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Impulse/force in pounds for the time frame

by waynexk8
Tags: frame, impulse or force, pounds, time
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waynexk8
#73
Jan27-12, 04:45 PM
P: 399
Quote Quote by douglis View Post
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?
You did say “speed” and speed is the rate/time at which an object covers distance, or the rate of change of position.

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
douglis
#74
Jan27-12, 05:14 PM
P: 148
Quote Quote by waynexk8 View Post

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
Everything is perfectly answered at my last post.I don't know what you can't understand but it's becoming ridiculous.
I'll follow sophiecentaur's advice and leave it.
waynexk8
#75
Jan27-12, 05:29 PM
P: 399
No sure if this is to me or all.

Quote Quote by Zula110100100 View Post
So it's more like force is the AMOUNT of "pushing", regardless of the time or distance.
Agree. However, for working things out you need to add in other quantities.

Quote Quote by Zula110100100 View Post
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.
Agreed.

Quote Quote by Zula110100100 View Post
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.
Time would have something to do with it.

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.

Quote Quote by Zula110100100 View Post
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]).
Agreed.

Wayne
waynexk8
#76
Jan27-12, 05:46 PM
P: 399
Quote Quote by sophiecentaur View Post
If you're the Expert now, on what Physics can and can't do, then I suggest you answer the question yourself.
Never said I was an expert.

Quote Quote by sophiecentaur View Post
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.
I have, and answer everything. Please show me what I have not answered ???

Quote Quote by sophiecentaur View Post
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.
I don’t understand why you say that. I asked a polite question, and “gave” 4 reasons why it could be wrong, and NOT once have you had a go at answering them. You seem to want to turn this into a mocking match; I simply do NOT want that. “Please” tell me why you can’t debate/answer my 4 reasons ??? You seem to be trying to mock me because I ask a question.

Quote Quote by sophiecentaur View Post
Does it, for one minute, strike you that your whole idea could just be flawed?
Hmm, not sure there, as it does take a certain force/strength to move an object so and so distance in so and so time frame, so there should be an answer.

Quote Quote by sophiecentaur View Post
Your resolute use of the nonsense expression "force / strength"
I explained that, however you do not say why you think this is wrong. A force is basically a push or pull, in this instance the force is my muscle strength. If you say why it’s wrong maybe I will agree.

Quote Quote by sophiecentaur View Post
and others,
What others please. Tell me and I will explain or learn from you.

Quote Quote by sophiecentaur View Post
demonstrates that you just don't really want to get to grips with the real stuff. Just why do you keep posting here?
I do want to learn, honest.

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
sophiecentaur
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Jan27-12, 05:46 PM
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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.
waynexk8
#78
Jan27-12, 07:13 PM
P: 399
Quote Quote by douglis View Post
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.
We all know you use energy to apply force, no one is debating this ???

Quote Quote by douglis View Post
I "bring up" the energy thing because higher energy usage(as in fast lifting) does NOT equate greater force application.
You seem to be tying yourself in knots, as you are constantly contradicting yourself.

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.

Quote Quote by douglis View Post
Apples and oranges.Only in Wayne's world those two mean the same thing.
I “never” once said they mean the same thing, odd thing to say ???


Quote Quote by douglis View Post
Your EMG states greater quadratic mean(RMS) as expected since in fast lifting the force has greater peaks.
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 ???

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.

Quote Quote by douglis View Post
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 ???

Quote Quote by douglis View Post
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
waynexk8
#79
Jan27-12, 07:33 PM
P: 399
Quote Quote by sophiecentaur View Post
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.
Please, just say what you do not get or understand and I will try to explain. I am not sure if I can say it short, as its “need” an explanation. As some things I need to explain, and I did my best trying to, what would you think or me if I left things out and told you later ???

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.

Quote Quote by sophiecentaur View Post
Admit it, though, you just want an unending chat and not an answer.
What I will admit, is I am sick of people not trying to give me an answer, because I seem to have proved them wrong, if you think I have not, then please post an answer and try and refute 1 to 4, is that too much to ask.




