Impulse/force in pounds for the time frame

[waynexk8]

This is all very interesting stuff but what's it doing on a General Physics Forum? The 'Physics' content is limited to the occasional use of words like Force and Acceleration. After acres and acres of talk about weight lifting - getting more and more Physiological with time there has emerged not a single paragraph to sum things up in strictly Physics terms - except for the very basics of Newton's Laws, which we were all taught when we were at School. Hardly surprising because Muscles are far too complex in their operation for simple analysis.

Ok, if you think you are right, why does the EMG state a higher average reading ? How and why do I move the weight 6 times further ? How and why do I use more energy ? how and why do I fail roughly 50% faster ?

And what equastions give you any proff ?

Wayne

 

[K^2]

Ok, if you think you are right, why does the EMG state a higher average reading ? How and why do I move the weight 6 times further ? How and why do I use more energy ? how and why do I fail roughly 50% faster ?

Muscles are funny that way. They require ATP just to maintain constant position under stress. Add to that the fact that you get better circulation while muscles move, and it seems quite reasonable.

 

[Dale]

Hi waynexk8, don't forget to respond to post 281, especially the 4 questions.

 

[douglis]

Thank you.

What I would like to learn more on is,

So stop your endless rambling and try to learn something.

Impulse, yes,

Not average force ? Why do you and D. keep on mentioning average force ? If the average force is the same for 1 rep at 1/1 and 100 reps at 1/1, we all know that the tension will be 99% times more on the muscles than the 1 rep at 1/1 thus the total/overall force must be higher it the tension is higher. I just don’t understand why you mention this, could you explain please ?

Wayne

With 100N

1)you do 1 rep with .5/.5
What's the average force?What's the impulse?

2)you do 6 reps with .5/.5
What's the average force?What's the impulse?

3)you do 1 rep with 3/3
What's the average force?What's the impulse?

 

[sophiecentaur]

Ok, if you think you are right, why does the EMG state a higher average reading ? How and why do I move the weight 6 times further ? How and why do I use more energy ? how and why do I fail roughly 50% faster ?

And what equastions give you any proff ?

Wayne

The only way that I could ever believe that the EMG is giving you accurate 'Work Done' information is if you tell me that, when you are stationary, the EMG reading is Zero.
EMG tells you something about how hard your muscles are working but nothing reliable about the effective work they are doing on the weights.
I know you are devastated that your expensive machine could possibly not be doing what you think it is but you may just have to get over that. It is a useful tool, no doubt, but not in the way you seem to think it is.

 

[waynexk8]

Sophiecentaur and D.

Could you please say why you cannot answer my question on my last posts ? I will number them this time. And if you don’t understand them, or anything part please say, but as this is a physicist forum, I don’t really see how they are complicated ?

dalespan, if you have the time, could you please anser ?

Dalespam wrote;
Impulse has the property that if you exert twice the force for the same amount of time you have doubled your impulse. So, for example, if you exert 100 lbs for 10 s and I exert 50 lbs for 10 s you have exerted twice the impulse that I have. Does this agree with your concept of "total/overall muscle force"?

Impulse also has the property that if you exert the same force for twice as long you have doubled your impulse. So, for example if you exert 100 lbs for 10 s and I exert 100 lbs for 5 s you have exerted twice the impulse that I have. Does this also agree with your concept of "total/overall muscle force"?

Question 1,
This is what I have been saying all along, I use exert 100 lbs for 10 s and you exert 80 lbs for 10 s I have exerted 25% more the impulse that I have.

Why is that so hard to understand ?



Question 2,
What forces do you think you have that can make up of balance out the higher propulsive forces of the fast in the studies ?

Let’s take the mean propulsive forces, slow 6.2mean in 10.9 seconds. Fast 45.3mean 2.8 seconds, now let’s divided the mean slow of 10.9 seconds by the fast 2.8 = 3.8, so now let’s divide the slow mean by 3.8 = 1.6.

Fast mean for 2.8 seconds = 45.3.

Slow mean for 2.8 seconds = 1.6.

The fast has nearly 3000% more mean propulsive force as in N's in the same time frame.

Please what forces have I left out that the slow has to make up or balance out these ?

http://www.jssm.org/vol7/n2/16/v7n2-16pdf.pdf


Question 3,

Which has the highest force, and which are you saying has the same force as the slow rep for the same time frame, 1 or 2 ?

I was just wondering and thinking, ARE you adding “all” the force, I mean with the fast there are NOT just force being exerted by the muscles, there are HUGE forces on them, for say .1 of a second x the 6 reps = high forces on the muscle for maybe .6 of a second. Have you added these in ?

I mean the peak force from the transition from negative to positive, the force on the muscle, NOT given out by the muscles ? We call them the MMMTs {Momentary Maximum Muscle Tensions} these forces “ON” the muscles can be as high as 140%

Have you added these on ?

Lift 1,
You lift 80% of the ground, up 1m and then down 1m all in 1 second, .5/.5

Lift 2,
You start at the top, lower the weight down 1m, and then lift it back up 1m all in 1 second, .5/.5

On lift 2, on the transition from negative to positive, there will be huge force on the muscles and coming from the muscles.

Which of the 2 forces on the 2 faster reps are you saying that’s the same as the slow rep for the same time frame ? As lift 1 and 2 use a different amount of force.



Question 4,
Fast reps person trys to use 100 pounds of force moving 80 pounds object for 6 seconds = 100/6

Slow reps person uses 80 pounds of force for 6 seconds = 80/6

How can 80/6 = or be as high as 100/6 ?
How can you trying to use 80 pounds of force for 6 seconds, be as high or the same as me using 100 pounds of force for 6 seconds ? HOW can 80 foirce be as high as 100 force ? As I always thought a 100 was 20 more then 80, or/and 25% more than 80, but you are saying that 80 = 100 ? Please for once explain how 80 = 100 ?

Wayne

 

[Dale]

dalespam, if you have the time, could you please anser ?

Hi waynexk8, yes, I could answer, but I won't at this time. If you stick it out, I will answer when I think you are ready.

If I came into your gym, having never done weightlifting, and asked you to show me how to do an Olympic snatch at twice my weight, would you teach me that right away or would you try to teach me the basics first?

For question 1 you are correct, it is 25% more impulse. For the rest of the questions we need to lay some more groundwork. Question 4 will be the next one we can address with a little more work.

 

[waynexk8]

Hi waynexk8, yes, I could answer, but I won't at this time. If you stick it out, I will answer when I think you are ready.

Ok that’s sounds more than fair.

Thank you very much dalespam,

If I came into your gym, having never done weightlifting, and asked you to show me how to do an Olympic snatch at twice my weight, would you teach me that right away or would you try to teach me the basics first?

Your right, I would never let you do that, you would have to do the basics first.

For question 1 you are correct, it is 25% more impulse. For the rest of the questions we need to lay some more groundwork. Question 4 will be the next one we can address with a little more work.

The odd part about this, as D would never even answer this, and I still don’t get why, unless he knows he’s wrong.

So could we all stay on question 1 for a while please.

So D. and S. it seems you worked the fast reps out, from the lifted that started on the ground, and was lifted up in .5 of a second and lowered in .5 of a second, = 1 second for the total lift, you then multiplied this by 6 = 6 seconds of lifting, as this is the same time frame as the one slow lift, 3 seconds up and 3 seconds down.

But you forgot to add in the peak forces, or the peak acceleration forces, the MMMTs, {Momentary Maximum Muscle Tension} the huge forces from the transition from negative to positive, and in 6 reps or 6 seconds there are 5 of these.

So I think, that if you add in the peak acceleration forces, too which are 25% higher, you x 25% by 5 = in 6 seconds the faster reps used 125% more impulse/force then the 1 slow rep in the same time frame. Could you comment please dalespam.

This that D. left out the most important forces, the peak acceleration forces, fits in with what he left out before, as he did the exact same this.

https://www.physicsforums.com/showthread.php?t=472526&page=9

Originally Posted by jarednjames
All this average acceleration stuff is non-sense. You are ignoring the fact that to get a rocket to accelerate at 1m/s2 uses significantly less energy than to get it to accelerate at 2m/s2. To increase the acceleration by even such a small amount requires a drastic energy increase.


douglis wrote;
O.K...let's start from here.
We agree that acceleration requires more energy and also the greater the acceleration the greater the energy.
So in this rocket example let's say that the rocket used force that caused initial acceleration equal with 2g(if we assume that was accelerated for 30sec) and then since the engine shut down used zero force for the final deceleretion(the last 30sec).
On average the force that was used produced average acceleration equal with g or else the average force was mg for 1 minute.
Exactly like the force that was used by the rocket that was standing still in the air.

