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Impulse/force in pounds for the time frame |
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| Feb23-12, 07:42 PM | #154 |
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Impulse/force in pounds for the time frameI did not say I knew nothing of physics, it’s more the physics theory I do not know. Ok, “please” sophiecentaur, one of the things that could have made me stroppy, is that I do not get why you or D. cannot answer this one, please “could” you ??? 1, I lift a weight up from a motionless start, up 500mm in .5 of a second. 2, A weight is being lowered under control, moving 500mm in .5 of a second, the mm before it hits the ground, I then have to try and lift it up 500mm in .5 of a second. 3, As of the acceleration components, when moving down with that acceleration, the weight will appear {not sure of the right physics saying for this sorry ???} to be heaver, or have more force than if it was motionless. So it is going to be heaver/harder to lift, thus I “will” need more force than the first lift, think we all agree there ??? 4, How much more force please. 5, If you can’t work out how much more force, and please you can add in your own accelerations for this if you need to. But if you can’t work it out, do you agree that the force needed would be about, and this is a very rough guess ??? 50% more force for .2 of a second, then the same force as the first lift ??? 6, When you two, or anyone else are compeering my fast lifts force, and your slow lifts force, which modal are you compeering your lift two, the lift started from a motionless start, or the lift that is being lowered first then lifted ??? I would say it’s the motionless lift, am I right or wrong please ??? I never said you did wrote anything wrong, I might have said I think you could have left something out, like not adding in all the variables, or kinology, or like in my 1 to 6 above. But what I don’t understand, and no one have tried to explain, is that when my peak acceleration forces are say a 100% or 100 pounds of force, how do you think that your 80% or 80 pounds of force can make up the 100 pounds of force, when its only 80 pounds, how can 80 ever be as high as a 100 ??? The only way the 80 pounds of force could make this extra force up, or the shall we say the 100 pounds of tension on the muscles to the 80 pounds of tension on the muscles, is if the 80 pounds of tension was on the muscle far far far longer, please do you agree there, if not, how does the 80 pounds of tension make up the higher 100 pounds of tension on the muscles, as a 80 pounds tension, can never be as high as the 100 tension, as how could 80 pounds feel like 80 pounds on the muscles ??? Its impulse, force with respect to time. A small force applied for a long time can 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. But in our case the reps, lifts are done for the same time frame. Ok, I imagine you know or have heard what a force plate/platform is ??? If I buy or have tests done, and that state the average and total or overall forces are higher in the fast, like the EMG states, what would you say then ??? http://en.wikipedia.org/wiki/Force_plate Will get back to the others tomorrow. Again, thank you for your time and help; I only want a friendly debate. Wayne |
| Feb24-12, 06:30 AM | #155 |
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It explains why you seem incapable of stating a question in less than ten paragraphs and incapable of understanding an answer that takes up a single line. Would you ever ever ever challenge a mathematician about something on the grounds that you know Maths but not the 'theory'? You do not believe that Physics cannot answer your question because you just don't know the Physics. Until you are prepared to learn some of the 'theory' you will never get this. Stick to arm waving. |
| Feb24-12, 06:26 PM | #156 |
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If I had four marbles on the table, and the mathematician took 2 away, and said I still have 4 there, then yes I would challenge his theory with my practical World experiments. Physics, or right/correct physics, is only right/correct physics, when the theory is proven with practical experiments, unless that, it stays an unproved theory. I am not arm waving, I have come here asking and hoping for answer to questions, so I ask quite simple questions like the below, but I get no answer, please I do not understand why you can’t answer the below. Not being sarcastic here, but it’s the only thing I can think of, you do understand what I am saying/asking below ??? If not please say, as its looks very straight forward to me, please tell me if the below is the wrong way to ask, or just say why you will not answer, as I do not get it. I can here to ask some friendly polite questions, and it seems more of a match to put me down, I just don’t understand this. 1, I lift a weight up from a motionless start, up 500mm in .5 of a second. 2, A weight is being lowered under control, moving 500mm in .5 of a second, the mm before it hits the ground, I then have to try and lift it up 500mm in .5 of a second. 