Impulse/force in pounds for the time frame

[sophiecentaur]

Calculate the peak force needed to throw a computer out of a window in sheer exasperation!

 

[waynexk8]

Can't ba answered if you don't know the acceleration, I'm afraid.

Will change to Metric.

Hmm, thought I had put enough information in, as if I am moving the weight 500mm at every .5 of a second, I decelerate for say the last 20% thought you would/could work this out, or am I doing it wrong ? Seems I was wrong interesting.

I am moving the weight on 1, at 1m/s. {m/s 1 meter per second}

I am moving the weight on 2, {which is now 83mm} in .5 of a second. Moving it at 83m/s. {83mm per second} This is basically going at a constant speed, as its accelerated to moving at 83m/s.

So are you saying that I could have different accelerations and decelerations ? This is as I thought, but D. {douglis} seems to think other. {that is right is it not D. ?}

Force=Mass X Acceleration
(Then add the weight, in this case)

Misunderstand that a little, so best say and ask.

Are you assuming constant force all the time?

Hmm, with a muscle it would not be a constant force, but let’s says it’s a machine moving the weights and the force is constant.

Wayne

 

[waynexk8]

Again that question.Define the value of force you're interested in.The peak...the average or the "total" force(integration of force in respect of time)?

Impulse force, total, overall, integration of force in respect of time.

For the peak force...we need more data.

Enough now for 1 and 3, as 2 will not be much higher than the weight.

For the average force...in both cases you start and end at rest.The net change in momentum is therefore zero,

How do you work out the net change in momentum/movement is zero ? As the net force in 1 = 500mm and in 2 = 166mm. So the net, total momentum/movement will be 500 and 83 ?

I start at rest and end in rest, but the distance move on 1 = 500mm and 2 = 83mm

which is equal to the net impulse delivered. Therefore, the average force is equal with the weight in both cases...80 pounds or 356N.

For the "total" force...again in both cases is equal with gravity's impulse for .5sec.So...356 X .5=153N*s.

See, I don’t get how you come to this, as the below, shows the total or overall force to be higher for the faster moving in less time ? And its not the peak. Please see the video stating that more overall, total force was used with the fast rep, but this time the more force was used in "less" time, go from 5 min.



Fast
P = 695
F = 579Newtons
V = 192

Slow
P = 649
F = 546Newtons
V = 161

Wayne

 

[waynexk8]

Calculate the peak force needed to throw a computer out of a window in sheer exasperation!

WaynesWorldphysics ?

Please sophiecentaur, could you work out what the Newont force is below is, peak, average or total ?

Go from 5 min.



Fast
P = 695
F = 579Newtons
V = 192

Slow
P = 649
F = 546Newtons
V = 161

This one may help.

http://www.youtube.com/watch?v=Ycu6y4EHfog&feature=related

Wayne

 

[waynexk8]

So that is your definition of 'Integration'? It is not the Mathematical definition that is used (and works) in Physics so there is no point in your using the term.

I will write to an EMG expert, and ask then on intergration and RMS.

I will also write to the makers of my machine, and ask them what actualy my machine reads out on the average.

Wayne

 

[waynexk8]

D. I just thought of somthing, I can video the machine and prove its NOT the peak force, but the average. Or you should be able to work it out here.

Take a look at the slow rep video, you will notice the peak force = ? was it a 190 somthing ? BUT the average was a 140 !

http://www.youtube.com/user/waynerock999?feature=guide#p/u/36/B8gtpp8ozvU

What do you say to that my friend D. ?

Wayne

 

[waynexk8]

How can the machine distinguish between situations where there are and there aren't antagonistic muscles at work?

If I am curling, arm flexion, then I will but the four pads on my biceps, as these are the prime moves.

Are you saying that your machine 'knows' when your arm is moving and when it's stationary? Does it know then you are holding, lifting and lowering the weights?

Yes the machine know when my arm is moving and when it's stationary, holding, lifting and lowering the weights. As the pads detect “all/every” of the electrical signals in my muscles the pads are on.

Here take a look at the makers video, go to 1.40min



Or I could make a video if you want.

So if they say the fast as a higher average, as a Physicist, thought you would be very interested, that is, if you do agree with D ? But to me he has not added in all the variables, like ground reaction muscle force and so forth, and left out Kinology and Biomechanics.

How can you expect to get a proper answer whilst you still insist on using the term force/strength? Do me a favour and look the individual words up. They describe entirely different aspects of Physics. Which one do you mean when you use that term? Why do you hang on to these deliberately nonsense terms instead of using the right one? You are saying black is white on every occasion.

Ok sorry, I will try the stick with force.

The very least you could do, if you really do want some sense, is to use the correct terms. Imagine you had a calculator and the + key sometimes gave you a - and the X key sometimes gave you a ÷. You would say it was rubbish, wouldn't you? Constantly using confusing terms is the equivalent to a dodgy calculator.

Ok get your point.

At least, you could do us all the courtesy of talking the right language. (The language that 11 year old kids are quite happy to learn to use in School.) Don't ask me to help you with this - just look up any word you want to use and see if it actually fits what you want to say. I get the impression that you are not looking anywhere else for your information but just want to be spoon fed by PF. This is unreasonable.

btw, I don't have to justify the Physics Equations in this context. They work fine for everyone but you so I am not out of step here. Give me a proper question (not ten pages of weight lifting jargon) and I can guarantee a good answer. Give me another 'Wayne' question and there will be no available answer from Physics.

Ok, I will try and use physics, if I can't will ask.

Wayne

 

[sophiecentaur]

WaynesWorldphysics ?

Please sophiecentaur, could you work out what the Newont force is below is, peak, average or total ?

1. If you can't tell me the acceleration (i.e. what is the variation of velocity with time during the lift? There are infinite possible combinations that will get the weight from bottom to top of the lift in a given time.) I cannot tell you the force. (Did you not read my F=mA formula?)

2. Because the weight starts and ends stationary, the Mean additional force must be zero and the mean force will be the weight.. (I think this had been pointed out several times already.)

3. What does "total force" mean? Do we add up the forces, measured every second, every tenth of a second, every 100th of a second? It's a nonsense concept as you can only validly add forces that operate at the same time.. Ask the makers of your machine for an answer. There is not a PF answer for you.

It matters not whether you are working in Imperial or Metric - all three questions are either nonsense of indeterminable.

 

[waynexk8]

Big thank you for staying with me.