Since in both cases the average acceleretion that the engined produced was equal I can't see how the fuels can be different.The fuels that the moving rocket spent when using 2g for 30 sec are exactly equal with the fuels that the standing rocket spent by using g for 1min.



Originally Posted by jarednjames
They most certainly are not.

Average acceleration doesn't apply to the fuel use because it isn't linear.

As previously, fuel use is exponential. If you double your speed, the fuel use more than doubles.

So to double your speed may half your travel time but it more than doubles fuel use.

1g for a minute may use 1L of fuel. 2g for 30 seconds will use 4L. 4g for 15 seconds will use 15L etc.

Obviously, it's not as straight forward as that but that's the gist of it. 1g for a minute does not equal 2g for 30 seconds.



douglis wrote;
Look...I tried to make an equivalent with weight lifting and I assumed that fuel expenditure is linear.If we don't make that assumption the discussion goes too far.


Originally Posted by jarednjames
In no case, ever, is fuel consumption linear. Weight lifting and rockets alike.

To make that assumption would completely change the physics and make it so that twice the speed with double the distance would give you the same energy use. Complete non-sense.



douglis wrote;
I insist that speed and distance are irrelevant and keep the nonsense comments for youself.
Forget for a moment the acceleration phases.Two rockets with different constant speeds will spend identical fuels for the same duration if you exclude the air resistance.

Again at the acceleration part now...do you believe that double force will result in more than double fuel expenditure?



Originally Posted by jarednjames
Constant speed yes, but it has to get to that speed which involves acceleration which, depending on the required velocity is different (as you are aware). To accelerate to 50mph does not take half the fuel it takes to accelerate to 100mph.

For simplicity you can ignore constant speed and simply have acceleration and deceleration.

If deceleration is due to gravity the rocket expends no fuel.

Leaving you only acceleration to consider for fuel use and the values for different accelerations (1m/s2, 2m/s2, 4m/s2 etc) do not have the same fuel use and it is not simply double fuel as you double acceleration.



Note that D. is trying to leave out the peak acceleration forces, just like he is doing now, and S. seems to be doing the same thing, but as they said, you HAVE to get to that speed, which must involve acceleration, and acceleration, involves more force, thus the peak acceleration forces.

D. got the energy used wrong over a year ago, and now gets the forces involved wrong, but only because he was looking at it the wrong way, and not adding all the forces in, easy done I suppose.


Wayne

 

[sophiecentaur]

Could anyone explain to me just what this thread is trying to do?
Wayne will, clearly, never take on board the Physics and is only interested in pointing out any inconsistency he can find in the (mostly) good Science. I can't think just what he really wants out of this - certainly not to learn any Physics. Yet the thread drags on and on. Can't it just be put to rest?

 

[Dale]

So could we all stay on question 1 for a while please.

So D. and S. it seems you worked the fast reps out, from the lifted that started on the ground, and was lifted up in .5 of a second and lowered in .5 of a second, = 1 second for the total lift, you then multiplied this by 6 = 6 seconds of lifting, as this is the same time frame as the one slow lift, 3 seconds up and 3 seconds down.

But you forgot to add in the peak forces, or the peak acceleration forces, the MMMTs, {Momentary Maximum Muscle Tension} the huge forces from the transition from negative to positive, and in 6 reps or 6 seconds there are 5 of these.

So I think, that if you add in the peak acceleration forces, too which are 25% higher, you x 25% by 5 = in 6 seconds the faster reps used 125% more impulse/force then the 1 slow rep in the same time frame. Could you comment please dalespam.

OK, question 1 was about impulse for constant forces, I don't think we are quite ready to discuss changing forces. If you would please go ahead and answer the four questions I posed earlier in post 281 for non constant forces. That way I can see where you are in your conceptualization of impulse.

 

[douglis]

But you forgot to add in the peak forces, or the peak acceleration forces, the MMMTs, {Momentary Maximum Muscle Tension} the huge forces from the transition from negative to positive, and in 6 reps or 6 seconds there are 5 of these.

Wayne

What kind of nonsense is that?
We talk about the average value of force for the exact reason that we recongnise that there're peak high and low fluctuations of force.Otherwise we would only talk about a constant velue of force.
BTW...the MMMTs are exctly the peak high of force.Those two are the same thing.The MMMTs are not an exotic and mysterious additional force.

Anyway...if I were you...I would shut up right now and I would take advantage of DaleSpam's superhuman patience to help you learn basic physics.

 

[waynexk8]

So stop your endless rambling and try to learn something.



With 100N

1)you do 1 rep with .5/.5
What's the average force?What's the impulse?

2)you do 6 reps with .5/.5
What's the average force?What's the impulse?

3)you do 1 rep with 3/3
What's the average force?What's the impulse?

Tell me please D. ?

Wayne

 

[waynexk8]

This thread is not to try and mock. The only reason for the above, as it seems, 1, you don’t understand the questions, which I find odd. 2, You do understand the questions, but will not answer as it will prove you wrong, that’s why you are not answering and trying to spoil the thread.

Could anyone explain to me just what this thread is trying to do?

We are debating which person uses the most total/overall force per time.

Let me explain one more time one of the questions you will not answer.

I use or try and use a 100 pounds of force for 6 seconds, that = 100 force x 6 seconds of time, let’s call that 100f/6t. You use 80 pounds of force for 6 seconds, that = 80 force x 6 seconds of time, let’s call that 80f/6t.

Fast rep = 100f/6t

Slow rep = 80f/6t

Please tell me, when we all agree that I use 100 force for 6 seconds, and you use 80 force for 6 seconds, how/why you think 80 is as high as 100 ? When 100 = 25% more than 80 and/or 20 more units than 80 ? And yes we both accelerate, and decelerate.

It’s like saying, I used 100 Joules of energy in 6 seconds, and you used 80 Jules of energy in 6 seconds, then you turn around and say you used the same amount of energy, “please” just tell me why you think this ?

Wayne will, clearly, never take on board the Physics

I AM a Man of physics, you have to believe in physics, it’s how the universe and life works or is explained, I can here to ask questions and to have them answered, I showed you that a EMG machines shows that the average force is higher, I showed you that you fail roughly 50% faster with these variables, I showed you that I move the weight 12m to your 2m, I showed you that you use more energy in the fast reps, this is all physics.

All you do is not answer.

and is only interested in pointing out any inconsistency he can find

Of course I will point out inconsistencies, and I would in all walks of life, and “if” you find any in what I say is you think wrong, please say.

in the (mostly) good Science.

What science are you thinking of, what do you think I have been showed ?

I asked a question, you and D. did/could not answer, dalespam did, I was right in what I said ?
For question 1 you are correct, it is 25% more impulse. Please how do you say that I did not say good science when I was right, it was not even a hard question, it was quite obvious

I can't think just what he really wants out of this - certainly not to learn any Physics. Yet the thread drags on and on. Can't it just be put to rest?

An answer, why can't we just debate and stay frendely.

Wayne

 

[waynexk8]

OK, question 1 was about impulse for constant forces, I don't think we are quite ready to discuss changing forces. If you would please go ahead and answer the four questions I posed earlier in post 281 for non constant forces. That way I can see where you are in your conceptualization of impulse.

Sorry I m issed this one, well not missed forgot to get back to it.

I am not saying that at all. The second law is correct; you just have some misunderstandings. We will get to those in a bit. Right now we need to focus on the progress that we have made and flesh out the concept of impulse.

OK, so now that we have confirmed that your concept of "total force" is the same as the standard physics concept of "impulse" I expect you to not write the words "total force" any more and to use the correct term "impulse". You may think that this is nitpicky, but there is a very important reason for doing this. Impulse has units of momentum, and not units of force, so the term "total force" is not only non-standard but incorrect. Something that does not have units of force cannot be any kind of force, let alone a "total force". Do you agree to this?

Ok, total/overall force = impulse, question first before I use it, impulse is the integral/whole of a force with respect to time ? I am sure it is, but best ask.

So, if you exert a constant 100 lb for 10 s then you have exerted an impulse of 1000 lb*s. And if you exert a constant 80 lb for 10 s then you have exerted an impulse of 800 lb*s. If you drew a force v time curve as I did above then each of these would be a simple straight flat line. And in both cases we would get the impulse by calculating the area under the curve (area of a rectangle is base times height which in this case is the time times the force).

