Can We Measure Deceleration in Weightlifting Without Specialized Equipment?

[waynexk8]

Hi all,

Without sophisticated equipment, {I have an EMG} when lifting weight up and down, as in weightlifting/bodybuilding, would there be any way to tell when/if the weight starts to decelerate ?

I have a Smith Machine, this is a machine, that the weight goes up and down on two rods, on bearings, see below.

With this machine, I am able to lift if I want open handed, {to which I down, and your lifts suffer} if wanted, if I lowed then lifted 50% of my 1RM, {Repetition Maximum} as explosively as I could, open handed, the weight at the end of the lift will move slightly out of my hands, {say the positive was 20in and the negative was 20in} this mean in my opinion that there was no deceleration when in contact with my hands, is this correct please ?

If I then lift 80% of my 1RM, in the exact same manner as above, but the weight because it was heaver did not leave my hands, {If you do the equations for a machine lifting 80% at 2ms, the weight would leave your hands for about 4in, but because of the biomechanical disadvantages and advantages, this does not happen with the muscles of the body} but still, does that not mean there is not deceleration on my behalf ? As I am trying to push the weight as explosively as I can, then after I push up explosively as I can, I then lower again. Next and more to the real, would there be a deceleration {I know there must be a deceleration, if I push up and then lower the weight} on my behalf ? Meaning if I push up as explosively as I can, 80% for 20in that will take say .3 of a second, will I use less force than the weight ? As if we go back and look at me pressing up 50% I would press up as hard as I can, thus the weight would move out of my hands, then, and only then would it start to decelerate, thus no deceleration on my behalf, then when it dropped into my hand I would then lower it, and repeat the lift.

The only problem here, I that I want to find if there is deceleration on my behalf lifting 80% closed handed ? As is not my weightlifting/bodybuilding lifting up and down, just like if I lifted 80% 20in, and then there was a steel plate to which I hit, then, I would think not deceleration on my behalf {Mind you I do not lift to a steel plate, thus, it’s basically the same} then lower.



Wayne

 

[Dale]

Without sophisticated equipment, {I have an EMG} when lifting weight up and down, as in weightlifting/bodybuilding, would there be any way to tell when/if the weight starts to decelerate ?

Not really. An EMG measures muscle activation. You would need an accelerometer or a good high speed video camera.

I have a Smith Machine, this is a machine, that the weight goes up and down on two rods, on bearings, see below.

Note, the rods will introduce some friction into the system, which will change some of the analysis that you and I have discussed in other threads.

With this machine, I am able to lift if I want open handed, {to which I down, and your lifts suffer} if wanted, if I lowed then lifted 50% of my 1RM, {Repetition Maximum} as explosively as I could, open handed, the weight at the end of the lift will move slightly out of my hands, {say the positive was 20in and the negative was 20in} this mean in my opinion that there was no deceleration when in contact with my hands, is this correct please ?

No, this is not correct. The weight will decelerate whenever the force on your hand is less than the force of gravity plus the friction force. The weight leaves your hand when the force is zero. So there is a considerable range of force where the weight is both in contact with your hand and decelerating.

This has nothing to do with wether you are lifting with an open or closed hand.

 

[waynexk8]

Not really. An EMG measures muscle activation. You would need an accelerometer or a good high speed video camera.

Maybe I could borrow an iPod, as that has an accelerometer, then dl this; http://liza.exelio.eu/

Would that be able to tell me when I was decelerating, as I am sure I don’t decelerate, well not in the normal decelerate, as the only way I think I decelerate, is because my arms cannot go any further, if you get what I am saying.

Note, the rods will introduce some friction into the system, which will change some of the analysis that you and I have discussed in other threads.

Yes totally agree, that the rods will introduce some friction, but I don’t see it damaging/altering the debate ?

No, this is not correct. The weight will decelerate whenever the force on your hand is less than the force of gravity plus the friction force.

Yes I understand that, that if I use less force than the weight that it will decelerate, but what if I went from using 100% to 99% then 98% and down to 85%, would not the weight still be accelerating ?