3, As of the acceleration components, when moving down with that acceleration, the weight will appear {not sure of the right physics saying for this sorry ???} to be heaver, or have more force than if it was motionless. So it is going to be heaver/harder to lift, thus I “will” need more force than the first lift, think we all agree there ??? 4, How much more force please. 5, If you can’t work out how much more force, and please you can add in your own accelerations for this if you need to. But if you can’t work it out, do you agree that the force needed would be about, and this is a very rough guess ??? 50% more force for .2 of a second, then the same force as the first lift ??? 6, When you two, or anyone else are compeering my fast lifts force, and your slow lifts force, which modal are you compeering your lift two, the lift started from a motionless start, or the lift that is being lowered first then lifted ??? I would say it’s the motionless lift, am I right or wrong please ??? I don’t understand, and no one have tried to explain, is that when my peak acceleration forces are say a 100% or 100 pounds of force, how do you think that your 80% or 80 pounds of force can make up the 100 pounds of force, when its only 80 pounds, how can 80 ever be as high as a 100 ??? The only way the 80 pounds of force could make this extra force up, or the shall we say the 100 pounds of tension on the muscles to the 80 pounds of tension on the muscles, is if the 80 pounds of tension was on the muscle far far far longer, please do you agree there, if not, how does the 80 pounds of tension make up the higher 100 pounds of tension on the muscles, as a 80 pounds tension, can never be as high as the 100 tension, as how could 80 pounds feel like 80 pounds on the muscles ??? Its impulse, force with respect to time. A small force applied for a long time can 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. But in our case the reps, lifts are done for the same time frame. Wayne |
| Feb24-12, 06:51 PM | #157 |
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LOOK, please let’s forget about that, as you did immediately say you was wrong when you was, I salute you for that, I will and do not want to rub it in every again, so sorry. No its not just a physics debate, it’s a World wide debate, and whatever needs to be added in, to answer the question, will be. Problem is, you do not answer questions. E-mail, sales.i@motion-labs.com Subject, Re: [From Website] to Edmund Cramp. E-mail, roren@prophysics.ch Subject, Re: Your email regarding EMG systems Then please try to answer the below, and if you say I run out of force because of I used up the energy fast, I know that, the question is sort of why did I use up more energy doing something faster ??? Did I use up more because I was using more force per unit of time, the same, or less. But the main issue is the force, the energy is “only” the supply, and we are on about the force, the energy supply. as in the fast rep fails at lifting first, that “must” mean, that the person doing the fast, used more muscle force per unit on time, if not why does the fast have no force to pick up the 80 pounds, but the slow does ??? Also, how does the fast move the weight 6 times further in the same time span, how does the fast use more energy in the same time span, if it’s not as you two say using more force up per unit of time. Wayne |
| Feb25-12, 01:02 AM | #158 |
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Only an instrument that measures what is happening to the weights themselves can measure the work involved and the forces involved. The EMG doesn't do this so it is not contradicting the Physics.
You are clearly not "trying to learn". Not even one small step at a time. Because you disregard every basic part of theory, preferring the 'but surely' argument and 'it stands to reason'. You dismissed my analogy concerning a mathematician by quoting an example where the Mathematician was not right and had no proper reasoning behind his result. But you argue against the RMS thing and that is totally Maths. |
| Feb25-12, 03:45 AM | #159 |
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I was wrong from biology point of view.From physics point of view I was perfectly right and you're saying nonsense.You can't possibly know if you spend more energy if you lift a weight 10 times up and down or you just hold it for 10 seconds. ![]() It's just a couple of us with superhuman patience trying to explain basic physics to you. The RMS is not the average.This is not offered for discussion....it's basic maths.You don't have to mail professors! ![]() Check the below numbers: -2, 5, -8, 9, -4 Their average is 0 but their RMS is 6.16. The RMS is NOT the average.It's the quadratic mean.You probably don't have a clue what that means so you have to trust the link and me. http://www.analytictech.com/mb313/rootmean.htm I'm sure by now that that's way beyond your intelligence. |
| Feb25-12, 06:23 AM | #160 |
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I stop here any physics related discussion with Wayne because obviously is a waste of time.
