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Impulse/force in pounds for the time frame |
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| Jan26-12, 01:18 PM | #69 |
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Impulse/force in pounds for the time frameDoes it, for one minute, strike you that your whole idea could just be flawed? Your resolute use of the nonsense expression "force / strength" and others, demonstrates that you just don't really want to get to grips with the real stuff. Just why do you keep posting here? |
| Jan26-12, 02:01 PM | #70 |
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You DON'T use force.You use energy to apply force.I "bring up" the energy thing because higher energy usage(as in fast lifting) does NOT equate greater force application. Apples and oranges.Only in Wayne's world those two mean the same thing. The force-energy relation is NOT linear.The force "balances out" while the energy NOT. |
| Jan26-12, 05:51 PM | #71 |
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Leave it, douglis, it ain't werf it.
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| Jan27-12, 04:45 PM | #72 |
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And the energy that supplies the muscles for the force/strength is basically calories, broken down into many chemicals. Question. Two Clones lift 80% of their 1RM, {Repetition Maximum or the most they can lift up one time} up 20 inch and down 20 inch. So .5/.5 means .5 of a second concentric, {up} and .5 of a second eccentric. {down} One repetition {rep for short} means one concentric and one eccentric, or once up and once down. Fast, 1 reps at .5/.5 = 6 seconds = distance the weight is moved = 240 inch. Slow, 1 rep at 3/3 = 6 seconds = distance the weight is moved = 240 inch. Or the below just for another example. However this example is a little false, but this needs to be told as it is part of the main of this debate. As if you did do 18 reps fast, you would not only do 3 reps with the slow = 18 seconds, but more like 5 to 6 reps = 3 to 36 seconds, as you always fail faster in the faster reps. This is because in “my” opinion, you have to use more overall or total force to move the weight faster and more distance in the same time frame. Total or overall force means in this instance, as you “only” {and this will apply to a machine or anything as well as human muscles} have a certain amount of available force to use, before it runs out, as you will hit muscular failure quite soon, meaning you cannot pick up the weight anymore as you have ran out of temporary force. So total or overall force means in this instance. As you only have a set amount of force to use in a set time, let’s call that 1000N. So which of the above and below use the most force, or the available force the fastest. Fast, 18 reps at .5/.5 = 18 seconds = distance the weight is moved = 720 inch. Slow, 3 reps at 3/3 = 18 seconds = distance the weight is moved = 140 inch. MY main opinions as to why you must use more total or overall force with the fast. 1, You fail faster, or hit momentary muscular failure faster on the faster reps, as you “have” to have used up all your temperedly force up and faster, and there is no question that you have not used up all your temperedly force up, because you have, as you hit momentary muscular failure, thus have no force left. 2, You use more energy, as in calories when doing the faster reps, why do you more energy doing “anything” faster, not just repping weight ??? In my opinion it’s because you’re using more force and/or more force faster. 3, You move the weight with the fast in scenario, 6 times the distance in the same time frame. You accelerate the weight far further then the slow moves it at a constant velocity. 4, EMG readings state more muscle activity in the faster. This machine is as reliable as a computer in adding equations. Have you heard of a force plate or platform ??? It’s a machine for taking many force reading, like ground force readings from runners, weightlifters and so on. I will either try to buy one or pay for some tests to be done, if this states the fast uses more force, what then, there will be 1 to 5. However I cannot see how anyone could argue with 1. My opinion is that the higher forces the fast puts out on the accelerations, when the fast is on its lower force and decelerations, the constant medium forces of the slow, cannot makeup or balance out the fast higher acceleration higher forces. They may add on paper, but that’s not real World practical actions which apply in World. Again, thank you for your time and help, and I have honestly come here to exasperate anyone, all I wanted was a nice quite polite debate. Wayne |
| Jan27-12, 04:45 PM | #73 |
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So your slow rep is more of a speed, or constant speed, so as don’t basically have a change in speed {other than the initial acceleration and closing deceleration, you change in speed is basically zero. “Please” you reason for this ??? As my faster reps have acceleration, and to accelerate an object is to change its speed, I have basically a constant change in speed on acceleration and deceleration. I ask again, your reason for this please. Now my questions, 1, If you cannot move the/a weight, as in the faster reps when you hit momentary muscular failure, this means you do not have sufficient force to move to move the weight, yes you do not have sufficient force or no you do have sufficient force??? 2, If you can move the/a weight, as in the slower reps when Clone 1 has hit momentary muscular failure, in the faster reps, this means you do have sufficient force to move to move the weight, yes you do have sufficient force to move to move the weight or no you don’t have sufficient force to move to move the weight??? 3, Most people would say, yes you do not have sufficient force to 1, and most people would say, yes you do have sufficient force to move to move the weight to 2. 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 |
| Jan27-12, 05:14 PM | #74 |
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I'll follow sophiecentaur's advice and leave it. |
| Jan27-12, 05:29 PM | #75 |
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No sure if this is to me or all.