OK, so now let's work a couple of problems to solidify the concept of impulse. Please show your work:

1) What is the impulse if you exert a 100 lb force for 5 s and then 80 lb for an additional 5 s?

900lb/10s

2) What constant force would give the same impulse as in 1) if exerted over 10 s?

90lb. 90 x 10 = 900

3) What is the impulse for the attached force vs time graph? (Hint: remember that the impulse is the area under the graph which is shaded in this graph and also remember that the area of a triangle is 1/2 base times height.)

Not sure why it’s a triangle, or the force curve is going down as the time goes on, it seems to me as you start using a 100 force, and then the force you use goes down and down the more seconds you are trying to apply an force ? Or is it a 100 force for 10 seconds 1000lb/10s ?

4) What constant force would give the same impulse as in 3) if exerted over the same amount of time?

If its 1000/10s it would be 100.

The graph is plotted in Mathematica using the following code:

y[t_] := 100 - 10 t;
Plot[y[t], {t, 0, 10}, Frame -> True, 
 FrameLabel -> {Style["Time (s)", Larger], 
   Style["Force (lb)", Larger]}, 
 PlotLabel -> Style["Force vs Time", Larger], Filling -> 0]

Slow rep,
So if we used 80 force for 10 seconds – off 1 second for the deceleration and call that second zero, it would be 800f/9s.

Fast reps,
So if we used 100 force for 10 seconds – off 1 second for the deceleration and call that second zero, it would be 1000f/9s. 20 more impulse in 10 seconds, or/and 25% more impulse in 10 secondsHowever, now we have left out the forces “on” the muscles, the forces from the transition from negative to positive {can you work this out ?} these forces for say a quarter of a second, can be as high as say ? 140% or more, so as there are 9 of these in 10 seconds, that = so and so force that we have to add onish, for so and so time, as this will put far far far more tension on the muscles, and this is the crux of this debate.

Wayne

 

[Dale]

Ok, total/overall force = impulse, question first before I use it, impulse is the integral/whole of a force with respect to time ? I am sure it is, but best ask.

Yes, impulse is the integral of a force with respect to time.

900lb/10s

This is incorrect, but I can see that you are heading in the right direction. Remember that division and multiplication are different operations, so 900lb/10s = 90 lb/s. This is not even in the correct units for impulse which is in units of force*time (lb*s) not force/time (lb/s). Do you see the difference?

So the way you calculate this is as follows:
(100 lb * 5 s) + (80 lb * 5 s) = (500 lb*s) + (400 lb*s) = 900 lb*s

Do you see the difference between 900 lb*s and 900lb/10s?

90lb. 90 x 10 = 900

Correct, but be sure to always put the units in correctly. The impulse is 900 lb*s and the force was applied for 10 s total. The constant force that would produce the same impulse is 90 lb since, as you said, 90 lb * 10 s = 900 lb*s.

Not sure why it’s a triangle, or the force curve is going down as the time goes on, it seems to me as you start using a 100 force, and then the force you use goes down and down the more seconds you are trying to apply an force ?

Yes, the person is starting exerting 100 lb force, but over the course of 10 s the force decreases linearly to 0 lb. I.e. somebody steadily getting weaker over the 10 s, unable to sustain the inital force of 100 lb.

Or is it a 100 force for 10 seconds 1000lb/10s ?

Remember the area of a triangle is 1/2 * base * height. In this case the triangle has a height of 100 lb and a base of 10 s, so the area (impulse) is:
1/2 * 10 s * 100 lb = 500 lb*s

If its 1000/10s it would be 100.

Based on the above, why don't you try this one again?

Do you understand the corrections I have made above, particularly the difference between division and multiplication and the correct use of the units?

 

[sophiecentaur]

. . . . . . .

We are debating which person uses the most total/overall force per time.

. . . . . . .

An answer, why can't we just debate and stay frendely.

Wayne

The problem is that "total/overall force per time" is not a quantity that is defined in Physics. You have continually ignored this and still use your own terms. You seem to want it all your own way. You want people to 'explain' real- physics things in your special, non-Physics terms (which in itself is logically impossible, of course) you leap on small inconsistencies in explanations as if that proves that your ideas must be right.

There is no meaning in the expression "total/ overall force" and adding "per unit time" doesn't help either. I can understand that you may want to relate knackeredness (units unspecified) to your various weightlifting situations but isn't that what your machine does?

We can discuss Force, Energy, Work, Momentum and a number of other things too but only when the words are assembled is a reasonable and meaningful order within a sentence. You have railroaded douglis and others into using your non-Physics because they seem to think they can lead you onto the straight and narrow path of real-Physics. I seriously doubt that you want to use that path. If you did, then you would be prepared, at least, to use the right words where they apply. The fact is that, if you actually were to try to do that, what you are being told regularly, would make sense to you and you might get somewhere with your quest.

You keep referring to a "debate". What debate? They tell you some universally held views about Physics and you describe endless weight lifting situations. The two 'sides' seem to be discussing two separate things.

I don't think you are acknowledging just how patient and tolerant people have actually been.

 

[waynexk8]

So stop your endless rambling and try to learn something.



With 100N

1)you do 1 rep with .5/.5
What's the average force?What's the impulse?

2)you do 6 reps with .5/.5
What's the average force?What's the impulse?

3)you do 1 rep with 3/3
What's the average force?What's the impulse?

You asked three question above, I said please tell me. Not only can you not answer my questions, you cannot answer your own, or did you put these for the physicists here to answer as you were asking them ?

Wayne

 

[sophiecentaur]

This question has been answered many times on this thread but you have refused to acknowledge it because you would have to climb down about everything you have been saying.
Instead of coming back with a further question, Wayne, just look at the 300+ posts and FIND the answer. (Clue: it is a very simple answer)

 

[waynexk8]

Yes, impulse is the integral of a force with respect to time.

We all agree on this then, that’s great.

if you exert a constant 100 lb for 10 s then you have exerted an impulse of 1000 lb*s. And if you exert a constant 80 lb for 10 s then you have exerted an impulse of 800 lb*s. If you drew a force v time curve as I did above then each of these would be a simple straight flat line. And in both cases we would get the impulse by calculating the area under the curve (area of a rectangle is base times height which in this case is the time times the force).

OK, so now let's work a couple of problems to solidify the concept of impulse. Please show your work:

1) What is the impulse if you exert a 100 lb force for 5 s and then 80 lb for an additional 5 s?

900lb/10s

This is incorrect, but I can see that you are heading in the right direction. Remember that division and multiplication are different operations, so 900lb/10s = 90 lb/s. This is not even in the correct units for impulse which is in units of force*time (lb*s) not force/time (lb/s). Do you see the difference?

SORRY, you got the / wrong, or not wrong, you misinterpreted it, maybe my fault, or I put it down wrong, maybe it was my mistake putting the divided sign, sorry there, thought you would have seen what I meant, as in this instance it’s not meant as divided, it just meant 900lb for 10 seconds. As we had 100lb force for 5s = 500lb for 5 seconds, then we had 80lb for 5 seconds = 400lb for 5 seconds. So we have 500 for 5 and 400 for 5, = 900 for 10, so I put 900lb/10s, maybe I should have put it like this; 900lb * 10s.

So the way you calculate this is as follows:
(100 lb * 5 s) + (80 lb * 5 s) = (500 lb*s) + (400 lb*s) = 900 lb*s

Do you see the difference between 900 lb*s and 900lb/10s?

Yes I see the difference, so the answer is as I said 900lb * 10s yes ?

Correct, but be sure to always put the units in correctly. The impulse is 900 lb*s and the force was applied for 10 s total. The constant force that would produce the same impulse is 90 lb since, as you said, 90 lb * 10 s = 900 lb*s.

Right.

Yes, the person is starting exerting 100 lb force, but over the course of 10 s the force decreases linearly to 0 lb. I.e. somebody steadily getting weaker over the 10 s, unable to sustain the initial force of 100 lb.

Agreed, BUT, in this, of the two examples we are looking at, the persons do not get weaker in the first example of the 6 seconds on the fast and slow reps. The fast uses exactly 100 force for the first 6 seconds on the fast, with doing 6 reps at .5/.5 = 6 seconds, as they are both using 80% or 80 pounds. The slow uses exactly 80 force for the first 6 seconds on the slow, with doing 1 reps at 3/3 = 6 seconds.