1. If you can't tell me the acceleration (i.e. what is the variation of velocity with time during the lift? There are infinite possible combinations that will get the weight from bottom to top of the lift in a given time.) I cannot tell you the force. (Did you not read my F=mA formula?)

Hmm, I thought I told you this on my other post ? Ok, this is where your help will have to come in, as I am not sure how to work this out. Can we call it for now a constant acceleration ? If so, on 1, the weight accelerates from rest to 400mm in 0.4 of a second, then decelerates the last 100mm in .1 of a second

2. Because the weight starts and ends stationary, the Mean additional force must be zero and the mean force will be the weight.. (I think this had been pointed out several times already.)

So as I push up with the force of the weight to move the weight, and then the weight cancels that force out, then you call the mean additional force {that my muscles creating force} must be zero ? If I am right, sort of get that, however, I have used forces, from a 100 to 1 pounds, and that’s all we are concerned about ? Or am I missing something.

My other point is, that I don’t think that when I am using far higher forces, like 100 pounds for the accelerations, that when I am then using slightly less force than 80 pounds for the decelerations, that when the slow rep is roughly using a constant 80 pounds, that the slow reps medium forces can NOT make up or balance out the very high force {the very high tensions that the very high forces have put on the muscles, as of the action reaction force thus tensions on the muscles} that the fast reps are putting out, that’s why the fast reps have used more energy and moved the weight 6 times further in my opinion.

Not sure you get what I say there, please say if you don’t, it’s like a person hits you with great force, and it hits you down, but it would take far more lower force hits and more “time” {but in this debate the time is the same} to hit you down with lower force hits. THIS is why you always fail at lifting the weight, or you hit momentary muscular failure faster with the faster reps, as they ARE doing more damage with the higher forces.

I am not saying the physics equations are wrong, it’s just they cannot tell the full story, as all the variables as liker the high force to medium force has not been worked out, that’s why the ENG states more average force used in the faster reps.

3. What does "total force" mean? Do we add up the forces, measured every second, every tenth of a second, every 100th of a second? It's a nonsense concept as you can only validly add forces that operate at the same time.. Ask the makers of your machine for an answer. There is not a PF answer for you.

It matters not whether you are working in Imperial or Metric - all three questions are either nonsense of indeterminable.

What I mean with total or overall force, is that if you lift say 80% of your 1RM, you can only lift it for a certain amount of times in a time frame at a certain speed. Let’s say you could lift it up and down 10 times at 1 second up and 1 second down, then at 20 seconds you could not lift it again, so you had in your muscles 20 seconds or 10 lifts in you at that rep speed, of force, after that your force was temporary no longer. Yes I know that sounds a bit daft, but that is actually what happens, and if you had lifted the same weight up and down in .5 of a second up and .5 of a second up, you would have most probably failed to lift the weight in 10 to 12 seconds. Meaning you have used up your temporary force up far faster.

So let’s “just” {please this is just an example to get my point over} say for an example you had 1000 forces to lift the weight, in the fast, you used up this force far far far faster, meaning if you both lift the weight for a set time, and do “not” lift until momentary muscular failure, the faster reps “must” be using up more force faster, as you fail faster lifting faster. Example of how the fast are using more force and faster, more energy used, more distance the weight has been moved, faster to muscular failure, the EMG states more muscle activity or muscle force. I know all the above sounds a bit complicated, but there is total since in there.

Thank you again for you time and help, not sure about the acceleration, hope to learn more on that.

Wayne

 

[douglis]

Big thank you for staying with me.

Hmm, I thought I told you this on my other post ? Ok, this is where your help will have to come in, as I am not sure how to work this out. Can we call it for now a constant acceleration ? If so, on 1, the weight accelerates from rest to 400mm in 0.4 of a second, then decelerates the last 100mm in .1 of a second

O.K. I'll try to help you understand once again.
Let's say in your above example the load is 800N(81.5kg) while your maximum force ability is 1000N.

For the first 0.4sec you're applying your Fmax (although in reality that's biomechanically impossible) so the net force is 1000-800=200N.
So your acceleration for those 0.4sec is a=F/m=200/81.5=2.45m/s^2

For the last 0.1sec you let the gravity decelerate the load so the net force is -800N
So your acceleration for those 0.1sec is a=F/m=-800/81.5=-9.81m/s^2 and obviously it's equal with g.

So as I push up with the force of the weight to move the weight, and then the weight cancels that force out, then you call the mean additional force {that my muscles creating force} must be zero ? If I am right, sort of get that, however, I have used forces, from a 100 to 1 pounds, and that’s all we are concerned about ? Or am I missing something.

My other point is, that I don’t think that when I am using far higher forces, like 100 pounds for the accelerations, that when I am then using slightly less force than 80 pounds for the decelerations, that when the slow rep is roughly using a constant 80 pounds, that the slow reps medium forces can NOT make up or balance out the very high force {the very high tensions that the very high forces have put on the muscles, as of the action reaction force thus tensions on the muscles} that the fast reps are putting out, that’s why the fast reps have used more energy and moved the weight 6 times further in my opinion.

As I proved you above you used positive acceleration equal with 2.45m/s^2 for the 80% and 4 times greater negative acceleration equal with -9.81m/s^2 for the last 20%.
That's why we're telling you the acceleration is offset by the deceleration.The average acceleration is always zero.
In the equation of force F=mg+ma the "ma" part is always zero so F=mg or else the force is equal with the weight.No Mean additonal force is required.The forces "make up","balance out" or call it whatever you like.

I don't believe it's possible to get a more simple explanation.

I am not saying the physics equations are wrong, it’s just they cannot tell the full story, as all the variables as liker the high force to medium force has not been worked out, that’s why the EMG states more average force used in the faster reps.

No...your EMG states that greater RMS is used in faster reps...not average force.

What I mean with total or overall force, is that if you lift say 80% of your 1RM, you can only lift it for a certain amount of times in a time frame at a certain speed. Let’s say you could lift it up and down 10 times at 1 second up and 1 second down, then at 20 seconds you could not lift it again, so you had in your muscles 20 seconds or 10 lifts in you at that rep speed, of force, after that your force was temporary no longer. Yes I know that sounds a bit daft, but that is actually what happens, and if you had lifted the same weight up and down in .5 of a second up and .5 of a second up, you would have most probably failed to lift the weight in 10 to 12 seconds. Meaning you have used up your temporary force up far faster.

