Impulse/force in pounds for the time frame

sophiecentaur

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.

douglis

No...the total/overall force(effect of force over time...impulse) is given by N*s...NOT by N.
The more N means greater peak force as they also state by themselves("Concentric strength(N) = Fmax in the push")

Yes...but you must also normalize the raw EMG data in order to integrate.Check the paragraph " materials and methods".It's described pretty well.The equation (1) shows that the integration is done for the normalized data.
http://jmbe.bme.ncku.edu.tw/index.ph...ewFile/635/839

Anyway...this discussion is meaningless.The fact is that the RMS is the 70% of the peak and naturally higher in fast lifting.End of story.

Those press up studies are perfectly designed...your mind is flawed.
For example....compare the durations and the Total Muscle Activations.

Slow push ups:Duration=101.2 sec.....TMA(triceps)=3145.29
Fast push ups:Duration= 84.2sec......TMA(triceps)=2138.91

The duration of the slow push ups is only ~20% greater but the Total Muscle Activation is ~47% greater.This is a direct PROOF that slow push ups have greater muscle activation per unit of time.
We have such a well designed study examining exactly what you're looking for and you're still around forums saying NONSENSE.

waynexk8

Back a little later.

This will please S. and D and the rest of the forum, some pure physics for a change.

Could anyone state the Newton’s needed for the below,

1,
From a still start, 80 pounds is moved up 20 inches, in .5 of a second, stops and reverses. Only the Newton’s for the forward.

2,
From a still start, 80 pounds is moved up 3.3 inches, in .5 of a second, stops and reverses. Only the Newton’s for the forward.

3,
80 pounds is being lowered under control at 20 inches in .5 of a second, then when still in full downward motion, it’s moved upward 20 inches, in .5 of a second, stops and reverses. Only the Newton’s for the immediate deceleration, {wherever this may be I would say the last 5%}
stop and forward/upward to stop.

Wayne

sophiecentaur

Can't ba answered if you don't know the acceleration, I'm afraid.
Force=Mass X Acceleration
(Then add the weight, in this case)

Are you assuming constant force all the time?

douglis

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)?

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

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

sophiecentaur

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

waynexk8

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. ???}


Misunderstand that a little, so best say and ask.

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

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

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

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

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.

http://www.youtube.com/watch?v=6clCe76uD-Q

Fast
P = 695
F = 579Newtons
V = 192

Slow
P = 649
F = 546Newtons
V = 161

Wayne

waynexk8

WaynesWorldphysics ???

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

Go from 5 min.

http://www.youtube.com/watch?v=6clCe76uD-Q

Fast
P = 695
F = 579Newtons
V = 192

Slow
P = 649
F = 546Newtons
V = 161

This one may help.

http://www.youtube.com/watch?v=Ycu6y...eature=related

Wayne

waynexk8

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/wayneroc...36/B8gtpp8ozvU

What do you say to that my friend D. ???

Wayne

waynexk8

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

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

http://www.youtube.com/watch?v=hG3EgCF90_I

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.

Ok sorry, I will try the stick with force.

Ok get your point.

Ok, I will try and use physics, if I cant will ask.

Wayne

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?)

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.

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

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.

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

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.

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.

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

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.
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

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.
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!