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
douglis
#80
Jan28-12, 02:10 AM
P: 148
Quote Quote by waynexk8 View Post
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
I didn't even read all the rest you wrote because you're repeating the same NONSENSE that has been answered dozens of times.

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.
waynexk8
#81
Jan28-12, 06:05 AM
P: 399
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
douglis
#82
Jan28-12, 07:53 AM
P: 148
Quote Quote by waynexk8 View Post
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.
That's an accelerometer.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&q...IrlXybsDwA_Pwg

The "total/overall force" can be found ONLY by using integrated electromyography(iEMG) like they did in the push ups studies.
waynexk8
#83
Jan28-12, 04:02 PM
P: 399
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;

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.
Not fully with you on that one sorry, or maybe reading you wrong. As when I tense up my arm with the pads on the moving muscles, lets say the biceps and foararm in the curl or arm flextion, if I just tence those muscles the reading on the machine tells me, if I just hold the weight half way up the machine tells me, and when I miove the weight up and down the machine tells me, and tells me the high signals, with my muscles are producing high force, and the lower end of the signals where my muscles are producing the lower force. So this is a net force, when you tense and when you lift.



sophiecentaur wrote;

as your machine would show, but the antagonistic muscles are producing equal and opposite force.
No, the antagonistic do “not” produce equal and opposite force in any barbell exercises, they produce very little force, maybe as little as 5 or 10% the agonist muscles do all the lifting and all the lowering. {actually the lowering or eccentric portion of the lift which the biceps and forearm does, {as well as the concentric} able a person to lower 40% more under control, say lower in 6 second, than the person can lift for their 1RM.

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.


sophiecentaur wrote;

So there's no direct connection between muscle activity and force produced.
Yes there is, it’s a direct comparison. As if I lift 30% then 80% the readings like the peak force and average force will be far higher.



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.
You only need the pads on the lifting muscles, as these are the ones producing the force for both up and down.

Yes the machine does do very complicated equations instantly all the time.



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?
Yes I understand the concept of work, if I do not move the weight no work has been done, but there is still muscle activity, energy and force being used.


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).
I find it odd you say that, but after reading the about you might differ.

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 ???

sophiecentaur wrote;

I can only suggest that you approach the manufacturers of your machine and ask them for their opinion.
These machines are/have been used in Hospitals and sports facilities for years, they are as efficient as your calculator, actually they are calculator/computers in another form. The machines are used as much as the everyday car; they are very well known and used.

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
sophiecentaur
#84
Jan28-12, 04:48 PM
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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.
waynexk8
#85
Jan28-12, 05:13 PM
P: 399
Quote Quote by douglis View Post
That's an accelerometer.
It a accelerometer, with built in velocity, power, force and other, the unit uses three-dimensional accelerometer. By measuring body movement in three dimensions (backwards and forwards, up and down, side to side), the body can be tracked in space. The speed (and change in speed) of movement can be used to calculate such things as flight time, number and speed of repetitions and power. Many physical parameters can then be calculated from these parameters.

Quote Quote by douglis View Post
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&q...IrlXybsDwA_Pwg
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 ???

Quote Quote by douglis View Post
The "total/overall force" can be found ONLY by using integrated electromyography(iEMG) like they did in the push ups studies.
integrated electromyography uses RMS, like my machine, I can set it for as many samples as I want. [b]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.

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
sophiecentaur
#86
Jan28-12, 05:20 PM
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Quote Quote by waynexk8 View Post
Do you know what Integrated means ??? Combining or coordinating separate elements so as to provide a harmonious, interrelated whole = RMS.
Wayne
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
#87
Jan28-12, 05:58 PM
P: 148
Quote Quote by waynexk8 View Post
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 ???
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")

integrated electromyography uses RMS, like my machine, I can set it for as many samples as I want. [b]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.
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.

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
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
#88
Jan29-12, 09:19 AM
P: 399
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
#89
Jan29-12, 09:41 AM
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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
#90
Jan29-12, 10:05 AM
P: 148
Quote Quote by waynexk8 View Post
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
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.


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