Wayne

This is the last time I bother to even read that particular question.

You "fail" in a weight lifting set when you have no longer enough energy to apply force equal with the weight.
"Failing" faster with a certain kind of lifting means that with this kind of lifting you spend energy at a higher rate...NOT that you use more force.In all cases the average force is equal with the weight.
It's the fluctuations of force that are more energy demanding.That's the only scientific answer you can get.

 

[waynexk8]

sophiecentaur I see you’re an Engineer, so hopefully you will be able to understand what I am saying, even thou a lot is not in pure physics talk, please, and I am not being sarcastic, as I know people find it hard the way I explain things, but please do you understand what I am trying to say in the below please ?

And please, if you agree with me or not, {AND YOU D.} do you understand what I am trying to say convey too you ? As this is very important, as it makes no difference if you agree or not now, only that you both and other members actually understand what I am “trying” to say ?

My other point is, that I don’t think that when I am using far higher forces, like 100 pounds for the accelerations, that when I am then using slightly less force than 80 pounds for the decelerations, that when the slow rep is roughly using a constant 80 pounds, that the slow reps medium forces can NOT make up or balance out the very high force {the very high tensions that the very high forces have put on the muscles, as of the action reaction force thus tensions on the muscles} that the fast reps are putting out, that’s why the fast reps have used more energy and moved the weight 6 times further in my opinion.

Not sure you get what I say there, please say if you don’t, it’s like a person hits you with great force, and it hits you down, but it would take far more lower force hits and more “time” {but in this debate the time is the same} to hit you down with lower force hits. THIS is why you always fail at lifting the weight, or you hit momentary muscular failure faster with the faster reps, as they ARE doing more damage with the higher forces.

I am not saying the physics equations are wrong, it’s just they cannot tell the full story, as all the variables as liker the high force to medium force has not been worked out, that’s why the EMG states more average force used in the faster reps.


Big thank you for staying with me.

Originally Posted by sophiecentaur
1. If you can't tell me the acceleration (i.e. what is the variation of velocity with time during the lift? There are infinite possible combinations that will get the weight from bottom to top of the lift in a given time.) I cannot tell you the force. (Did you not read my F=mA formula?)

Hmm, I thought I told you this on my other post ? Ok, this is where your help will have to come in, as I am not sure how to work this out. Can we call it for now a constant acceleration ? If so, on 1, the weight accelerates from rest to 400mm in 0.4 of a second, then decelerates the last 100mm in .1 of a second

Originally Posted by sophiecentaur
2. Because the weight starts and ends stationary, the Mean additional force must be zero and the mean force will be the weight.. (I think this had been pointed out several times already.)

So as I push up with the force of the weight to move the weight, and then the weight cancels that force out, then you call the mean additional force {that my muscles creating force} must be zero ? If I am right, sort of get that, however, I have used forces, from a 100 to 1 pounds, and that’s all we are concerned about ? Or am I missing something.

My other point is, that I don’t think that when I am using far higher forces, like 100 pounds for the accelerations, that when I am then using slightly less force than 80 pounds for the decelerations, that when the slow rep is roughly using a constant 80 pounds, that the slow reps medium forces can NOT make up or balance out the very high force {the very high tensions that the very high forces have put on the muscles, as of the action reaction force thus tensions on the muscles} that the fast reps are putting out, that’s why the fast reps have used more energy and moved the weight 6 times further in my opinion.

Not sure you get what I say there, please say if you don’t, it’s like a person hits you with great force, and it hits you down, but it would take far more lower force hits and more “time” {but in this debate the time is the same} to hit you down with lower force hits. THIS is why you always fail at lifting the weight, or you hit momentary muscular failure faster with the faster reps, as they ARE doing more damage with the higher forces.

I am not saying the physics equations are wrong, it’s just they cannot tell the full story, as all the variables as liker the high force to medium force has not been worked out, that’s why the EMG states more average force used in the faster reps.

Originally Posted by sophiecentaur
3. What does "total force" mean? Do we add up the forces, measured every second, every tenth of a second, every 100th of a second? It's a nonsense concept as you can only validly add forces that operate at the same time.. Ask the makers of your machine for an answer. There is not a PF answer for you.

It matters not whether you are working in Imperial or Metric - all three questions are either nonsense of indeterminable.

What I mean with total or overall force, is that if you lift say 80% of your 1RM, you can only lift it for a certain amount of times in a time frame at a certain speed. Let’s say you could lift it up and down 10 times at 1 second up and 1 second down, then at 20 seconds you could not lift it again, so you had in your muscles 20 seconds or 10 lifts in you at that rep speed, of force, after that your force was temporary no longer. Yes I know that sounds a bit daft, but that is actually what happens, and if you had lifted the same weight up and down in .5 of a second up and .5 of a second up, you would have most probably failed to lift the weight in 10 to 12 seconds. Meaning you have used up your temporary force up far faster.

So let’s “just” {please this is just an example to get my point over} say for an example you had 1000 forces to lift the weight, in the fast, you used up this force far far far faster, meaning if you both lift the weight for a set time, and do “not” lift until momentary muscular failure, the faster reps “must” be using up more force faster, as you fail faster lifting faster. Example of how the fast are using more force and faster, more energy used, more distance the weight has been moved, faster to muscular failure, the EMG states more muscle activity or muscle force. I know all the above sounds a bit complicated, but there is total since in there.

Thank you again for you time and help, not sure about the acceleration, hope to learn more on that.


Wayne

 

[sophiecentaur]

Wayne
You still don't get it, do you?
Whatever you 'feel' in your arms and whatever you think that machine is telling you, my points 1,2 and 3 still apply.

1. You can't say it's accelerating constantly all the time. If it were, then it would still be moving at the top. It isn't because, of course, it stops at the top. And, in practical terms, you have no way of knowing without measuring the velocity at very short intervals over the whole of a lift.

2. For the weight to start at zero velocity and to end at zero velocity then the average force Has to be weight. This may annoy or confuse you and it may not be what you think the machine is telling you but is true beyond any shadow of a doubt. You could ask Sir Isaac Newton himself and he would tell you the same.

3. You actually don't know what you mean when you talk of "total force" because you are after an arm waving description of what it feels like to do a lift. I really don't know why you won't accept (from not just me but others, too) that your idea can't be stated in Physics.