But if I lifted a weight, or a machine lifted a weight, 50% and then 80% and the weight was in lift 1 and 2, lifted with 100% force to full stretch, would not it be the length of my arm reach or the machine reach that at that moment makes for the immediate deceleration/stop ? If so, then my force, {forget about the first acceleration for now} for 99% of the, sorry 100% of the lift would be 100% or as close as I could get to using my maximum force for that lift.

What I am saying, that if I am lifting 80% the way I do, as explosively/fast as I can, I don’t think there is in the first several rep, much deceleration at all. Could you maybe do a lift like this somehow and see what I mean, not sure if you could do/try this, or shall I video me doing a few lifts on the Smith machine maybe ? As if I am lifting 80% with 100% I don’t see on the first several reps, if I am trying to move the weight up as fast as I can, how there can be much/any deceleration. Take a shot putter, now there is no deceleration on his behalf, this is what I am saying, the deceleration on the putt come after it leaves his hand, so what if he putted as usual, but at the instant that the putt left his hand, a steel plate appeared ?

The weight leaves your hand when the force is zero.

Right, as then there is no force pushing it.

So there is a considerable range of force where the weight is both in contact with your hand and decelerating.

But has it got to be like this, as I said above about the shot putter and so forth. I don’t see why I can’t, or more why a machine could not push up for full range with 100% force ? I don’t see a reason when a person, or more to the point a machine has to use less force and decelerate ?

This has nothing to do with wether you are lifting with an open or closed hand.

It must have something to do with if I have open hands or not, as if I repped 20% with open hands, it would leave my hand well before I had finished the rep, or ROM {Range Of Motion} but if I had closed hands, I would still be exerting force on the weight, to which I would not open handed. What I am doing when closed hands, is taking more of the acceleration force, propulsive forces on my muscles.

So if I was to accelerate a weight of 80% for a whole .5 of a second, to a person moving a weight at a constant very slow speed of .5 of a second, would not I be using more force ?

Wayne

 

[Dale]

Yes I understand that, that if I use less force than the weight that it will decelerate, but what if I went from using 100% to 99% then 98% and down to 85%, would not the weight still be accelerating ?

You need to learn to be more specific. What is 100% in this case? What is the weight?

As I said above, the weight will decelerate any time the force on the hand is less than the force of gravity plus the friction force. There is not enough information in "using 100%" to determine what happens. Provide the required information and try the calculation yourself.

Take a shot putter

Please stick with one scenario.

So if I was to accelerate a weight of 80% for a whole .5 of a second, to a person moving a weight at a constant very slow speed of .5 of a second, would not I be using more force ?

80% of what? How much force are you using to accelerate? Do you mean "compared to a person" or "towards a person". 0.5 s is a time, not a speed. More force than what?

 

[waynexk8]

Just thought would say, will have to get back to your posts and PM’s tomorrow.

Wayne

 

[waynexk8]

You need to learn to be more specific. What is 100% in this case? What is the weight?

Sorry.

The weight is 80 pounds, too which that is 20% off my 1RM, {Repetition Maximum} so let’s just say I could use 100% or 100 pounds, if I then went from using 100 pounds to 99 pounds then 98 pounds and down to 856 pounds, would not the weight still be accelerating ?

As I said above, the weight will decelerate any time the force on the hand is less than the force of gravity plus the friction force. There is not enough information in "using 100%" to determine what happens. Provide the required information and try the calculation yourself.

Do you mean that if the force is a little under 80 pounds, then the weight is slowing/deceleration down ? However, I could push up with from 100 to 81 pounds for the full rep/push, then there would be no deceleration on my part, as the only reason the weight would decelerate and stop, would be because my arm had reached its stretching length, and as I said, as of the biomechanical disadvantages and disadvantages, thought the ROM {range Of Motion} the weight would repped/pushed up with a constant force, thus it would not leave the hand. So basically as I said, If I could push up with from 100 to 81 pounds for the full rep/push, then there would be no deceleration on my part, as the only reason the weight would decelerate and stop, would be because my arm had reached its stretching length, there would just be an extra impulse/jolt on the bones and muscles as full ROM was reached.

Please stick with one scenario.

Ok.

80% of what? How much force are you using to accelerate? Do you mean "compared to a person" or "towards a person". 0.5 s is a time, not a speed. More force than what?