That was my initial point years ago, both have a maximum of 100 pounds force, and both use 80% 80 pounds for the reps, and so the fast uses 100 pounds of force to move the 80 pounds up and down for 6 seconds, and the slow uses just 80 pounds of force to move the 80 pounds up and down for 6 seconds.

1,
Fast uses 100 pounds of force on 80 pounds for 6 seconds.

2,
Slow uses 80 pounds of force on 80 pounds for 6 seconds.

3,
Fast uses 25% or/and 20 pounds more force for 6 seconds, “HOW” can the lower force of 80 pounds used for 6 seconds be as high as 100 pounds used for 6 seconds ?

Second example was, fast does 24 reps with the same weight and force as before, at .5/.5 = 24 seconds, then hits momentary muscular failure. Slow does 4 reps with the same weight and force as before, at 3/3 = 24 seconds, and stops, but has “not” modernity muscular failure.

Remember the area of a triangle is 1/2 * base * height. In this case the triangle has a height of 100 lb and a base of 10 s, so the area (impulse) is:
1/2 * 10 s * 100 lb = 500 lb*s

Based on the above, why don't you try this one again?

Right get you, we are only looking at the half triangle. But you just told me the answer ? As you say, you half the ten seconds, = 5 seconds at 100 pounds, then it’s a 100 x 5 = 500.

Do you understand the corrections I have made above, particularly the difference between division and multiplication and the correct use of the units?

Yes I understand the graph now; however I did understand the other examples, but put them down a little different than I should then.

So could we clear this up a little more please.

1,
Fast rep = 100 force x 10 seconds = 1000lb * 10s.

2,
Slow rep = 80 force x 10 seconds = 800lb * 10s.

3,
So I was right all along ! So what are D and S. on about ?

Wayne

 

[sophiecentaur]

Read my last post again.
You are thrashing about all around the topic without taking anything on board, it seems. You have been given answers which involve the strict definitions of the quantities involved. You are choosing not to read them properly and ignore the vital differences between what you read and how you choose to interpret. If you want to join the club then you have to obey the rules.

 

[waynexk8]

This question has been answered many times on this thread but you have refused to acknowledge it because you would have to climb down about everything you have been saying.

No, you will have to climb down from all you said, you was wrong, I was right, please explain what you are thinking ? How and why do you think you were right, and how do you think I was wrong ?

1,
As I said, the EMG proved me right,

2,
Moving the weight 10m more in the same time frame proved me right,

3,
Using more energy in the fast proved me right,

4,
As you fail with the fast 50% faster proved me right,

5,
I used 100 force moveing 80 for 6 seconds, you used 80 force moving 80 for 6 seconds, 100 is more than 80, the 100 = 20 more than 80, 100 = 25% more than 80, that proved me right.

I could go on, but I don't understand what you are now on abouyt ?

Instead of coming back with a further question, Wayne, just look at the 300+ posts and FIND the answer. (Clue: it is a very simple answer)

Unsure what you mean now ? As some of you said I was wrong, but I was not ? You and D. and some others said that in the examples/s, that both rep speeds used the same, as I called then overall/total force, but as I know I should have called it impulse, BUT both rep speeds over the same time frame did not use the same impulse, the faster reps used “more” impulse, just as I said all along.

Several of you were wrong, and I was right, it was so simple I don’t understand how D. and some others over at BB.com and a few other sites, could even think I was wrong. But the best part about this as I have learned a very lot.

Over 5 years ago I started doing the heavier weights with the very explosive reps, added about 40 of pounds to my bodyweight, 30 muscle and 10 fat, in the last year or more added another 8 pounds, and lost some fat. No one will add much doing slow reps, as you come to sticking points in about 6 months, sorry going on.

Wayne

 

[sophiecentaur]

Your EMG machine agrees fine with what you feel when you do exercise and it helps you improve your performance. That's fair enough. But it doesn't tell you anything about Work-Done-On the weights or the true average force during a cycle. It just doesn't Care. It doesn't even appear to Claim to. EMG speak is not Physics. I have no argument with EMG and I think that is because I understand what they are saying and how it sits with respect to real Physics.
I don't think you have understood the glaring difference between the two worlds. It suits you not to because you like this pseudo Physics of yours.
I think I've said this before but go and enjoy your body building. It sounds great fun (and hard work) and you are successful. Isn't that enough?
If you want to do Physics then take a college course and learn it properly. That's fun too - and hard work, because you have to actually learn some mental and verbal discipline.

 

[douglis]

Unsure what you mean now ? As some of you said I was wrong, but I was not ? You and D. and some others said that in the examples/s, that both rep speeds used the same, as I called then overall/total force, but as I know I should have called it impulse, BUT both rep speeds over the same time frame did not use the same impulse, the faster reps used “more” impulse, just as I said all along.

Several of you were wrong, and I was right, it was so simple I don’t understand how D. and some others over at BB.com and a few other sites, could even think I was wrong.

Wayne

Oh God...just follow DalleSpam's lessons.He's trying to teach you the meaning of impulse and to show you why it's the SAME regardless the rep speed.

 

[waynexk8]

D. “please” answer your questions ? You not only can not answer and debate against me, you can not even answer you own question, you also can NOT even say what you point is on them ?

With 100N

1)you do 1 rep with .5/.5
What's the average force?What's the impulse?

2)you do 6 reps with .5/.5
What's the average force?What's the impulse?

3)you do 1 rep with 3/3
What's the average force?What's the impulse?

Oh God...just follow DalleSpam's lessons.He's trying to teach you the meaning of impulse and to show you why it's the SAME regardless the rep speed.

What are you talking about now, read the below.

DaleSpam wrote; if you exert a constant 100 lb for 10 s then you have exerted an impulse of 1000 lb*s. And if you exert a constant 80 lb for 10 s then you have exerted an impulse of 800 lb*s.

The only problem in the above, is that the slow may exerted a constant {lets just say constant for now, as its quite close}force for say 80% of the rep, say 10% for acceleration, and 10% for deceleration. However the fast will accelerate for say 80% and decelerate for 20%

Also we have “still” left out the force “on” the muscles from the transition from negative to positive, as in the above we have only worked out the force exerted by the muscles.


1,
Fast rep = 100 force x 10 seconds = 1000lb * 10s.

2,
Slow rep = 80 force x 10 seconds = 800lb * 10s.

3,
So I was right all along ! So what are D and S. on about ?

Your EMG machine agrees fine with what you feel when you do exercise and it helps you improve your performance. That's fair enough. But it doesn't tell you anything about Work-Done-On the weights or the true average force during a cycle. It just doesn't Care. It doesn't even appear to Claim to. EMG speak is not Physics. I have no argument with EMG and I think that is because I understand what they are saying and how it sits with respect to real Physics.
I don't think you have understood the glaring difference between the two worlds. It suits you not to because you like this pseudo Physics of yours.
I think I've said this before but go and enjoy your body building. It sounds great fun (and hard work) and you are successful. Isn't that enough?
If you want to do Physics then take a college course and learn it properly. That's fun too - and hard work, because you have to actually learn some mental and verbal discipline.

An EMG works by physics, it does NOT just agree with what I feel, as the funny thing here is, the very slow reps are very hard physically and mentally, as of the lactic acid build up they feel harder give more pain, so thus shoots your debate.

I would “NEVER” talk pseudo Physics, now I know total/overall is impulse I say that, I am a Man of science, take a look at my other videos of my wrought Iron, these were designed and fabricated by me, I walk though life with science not against it. You again fail to answer any of my questions or debate against what I say, or put up any equations or debate for your way of thinking ? Just take 4, it should be quite obvious that you use more force, as there is more tension being put on the muscles per unit of time as you fail faster, thus more tension more impulse/force.

2,
Moving the weight 10m more in the same time frame proved me right,

3,
Using more energy in the fast proved me right,

4,
As you fail with the fast 50% faster proved me right,

5,
I used 100 force moving 80 for 6 seconds, you used 80 force moving 80 for 6 seconds, 100 is more than 80, the 100 = 20 more than 80, 100 = 25% more than 80, that proved me right.

6,
a,
Fast rep = 100 force x 10 seconds = 1000lb * 10s.

b,
Slow rep = 80 force x 10 seconds = 800lb * 10s.


More later.


Wayne

 

[douglis]

What are you talking about now, read the below.