What you could find out (but only by measurement) would be the maximum force, the total work done on the weights (thats N lifts times weight times height) and the Power developed (dividing the total work done by the time for the whole exercise).

I don't understand why that isn't enough.
All the other acres and acres of figures you have written on these pages have been pretty much wasted. I'd bet that no one has actually read more than a few sample lines of it and then given up.

How can you be so sure that your question is reasonable when, in the same breath, you admit to not knowing much real Physics?

I have seen some of the movies on your website and I am really impressed by your dedication to your sport. You are clearly a bit of an expert on the practical aspects of it and your advice on how to do the exercises and how to body build is, no doubt, correct. But, when it comes to the Physics of the situation, you just have to be much more rigorous and 'go along with the rules'. Those rules say that you are on a fruitless quest and have been talking mostly rubbish. Why are you bothering and why can't you just accept what you are being told? Do you just like a friendly chat - is that what it's all about?

That is a DEAD PARROT!

 

[waynexk8]

O.K. I'll try to help you understand once again.

Ok thx.

Let's say in your above example the load is 800N(81.5kg) while your maximum force ability is 1000N.

Right.

For the first 0.4sec you're applying your Fmax

Right.

(although in reality that's biomechanically impossible)

Right.

so the net force is 1000-800=200N.

This is where you sort of lose me. As you saying ? The net force on me and the weight = 200N ? If so I am still with you, but why say the net force on my muscles and the weight ? AND WHY take 800 from 1000 ? As we are only concerned with the weight/force on me, my muscles.

So am I not right in saying what physics states, is the net force on me, my muscles is only the weight = 800N ? But that’s “only” when you are holding the weight of moving it very slow.

Now here’s where it get a little more complicated. As if you agree that there is 800N pushing down and me pushing up with 800N, the moment I try too accelerate it as fast as possible I will be pushing up with a 1000N, but not only that, but the weight of 800N will not only be pushing back down with just the 800N, but as of the acceleration components, the weight will pushing down more than 800N as of the action reaction law ?

So you you weight me and the weight on the scales, it could register ? 400 pounds, but the faster I try to accelerate, the more weight/force is pushing back onto my muscles as the reading on the scales goes up and up the faster I try to accelerate.

So your acceleration for those 0.4sec is a=F/m=200/81.5=2.45m/s^2

Right.

For the last 0.1sec you let the gravity decelerate the load so the net force is -800N

Not sure on you here, are you saying that only gravity does the work and force on the weight for the last 0.1 of a second ? If so I do not agree there, as I have to keep pushing for the weight to keep moving. I decelerate the weight, if the weight was heavier or lighter, or if gravity had more force or less force, I my muscles decelerate the weight, I am in full control, especially on 80% if we were talking of 20%, then the accelerating components will be helping moving the weight after a certain speed/force is applied to the weight, as if I immediately stopped, the weight would keep moving.

So your acceleration for those 0.1sec is a=F/m=-800/81.5=-9.81m/s^2 and obviously it's equal with g.

Sorry you lose me a little there ? Equal with g ? I know what g is, but not sure why you are saying equal with it.

As I proved you above you used positive acceleration equal with 2.45m/s^2 for the 80% and 4 times greater negative acceleration equal with -9.81m/s^2 for the last 20%.
That's why we're telling you the acceleration is offset by the deceleration.The average acceleration is always zero.

Right get what you’re saying here, and have done all along, your sort of saying, and I know this is different, but want you to understand that I understand. You’re saying that when my accelerations are higher than your forces, that your forces make up or balance out when my forces are on the decelerations ? BUT what I am saying, is that it may look that way on paper, but how do you think your 80 medium force, can make up for/to my 100 force ? I am saying to do this; you would need FAR more time using your 80 force for the tension to build up on the muscles.

As “IF” you did make up or balance this force up, thus make up or balance this opposite action reaction tension on the muscles, HOW and WHY do I use “more” overall/total energy ? How/why do I move the weight “more” overall/total distance ? How and why do I fail roughly 50% faster “more” ? How and why does the EMG read out far more muscle activity, force on the faster rep.

Four definite things prove you “CAN’T” be making up the force, see my point ?

Don’t you get, that your measure or force does and can never equally my higher force ! Thus you have to need more time to let your force put more tension on the muscles. A small force applied for a LONGER TIME produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important.

In the equation of force F=mg+ma the "ma"

You lost me on ma, I need to know what that is for me to understand you point.

part is always zero so F=mg or else the force is equal with the weight.No Mean additonal force is required.The forces "make up","balance out" or call it whatever you like.

I don't believe it's possible to get a more simple explanation.

This is the last time I bother to even read that particular question.

You "fail" in a weight lifting set when you have no longer enough energy to apply force equal with the weight.

You can NOT just say it’s the energy that fails or runs out, that is non scientific, very biased and unfair.

You have no temporary energy left and NO temporary force left. As we are not in this case taking about energy are we ? So if I cannot lift 400 pounds above my head, you will say I have not enough force, right ? And if I can lift up 200 pounds 30 times you will say in the end you did not have enough force left right ? Well actually I did not have {and I am being scientific, non biased and fair here} force “and” energy left, you can’t have one without the other, you “HAVE” to include force and energy when you say I hit momentary muscular failure faster than you.

I ran out of temporary force because I used up all my temporary energy, or/and I used up all my temporary energy because I used up all my temporary force. BOTH are used, and used up. I STILL have energy and force left, but I need to recuperate to get them both back.

"Failing" faster with a certain kind of lifting means that with this kind of lifting you spend energy at a higher rate...NOT that you use more force.In all cases the average force is equal with the weight.

Ok if you want to think backwards, tell me “why” I use up my energy faster ? Or with this kind of lifting why do you think you spend energy at a higher rate ?

More acceleration/force ? if not what and why ?

It's the fluctuations of force that are more energy demanding.That's the only scientific answer you can get.

But this is what I “have” be saying all along, the high fluctuations of force that are more energy demanding, meaning if they are more demanding as you just said, means your forces can not make up or balance out, right.

Wayne

 

[waynexk8]

Wayne
You still don't get it, do you?
Whatever you 'feel' in your arms and whatever you think that machine is telling you, my points 1,2 and 3 still apply.

1. You can't say it's accelerating constantly all the time. If it were, then it would still be moving at the top. It isn't because, of course, it stops at the top. And, in practical terms, you have no way of knowing without measuring the velocity at very short intervals over the whole of a lift.