I am lifting up 80 pounds with a constant force of 100 pounds, for 20 inch, that will take .5 of a second. Person 2 is lifting the up 80 pounds with a constant force of 80 pounds, for 3.3 inch, that will take .5 of a second, the only deceleration on both parts, will be instantaneously, as both arms had reached out full ROM. Which please will use the more impulse.

Wayne

 

[Dale]

The weight is 80 pounds, too which that is 20% off my 1RM, {Repetition Maximum} so let’s just say I could use 100% or 100 pounds, if I then went from using 100 pounds to 99 pounds then 98 pounds and down to 856 pounds, would not the weight still be accelerating ?

Neglecting the friction force, yes.

Do you mean that if the force is a little under 80 pounds, then the weight is slowing/deceleration down ?

Yes.

However, I could push up with from 100 to 81 pounds for the full rep/push, then there would be no deceleration on my part

Neglecting friction, yes. However, even ideally you cannot exert a constant force for very long. As soon as you get to the end of your reach, the force must change. It can change to 0, if you are throwing the weight. It can change to some force less than the weight if you are gently decelerating it. Or it can even change to a negative force if you are almost throwing it but not letting go.

I am lifting up 80 pounds with a constant force of 100 pounds, for 20 inch, that will take .5 of a second.

This is impossible.

You can lift 70 lb with a constant force of 100 lb for 20 in, that will take .5 s.
You can lift 80 lb with a constant force of 113 lb for 20 in, that will take .5 s.
You can lift 80 lb with a constant force of 100 lb for 12 in, that will take .5 s.
You can lift 80 lb with a constant force of 100 lb for 20 in, that will take .6 s.

Person 2 is lifting the up 80 pounds with a constant force of 80 pounds, for 3.3 inch, that will take .5 of a second, the only deceleration on both parts, will be instantaneously, as both arms had reached out full ROM. Which please will use the more impulse.

Assuming that you adjust your numbers to something possible, then the question is, which is going faster at the end? If the instantaneous decelerations at the end both bring the weights to a stop then the impulse is the same. If they are throwing the weights then the one that throws with a higher velocity has the higher impulse. The "instantaneous" force contributes to the impulse.

Remember the equation that relates impulse and velocity:
m \Delta v = I_{lift}-I_{grav}

 

[waynexk8]

Neglecting the friction force, yes.

Yes we are negtectiing friction, air resistance. Yes this is what I always said.

Yes.

Right.

Neglecting friction, yes.

Glad you said that, as some seemed to think that had to be a deceleration. Hmm, just looked at your last answer, let’s get this right, so if I move 80% of my 1RM, with full acceleration to full ROM, and you move 80% of your 1RM, with a very slow speed for the same time frame, are you saying that both have the same impulse, or that I the person moving the weight at full acceleration for the whole time has the higher impulse ?

However, even ideally you cannot exert a constant force for very long. As soon as you get to the end of your reach, the force must change. It can change to 0, if you are throwing the weight. It can change to some force less than the weight if you are gently decelerating it. Or it can even change to a negative force if you are almost throwing it but not letting go.

Yes.

This is impossible.



You can lift 70 lb with a constant force of 100 lb for 20 in, that will take .5 s.
You can lift 80 lb with a constant force of 113 lb for 20 in, that will take .5 s.
You can lift 80 lb with a constant force of 100 lb for 12 in, that will take .5 s.
You can lift 80 lb with a constant force of 100 lb for 20 in, that will take .6 s.

Ok, as I always said, these number were not spot on, just a rough reference, let’s go for the 80 lb with a constant force of 100 lb for 20 in, that will take .6 s.

Assuming that you adjust your numbers to something possible, then the question is, which is going faster at the end? If the instantaneous decelerations at the end both bring the weights to a stop then the impulse is the same. If they are throwing the weights then the one that throws with a higher velocity has the higher impulse. The "instantaneous" force contributes to the impulse.