The only problem in the above, is that the slow may exerted a constant {lets just say constant for now, as its quite close}force for say 80% of the rep, say 10% for acceleration, and 10% for deceleration. However the fast will accelerate for say 80% and decelerate for 20%

Also we have “still” left out the force “on” the muscles from the transition from negative to positive, as in the above we have only worked out the force exerted by the muscles.


1,
Fast rep = 100 force x 10 seconds = 1000lb * 10s.

2,
Slow rep = 80 force x 10 seconds = 800lb * 10s.

3,
So I was right all along ! So what are D and S. on about ?

Wayne

...he just gave you an example on how to calculate the impulse when the force is constant...that doesn't mean that those values represent fast and slow reps!

D. “please” answer your questions ? You not only can not answer and debate against me, you can not even answer you own question, you also can NOT even say what you point is on them ?

With 100N

1)you do 1 rep with .5/.5
What's the average force?What's the impulse?

2)you do 6 reps with .5/.5
What's the average force?What's the impulse?

3)you do 1 rep with 3/3
What's the average force?What's the impulse?

Here's what required 21 pages of nonsense!

1)avg force=100N impulse=100N*s

2)avg force=100N impulse=600N*s

3)avg force=100N impulse=600N*s

So the big mystery is now solved.
6 reps with .5/.5 have the same impulse with 1 rep with 3/3!

 

[Dale]

SORRY, you got the / wrong, or not wrong, you misinterpreted it, maybe my fault, or I put it down wrong, maybe it was my mistake putting the divided sign, sorry there, thought you would have seen what I meant, as in this instance it’s not meant as divided, it just meant 900lb for 10 seconds. As we had 100lb force for 5s = 500lb for 5 seconds, then we had 80lb for 5 seconds = 400lb for 5 seconds. So we have 500 for 5 and 400 for 5, = 900 for 10, so I put 900lb/10s, maybe I should have put it like this; 900lb * 10s.

Oh my. 90 lb for 10 s is 900 lb*s, not 900 lb for 10 s. 100 lb for 5 s is 500 lb*s not 500 lb for 5. You should not put 900 lb * 10 s (9000 lb*s) because that is off by a factor of 10 from the 90 lb * 10 s (900 lb*s) specified in the problem.

Here are a few sites on doing calculations with units:
http://en.wikipedia.org/wiki/Units_of_measurement#Calculations_with_units
http://www.chem.ox.ac.uk/teaching/Physics%20for%20CHemists/Units.html
http://www.chem.ox.ac.uk/teaching/Physics%20for%20CHemists/Units/Calculations.html

You NEED to know this if you want to do any physics at all.

Yes I see the difference, so the answer is as I said 900lb * 10s yes ?

No, the answer is 900 lb*s which is not 900 lb * 10 s = 9000 lb*s.

Agreed, BUT, in this, of the two examples we are looking at

The purpose of this is not to provide an accurate force profile for weight lifting, but just to teach you how to calculate impulse in general by looking at a force v. time diagram. I am not sure that you have that correct yet.

The fast uses exactly 100 force for the first 6 seconds on the fast

This is physically impossible for a human to do for reasons that we will get to next.

Right get you, we are only looking at the half triangle. But you just told me the answer ? As you say, you half the ten seconds, = 5 seconds at 100 pounds, then it’s a 100 x 5 = 500.

OK, here is another place where you really need to specify the units. 1/2 * 100 lb * 10 s = 100 lb * 5 s = 500 lb*s. So the question is what constant force (units of lb), applied for 10 s, gives 500 lb*s impulse?

Yes I understand the graph now; however I did understand the other examples, but put them down a little different than I should then.

I am not sure that you do, the correct use of units is not an optional part of physics. It looks like you may not be familiar with using units. Please go over the links above and let me know if it makes sense to you. If so, then I will go over some new impulse problems. As it is I cannot tell if you are actually understanding the concept of impulse or if you just think that you understand it.

So I was right all along ! So what are D and S. on about ?[/b]

No, douglis and sophiecentaur are correct. I am trying to teach you step-by-step why they are correct.

The bottom line is that there is not a simple "external" physics quantity (that I know of) which corresponds well to the internal biological state of muscle fatigue. You seem to think that the impulse is such a quantity, so I am trying to teach you the properties that impulse actually has so that you can understand that it does not have the properties you think.

 

[waynexk8]

...he just gave you an example on how to calculate the impulse when the force is constant...

Yes we all know it’s an example.

that doesn't mean that those values represent fast and slow reps!

“WHAT” do you mean, if you use 80 it is a SLOW rep, and if you use a 100 it’s a fast rep, it does and can only represent fast and slow reps. What ELSE do you think it represents ? I mean we all agree that the slow uses 80 and the fast 100.

1,
Fast rep = 100 force x 10 seconds = 1000lb * 10s.

2,
Slow rep = 80 force x 10 seconds = 800lb * 10s.

Here's what required 21 pages of nonsense!

1)avg force=100N impulse=100N*s

2)avg force=100N impulse=600N*s

3)avg force=100N impulse=600N*s

So the big mystery is now solved.
6 reps with .5/.5 have the same impulse with 1 rep with 3/3!

This is a physics forum; you “NEED” to say how you came to those numbers. And then when you do, if you could then add the forces “ON” the muscles from the transition from negative to positive.

You never and need to explain your way of thinking.

Please explain from me using 125 on a 100 do you get that the average force = 100 and then explain when you use 100 on a 100 that your average force = 100.

Tell me something then, “HOW” am I ONLY producing 100N ? When moving a 100N, I am using as we both agreed, 25% more than you, this = I am using 125N NOT 100N, you are the one using a 100N, I use more force than you, you HAVE to use more force to move the same weight in the same time frame 10m more.

Wayne

 

[waynexk8]

Oh my. 90 lb for 10 s is 900 lb*s, not 900 lb for 10 s. 100 lb for 5 s is 500 lb*s not 500 lb for 5. You should not put 900 lb * 10 s (9000 lb*s) because that is off by a factor of 10 from the 90 lb * 10 s (900 lb*s) specified in the problem.

Correct, but be sure to always put the units in correctly. The impulse is 900 lb*s and the force was applied for 10 s total. The constant force that would produce the same impulse is 90 lb since, as you said, 90 lb * 10 s = 900 lb*s.

Sorry that was as I how you will see just writing, or more like a hurrying mistake, as you x the 90 by 10 = 900, I was just in a hurry sorry.

Yes, 90lb for 10 s is 900lbs

Here are a few sites on doing calculations with units:
http://en.wikipedia.org/wiki/Units_of_measurement#Calculations_with_units
http://www.chem.ox.ac.uk/teaching/Physics%20for%20CHemists/Units.html
http://www.chem.ox.ac.uk/teaching/Physics%20for%20CHemists/Units/Calculations.html

You NEED to know this if you want to do any physics at all.

Will take a look at the sites, thx.

No, the answer is 900 lb*s which is not 900 lb * 10 s = 9000 lb*s.

Yes sorry did it again, as you say it’s, 100lb x 5s = 500lb + 80 lb x 5s = 400lb 400lb + 500lb = 900lb

The purpose of this is not to provide an accurate force profile for weight lifting, but just to teach you how to calculate impulse in general by looking at a force v. time diagram. I am not sure that you have that correct yet.

Yes I do get you; I was just in a hurry.

This is physically impossible for reasons that we will get to next.

YES I do understand that, one, it’s something you might not know, but the force/velocity curve states you cannot use 100% force on a weight as light as 80% to use 100, you need to be trying to move 100% but I don’t think you meant that.

As let’s just say I was moving 80% weight with a 100% force. I would be trying to use a 100% until the deceleration, and on the deceleration you will be using less force than the weight of 80%. I think that’s what you were getting at. This is also true with the slow rep, as when trying to move 80% weight with 80% force, you still must use less force than the 80%.

OK, here is another place where you really need to specify the units. 1/2 * 100 lb * 10 s = 100 lb * 5 s = 500 lb*s. So the question is what constant force (units of lb), applied for 10 s, gives 500 lb*s impulse?

Not get you there sorry. This is what it looks like to me the way you wrote it this time, you wrote, and this s what I think it says; 1/2 * 100 lb * 10 s = 100 lb * 5 s = 500 lb*s. 1/2 x 100 lb = 50 x 10 s = 500 lb x 5 s = 2500 lbs.