Right.

2. For the weight to start at zero velocity and to end at zero velocity then the average force Has to be weight. This may annoy or confuse you and it may not be what you think the machine is telling you but is true beyond any shadow of a doubt. You could ask Sir Isaac Newton himself and he would tell you the same.

Ah, but that’s what I am not saying is not true, meaning you’re not understanding what I am trying to say, I am saying the above is right, I always have.

I am saying that my higher acceleration forces, and they are higher and always higher than the slow, as my higher force on the transition from negative to positive could be as high as 140% but my normal higher forces are say a 100 to your 80, thus higher.

My point is, when you push out a 100 force there is a 100 force/tension on the muscles, right ? Now you only push out a 80 force there is a 80 force/tension on the muscles, right ?

You have never put a 100 force/tension on the muscle, right ? So why and how do you think that your 80 force/tension can equal a 100 ? As 80 = 80 not a 100. It would take far far far longer in time for your 80 to make up the force/tension on the muscles/object

I am saying I am doing more damage with my higher forces, and your lower forces can NOT do the same damage in the same time frame, that’s why the below happens;

1,
I use more energy,

2,
I move the weight more distance,

3,
My muscles fail faster in the faster reps, proving they must be doing more damage,

4,
EMG agrees.

5,
I am NOT saying your questions are wrong, I am just stating my higher impulse, my impulse force with respect to time, YOUR SMALL FORCE applied for a long time can produce then same momentum change as MY LARGE FORCE applied briefly, because it is the product of the force and the time for which it is applied that is important.

Please if you do not agree, all I would like is that you do/can understand what I am just trying to say ? I am not at this moment in time concerned who is right or wrong, just that you see my point, as it’s quite frustrating when something argues over you, but they are not actually seeing my point, as they keep repeating what I agree with.

No time for the below its late here.

3. You actually don't know what you mean when you talk of "total force" because you are after an arm waving description of what it feels like to do a lift. I really don't know why you won't accept (from not just me but others, too) that your idea can't be stated in Physics.

What you could find out (but only by measurement) would be the maximum force, the total work done on the weights (thats N lifts times weight times height) and the Power developed (dividing the total work done by the time for the whole exercise).

I don't understand why that isn't enough.
All the other acres and acres of figures you have written on these pages have been pretty much wasted. I'd bet that no one has actually read more than a few sample lines of it and then given up.

How can you be so sure that your question is reasonable when, in the same breath, you admit to not knowing much real Physics?

I have seen some of the movies on your website and I am really impressed by your dedication to your sport. You are clearly a bit of an expert on the practical aspects of it and your advice on how to do the exercises and how to body build is, no doubt, correct. But, when it comes to the Physics of the situation, you just have to be much more rigorous and 'go along with the rules'. Those rules say that you are on a fruitless quest and have been talking mostly rubbish. Why are you bothering and why can't you just accept what you are being told? Do you just like a friendly chat - is that what it's all about?

That is a DEAD PARROT!

Wayne

 

[douglis]

This is where you sort of lose me. As you saying ? The net force on me and the weight = 200N ? If so I am still with you, but why say the net force on my muscles and the weight ? AND WHY take 800 from 1000 ? As we are only concerned with the weight/force on me, my muscles.

I stopped here.It was my mistake to try to make my point with physics equations.It's obvious that you can't follow them.That's not a bad thing but I don't have the time and the patience to explain every single line that I write.

I suggest to leave out physics.We have two push ups studies that examine exactly you're looking for.In physics there's no such thing as "total force" but in biomechanics there's a magnitude called Total Muscle Activation.
Let's discussed them via mail because obviously they're unrelated with physics.

You can NOT just say it’s the energy that fails or runs out, that is non scientific, very biased and unfair.

What makes you think that you're able to judge what's scientific and what's not?
But even common sense tell us that energy runs out and force is applied.Force doesn't run out.It's either applied or not.

You have no temporary energy left and NO temporary force left. As we are not in this case taking about energy are we ? So if I cannot lift 400 pounds above my head, you will say I have not enough force, right ? And if I can lift up 200 pounds 30 times you will say in the end you did not have enough force left right ? Well actually I did not have {and I am being scientific, non biased and fair here} force “and” energy left, you can’t have one without the other, you “HAVE” to include force and energy when you say I hit momentary muscular failure faster than you.

I ran out of temporary force because I used up all my temporary energy, or/and I used up all my temporary energy because I used up all my temporary force. BOTH are used, and used up. I STILL have energy and force left, but I need to recuperate to get them both back.

You can't lift 400 pounds above your head because you don't have the ability to apply that force.
When you "fail" after you have lifted 200 pounds 30 times,even though you have the ability to apply that force,you have no longer the required energy to apply that force.You run out of energy.You didn't run out of force.

Ok if you want to think backwards, tell me “why” I use up my energy faster ? Or with this kind of lifting why do you think you spend energy at a higher rate ?

More acceleration/force ? if not what and why ?

Not more acceleration...neither more force.The mean force is the weight which means "the forces balance out...make out" or whatever.Accept that fact.
The generation of force(fluctuations are generation/degeneration cycles) is more energy demanding than the maintenance of force(as if holding the weight).There're biology studies that prove that.
That's a perfectly scientific answer.

 

[sophiecentaur]

Please if you do not agree, all I would like is that you do/can understand what I am just trying to say ? I am not at this moment in time concerned who is right or wrong, just that you see my point, as it’s quite frustrating when something argues over you, but they are not actually seeing my point, as they keep repeating what I agree with.

No time for the below its late here.




Wayne

The point you are trying to make is not to do with Physics. It's to do with what you feel in your muscles and how you are trying to fit it to some kind of 'Mechanical Science', that you've invented. You have absolutely no idea of what your muscles are doing at the time they're doing it. There is a delay, corresponding to a large fraction of the time of one lift, in what you do and your consciousness of what you did. That's no basis for an experiment because the evidence is fundamentally flawed.