You seem to be saying that full acceleration to just holding a weight or moving it very slowly, uses the same impulse ? I do not agree with that, F=ma is the force experienced by an accelerating object, and according to Newton’s laws of motion, force is equal to mass into acceleration. Now acceleration is the rate of change of velocity. This is called impulse which is the product of force and the duration for which it is applied. I am not talking off any deceleration here on my part, as I am not debating the time when the weights are at zero velocity/speed/acceleration/movement, I do not want to add in or take off anything after the weight has stopped moving, I am only concerned with the movement.

F = ma. A weight of 200kg with an Acceleration of 20m/s is 200 x 20 = 4000N, 40m/s is 200 x 20 = 8000N, if they both came to an instantaneously stop at the same time frame, 1 second, the N’s would still be 4000N and 8000N. The acceleration of the weight is directly proportional to the net force. Impulse is the hitting the barbell with my hand, the impulse in this situation is the average force exerted by the hand {arm and all muscles involved} multiplied by the time the hand and barbell were in contact. So the impulse given to the barbell by my hand causes a change in momentum/movement of the weight. As you told me, Impulse is the force over time, it is measured in Newton ,seconds, so a force of one Newton applied for one second will change the momentum/movement, a force of two Newton's applied for one second will change the momentum even more, and that’s what I have done, I have used more Newton’s in the same time frame, I am not interested in the time “AFTER” the rep, the time when momentum/movement is zero, “ONLY” when there is motion

A little problem, say your muscles have a tearing strain of a 100 pounds of force, so if I lift 80 pounds with a 100 pounds of force, in let’s say .3 of a second my muscles will tear. But if you lift the 80 pounds with 80 pounds of force, the muscles would not tear until they were totally fatigued, and that could take up to 30 seconds or more. So here is why I am saying that the forces and average forces, and the impulses and average impulses that I have produced in a very short time, cannot be made up or balanced out by the lower forces and average forces, and the lower impulses and average impulses in the same time frame. My first impulses are 100, 100, 100, 100, yours are 80, 80, 80, 80, my average is higher than yours, that’s why I fail faster, as I fail faster, you cannot average the other forces and impulse forces out, the maybe the only scenario that we can work out, is me failing 50% faster than you, and in that time I have used more force, and more impulse force yes ? That’s why I use more energy and the EMG reads higher. Is not F=ma, where F is the force applied, m is the mass of the body and ‘a’ is its acceleration, acceleration is rate of change of velocity, so the weight and its velocity is its momentum, so the change in momentum of a body is a product of the force applied on it for some duration. So force effects the acceleration of an object {f=ma} while Impulse simply sets the velocity of an object. This means that same change in momentum/movement can be induced on a weight with a large force operating over a small time period and a small force operating over a long period of time. So the effect of a force is not just the amount of the force but also dependent upon the duration for which it is being applied.

Remember the equation that relates impulse and velocity:
m \Delta v = I_{lift}-I_{grav}

Yes impulse and velocity are related.

Wayne

 

[Dale]

Wayne, there are very few important details:

1) how much is the weight
2) how long does it take
3) what is the change in velocity (in our previous discussions I had always assumed this to be 0, but with your discussion about throwing weights it is no longer certain)

Nothing else matters. With that information you can determine the impulse, without that information you cannot. If those three factors are kept constant then none of the other details about the force matter, the impulse is the same.

so if I move 80% of my 1RM, with full acceleration to full ROM, and you move 80% of your 1RM, with a very slow speed for the same time frame, are you saying that both have the same impulse, or that I the person moving the weight at full acceleration for the whole time has the higher impulse ?

No way to tell from your description. Weight and change in velocity are missing.

F = ma. A weight of 200kg with an Acceleration of 20m/s is 200 x 20 = 4000N, 40m/s is 200 x 20 = 8000N, if they both came to an instantaneously stop at the same time frame, 1 second, the N’s would still be 4000N and 8000N.

The weight is 200 kg in both cases, the duration is 1 s in both cases, and the change in velocity (0 m/s) is the same in both cases, therefore the impulse is the same in both cases (~2000 Ns). What you are neglecting is the negative impulse from the sudden deceleration which is -2000 Ns for one and -6000 Ns for the other.

the impulse in this situation is the average force exerted by the hand {arm and all muscles involved} multiplied by the time the hand and barbell were in contact.