Until I get that right, I can’t work out the constant; mind you the fast rep is not constant ? It’s more or less acceleration for 80% so the fast and slow look likes this roughly.

Fast = 100 force for 8 seconds = 800, we need to work out the deceleration force, then add on the force not from the muscles, but on the muscles from the several transitions from negative to positive

Slow = 80 force for 8 seconds = 640; we need to work out the deceleration force.

I am not sure that you do, the correct use of units is not an optional part of physics. It looks like you may not be familiar with using units. Please go over the links above and let me know if it makes sense to you. If so, then I will go over some new impulse problems. As it is I cannot tell if you are actually understanding the concept of impulse or if you just think that you understand it.

No, douglis and sophiecentaur are correct. I am trying to teach you step-by-step why they are correct.

But they have no theory evidence or proof or practical, I have both and then cannot, not only not counter it, but can’t even answer it.

The bottom line is that there is not a simple "external" physics quantity (that I know of) which corresponds well to the internal biological state of muscle fatigue. You seem to think that the impulse is such a quantity, so I am trying to teach you the properties that impulse actually has so that you can understand that it does not have the properties you think.

Seems like this is a far better debate, and thank you.

Only, you don’t seem to say why you think you use the same impulse in the same time frame using the same weight ? I have stated these before, or would you like to stay on the above first ? Let us recap.

Also we have “still” left out the force “on” the muscles from the transition from negative to positive, as in the above we have only worked out the force exerted by the muscles.


1,
Fast rep = 100 force x 10 seconds = 1000lb for 10s. – off something for decelerations.

2,
Slow rep = 80 force x 10 seconds = 800lb for 10s. - off something for decelerations.


Also we have “still” left out the force “on” the muscles from the transition from negative to positive, as in the above we have only worked out the force exerted by the muscles.


When can we add these on ?

Question 1,
This is what I have been saying all along, I use exert 100 lbs for 10 s and you exert 80 lbs for 10 s I have exerted 25% more the impulse that I have. Why when we agree that you have used 80 and me a 100 do you think after that 80 = 100 ?




Question 2,
What forces do you think you have that can make up of balance out the higher propulsive forces of the fast in the studies ?

Let’s take the mean propulsive forces, slow 6.2mean in 10.9 seconds. Fast 45.3mean 2.8 seconds, now let’s divided the mean slow of 10.9 seconds by the fast 2.8 = 3.8, so now let’s divide the slow mean by 3.8 = 1.6.

Fast mean for 2.8 seconds = 45.3.

Slow mean for 2.8 seconds = 1.6.

The fast has nearly 3000% more mean propulsive force as in N's in the same time frame.

Please what forces have I left out that the slow has to make up or balance out these ?

http://www.jssm.org/vol7/n2/16/v7n2-16pdf.pdf


Question 3,

Which has the highest force, and which are you saying has the same force as the slow rep for the same time frame, 1 or 2 ?

I was just wondering and thinking, ARE you adding “all” the force, I mean with the fast there are NOT just force being exerted by the muscles, there are HUGE forces on them, for say .1 of a second x the 6 reps = high forces on the muscle for maybe .6 of a second. Have you added these in ?

I mean the peak force from the transition from negative to positive, the force on the muscle, NOT given out by the muscles ? We call them the MMMTs {Momentary Maximum Muscle Tensions} these forces “ON” the muscles can be as high as 140%

Have you added these on ?

Lift 1,
You lift 80% of the ground, up 1m and then down 1m all in 1 second, .5/.5

Lift 2,
You start at the top, lower the weight down 1m, and then lift it back up 1m all in 1 second, .5/.5

On lift 2, on the transition from negative to positive, there will be huge force on the muscles and coming from the muscles.

Which of the 2 forces on the 2 faster reps are you saying that’s the same as the slow rep for the same time frame ? As lift 1 and 2 use a different amount of force.



Question 4,
Fast reps person trys to use 100 pounds of force moving 80 pounds object for 6 seconds = 100/6

Slow reps person uses 80 pounds of force for 6 seconds = 80/6

How can 80/6 = or be as high as 100/6 ?
How can you trying to use 80 pounds of force for 6 seconds, be as high or the same as me using 100 pounds of force for 6 seconds ? HOW can 80 foirce be as high as 100 force ? As I always thought a 100 was 20 more then 80, or/and 25% more than 80, but you are saying that 80 = 100 ? Please for once explain how 80 = 100 ?

Wayne

 

[sophiecentaur]

I would “NEVER” talk pseudo Physics, now I know total/overall is impulse I say that, I am a Man of science, take a look at my other videos of my wrought Iron, these were designed and fabricated by me, I walk though life with science not against it. You again fail to answer any of my questions or debate against what I say, or put up any equations or debate for your way of thinking ? Just take 4, it should be quite obvious that you use more force, as there is more tension being put on the muscles per unit of time as you fail faster, thus more tension more impulse/force.

Your Physics really is pseudo Physics because you neither use the right terms, nor do you obey the rules for discussing the relationships between the quantities. To be dealing in Real Physics you have to be doing both those things. The fact is that you are not prepared to state a question in any way other than in the form of an endless story about an imaginary weight lifting session. Physics goes beyond the specific and reduces things to the simplest possible form - usually with Maths. This is what you will not do so you are not actually taking part in real Physics at all. You are not a "man of Science". You are a craftsman and an accomplished expert(?) on lifting weights. They are not Science. There is no value judgement in my statement but you prove in everything you write that you are not really interested in the true scientific approach. If you were, you would have modified what you say and how you say it, by now and taken on board what you've been (so patiently) told.
You wouldn't recognise a correct equation if it got up and bit you on the leg - you have proved this by ignoring / rejecting the correct answers you have been given time after time.
I say again: Debate? What Debate? It's just two or three people banging their heads against a brick wall.

 

[Dale]

Sorry that was as I how you will see just writing, or more like a hurrying mistake, as you x the 90 by 10 = 900, I was just in a hurry sorry.

Yes, 90lb for 10 s is 900lbs

No, 90 lb for 10 s is 900 lb*s, not 900 lb!

Yes sorry did it again, as you say it’s, 100lb x 5s = 500lb + 80 lb x 5s = 400lb 400lb + 500lb = 900lb

No, no.
100 lb * 5 s = 500 lb*s
80 lb * 5 s = 400 lb*s

While it is true that
400 lb + 500 lb = 900 lb
it is completely irrelevant to the problem. What is relevant to the problem is:
400 lb*s + 500 lb*s = 900 lb*s

Yes I do get you; I was just in a hurry.

Please read those links I gave you very carefully and ask questions. It is clear that you do not get the units, and this is very important in physics.

1/2 x 100 lb = 50 x 10 s

Look at the units I bolded. Pounds and seconds are different types of quantities. One is a unit of force and the other is a unit of time. There is simply no way whatsoever to make any number of pounds equal to any number of seconds.

I have stated these before, or would you like to stay on the above first ?

We need to stay on the above until you get the units of force (lb), time (s), and impulse (lb*s) straightened out. Again, please read those links and come back with questions.

PS you may think that this is being overly picky, but it is not. I still remember it vividly. My very first lecture of my very first physics class, the professor spent the entire class going over units and emphasizing the importance of always correctly handling the units. He claimed that checking your units was the easiest way to catch the large majority of physics mistakes, and my decades of physics experience since has proven him right again and again. I cannot overemphasize the importance of correct units if you want to do physics at all.

 

[Dale]

Hi waynexk8,

After you read the links try the following problems:
1) 100 lb * 5 ft =
2) 20 ft*lb / (2 ft) =
3) 32 ft/s² * 10 s =
4) 5 ft/s * 320 lb / (32 ft/s²) =

Do not rush or hurry this, make sure that you get the units correct. This is important.

 

[waynexk8]

No, 90 lb for 10 s is 900 lb*s, not 900 lb!

Yes, that is what I said 900lbs not 900lb all I left out is the s, 900lb = 900lb + 900lbs = 900lbs, they are both the same 900lbs = 900lbs, 900lb = 900lb. It’s the same as 1 stone = 1 stone or 1 stones = 1 stones, there is no difference between 9 stone and 9 stones. 9 stone = 9 stone and 9 stones = 9 stones. How can you say 9 stone and 9 stones = different ?

No, no.
100 lb * 5 s = 500 lb*s
80 lb * 5 s = 400 lb*s

That’s what I said; 100 x 5 = 500, and 80 x 5 = 400. We both said the same thing ?