You must have spent hours and hours on this and it really isn't a good way to learn anything about a subject.
This is not uncommon. Many people write posts about their pet approach to some topic and insist on having it explained in their terms only. It so often fails because they are not prepared to start from scratch and get the subject properly sorted out. Douglis has very patiently, point by point, taken you through your problem and you just reply with remarks like:
"Sorry you lose me a little there ? Equal with g ? I know what g is, but not sure why you are saying equal with it."
His conclusion would be absolutely blindingly obvious to you if you had taken the trouble of approaching this from the beginning instead of doggedly insisting that your view is the right one. Science doesn't necessarily 'make sense' when you try to reconcile it with ideas which, themselves, don't make sense.
You could have completed a whole GCSE Physics Course in the time you have wasted trying to justify your misplaced views on these threads. If you had, then you would have no trouble understanding this really elementary bit of Classical Science. You would also see how simple Physics can't help you with this 'undefined' question that you keep trying to ask.

As I hinted before, I really question your motives here. I suspect you are just attention seeking.
I'll bet you haven't read all the words in this post, too.

 

[waynexk8]

3. You actually don't know what you mean when you talk of "total force" because you are after an arm waving description of what it feels like to do a lift. I really don't know why you won't accept (from not just me but others, too) that your idea can't be stated in Physics.

Ok I will be able to except that the total or overall force will not be able to be equated by physics, I just find it odd, as this the below F = Ma seems so easy, but then when I ask for the force for 80 pounds moving 20 inch in .5 or a second, and 80 pounds moving 3.3 inch in .5 of a second, it can’t be done ?

In total or overall force, I mean the muscles only have a temporary limited amount of force before its used up, so let’s say I can lift 200 pounds 10 x at 1/1 = 20 seconds, then I could not lift it a full rep again, but maybe half a rep, but let’s forget about the half a rep, could not we work out how much force I have, or much I have used ?

F = Ma.
200kg and an Acceleration of 20m/s or 10m/s then we can work out the Force pushing the car by multiplying the weight by the Acceleration like this,

Fast rep,
200 x 20 = 4000N.

Slow rep,
200 x 10 = 2000N

Yes I know for my reps there will be a deceleration, but don’t see how the above 2000N can make up or balance out the 4000N ?

PLEASE, could explain this too me. So this means that I would have to use a force of 4000N to be able to move the 200kg 20m in 1 second, so that means I would have to apply the 4000N for 1 second only for it to move 20m in 1 second ? And what ?

What you could find out (but only by measurement) would be the maximum force, the total work done on the weights (thats N lifts times weight times height) and the Power developed (dividing the total work done by the time for the whole exercise).
I don't understand why that isn't enough.

Ok Maximum force would be ok for a start, how would you do that please ? And total work done on the weights.

Believe it or not I can work the power out quite easy, that’s one of the first things that got me back interested in physics. I know this is like 1 and 1 too you, but hope I am doing it right here ?

Calculate how much power I would be used on both rep speeds. Distance weight moved 1.85 M.

Determine the force we will need to figure out what the weight of the barbell is (W = mg = 91 kg x 9.81 m/s = 892 kg.m/s or 892 N). Now, if work is equal to Force x distance then, U = 892 N x 1.85 m = 1650 Nm.

We can calculate that lifting a 200 lb barbell overhead a distance of 1.85 m required 1650 J of work.

Let us only add up the positive part of the lift.

The concept of power however, takes time into consideration. If for example, it took .5 seconds to complete the lift, then the power generated is 1650 J divided .5 s = 3300 J/s.

If it took 2 seconds to complete the lift, then the power generated is 1650 J divided 2 s = 825 J/s.

Slow set,
825 x 6 reps = 4950Joules.

Fast set,
3300 x 25 reps = 82500Joules

All the other acres and acres of figures you have written on these pages have been pretty much wasted. I'd bet that no one has actually read more than a few sample lines of it and then given up.

How can you be so sure that your question is reasonable when, in the same breath, you admit to not knowing much real Physics?

I have seen some of the movies on your website and I am really impressed by your dedication to your sport. You are clearly a bit of an expert on the practical aspects of it and your advice on how to do the exercises and how to body build is, no doubt, correct.

Big thank you for that as at first I thought you was trying to make out I was a crank, as imagine you going to a place where no one knows you, and they say you know knowing of physics and don’t know what you’re talking about. But I have been training for nearly 40 years, and the Wrought Iron is my business, thanks again for saying that, as was getting a bit up tight.

But, when it comes to the Physics of the situation, you just have to be much more rigorous and 'go along with the rules'. Those rules say that you are on a fruitless quest and have been talking mostly rubbish. Why are you bothering and why can't you just accept what you are being told? Do you just like a friendly chat - is that what it's all about?

That is a DEAD PARROT!

Ok yes I will have to go along with the rules, but I know I go on about it, but the EMG ? Or do you know what a force plate is ? I imagine so, what if I bought one, or got someone to do some tests for me, and just say, and please just say, and I am not saying you are, but let’s just say that like the EMG the force plate readings come back stating the fast do have a higher average force reading, what would you do, would you be very interested, and very interested to find out not how the physics equations were wrong, as they are not, but would you be very interested to find out what variables you left out ?

This debate is HUGE on the internet. Over several different forums, by 1000s of different people, and lots or separate debates, and it’s been going on for well over ten years.

Not that well, back to answer the rest tomorrow.

Wayne

 

[sophiecentaur]

The basic rules of all this are so simply stated that they don't justify all this 'chat'. All the blather is just clouding the issue.
I shall not bother to repeat, yet again, the very few relationships and definitions.
All I can say is that, if you cannot state your question in just a few words with no numbers and just the right terms, used correctly, there isn't an answer. No one wants to read another list of numbers and jargon terms. We have established that it doesn't get us anywhere.

 

[waynexk8]

This is what I want to do, but not on a jump, but 6 reps at .5/.5 = 6 seconds and 1 rep at 3/3 = 6 seconds. And find the average force used, and the peak, but the total or overall force used.

http://people.brunel.ac.uk/~spstnpl/Publications/VerticalJump(Linthorne).pdf

Wayne

 

[douglis]

This is what I want to do, but not on a jump, but 6 reps at .5/.5 = 6 seconds and 1 rep at 3/3 = 6 seconds. And find the average force used, and the peak, but the total or overall force used.
Wayne

1)The average force is the weight.
2)The peak force requires more data.If you use an accelerometer you can find the peak acceleration and then the peak force.
3)Total or overall force doesn't exist in physics.What you're looking for is the integrated EMG or maybe you can describe it with gravity's impulse which is the same in both cases of your example(weight X 6 seconds).

 

[sophiecentaur]

He has been told this many times but just does not choose to take it on board. Until he does, we are all wasting our time.

 

[waynexk8]

Just composing my next post.