Yes. And the average force is ALWAYS equal to the weight if the barbell starts at rest and stops at rest.

I am not interested in the time “AFTER” the rep, the time when momentum/movement is zero, “ONLY” when there is motion

In a rep the weight starts at rest at a low point, is lifted upwards to a high point where it is again at rest (at least momentarily), and is then brought back down to the same low point once again at rest. Correct? I have never discussed the time before nor after a rep, but this is my understanding of what a rep is.

I would prefer to return to discussing weightlifting and reps like normal, but we can continue talking about throwing weights if you prefer.

, that’s why I fail faster, as I fail faster, you cannot average the other forces and impulse forces out,

Yes, not only can you, physics demands it. Irrevocably and invariably. If those three important factors are held constant, then everything that happens in the middle must unavoidably average out. No amount of EMG readings or anecdotes about fatigue can change this simple physical fact.

Is not F=ma, where F is the force applied, m is the mass of the body and ‘a’ is its acceleration, acceleration is rate of change of velocity, so the weight and its velocity is its momentum, so the change in momentum of a body is a product of the force applied on it for some duration.

Yes, exactly. That quantity, the force applied for some duration, is the impulse.

This means that same change in momentum/movement can be induced on a weight with a large force operating over a small time period and a small force operating over a long period of time. So the effect of a force is not just the amount of the force but also dependent upon the duration for which it is being applied.

Yes, which is why impulse includes both.

 

[douglis]

you cannot average the other forces and impulse forces out, the maybe the only scenario that we can work out, is me failing 50% faster than you, and in that time I have used more force, and more impulse force yes ? That’s why I use more energy and the EMG reads higher.
Wayne

Here we go again!

NO...either you lift the weight up and down for a minute or you just hold it you use the same impulse and the same force per second.
It's irrelevant how fast you fail or what your EMG reads.

 

[Dale]

you cannot average the other forces and impulse forces out,

Hi waynexk8, you have made this comment enough times that I think that we need to address it head-on. Here is the proof that your claim is incorrect. Please walk through the proof and ask questions. This is something that you need to understand, so let's take the time to do it correctly.

Here is the mathematical proof, it relies only on 3 assumptions:
1) Newton's second law: F_{net}=ma
2) Free weights have only two forces acting on them, gravity and lift from the person: F_{net}=F_{grav}+F_{lift}
3) The lift starts and stops at rest: v_i=v_f=0

I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} ma \; dt = m \int_{t_i}^{t_f} \frac{dv}{dt} \; dt = m (v_f-v_i) = 0
I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} F_{lift} \; dt + \int_{t_i}^{t_f} F_{grav} \; dt = I_{lift} + \int_{t_i}^{t_f} mg \; dt = I_{lift} + mg(t_f-t_i) = 0
I_{lift} = -mg(t_f-t_i)

So the impulse of the lift depends only on the weight (mg) and on the time (t_f-t_i), not on the speed of the lift.

 

[waynexk8]

Here we go again!

NO...either you lift the weight up and down for a minute or you just hold it you use the same impulse and the same force per second.
It's irrelevant how fast you fail or what your EMG reads.

If, and I do mean if, if you’re going to debate on a physics forum, please "try" and state/show your case for being right, and why you think that these are irrelevant here.

1,
The faster you move the weight, the faster you fail.
And more energy used.
You need to explain that I don’t use more impulse over the same time as you, but why I do use more energy, I say you use more energy, and you use more energy the more impulse you use, and this is a proven fact. As the faster you move the weight the more energy you use, and it's directly proportional. Example, if I use 200 calories {energy} to move a weight in 5 seconds, and then use 300 calories to move a weight in 5 seconds, the time I used 200 calories to move the weight, the weight might move 200m, and then the time I used 300 calories to move the weight, the weight would then move 300m, and as you should know, you HAVE to use a higher force and a higher/more impulse to move a weight further in the same time frame. The acceleration a of the weight is directly proportional to the net force and proportional to the mass. F = ma, so more full acceleration for .5 of a second, you HAVE to use more force, I am not talking about the time before acceleration, or after acceleration. The force at work on a car as it starts a race, f the car has a Mass of 500 pounds and an acceleration of 10m/s work out the Force pushing the car by multiplying the Mass by the Acceleration like this 500 x 10 = 5000N, if it’s moving SLOWER, 5m/s its 500 x 5 = 2500N. Please not that 5000 = more than 2500. If I accelerate a weight for 1 second using 10N and you move a weight for 1 second using 5N, I have used 10N for 1 second, you have used 5N for 1 second, how is 5 as high/more than 10 ?