What's the point anyway?The below study is exactly what he's looking for. Take a look again: The fast push ups failed at 81.2 sec while the slow push ups failed at 101.2 sec.It's obvious that the fast push ups had greater rate of energy expenditure(see table 1). BUT the Total Muscle Activation(for the pectoralis major for example) for the fast push ups is 2114.23 while for the slow push ups is 3121.81(see table 3). So the muscle activation per second for the fast push ups is 2114.23/81.2=26.04 and the muscle activation per second for the slow push ups is 3121.81/101.2=30.85 http://jmbe.bme.ncku.edu.tw/index.ph...ewFile/635/839 Conclusion: greater rate of energy expenditure does NOT equate greater muscle tension per unit of time.In fact,in the case of push ups,the exact opposite is true. End of discussion. |
| Feb25-12, 10:58 AM | #161 |
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But it’s the muscles forces we are after, or the tensions on the muscles, that’s the forces from the muscles forces onto the weights, with the opposite reaction forces creating tensions on the muscles, and as the EMG takes the muscle activity = muscle force, more muscle activity = more muscle force, yes ??? When you say work, do you mean the mechanical work, the product of a force times the distance through which it acts, work of the force, we know the fast does more work, or more work of the force, so is not that more force if it’s done more work, work of the force ??? Another thing, I have not been explained what RMS is ??? But then again, I could say that you are now disagreeing with people who make and work with EMG machines, and two of these stated to measure the total or overall force output of the muscles, RMS is about the best. "Please" could you ansner this ??? If not please state why. 1, I lift a weight up from a motionless start, up 500mm in .5 of a second. 2, A weight is being lowered under control, moving 500mm in .5 of a second, the mm before it hits the ground, I then have to try and lift it up 500mm in .5 of a second. 3, As of the acceleration components, when moving down with that acceleration, the weight will appear {not sure of the right physics saying for this sorry ???} to be heaver, or have more force than if it was motionless. So it is going to be heaver/harder to lift, thus I “will” need more force than the first lift, think we all agree there ??? 4, How much more force please. 5, If you can’t work out how much more force, and please you can add in your own accelerations for this if you need to. But if you can’t work it out, do you agree that the force needed would be about, and this is a very rough guess ??? 50% more force for .2 of a second, then the same force as the first lift ??? 6, When you two, or anyone else are compeering my fast lifts force, and your slow lifts force, which modal are you compeering your lift two, the lift started from a motionless start, or the lift that is being lowered first then lifted ??? I would say it’s the motionless lift, am I right or wrong please ??? Wayne |
| Feb25-12, 11:57 AM | #162 |
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I am not disagreeing with what the EMG people are saying. The problem is that you do not seem to understand what they are saying and how it applies to your question.
Why should you need to have RMS explained any more. It isn't an explanation that you need. What you need to do is to use the definition to work out some numbers yourself. Douglis has given you a perfect worked example. I cannot be bothered to answer that question because it is just a smoke screen to protect you from going to the trouble of working stuff out for yourself. |
| Feb25-12, 12:22 PM | #163 |
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Then do another sherach, on room calorimetry. the Energy Expenditure and Nutrient Oxidation will and have been established in a room calorimetry. Many Statistics will be measured eg; Heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, to determine the effects of exercise at different intensities. On every occasion when an activity has been practiced at a faster rate, the heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, has been higher, therefore energy expenditure, nutrient oxidation will be higher. The increase in the rate of energy expenditure/cost will be due the increase in the intensity/force of muscle activity. When a muscle is subjected to a greater acceleration, speed or velocity, EMG reading will go up, when muscle uses more force, more force will equal more acceleration, speed, velocity, equals more energy. jarednjames wrote;Let's say it takes 10N to lift the weight in 1 second (a = f/m), if you apply 20N it would only take 0.5 seconds and so on. But the force applied per unit time is the same. The difference comes when you look at energy. The kinetic energy of moving the weight (100lbs) at an average velocity of 1m/s is 22 joules, but the KE of moving the weight at 2m/s is 90 joules. So the higher the average velocity of the weight, the more the energy use increases - which is why you get tired quicker. The faster you move something, the more energy it takes. As you can see, simply doubling the velocity (which would halve the rep speed) requires over four times more energy. Each time you double speed of the reps, and as such halve the time for the reps, you are increasing the energy requirement in this manner. That is why you're using more energy. * All figures are for guidance only. jarednjames wrote; To reduce the time to move an object a distance of 1m, you have to increase the force. By increasing the force, the acceleration becomes larger: a = f/m therefore if you double to force to 2f you get 2a = 2f/m. Let's say we have an object that is 1kg. To move that 1m in 1s (1m/s) requires a KE of 0.5mv2 = 0.5*1*1 = 0.5 Joules of energy. Now, to move it 100m in 1s (100m/s) requires a KE of 0.5*1*10000 = 5000 Joules. So in the first case I need a tiny amount of energy, in the second I need a huge amount in comparison. For your case to move that 1m in 1s (1m/s) requires a KE as above (0.5 Joules). Now, to move it 1m in 0.5s (2m/s) requires a KE of 0.5*1*4 = 2 Joules. So again you can see how simply halving the time of the repetition requires you to use more energy to complete it. The time applied is considered in the velocity figure. For you to move the weight 1 rep in 1s requires 0.5 