As the more compressed the spring got the more force needed, and vice versa. Also, if you were taking about applying muscular force, then time again would come into it. Wayne |
| Jan27-12, 05:46 PM | #76 |
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Sophiecentaur, you don’t know me and I don’t know you, but I can honestly say I did or do not want it to turn out like this, all I want is a polite friendly debate, also I do not get why you do not try and debate/answer my 4 reasons, “OR” answer this question to which I asked before and is full physics. Can you have more power and the same force in the same time frame ??? Please again Sophiecentaur, thank you for your time and help, and sorry if IO upset you, it was/is not intended. Wayne |
| Jan27-12, 05:46 PM | #77 |
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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.
Admit it, though, you just want an unending chat and not an answer. |
| Jan27-12, 07:13 PM | #78 |
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1, We all agree that you have to use more energy the faster you do anything in the same time frame. 2, When you do something faster in the same time frame you “have” to have high/faster/longer accelerations, these high/faster/longer accelerations “must” have higher forces, as you can’t have high/faster/longer accelerations without initial higher forces for the acceleration. But you say, I quote; higher energy usage(as in fast lifting) does NOT equate greater force application. 3, So do you think it’s “just” a coincidence that when you have high/faster/longer accelerations to a constant speed, that the energy needed for the high/faster/longer accelerations goes up ??? Or do you think it’s because the force “has” to go up for the high/faster/longer accelerations ??? 4, Tell me, as of several months back you stated that you do not use more energy doing anything fast, you then over the debate got convinced you were wrong by someone from this forum, then agreed that you were wrong, and admitted that you have to use more energy when doing anything fast, and in our case lifting weight. Could you please state why you have to use more energy ??? Is it a, when the accelerations and force go up, or b, on the deceleration when the force goes down ??? 5, What/why does equate greater energy application in the faster reps, is its not force ??? First on a physics forum, you “NEED” to “state/say” WHY you “think” higher energy usage does not equate greater force application. The RMS also known as the quadratic mean, is a statistical measure of the magnitude of a “varying” quantity. So YOU state that my high forces and my low forces are all balanced out by your medium forces. So if the EMG takes the highs and the lows on my high and low forcers, as you claim they should all balance out, but they do NOT do they ??? RMS It is especially useful when varieties are positive and negative. It can be calculated for a series of “discrete” values or for a “continuously” “varying” function. The name comes from the fact that it is the square root of the mean of the squares of the values. It is a special case of the generalized mean/average. [QUOTE=douglis;3729130]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. 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 |
| Jan27-12, 07:33 PM | #79 |
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Question short, I lift a weight 1000mm in .5 of a second, you lift a weight 166mm in .5 of a second, which uses the most force, not the peak force used, but the total or overall force. Question long as I think it needs more of an explanation as I know more about it. Two Clones lift 80% of their 1RM, {Repetition Maximum or the most they can lift up one time} up 20 inch and down 20 inch. So .5/.5 means .5 of a second concentric, {up} and .5 of a second eccentric. {down} One repetition {rep for short} means one concentric and one eccentric, or once up and once down. Fast, 1 reps at .5/.5 = 6 seconds = distance the weight is moved = 240 inch. Slow, 1 rep at 3/3 = 6 seconds = distance the weight is moved = 240 inch. Or the below just for another example. However this example is a little false, but this needs to be told as it is part of the main of this debate. As if you did do 18 reps fast, you would not only do 3 reps with the slow = 18 seconds, but more like 5 to 6 reps = 3 to 36 seconds, as you always