2,
You need to explain why a very sophisticated bit of equipment, that has a computer that adds up things very fast and accurate, is as you think wrong ?

3,
Power,
At least you can understand that the faster moving uses more power.

Wayne

 

[waynexk8]

Hi waynexk8, you have made this comment enough times that I think that we need to address it head-on. Here is the proof that your claim is incorrect. Please walk through the proof and ask questions. This is something that you need to understand, so let's take the time to do it correctly.

Here is the mathematical proof,

I would say there are two proofs, paper equations and then a real World practical test/study, problem is, the real World practical test/study, as in energy, hitting failure and the EMG does not agree with equations, the equations are not wrong, just the way of doing these equations are wrong, something will be needed to be changed. It’s like when we did equation to see how far a weight would move if we lifted at a certain speed and stopped, it was 3 inch, but when we did the test the weight did not move, the equations were not wrong, it was because of the muscles biomechanical advantages and disadvantages thought-out the ROM, that they could not push will the full equations force for the whole rep. I this debate, there could be something completely different that has to be change, and nothing either to do with the human body. if the car has a Mass of 500 pounds and an acceleration of 10m/s work out the Force pushing the car by multiplying the Mass by the Acceleration like this 500 x 10 = 5000N, if it’s moving SLOWER, 5m/s its 500 x 5 = 2500N. Please not that 5000 = more than 2500. If I accelerate a weight for 1 second using 10N and you move a weight for 1 second using 5N, I have used 10N for 1 second, you have used 5N for 1 second, how is 5 as high/more than 10 ?

it relies only on 3 assumptions:
1) Newton's second law: F_{net}=ma
2) Free weights have only two forces acting on them, gravity and lift from the person: F_{net}=F_{grav}+F_{lift}
3) The lift starts and stops at rest: v_i=v_f=0

In this debate, we are not debating anytime before or after the has started to move.

I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} ma \; dt = m \int_{t_i}^{t_f} \frac{dv}{dt} \; dt = m (v_f-v_i) = 0
I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} F_{lift} \; dt + \int_{t_i}^{t_f} F_{grav} \; dt = I_{lift} + \int_{t_i}^{t_f} mg \; dt = I_{lift} + mg(t_f-t_i) = 0
I_{lift} = -mg(t_f-t_i)

So the impulse of the lift depends only on the weight (mg) and on the time (t_f-t_i), not on the speed of the lift.[/QUOTE]

I cannot pretend to know I understand the equations, but if they were right, why do I fail faster using a faster velocity, why do I use more energy {calories} why does the EMG state I use more muscle activity, and more muscle activity = more force.

Here is why I fail faster, I fail faster because the faster I move the move tension I put on the muscles, thus the more strain on them, if the slower moving reps were to equal out as the equation seems to same, it says that’s the forces average out, if the forces averaged out, then the slow repping person would average out their forces thus tension on the muscles thus strain, and fail the same time as the faster moving person, but this does not happen, as its like this, 100, 100, 100, 100, 000 fast, 80, 80, 80, 80, 80, slow, 100 is higher than 80, 100 is higher than 80 force 80% of the whole time of the rep, so how can 80 done for just 20% make this up ? In my opinion it does and cannot, if it did make this force up, then in the real would practical experiment as in lifting weight and seeing who hits momentary muscular failure faster, both would hit momentary muscular failure at the same time, because the slow would equal out balance out all the forces. But both reps do not hit failure the same time, the fast hits failure 50% faster, thus how can the forces be the same ? How can they be the same if both don’t hit failure the same time ? Here is another reason why, 100, 100, 100, 100, 000 fast, 80, 80, 80, 80, 80, note the fast does have to more force, and more impulse force the first 80% of the time frame, forget the last 20% for now, see it ?

