBxS-&sig=AHIEtbT6QUrHzT2mMWFVWF8f8ikRAXeHCA

Also, if you work out in physics how far a barbell will move if you let go when pressing it, and your using 80% for 15 inch, is does NOT go what physics say, as of the biomechanical disadvantages and biomechanical advantages of the body. But it would move the 3 inch if a machine pushed with a constant force of 80% for that force and time.

Jeff Pinter wrote, and he did say I could use this.

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

So as I said before, could we not keep this to a machine moving the weight please ?

Big thanks for your time and help DaleSpam, and the rest, thank you.

Now its too late and I have no time to read or answer the others.

Wayne

 

[sophiecentaur]

So as I said before, could we not keep this to a machine moving the weight please ?

Wayne

Absolutely. But you still are not accepting that the energy that a machine needs to supply can be near ZERO, the average force will be ZERO and the net Impulse will also be ZERO.
You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate.
But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?

 

[douglis]

Either it's a muscle or a machine...one thing will always be for sure.
The muscle's/machine's impulse will always be equal with gravity's impulse regardless if the lifting is fast or slow for the same duration.The net impulse is always zero.

 

[Dale]

That’s an odd question.

Yes, I know that. So if you are lifting a weight up 500mm in 3 seconds, and going to reverse the direction, as in weightlifting, lifting a weight up and down. First you have to accelerate, then “try” to move it with a constant force, and then decelerate the weight.

OK, I am glad you understand that. I apologize that you think it was an odd question, but from several of your previous comments it was not clear that you understood that point. I am glad that you do and it was just a miscommunication.

So, let's take the scenario of a machine or a person lifting a 50 kg weight up and down 500 mm in 3 s as smoothly as possible. I generated the attached plot to represent a typical position over time graph for such a lift. Look at it and see if it seems like a reasonable approximation to you.

The plot is generated in Mathematica using the following code:

y[t_] := -.25 Cos[2 \[Pi] t/3] + .25;
Plot[y[t], {t, 0, 3}, Frame -> True, 
 FrameLabel -> {Style["Time (s)", Larger], 
   Style["Height (m)", Larger]}, 
 PlotLabel -> Style["Position of Weight During Lift", Larger]]

 

[waynexk8]

Absolutely.

Great. Said this as the muscle will start to complicate things with their biomechanical advantages and disadvantages thought-out the ROM.

But you still are not accepting that the energy that a machine needs to supply can be near ZERO,

You lost me there ? If the machine is powered by the energy, diesel or electricity, to move 80% of its maximum force up and down, will take a certain amount of this energy, but as we all agree, that the faster/further an object moves in the same time frame the more energy is needed, there is no debate there, or is there ? As something, makes for the more energy used, and I say it’s the higher forces of the accelerations that cannot be made up or balanced out by the more constant forces of the slower repetitions, when the fast is on the deceleration.

Run 1 mile = 3 minutes, walk 1 mile 9 minutes, both use the same energy. Run 3 minutes, walk 3 minutes the run uses more energy, this is what I have said all along.

the average force will be ZERO and the net Impulse will also be ZERO.

Not sure I understand that, as if I lift 80% of my 1RM, {hope we all know what 1RM means by now, if anyone does not, I have explained many times, it’s the 1 repetition maximum a person can lift} I will try to use 100% force thought the ROM/lift, the only time I will use at the beginning and end, but they do not count, as it’s the lifting we are on about, not the times after the lifting.

But thought we all agreed average means nothing in this debate, as if the average force is the same for 1 repetition and a 100 repetitions for the same speed, it means nothing.

You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate.

Right, sure we all agree on this, however, as I lift the same weight 6 times further in the same time frame, using 100% force to the slow using 80% force, how can the total overall muscle activity or force be the same, you might say again, but in physics you can’t measure total overall muscle force, well try this.

Set our machine to lift 80% up and down for 60 seconds set at a lifting force of 80% of the machines maximum moving force, and then set the machine to move 80% up and down for 60 seconds set at a lifting force of 100%

Fast, the machine lifted with a set force of 100% efficiency for 60 seconds.

Slow, the machine lifted with a set force of 80% efficiency for 60 seconds.

The fast lift with 20% more force efficiency for 60 seconds

But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?

Said this as the muscle will start to complicate things with their biomechanical advantages and disadvantages thought-out the ROM.

There has to be a reason, why the faster makes for the more energy used, I said why I think, no one else does, D, mentions, biological, that’s no answer at all, it’s says nothing.

Wayne

 

[waynexk8]

Absolutely. But you still are not accepting that the energy that a machine needs to supply can be near ZERO, the average force will be ZERO and the net Impulse will also be ZERO.
You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate.
But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?

