- #71
sophiecentaur
Science Advisor
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Leave it, douglis, it ain't werf it.
sophiecentaur said:@Wayne
I just read your long post. Your insistence on using terms wrongly can only be regarded as arrogant - when you have been told so many times what the right terms are. You equate force / strength / energy at every opportunity which makes the accompanying reams of stuff meaningless.
sophiecentaur said:What's the point in saying that you're sorry if you continue to say the same old rubbish.
sophiecentaur said:You seem to want to pick holes in the explanations given by Douglass and others when they / we have no real idea what question you are actually asking.
sophiecentaur said:If you say that reading a Physics textbook is beyond you then how could you think that you would understand and be qualified to argue with what you read here?
sophiecentaur said:I should just turn on your EMG /ESP /SPQR machine and enjoy reading the figures it shows you. That's what you paid your money for. Aim at making them higher (or lower) and see if you feel fitter / stronger as a result. Enjoy things at that level. No one will argue with you or get exasperated.
douglis said:Great!For the first time you used some physics!
So what's the change in SPEED when you start and end at rest as it's done when you lift a weight?
waynexk8 said:So this means you use up your temporary force faster on the faster reps, meaning you are using more total or overall force up. And faster, right ?
Wayne
Zula110100100 said:So it's more like force is the AMOUNT of "pushing", regardless of the time or distance.
Zula110100100 said:Take for example when you press your hands together, there is a certain force, even though there is no distance, and the time has nothing to do with the force, except that it might decrease as you get tired, and will certainly vary somewhat over time.
Zula110100100 said:Consider if you took some dynamic(changing over time) situation and you had a spring in between something apply force and something being moved, you could take a snapshot at any time and determine the force being applied by measuring the spring(assuming you knew it's equilibrium length and spring constant), so time has nothing to do with it.
Zula110100100 said:An impulse has doesn't mean a sudden impact, and doesn't mean its "high"(especially as that is relative to what you are talking about[an elephant can probably handle much higher impulses than a mouse]). You can have an impulse of .0001N*S, taking place over an hour(theoretically), What makes it an impulse is that it is force taking place over time, not just the force we would see in a snapshot, but the measure of that "amount of pushing" for some "amount of time". The same way that acceleration would have a different meaning that acceleration for some amount of time(which is a change in velocity, [itex]/delta v[\itex]).
sophiecentaur said:If you're the Expert now, on what Physics can and can't do, then I suggest you answer the question yourself.
sophiecentaur said:I have just read your comments on Douglis's post and it is clear that you don't even read his sentences to the end.
sophiecentaur said:I have to conclude that you find us all to be totally incompetent in the field of Physics so I suggest you go and find a Forum in which the contributors know enough of the right sort of Physics to satisfy you.
sophiecentaur said:Does it, for one minute, strike you that your whole idea could just be flawed?
sophiecentaur said:Your resolute use of the nonsense expression "force / strength"
sophiecentaur said:and others,
sophiecentaur said:demonstrates that you just don't really want to get to grips with the real stuff. Just why do you keep posting here?
douglis said:For God's sake...read my whole sentence and try to make your brain work.
You DON'T use force.You use energy to apply force.
douglis said:I "bring up" the energy thing because higher energy usage(as in fast lifting) does NOT equate greater force application.
douglis said:Apples and oranges.Only in Wayne's world those two mean the same thing.
douglis said:Your EMG states greater quadratic mean(RMS) as expected since in fast lifting the force has greater peaks.
douglis said:Here it is again!Are you able to understand the difference of these three?
I never said I do not know the difference ? Why would you think or say that ? I said, I quote; I fail faster in the faster reps, thus I “have” used up my temporary energy/force/strength.
I fail in the faster reps because I have used my temporary energy up, as in the muscles energy ATP glucose and creatine. And used my temporary force up, push or pull. And used up my strength push or pull.
douglis said:The acceleration is always exactly balanced by the deceleration regardless the length of each phase.That's why the average force is always equal with the weight.
Why do you “think” its balanced, you cannot say, if it was balanced, why is you energy and distance not made up or balanced out in your slower reps, if you claim its balanced out ?
douglis said:NONSENSE.
The force-energy relation is NOT linear.The force "balances out" while the energy NOT.
Again you state something with “nothing” to back it up ? You need to try and explain why the faster reps use more energy, you need to say why and how you think the forces balance out make up but the energy does not ? I say why, but you can not ?
So what if I buy or get tests done on a force plate, and they state what I claim, will you accept your physics equations are missing variables then ?