While it is true that
400 lb + 500 lb = 900 lb

it is completely irrelevant to the problem. What is relevant to the problem is:
400 lb*s + 500 lb*s = 900 lb*s[/quote]

Sorry, not sure what you are getting at, 900lbs = 900, 900lb = 900, what are you saying thinking is different ?

Please read those links I gave you very carefully and ask questions. It is clear that you do not get the units, and this is very important in physics.

Yes I will, but 100 or what ever = 100s unless you state otherwise. How can a 100lb = more than a 100lbs ?

Look at the units I bolded. Pounds and seconds are different types of quantities. One is a unit of force and the other is a unit of time. There is simply no way whatsoever to make any number of pounds equal to any number of seconds.

What do you mean ? If I use 1 pound of force for 1 second, and the use 1 pound of force for 3 seconds, of “course” I have used the same amount of force for longer, as in 1 second first, then 2 seconds.

Look, if I use 1 pound of force for 1 second, then use 1 pound of force for 100 seconds, of course I have used more force, or the same force for a longer time frame. How can you think different ? I

We need to stay on the above until you get the units of force (lb), time (s), and impulse (lb*s) straightened out. Again, please read those links and come back with questions.

Yes will read them.

But I don’t understand you, I use 1 pound of force for 1 second, its more than using 1 pound of force for half a second, how can you think different ?

PS you may think that this is being overly picky, but it is not. I still remember it vividly. My very first lecture of my very first physics class, the professor spent the entire class going over units and emphasizing the importance of always correctly handling the units. He claimed that checking your units was the easiest way to catch the large majority of physics mistakes, and my decades of physics experience since has proven him right again and again. I cannot overemphasize the importance of correct units if you want to do physics at all.

I totally agree with you, but not sure what you are getting at, if you explain, I might see you point.
But if you use or try and use 80 pounds of force for 6 seconds and I use or try and use 100 pounds of force for 6 seconds both moving 80% of the maximum force we can move, HOW can you think or say that 80 = 100 ?

Wayne

 

[waynexk8]

Hi waynexk8,

After you read the links try the following problems:
1) 100 lb * 5 ft =
2) 20 ft*lb / (2 ft) =
3) 32 ft/s² * 10 s =
4) 5 ft/s * 320 lb / (32 ft/s²) =

Do not rush or hurry this, make sure that you get the units correct. This is important.

Not sure what your symbols mean.

Does a 100 lb * 5 ft = 100 lb x 5 ft ? But what does that mean or say ?
Does 20 ft*lb / (2 ft) mean 20 ft*lb divided by (2 ft) ? and what does that mean ?
Does 32 ft/s² * 10 s = mean 32 ft/s² x 10 s and what does that mean ?
Does 5 ft/s * 320 lb / (32 ft/s²) = 5 ft/s x 320 lb / (32 ft/s²) and what does that mean ?


All I am saying is, I try and use 100 pounds of force moving 80 pounds for 6 seconds, the fast, and then you try and use 80 pounds of force moving 80 pounds for 6 seconds the slow, we both agree on this, then you try and turn around and say you used 80 pounds of force for 6 seconds, but your 780 pounds of force somehow = 100 pounds of force ? To me this does and cannot make sense, how can 80 pounds of force used for 6 seconds = 100 pounds of force used for 6 secounds ?

Wayne

 

[Dale]

Not sure what your symbols mean.

Does a 100 lb * 5 ft = 100 lb x 5 ft ?

Yes. Since x is often used as a variable the general practice is to use * for multiplication. So 100 lb * 5 ft means 100 pounds multiplied by 5 feet.

Given that explanation and given the links I posted, please go back and work the problems in 332. Don't worry about what they mean, just see if you can get the right answers.

 

[Dale]

Yes, that is what I said 900lbs not 900lb all I left out is the s, 900lb = 900lb + 900lbs = 900lbs, they are both the same 900lbs = 900lbs, 900lb = 900lb. It’s the same as 1 stone = 1 stone or 1 stones = 1 stones, there is no difference between 9 stone and 9 stones. 9 stone = 9 stone and 9 stones = 9 stones. How can you say 9 stone and 9 stones = different ?

No, you are completely missing the point and consistently getting the units wrong "lb*s" is not the plural of pound (a unit of force), it is a unit of impulse, the "pound second". It is literally 1 pound times 1 second. Your answers above are all wrong because of this critical mistake. 10 lb*s is not at all the same as 10 lb. They are incompatible units. Please read the links and work the problems.

That’s what I said; 100 x 5 = 500, and 80 x 5 = 400. We both said the same thing ?

No, we said very different things. Your units were all wrong. We did not say the same thing any more than 10 miles is the same thing as 10 days.

Do you understand the difference between miles and days? That is the difference between lb and lb*s. They are different units.

Sorry, not sure what you are getting at, 900lbs = 900, 900lb = 900, what are you saying thinking is different ?

The units are different. 900 lb is almost half a ton, it weighs a lot. 900 is a pure number, it weighs nothing. 900 lb ≠ 900.

How can you possibly think that a pure number is the same thing as nearly half a ton? Have you ever heard the expression "comparing apples and oranges"? That is what you are doing wrong, and you are doing it very consistently wrong. We really need to address this before proceeding.

 

[waynexk8]

No, you are completely missing the point and consistently getting the units wrong "lb*s" is not the plural of pound (a unit of force), it is a unit of impulse, the "pound second". It is literally 1 pound times 1 second. Your answers above are all wrong because of this critical mistake. 10 lb*s is not at all the same as 10 lb. They are incompatible units. Please read the links and work the problems.

“RIGHT” get you now, it’s not the plural, it’s the pound second, will rewrite and reanswer the post for you later, and thx, don’t think I have enough time now.

No, we said very different things. Your units were all wrong. We did not say the same thing any more than 10 miles is the same thing as 10 days.

Do you understand the difference between miles and days? That is the difference between lb and lb*s. They are different units.

The units are different. 900 lb is almost half a ton, it weighs a lot. 900 is a pure number, it weighs nothing. 900 lb ≠ 900.

How can you possibly think that a pure number is the same thing as nearly half a ton? Have you ever heard the expression "comparing apples and oranges"? That is what you are doing wrong, and you are doing it very consistently wrong. We really need to address this before proceeding.

So let’s get this right for later, when you ask me something, and you say 100lb*s you mean you have applied a force of 100lb for 1 second ?

Wayne

 

[Dale]

So let’s get this right for later, when you ask me something, and you say 100lb*s you mean you have applied a force of 100lb for 1 second ?

What we mean is that we have applied an impulse of 100 lb*s. That could be a force of 100 lb for 1 s as you said, or it could be a force of 50 lb for 2 s or a force of 25 lb for 4 s or a force of 10 lb for 10 s ...

 

[waynexk8]

One quick one please, for all.

If I hold the 80lb half way up {forget the acceleration/force it took to get there, say someone handed it to me.} for 6 seconds, are you all saying I used the same impulse/force ? As you might or might not know, if you did a fast, then slow rep, and the static hold all the momentary muscular failure, you would be able to hold the weight stationery far far far longer.

Wayne

 

[sophiecentaur]

Do you mean force or do you mean impulse? What "man of Science" would ignore the distinction? This is ridiculous.
You do not "use" a force. You apply a force. When you do it with your muscles, it involves energy expenditure. When you rest the weight on a table, no energy is involved. A table never gets tired, does it?
Muscle failure is not a Physics phenomenon. It's a physiological thing.

 

[waynexk8]

DaleSpam wrote; if you exert a constant 100 lb for 10 s then you have exerted an impulse of 1000 lb*s. And if you exert a constant 80 lb for 10 s then you have exerted an impulse of 800 lb*s.

With the above, its seems you are saying that the fast rep = an impulse of 1000lb*s, slow rep = an impulse of 800lb*s. But then you say no, and that I/that is not right and S. and D. are ?

The only problem in the above, is that the slow may exerted a constant {lets just say constant for now, as its quite close}force for say 80% of the rep, say 10% for acceleration, and 10% for deceleration. However the fast will accelerate for say 80% and decelerate for 20%

Also we have “still” left out the force “on” the muscles from the transition from negative to positive, as in the above we have only worked out the force exerted by the muscles.


1,
Fast rep = 100 force x 10 seconds = 1000lb * 10s.

2,
Slow rep = 80 force x 10 seconds = 800lb * 10s.