In the meantime ou might enjoy the below from some recent communications I had from about the best known muscle physiologists, Roger M. Enoka, Ph.D. Professor and Chair Department of Integrative Physiology University of Colorado And one of his colleagues Per Aagaard Professor, PhD Institute of Sports Science and Clinical Biomechanics University of Southern Denmark.


Rogers Neuromechanics of Human Movement is out on [Audiobook] (CD-ROM)

https://www.amazon.com/dp/0736002510/?tag=pfamazon01-20

The number of muscle fibers activated to lift a weight depends on two factors:

(1) the amount of weight; and (2) the speed of the lift. Although more muscle fibers are activated during fast lifts, they are each generating MORE force. We know this because the rate at which the muscle fibers are activated by the nervous system increases with contraction speed.

Although your question seems relatively straight forward, it is not. Despite the popularization of the terms slow and fast muscle fibers, the characteristics of muscle fibers are not so black and white. Human muscle fibers are often classified as types I, IIa, and IIx.

This distinction is NOT based on contraction speed (slow or fast) but is based on the activity of an enzyme that is related to contraction speed. When the enzyme activity is assessed with an histochemical stain, the fiber types appear quite distinct: black, grey, and white.

When the enzyme activity is quantified, however, there is a continuous distribution of enzyme activity across the population. Furthermore, muscle fiber size (a measure of force capacity) varies continuously across the population and in some cases type I ("slow") fibers are actually the biggest.

I do not know how much work is performed by the different fiber types in the two scenarios you describe. I don't think this has been measured. The closest muscle physiologists have come to answering your question is to measure the size of muscle fibers in individuals who perform different types of training.

The most common finding is that it is the intermediate fiber type, the fast muscle fiber (type IIa) that experiences the biggest increase in size (strength) in individuals who perform conventional weight lifting (heavy loads,) and body building (lighter loads, fast/explosive reps) training. Neither type of training appears to have a significant effect on the size of types I and IIx fibers.

Cheers.

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

When a given load is lifted very fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.

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

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

The all-or-nothing principle only refers to the discharge of action potential by a motor neuron; either it discharges an action potential or it does not. It is not correct to apply this principle to the force generated by a muscle fiber, which depends on the action potential-mediated level of calcium within the fiber.

Each muscle fiber action potential releases a certain amount of calcium from the storage site (sarcoplasmic reticulum) that enables the contractile proteins to interact and produce force. The amount of calcium released by a single action potential is less than that required to produce maximal muscle fiber force. Consequently, the force produced by a muscle fiber depends on action potential rate.

Cheers.

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


Roger M. Enoka, Ph.D. Wrote;
One important point to emphasize in these types of discussions is the concept of slow and fast twitch muscle fibers. Unfortunately, this terminology is misleading because there are not two (or three) types of muscle fibers; rather, there is a continuous distribution in every muscle from the fibers with slow contractile kinetics through to those with fast kinetics.

Because there are not distinct types of muscle fibers, it is not possible to design an exercise program that stresses either "fiber type".

A more appropriate functional distinction between muscle fibers is the force at which the motor units are activated during a muscle contraction, which is known as recruitment threshold.

Motor units with low recruitment threshold can be either slow or fast twitch, whereas motor units with high recruitment thresholds are all fast twitch. But, recruitment thresholds decrease with contraction speed so that all motor units in a muscle are activated when rapid contractions are performed with loads 40% of maximum.

The force that a muscle must exert to move a load depends on two factors: the mass of the load and the amount of acceleration imparted to the load. The number of muscle fibers recruited during the lift increases with the speed the lift.

The rate at which any motor unit, low or high threshold, can discharge action potentials is not maximal during slow contractions. As contraction speed increases, so does discharge rate for all motor units.

Hi Roger,
The part on recruitment threshold, is a tricky one to get your head around. I think it means the faster you lift, the muscle fibers lowers their activation recruitment force, so that more can be recruited faster, and are thus recruited faster, as more are needed faster.

Am I right or half right or wrong ?

Hi Wayne,

You were right.

Cheers.


Wayne

 

[sophiecentaur]

Fascinating stuff but what has it to do with the quasi Physics that you want to apply to the problem, Wayne?
Where does he mention average or total force? This guy knows what he's talking about and you should pay attention to what he says. He talks of muscle fibre activity and energy - which is just what I should have expected - but none of your old rubbish.

 

[waynexk8]

1)The average force is the weight.

Get back to that one.

2)The peak force requires more data.If you use an accelerometer you can find the peak acceleration and then the peak force.

We all agree that the peak will/must be higher in the faster reps.

3)Total or overall force doesn't exist in physics.

I am sure it must ? As they work the average ground reaction force of exercises out on a force plate ?

What you're looking for is the integrated EMG or maybe you can describe it with gravity's impulse which is the same in both cases of your example(weight X 6 seconds).

Why do you seem to think you are such an expert on EMG now ? As you did not know anything before, and why do you keep say my EMG is RMS ?

D. you said my EMG showed only the peak forces, I said it measured the total or overall muscle activity/ force output. I showed you this before but you did not comment, could you please, I can video the machine and prove its NOT the peak force, but the average. Or you should be able to work it out here.

Take a look at the slow rep video, you will notice the peak force = ? Was it a 190 something ? BUT the average was a 140 !

http://www.youtube.com/user/wayneroc...36/B8gtpp8ozvU

What do you say to that my friend D. ?

Wayne

 

[sophiecentaur]

Wayne. As you do not know how that EMG machine works, you cannot insist about what it tells you.
If you insist "it must" it's up to you to prove it. Our physics is quite self consistent at this level and needs no further justification. It's innocent until YOU prove it guilty.

 

[waynexk8]

Wayne. As you do not know how that EMG machine works, you cannot insist about what it tells you.
If you insist "it must" it's up to you to prove it. Our physics is quite self consistent at this level and needs no further justification. It's innocent until YOU prove it guilty.

Fascinating stuff but what has it to do with the quasi Physics that you want to apply to the problem, Wayne?
Where does he mention average or total force? This guy knows what he's talking about and you should pay attention to what he says. He talks of muscle fibre activity and energy - which is just what I should have expected - but none of your old rubbish.

sophiecentaur, ok it’s up to me to provide physics is not wrong, but I think it’s not adding in all the variables. As I think the medium forces of the slow, cannot make up or balance out the higher forces of the faster when they are on the decelerations. And in my opinion this is what the machies adds in, the ground reaction higher peck force.