OK, I am glad you understand that. I apologize that you think it was an odd question, but from several of your previous comments it was not clear that you understood that point. I am glad that you do and it was just a miscommunication.

K great, thank you.

So, let's take the scenario of a machine or a person lifting a 50 kg weight up and down 500 mm in 3 s as smoothly as possible. I generated the attached plot to represent a typical position over time graph for such a lift. Look at it and see if it seems like a reasonable approximation to you.

The plot is generated in Mathematica using the following code:

y[t_] := -.25 Cos[2 \[Pi] t/3] + .25;
Plot[y[t], {t, 0, 3}, Frame -> True, 
 FrameLabel -> {Style["Time (s)", Larger], 
   Style["Height (m)", Larger]}, 
 PlotLabel -> Style["Position of Weight During Lift", Larger]]

Yes that look quite ok, I think. We are lifting 80% are we not, or close too ?

Wayne

 

[waynexk8]

This demonstrates well how you have got this wrong from beginning to end. Your muscles are not a machine. I can design a machine that uses NO energy whilst doing 'reps' at any rate and for any length of time (except to overcome some friction - and we could reduce this to an arbitrarily small amount).

How do you work that out ? What machine can lift say 80% up, and then lower it down using very little energy, and what energy is this ?

The way the forces vary as the weights accelerate will depend on the rate

Right, the faster you try a move/accelerate the weight, the more action reaction forces go back on your muscles as tension, move/accelerate very fast very high forces, = very high tensions on the muscles, move very slow, very low force, = very low tensions on the muscles, move very slow, very low force.

but the mean velocity is zero for each cycle, and so is the energy transfer.

Not sure what this has to do with the debate, as we need to know about the velocities themselves, that are produced by the person/machine creating these forces, as I just said, the faster you try a move/accelerate the weight, the more action reaction forces go back on your muscles as tension.

btw, when you describe your machine as having a certain "lifting force" you do not specify for how long or over what distance this force is applied - nor what happens on the way down; the model is incomplete.

Ok sorry there.


The machine has a lifting force of 100 pounds, it can lift 100 pounds up 1m in .5 of a second, and any speed slower, and can lower it 1m in .5 of a second, and any speed slower, for an unlimited time, for debates sake, the weight the machine will use will be 80% that’s 80 pounds.


The machine first lift the 80% up 1m in 3 seconds and lowers it 1m in 3 seconds, 1 time, = 2m in 6 seconds.


Then the machine lifts the 80% up 1m in .5 of a second, and lowers it 1m in .5 of a second, 6 times = 12m in 6 seconds.

On the way down it lowers it in control 1m in .5 of a second, I say in control, as if let to drop, it would get to the ground far faster. Also, as of the acceleration components, when the weight is being lowered, the faster it’s lowered the more force it puts on the machine/muscles, like in the faster it hits the ground the more impact force.

All this machine needs to consists of is a wheel, with the weights hung on the periphery. You could modify it, if you like, so that the weights are on a crank and push rod so that they are constrained to go just up and down. Over one cycle of operation, ZERO work has been done on the weights and Zero energy is expended by the energy source. All that is necessary is to start the machine up and bring it to the desired speed. This (kinetic) energy could all be reclaimed at the end of the exercise.
Perhaps this will help you to see that there is absolutely no parallel between what your muscles are doing and a simple mechanical model. Anything you may 'think' that your muscles are doing is entirely subjective. However many times you repeat a story about reps and rates and what you reckon you can tell us about what your muscles feel like, there is no direct correspondence between your body and simple mechanics. I believe that is what you have been trying to show all along.
__________________

Wow that sounds interesting, but its far too late to comment on that now, will get back to it tomorrow.


Wayne

 

[waynexk8]

To all the other that have posted here, I will get back to them tommorrow, and thanks for your time and help.

Wayne

 

[sophiecentaur]

How do you work that out ? What machine can lift say 80% up, and then lower it down using very little energy, and what energy is this ?


Wayne

Plus all the rest
That comment shows that yo don't get the mosgt basic part of all this thread and others.

The machine doesn't need to use use "very little energy" on the way down. IT GETS ALL THE ENERGY BACK! Unlike your muscles, which don't have Energy Recovery. So the two cannot be compared.
You insist that this problem can be solved your way and you have the nerve to hang onto the idea in the face of people who know much more basic Physics than you. The only hope you have is to do a Physics course at some level which may help you understand what you need to know in order to grasp how crazy your idea is.
If someone told you that swimming the Pacific is a no no, how long would you not believe them?

 

[douglis]

sophiecentaur...I think that he first needs to grasp more simple things.For example...he doesn't seem to understand the meaning of average force.
Otherwise he wouldn't say nonsense like "...the forces don't make up..." or "...average means nothing in this debate...".
More important he would understand that same average force equates same impulse(what he calls "total/overall force") when is applied for the same duration.