Wayne
sophiecentaur said:There you go again with acres of unreadable (and unread) rambling. State your question like everyone else does: in a couple of lines, using the right terms (which you know by now). No one will make sense of that old ramble above.
sophiecentaur said:Admit it, though, you just want an unending chat and not an answer.
waynexk8 said:I have higher peak, and lower lows, please do not forget about the lower lows, as the EMG WILL take these reading to, and average them all up, or do you want to try and forget about my lows when its convenient ?
Wayne
waynexk8 said: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.
sophiecentaur wrote;
You claim that your machine tells you the force involved (in N) but then say that it reads electrical activity in μV. When you 'tense up' your arm, there is no net force at all (it stays still, in its original position) but there is loads of muscle activity.
sophiecentaur wrote;
as your machine would show, but the antagonistic muscles are producing equal and opposite force.
sophiecentaur wrote;
So there's no direct connection between muscle activity and force produced.
sophiecentaur wrote;
That is unless you have electrodes on every muscle group and the machine can do some complicated 'addition' of the effects of all the muscles.
sophiecentaur wrote;
You are still after some relationship between that muscle activity and the measurable work done on a weight when lifting it. But if it's possible to have loads of muscle activity and Zero work done, then there clearly is not one. Can you not accept that?
sophiecentaur wrote;
There is really no more to be said on the topic (except for another acre of figures about rep rates and pounds lifted).
sophiecentaur wrote;
I can only suggest that you approach the manufacturers of your machine and ask them for their opinion.
douglis said:That's an accelerometer.
douglis said:From the change in speed estimates the peak value of the acceleration and then from the equation F=mg+ma finds the peak value of force(Fmax).
Check their site..."Concentric strength(N) = Fmax in the push"
https://docs.google.com/viewer?a=v&...Vseg--&sig=AHIEtbSVGzCSQRrN1crmIrlXybsDwA_Pwg
douglis said:The "total/overall force" can be found ONLY by using integrated electromyography(iEMG) like they did in the push ups studies.
waynexk8 said:Do you know what Integrated means ? Combining or coordinating separate elements so as to provide a harmonious, interrelated whole = RMS.
Wayne
waynexk8 said:Not sure if I get you there, as the fast produced more Newtons, yes ? So more Newtons is more total or overall force, right ? As how can more N be the same ?
integrated electromyography uses RMS, like my machine, I can set it for as many samples as I want. You have to use RMS to perform the integration. Do you know what Integrated means ? Combining or coordinating separate elements so as to provide a harmonious, interrelated whole = RMS.
I showed you the those press up studies were flawed, in that it was total muscle activity they took, and as the slow went on for longer.
Wayne
waynexk8 said: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.
Wayne
sophiecentaur said:Can't ba answered if you don't know the acceleration, I'm afraid.
sophiecentaur said:Force=Mass X Acceleration
(Then add the weight, in this case)
sophiecentaur said:Are you assuming constant force all the time?
douglis said: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)?
douglis said:For the peak force...we need more data.
douglis said:For the average force...in both cases you start and end at rest.The net change in momentum is therefore zero,
douglis said: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 said:Calculate the peak force needed to throw a computer out of a window in sheer exasperation!
sophiecentaur said: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.
sophiecentaur said:How can the machine distinguish between situations where there are and there aren't antagonistic muscles at work?
sophiecentaur said: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?
sophiecentaur said: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.
sophiecentaur said: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.
sophiecentaur said: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.
waynexk8 said:WaynesWorldphysics ?
Please sophiecentaur, could you work out what the Newont force is below is, peak, average or total ?
sophiecentaur said: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?)
sophiecentaur said: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.)
sophiecentaur said: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 said: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.
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.
Wayne
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?)
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.)
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.
douglis said:O.K. I'll try to help you understand once again.
douglis said:Let's say in your above example the load is 800N(81.5kg) while your maximum force ability is 1000N.
douglis said:For the first 0.4sec you're applying your Fmax
douglis said:(although in reality that's biomechanically impossible)
douglis said:so the net force is 1000-800=200N.
douglis said:So your acceleration for those 0.4sec is a=F/m=200/81.5=2.45m/s^2
douglis said:For the last 0.1sec you let the gravity decelerate the load so the net force is -800N
douglis said: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.
douglis said: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.
You lost me on ma, I need to know what that is for me to understand you point.douglis said:In the equation of force F=mg+ma the "ma"
douglis said: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.
douglis said:"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.
douglis said:It's the fluctuations of force that are more energy demanding.That's the only scientific answer you can get.
Right.sophiecentaur said: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.
sophiecentaur said: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.
sophiecentaur said: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 said: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.
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.
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 ?