Joules - that is the energy you must provide to do it. For you to move the weight 1 rep in 0.5s requires 2 Joules - again, that is the energy you must provide to do it. If you do not provide that energy, you can't complete the rep in the required time. 1, As when using 80% of your 1RM {Repetition Maximum} when you fail to lift the weight again, or you hit momentary muscular failure 50% faster on the faster reps, this must surely mean you have used up your temporary force faster, it can’t mean you have used up your temporary force slower, or you would still have force left, but you don’t, you don’t have any force left, so that can mean one think only, that on the fast you are using more force per unit of time. I use more energy in the same time frame; I move the weight far far far further in the same time frame. And if you say I run out of force because of I used up the energy fast, I know that, the question is sort of why did I use up more energy doing something faster ??? Did I use up more because I was using more force per unit of time, the same, or less. But the main issue is the force, the energy is “only” the supply, and we are on about the force, the energy supply. as in the fast rep fails at lifting first, that “must” mean, that the person doing the fast, used more muscle force per unit on time, if not why does the fast have no force to pick up the 80 pounds, but the slow does ??? Also, how does the fast move the weight 6 times further in the same time span, how does the fast use more energy in the same time span, if it’s not as you two say using more force up per unit of time. If not please explain your way of thinking, please for once answer. 2, I don’t understand, and no one have tried to explain, is that when my peak acceleration forces are say a 100% or 100 pounds of force, how do you think that your 80% or 80 pounds of force can make up the 100 pounds of force, when its only 80 pounds, how can 80 ever be as high as a 100 ??? The only way the 80 pounds of force could make this extra force up, or the shall we say the 100 pounds of tension on the muscles to the 80 pounds of tension on the muscles, is if the 80 pounds of tension was on the muscle far far far longer, please do you agree there, if not, how does the 80 pounds of tension make up the higher 100 pounds of tension on the muscles, as a 80 pounds tension, can never be as high as the 100 tension, as how could 80 pounds feel like 80 pounds on the muscles ??? Its impulse, force with respect to time. A small force applied for a long time can 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. But in our case the reps, lifts are done for the same time frame. If not please explain your way of thinking, please for once answer. RMS or quadratic mean will used in the situations, where it is the square of the values that matters; an electrical current squared will be proportional to the power. A quadratic mean, RMS should be used for periodic data (i.e., over time, a graph describes a sine curve), or when both positive and negative data are included, and what you seek is an "average distance from zero" for the various data point. The quadratic mean, a statistical measure of the magnitude of a varying quantity. It is especially useful when varieties are positive and negative. You times it by two; you're multiplying a number by itself. Five squared is 25, five times five is 25. The problem is, why are you two not adding in the negative force ??? Wayne |
| Feb25-12, 12:26 PM | #164 |
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This should interest you, see table 2. Force-velocity, impulse-momentum relationships: http://www.jssm.org/vol7/n2/16/v7n2-16pdf.pdf Wayne |
| Feb25-12, 05:21 PM | #165 |
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I have read the paper an I think I can see your problem. I find little to disagree with what is written.
They are not considering the whole cycle (lift/lower) when they refer to mean force. They are not only discussing free lifts but machines that present RESISISTIVE loads. Everything changes in that case because you are not just changing gravitational potential energy in that case but work is being done in overcoming friction. If you had read what they say then you would not think they are disagreeing with established physics at all. You did not understand what you were arguing about because you insisted on giving details instead of condensing your questions into something meaningful. |
| Feb26-12, 06:42 AM | #166 |
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Hi waynekx8, I didn't notice that you were back.
It looks like you are still confusing "work done" and "energy expended" as we discussed last year: http://www.physicsforums.com/showthr...15#post3190515 |
| Feb26-12, 09:26 AM | #167 |
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| Feb26-12, 10:09 AM | #168 |
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So the whole thread is GIGO, as usual. And I wonder why they use a term like "propulsive force" which has no strict definition faik.
You need to read more carefully, Wayne. |
| Feb27-12, 01:17 PM | #169 |
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Will be back later to get back to some of the old questions I have missed.
Here D, I double checked, what do you say now on this and the other mails regarding RMS and EMG ??? I understand why they say to usde RMS, and the other methords, do you now understand why you use ROM ??? Hi Wayne, There are a few research papers floating around that suggest that the RMS envelope of EMG activity approximates to the force exerted by a muscle. There are several different methods of integrating and averaging muscle activity - all produce estimates of muscle force, the RMS method appears to be slightly more favored but I don't believe that there's any solid proof that it's any more accurate than any other method - the different methods are discussed in the EMG section of the Craik and Otis book which goes into a reasonable amount of detail while remaining understandable. Regards, Edmund Cramp -- << Personal contact information removed by Moderators >> Wayne |
| Feb27-12, 02:42 PM | #170 |
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Wayne still doesn't understand that answer.
I notice he doesn't ever quote from a long correspondence with any of his other sources. |
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