The acceleration a of the weight is directly proportional to the net force and proportional to the mass . F = ma, so more full acceleration for .5 of a second, you HAVE to use more force, I am not talking about the time before acceleration, or after acceleration. If I use 10N of force for .5 of a second, and you use 3N of force for .5 of a second, I have used more force, and more impulse !

Newton's 2nd Law and the dynamics of Aristotle. According to Newton, a force causes only a change in velocity, {an acceleration} it does not maintain the velocity as Aristotle held.

So how can/is the object moving ? We will say it’s a weight, I acceleration it using force, I move it at a constant velocity using force, I deceleration at with force, if I stopped using force, the weight would start accelerant until it hit the Earth, if it was dropped from a great height, it would accelerate until the air pressure was the same, then it would move at a constant speed. However if I was pushing a weight up, UI would HAVE to use force for both acceleration and constant speed, looks like both were right ? If you have a 80 pound weight, and it’s traveling at a constant speed of 50m/s, Newton's second law, says there should be no force because there is no acceleration, but there is forces, the force of gravity, the force of friction, the force of momentum/movement, motion, a force pushed moved the weight, thus now the force moving the weight is the weight/mass of the weight itself, gravity and friction. As if it hits another object it will move the object, so there is force there but according to F=ma there is no force.

Sorry and thank you for your help agai8n dalespam, but it’s a little last for your other post sorry.


Wayne

 

[Dale]

I would say there are two proofs, paper equations and then a real World practical test/study,

I agree. That is why scientists continuously put the paper equations to the most careful and rigorous experiments that they can imagine, and great honors are bestowed on scientists which prove equations wrong and come up with better ones. The particular equations that we are discussing here have passed the most exacting tests imaginable for over 3 centuries of the most brilliant experimentalists. The real world tests completely agree with the equations I posted.

problem is, the real World practical test/study, as in energy, hitting failure and the EMG does not agree with equations, the equations are not wrong, just the way of doing these equations are wrong

No, the way of doing the equations is correct. If you disagree then please point out EXACTLY where I made my mistake.

,If I accelerate a weight for 1 second using 10N and you move a weight for 1 second using 5N, I have used 10N for 1 second, you have used 5N for 1 second, how is 5 as high/more than 10 ?

10 N*s is indeed more impulse than 5 N*s, nobody is disputing that.

What you don't seem to realize is that after you have applied 10 N*s and I have applied 5 N*s your weight is moving faster than mine. Over the course of the next second, if we both bring our weights to rest, then you must use LESS force than me such that our average force is the same.

In this debate, we are not debating anytime before or after the has started to move.

Good to hear. I certainly have never mentioned anything about any time before the weight has started to move or after it had stopped.

I cannot pretend to know I understand the equations, but if they were right, why do I fail faster using a faster velocity, why do I use more energy {calories} why does the EMG state I use more muscle activity, and more muscle activity = more force.

The impulse is the same for fast or slow, the energy is higher for fast than for slow, therefore, the energy does not depend only on the impulse.

The impulse is the same for fast or slow, the EMG is higher for fast than for slow, therefore, the EMG does not depend only on the impulse.

The impulse is the same for fast or slow, you fail sooner for fast than for slow, therefore the failure time does not depend only on the impulse.

...

if the forces averaged out, then the slow repping person would average out their forces thus tension on the muscles thus strain, and fail the same time as the faster moving person

That would be correct IF the failure depended only on the average strain, therefore the failure does not depend only on the average strain.

, but this does not happen, as its like this, 100, 100, 100, 100, 000 fast, 80, 80, 80, 80, 80, slow, 100 is higher than 80, 100 is higher than 80 force 80% of the whole time of the rep, so how can 80 done for just 20% make this up ?

Let's say that the weight is 75 lb and you lift for 1 s at 100 lb and I lift for 1 s at 80 lb. At the end of that second your weight is moving 11 ft/s and my weight is moving just 2 ft/s. Now, let's say we want to decelerate the weight in 0.5 s, in order to do that, you must use a force of 25 lb, and I must use a force of 65 lb. Any other force and the weight will not stop, i.e. the rep will not be at an end.
For me: 80 lb * 1 s + 65 lb * 0.5 s = 112.5 lb*s
For you: 100 lb * 1 s + 25 lb * 0.5 s = 112.5 lb*s

In my opinion it does and cannot, if it did make this force up, then in the real would practical experiment as in lifting weight and seeing who hits momentary muscular failure faster, both would hit momentary muscular failure at the same time, because the slow would equal out balance out all the forces.