But for the moment could you please clear up and explain what you think RMS, {Root mean square}
means please after looking at these two very short videos.

Slow test.
http://www.youtube.com/user/waynerock999?feature=guide#p/u/36/B8gtpp8ozvU

Fast test.
http://www.youtube.com/user/waynerock999?feature=guide#p/u/37/pd0ZAm_2ioY

As if you look at the low rest reading you can see {not saying this is at all the exact lowest reading} on the slow, its 70, and highest 196, average = 133, machines states 140. As if you look at the low rest reading on the fast, its 155, and highest 226, average = 190, machines states 196. So it can not be the pack muscle activity, it’s the average muscle activity, force output.

By the way this weight was far too light.

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


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

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

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

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

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

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

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

Wayne

 

[waynexk8]

The point you are trying to make is not to do with Physics. It's to do with what you feel in your muscles and how you are trying to fit it to some kind of 'Mechanical Science', that you've invented.

Here is another way of seeing why in my opinion the faster peak force made in the accelerations of the faster reps can not be made up by the mediam forces of slower reps when the faster reps are on the lower forces in their deccelations. As in my opinion the higher peak forces of the faster reps, will do more damage

Could I just for thins post, post some explanations/scenarios of what I mean in laymans terms when I mean that the fast reps will do more damage, thus use more overall or total force.

You drive {ONE SLOW REP = 6 MINUTES} a 80 pound car over a bridge that has a length of 6 miles and 101 pound breaking force {the car is the force created by the muscles, and the bridge in the muscles, and the breaking force in the breaking force, or the force that would tear the muscles, a tear means the muscles have taken more than they can stand, and are totally damaged} You drive your car over in 6 minutes and all is fine with the bridge {muscles} structurally.

I on the other hand, I drive {SIX SLOW REPS = 6 MINUTES} my 6 cars at a 100 pound each over the 6 mile bridge one at a time at 1 minute each, the first car generates a force very close to the breaking force, so do the second and third cars, but the time the fourth, fifth and sixth cars have gone over the bridge, they will most probably caused enough damage that the bridge breaks, why ? Because each was generating MORE force.

Say a Bird flapped his wings once very slowly for 30 minutes, he would not fly, flapped his wing very fast for 30 seconds, he would be flying very high and far. Thus is that not more total or overall force used ? As the Bird has moved his bodyweight far far far further then the slow flapping Bird, thus more force, more acceleration, more velocity, more energy, more disstyance.

My point, the peak force of the faster reps as of the accelerations, are far far far higher than those of the slow reps medium forces, thus in the peak forces small time frame but done 6 times as 6 reps, will put more tension on the muscles, causing them to tear.

Just imagine an imaginary rubber band going across your room in front of you. You roll a light weight over it in 6 seconds, the rubber band bend/dip slightly, as its having a constant light force on it, now roll along it a heaver weight, but each weight is rolled over 6 times in the same time frame as the lighter weight was rolled over once. On each of the fast rolling weight, you see a far far far larger bend/dip. That’s because there is more overall/total force on the rubber band, thus more tension, damage is being done.

If I arm wrestled you, and you put 80 pounds of force and I put 100 pounds of force, I would beat you the 6 times I went again you, and the time frame would be the same. Even if I decelerated for the last 30% then put the 100 pounds of force on again, I would still beat you on my second pull/push of force.

Is not there a way to work out my higher impulse impact forces ?

 

[douglis]

Why do you seem to think you are such an expert on EMG now ? As you did not know anything before, and why do you keep say my EMG is RMS ?

Well...you don't have to be an expert.It's not quantum mechanics.Every manual states that rectifies the raw EMG with the RMS method.
We all learned at school that the RMS is the 70% of the positive peak value.

D. you said my EMG showed only the peak forces, I said it measured the total or overall muscle activity/ force output. I showed you this before but you did not comment, could you please, I can video the machine and prove its NOT the peak force, but the average. Or you should be able to work it out here.

Take a look at the slow rep video, you will notice the peak force = ? Was it a 190 something ? BUT the average was a 140 !

http://www.youtube.com/user/wayneroc...36/B8gtpp8ozvU

What do you say to that my friend D. ?

Wayne

No...you didn't meausure any "total or overall muscle activity/ force output".You also did NOT measure the peak.You meusured the RMS(70% of the positive peak).
The Total Muscle Activation is meusured by the integrated EMG which is the area below the EMG curve.

You drive {ONE SLOW REP = 6 MINUTES} a 80 pound car over a bridge blah...blah...blah

All these "explanations/scenarios"() are the effect of the peak force.

 

[sophiecentaur]

Acres of blather, yet again, Wayne. I, and everyone else, understand what RMS means; it is very well defined. What your machine does in order to produce the "RMS" figures it displays is a complete mystery to both you and me. We can draw no conclusions.
Btw, douglis, RMS is only your value of 0.7ish in the case of sinusoidal situations. RMS value depends totally on the time profile and, to measure it accurately, you need a good number of samples over the cycle. A weight lifting cycle will be far from sinusoidal.

 

[waynexk8]

Sorry I have been very busy.

One thing, could we possibly answer my original question, before I come back on the rest, as I am not sure if I am getting my point over, As I have and always been on about the extra force, the ground, or in our case the more muscles reaction force, thus more tension on the muscles, that is more from the peak acceleration force from the transition from negative to positive, the MMMT {Momentary Maximum Muscles Tensions}. HOPE this question. Newton’s Third Law, is more of a physics equation you can easy work out and get your teeth into.

I think/hope we will all agree, in that if you lift a weight up, then immediately lower, then in a completely different lift, you lower, then immediately lift the weight, there will be more force needed on the second lift, as the moment I lifted the first 80 pounds, there would be just 80 pounds to lift, however, if 80 pounds is traveling down 20 inch in .5 of a second, the acceleration components will appear to make the 80 pounds, far far far heaver that it actually is. Could we please try and work out how much more heavy the weight would be, or register on a scales if you let it drop, and then the extra force that is needed on the second lift ? Say in % ways.

A,
You lift 80 pounds up, from a still start, to 20 inch in .5 of a second; the weight is moving at 40 inch per second, and you then immediately lower the weight back down in .5 of a second.

B.
You lower 80 pounds down 20 inches; you immediately lift the weight back up 20 inch.

Thank you all for your time and help.

Wayne