fail faster in the faster reps. This is because in “my” opinion, you have to use more overall or total force to move the weight faster and more distance in the same time frame. Total or overall force means in this instance, as you “only” {and this will apply to a machine or anything as well as human muscles} have a certain amount of available force to use, before it runs out, as you will hit muscular failure quite soon, meaning you cannot pick up the weight anymore as you have ran out of temporary force. So total or overall force means in this instance. As you only have a set amount of force to use in a set time, let’s call that 1000N. So which of the above and below use the most force, or the available force the fastest. Fast, 18 reps at .5/.5 = 18 seconds = distance the weight is moved = 720 inch. Slow, 3 reps at 3/3 = 18 seconds = distance the weight is moved = 140 inch. MY main opinions as to why you must use more total or overall force with the fast. 1, You fail faster, or hit momentary muscular failure faster on the faster reps, as you “have” to have used up all your temperedly force up and faster, and there is no question that you have not used up all your temperedly force up, because you have, as you hit momentary muscular failure, thus have no force left. 2, You use more energy, as in calories when doing the faster reps, why do you more energy doing “anything” faster, not just repping weight ??? In my opinion it’s because you’re using more force and/or more force faster. 3, You move the weight with the fast in scenario, 6 times the distance in the same time frame. You accelerate the weight far further then the slow moves it at a constant velocity. 4, EMG readings state more muscle activity in the faster. This machine is as reliable as a computer in adding equations. Let’s change this around for a minute, and YES I know this is different, but let’s say you ran up to the shop and back, which is uphill, 100m away, as fast as you could 6 times in 300 seconds = 1200m, most people would either not make it, or be totally exhausted. You then walk very slowly to the shop and back one time in 300 seconds = 200m, do you honestly think you have used the same overall or total force, or the same overall or total muscle force ??? I say again, I have not started this mocking type match; I have been polite and only asked questions, and said if anyone does not understand them to ask. All I want is some kind of answer. Wayne |
| Jan28-12, 02:10 AM | #80 |
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The above is the only one that worth answering. The RMS takes the square root of the mean of the squares of the values.By taking the squares of the values you turn the negative values in to positives. I know you have problems with basic maths but you must have been taught that in high school.For example...-2^2=4 not -4. So...your EMG turns the "lower lows" in to "higher peaks".The RMS is NOT the technical mean that we've been discussing so far.....it's the quadratic mean.....huge difference. The RMS(quadratic mean) is exactly the 70% of the peaks....the positive peaks and also the negatives that turned in to positive. Since fast lifting has greater peaks will definitely have higher RMS. Your EMG maybe fine for comparing the activation of a muscle group between different exercises but is NOT offered for a fast vs slow comparison. That comparison could only be possible if you'll find the integrated EMG(iEMG) like they did in the push ups studies(and found greater iEMG for the slow push ups!!!) or as I did at the graphs I mailed you. |
| Jan28-12, 06:05 AM | #81 |
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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 = 579 V = 192 Slow P = 649 F = 546 V = 161 Wayne |
| Jan28-12, 07:53 AM | #82 |
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Check their site...."Concentric strength(N) = Fmax in the push" https://docs.google.com/viewer?a=v&q...IrlXybsDwA_Pwg The "total/overall force" can be found ONLY by using integrated electromyography(iEMG) like they did in the push ups studies. |
| Jan28-12, 04:02 PM | #83 |
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sophiecentaur, I was woundering why you did not replly to this post ??? As without sounding sarcastic, you got the force produced by the agonist and antagonist wrong, thus i think we need to clear thins up, and the other things.