Wayne

 

[Dale]

If I hold the 80lb half way up {forget the acceleration/force it took to get there, say someone handed it to me.} for 6 seconds, are you all saying I used the same impulse/force ?

If you hold 80 lb stationary for 6 s then you have exerted an impulse of 480 lb*s. That is the same impulse as 40 lb for 12 s or 160 lb for 3 s.

 

[waynexk8]

Hi waynexk8,

After you read the links try the following problems:
1) 100 lb * 5 ft =
2) 20 ft*lb / (2 ft) =
3) 32 ft/s² * 10 s =
4) 5 ft/s * 320 lb / (32 ft/s²) =

Do not rush or hurry this, make sure that you get the units correct. This is important.

Right, I have/am having a go of this, but I think it could be best if you did one calculation, as I am not sure what I am supposed to be calculating, on 1, it can be 100 x 5 = 500, and if you put 100 lb * 5 ft = on Google, it gives you, 69.1274772 m kg, but I don’t know if that’s right or how it came to that, will have another read and think, but thought best post this now.

Wayne

 

[waynexk8]

Going to have to come back to the other.

If you hold 80 lb stationary for 6 s then you have exerted an impulse of 480 lb*s. That is the same impulse as 40 lb for 12 s or 160 lb for 3 s.

Ok, so could you say the following please, all with 80 pounds for 6 seconds. But please remember the person/machine can use up too 100 pounds or force anytime. And “please add in the forces on the muscles, as you did say that the rep stating with an eccentric going into the concentric gives out a higher force/impulse

1,
Static hold = 480lb*s.

2,
Fast reps =

3,
Slow rep =

As what I would say is; impulse is the integral of a force with respect to time. A small force applied for a long time can produce the same momentum/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 what you 3 seem to be saying; A small force applied for the same time can produce the same momentum/momentum change as a large force applied for the same time.

Wayne

 

[douglis]

Going to have to come back to the other.

Ok, so could you say the following please, all with 80 pounds for 6 seconds. But please remember the person/machine can use up too 100 pounds or force anytime. And “please add in the forces on the muscles, as you did say that the rep stating with an eccentric going into the concentric gives out a higher force/impulse

1,
Static hold = 480lb*s.

2,
Fast reps =

3,
Slow rep =


Wayne

1,
Static hold = 480lb*s.

2,
Fast reps = 480lb*s.

3,
Slow rep = 480lb*s.

In all three cases the change in momentum is zero hence the average muscle force is equal with the weight.Impulse equals average force X time.In all cases is equal with 480lb*s.

The last five years you've been explained that half a billion of times by at least twenty different people.Even if you had started from an absolute zero physics level you could have become a physics professor by now...if you only had shut up and listen to them.
But no...you're not here to learn.Your obsession kept you to the absolute zero level.What a waste of time...

 

[waynexk8]

1,
Static hold = 480lb*s.

2,
Fast reps = 480lb*s.

3,
Slow rep = 480lb*s.

In all three cases the change in momentum is zero hence the average muscle force is equal with the weight.Impulse equals average force X time.In all cases is equal with 480lb*s.

I thought you were going to say that.

What do you mean the momentum/movement is zero, its zero in the start hold, but with the slow there is 2m of momentum/movement, and the fast there is 12m momentum/movement.

Say I just repped up 1m would you still say there was zero momentum/movement ? And if so why ?

Why do you not answer questions that puts your little theory as wrong ? Like the below ? Why when you stated yourself the only way to sort this out is EMG, then when EMG states you wrong, you think paper equations written on paper are better, It’s like saying you said we can’t get Men to the Moon, I than go and send Men to the Moon, but you still say I am wrong, even thou I send Men to the Moon ?

Don’t you understand that when muscles are at accelerations, and have to be using more force, that they are under more strain, TENSIONS? And the only way your slower lower force reps can make up this stain, is if they do the reps for longer than the faster reps, this IS proved in that I fail 50% faster, I fail faster because I HAVE put more tension on the muscles per unit of time, thus I have put out more force per unit of time.

I NEED to explain that better tonight, and I don’t think you or the physicists are seeing this.

Let me ask another question please all.

1,
Static hold done for 5 seconds, force used ? =

2,
Static hold done for 10 seconds force used ? =

3,
Fast reps done for 5 seconds, force used ? =

4,
Fast reps done for 10 seconds, force used ? =

5,
Slow reps done for 5 seconds, force used ? =

6,
Slow reps done for 10 seconds, force used ? =

The last five years you've been explained that half a billion of times by at least twenty different people.Even if you had started from an absolute zero physics level you could have become a physics professor by now...if you only had shut up and listen to them.
But no...you're not here to learn.Your obsession kept you to the absolute zero level.What a waste of time...

IF you “think” you are right, you need to answer, address and then “try” and counter when real world experiments prove you little theory wrong, as of yet you have not, a “theory” and that’s “all” you have, “NEEDS” to be proven with real World practical experiments/tests, like I have done, as of yet you have not, you hold no ground at the moment, all you have is a theory on paper, which has by me “^” times be proven “WRONG”.

1,
As I said, the EMG proved me right,

2,
Moving the weight 10m more in the same time frame proved me right,

3,
Using more energy in the fast proved me right,

4,
As you fail with the fast 50% faster proved me right,

5,
I used 100 force moving 80 for 6 seconds, you used 80 force moving 80 for 6 seconds, 100 is more than 80, the 100 = 20 more than 80, 100 = 25% more than 80, that proved me right.

I could go on.

Wayne

 

[douglis]

I thought you were going to say that.

What do you mean the momentum/movement is zero, its zero in the start hold, but with the slow there is 2m of momentum/movement, and the fast there is 12m momentum/movement.

Say I just repped up 1m would you still say there was zero momentum/movement ? And if so why ?

Wayne

Not zero momentum...zero CHANGE in momentum.That's what the impulse represents.

For all the rest I don't have the patience.Just read carefully DaleSpam posts.

 

[Dale]

Right, I have/am having a go of this, but I think it could be best if you did one calculation, as I am not sure what I am supposed to be calculating

OK, here is a step-by-step detailed example of how to calculate a quantity with units in physics:
5 s * 64 lb / (32 ft/s) =
(5 * 64/32) * (s*lb/(ft/s)) =
(5 * 2) * (s*lb*s/ft) =
10 lb*s²/ft

 

[Dale]

As what I would say is; impulse is the integral of a force with respect to time. A small force applied for a long time can produce the same momentum/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.

This is correct.

But what you 3 seem to be saying; A small force applied for the same time can produce the same momentum/momentum change as a large force applied for the same time.

None of us have ever said this. The reason that you believe that we have said it is because of a misunderstanding you have of impulse for time-varying forces and/or a misunderstanding of the forces over a fast vs a slow rep. That is what I am trying to teach you, but we have not gotten to the time-varying part yet.

It would be helpful if you would just focus on the new questions I have asked of you recently instead of repeating your old questions every couple of posts. I am well aware of your old questions, and I am trying to get you to the point where you will understand the answers as quickly as possible. Please focus on making forward progress rather than continually looking back and rehashing the scenario. It is a waste of your time since it provides no new information to me and prevents you from taking the time to learn from what I have already posted.

 

[waynexk8]

Back later to answer the post/s.

Let me try and convince you this way that you are wrong, or that your equations work on paper, but not in practice.

I am going to lift 90% of my 1RM, in this case my 1RM = 100 pounds. So 90% = 90 pounds, you’re going to try and lift 90% of your 1RM which is also a 100 pounds.

1,
I apply my 1RM force, as this is a fast rep, I lift the 90 pounds up and down say 3 times in 6 seconds, I have applied 100 pounds of force for 6 seconds, well not quite that much as my force will have been diminishing But I have overcome gravity for 6 seconds lifting the 90 pounds

2,
You apply 80% or your 1RM as this is a slow, you cannot lift the weight at all, in 6 seconds you have applied 80 pounds of force only. You have not overcome gravity in the 6 seconds, and you have not lifted the weight.

3,
I lifted the weight 3 times in 6 seconds, I thus used more {nearly said it again, the total/overall force, but let’s be correct.} impulse than you in 6 seconds as I overcome gravity and lifted the weight up and down 3 times. THUS putting MORE tensions on the muscles.

Or do you think you put the same tensions on the muscles not lifting a weight as you have not, are not used enough force ?

Wayne

 

[Dale]

This scenario is not in dispute. We all agree that the impulse is greater for 1 than for 2.