http://www.physicsforums.com/showpos...9&postcount=63 Could you go from here please; So no the antagonistic muscles are not producing equal and opposite force. As the biceps are the curling or flexing muscles, and the triceps are the extending muscles. Yes the machine does do very complicated equations instantly all the time. What about this question then ??? I am sure you can talk or understand scenarios outside of physics ??? Like other branches of physics, kinology, biomechanics. Or, just thought of this, let’s say your back at university, and you have to do a practical test and scenario for your PhD, and the Lecturer asks you the below, and you have to answer. Or could you just have a go for me, or suggest another way I can ask it, but as a physics adviser, I thought you would like to try and work out how the equations do not add up in the real World tests/experiments as in below ??? Please ??? As I don’t see how you can step outside the box, as I thought all physics should be tested in the real World after they have been calculated on paper. The Question. You always fail faster about 50% faster in the faster repetitions, so if you fail faster, you “HAVE” used up your temporary force/strength, as you hit muscular failure faster thus you cannot lift the weight anymore. To me, if both Clones started the fast and slow lifting at the same time, as the slow Clone are still lifting the weights when the fast Clone has hit muscular failure and unable to lift the weight anymore, this “is/seems irrefutable, or categorically right to me, and I am not trying to sound smug or anything here, but if the slow Clone is still moving the weight, then the fast Clone “has” used up more overall or the total force/strength they had faster, if you see my point, please do you see what I am saying here ??? http://medical-dictionary.thefreedic...ectromyography Originally Posted by sophiecentaur They may well be more prepared to speak you language as it is in their interest to sell as many of those machines as they can. I think they will tell you that the machine gives a good indication of how much energy the muscles are transferring and / or the forces. I have no problem with that (the neurological application of the machine seem very worth while). They may even launch into some link between that and the mechanical work done. That will make you happy. Great, but it won't make the Science any more valid. IF, you want to prove your physics equations right, you would need to “try” and answer the question above, and say why the below happens in the real World, as I am in contact; New Scientist Magazine, Physicstoday Magazine and Physics World Magazine. If you could be the first to solve the puzzle of how and why the equations and real World tests on this Phenomena is, maybe you could be in these Mags. Thank you for your time and help. Wayne |
| Jan28-12, 04:48 PM | #84 |
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How can the machine distinguish between situations where there are and there aren't antagonistic muscles at work? 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?
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. 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. 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. |
| Jan28-12, 05:13 PM | #85 |
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Calculate the integrated EMG envelope on- and off-line. The integration function incorporates an RMS (Root Mean Square) feature set to operate over a user-specified number of samples. Adjust the RMS time constant by increasing or decreasing the number of samples used to perform the integration. The number of samples used in the RMS integration divided by the sample rate is proportional to the time constant of the integration. http://www.biopac.com/researchApplic...&AF=61&Level=3 Do a search for integrated below, you will find RMS is part or integrated, as RMS has to be used, as RMS = Averaged or root-mean-square (RMS) Integrated EMG (iEMG) important for quantitative EMG relationships (EMG vs. force, EMG vs. work) best measure of the total muscular effort, useful for quantifying activity for ergonomic research, various methods: mathematical integration (area under absolute values of EMG time series) root-mean-square (RMS) times duration is similar but does not require taking absolute values. http://uk.search.yahoo.com/r/_ylt=A7...311/emg-p2.pps Why do you refer to RMS on my machine ??? Show me the machine they used in the press up ??? 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 |
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