Your opinion is wrong, as proven mathematically, and the real world careful controlled experiments confirm the math. The average force is the same for fast and slow. Momentary muscular failure occurs sooner for fast than for slow. Therefore momentary muscular failure is not due only to the average force.

If I use 10N of force for .5 of a second, and you use 3N of force for .5 of a second, I have used more force, and more impulse !

Yes. And your weight is moving faster than mine. Since the rep doesn't end until the weight is stopped again, this is not a complete description of a rep.

waynexk8, I really think that you should look at the mathematical proof I posted and try to understand it. You are just shrugging your shoulders and ignoring it instead of taking the opportunity to learn something. I cannot help if you don't tell me what part of it you do not understand. Do you understand the 3 assumptions?

 

[waynexk8]

You hit a Snooker ball with a force of 10N, the cue comes in contact with the ball for .1 of a second, it moves 1m hits another Snooker ball, the original Snooker ball stops, and the second Snooker ball goes 1m. You hit a Snooker ball with a force of 5N, the cue comes in contact with the ball for .1 of a second, it moves 1m hits another Snooker ball, the original Snooker ball stops, and the second Snooker ball goes 500mm. This will be the same as your muscles moving a weight, as the reaction, reaction, will then be the muscles moving the weight, and the reaction, reaction from the force and impulse will be the tension on the muscles.

I do not get why/how you are saying that both hits have the same force and impulse, average force and average impulse, when I stated the Ns were different but the time frame the same. As everything, the N, the time, and the distance the second Snooker ball moves points/says that more force and more impulse, more average force and more average impulse was used, there is nothing saying both were the exact same. Back later.

Wayne

 

[douglis]

If, and I do mean if, if you’re going to debate on a physics forum, please "try" and state/show your case for being right, and why you think that these are irrelevant here.

Wayne

I have stated so many times why I think your examples are irrelevant that I'm sure it would be a waste of time to do it once again.But what the hell...

I showed you studies that found that the greater fluctuations of the same average muscle force are more energy demanding.
For example,if the applied force varies from 0 to 10N in a second requires more energy than an applied force that varies from 4 to 6N in a second.Despite the fact that impulse and the average force is identical.
That's the ONLY reason that fast reps lead to failure faster.They're more energy demanding(due to their greater fluctuations of force) even though their average force per second is identical with slow reps.

You hit a Snooker ball with a force of 10N, the cue comes in contact with the ball for .1 of a second, it moves 1m hits another Snooker ball, the original Snooker ball stops, and the second Snooker ball goes 1m. You hit a Snooker ball with a force of 5N, the cue comes in contact with the ball for .1 of a second, it moves 1m hits another Snooker ball, the original Snooker ball stops, and the second Snooker ball goes 500mm. This will be the same as your muscles moving a weight, as the reaction, reaction, will then be the muscles moving the weight, and the reaction, reaction from the force and impulse will be the tension on the muscles.

Your example proves that you didn't even read or you didn't understand(most probably) the assumptions that DaleSpam is talking about.

We're talking about cases where the load starts and stops at rest hence there's zero change in momentum.In your example,in the first case the one ball hits the other with double speed than the second case(I assume that the 10N and 5N are average values) and in .1sec the balls stop in both cases.
So the first case has double change in momentum than the second.Nothing like our weight lifting example where the change in momentum is always zero.

 

[Dale]

You hit a Snooker ball ...

I do not get why/how you are saying that both hits have the same force and impulse, average force and average impulse

I have never said that, nor has anyone else. Please focus on the weightlifting scenario and stop bringing in irrelevant scenarios. They only distract you and the respondents.

If you really wish to discuss Snooker, then open a separate thread and keep the two topics separate.

 

[waynexk8]

Back in full tomorrow, and the pm’s, SORRY for not getting back to the older, will do first thing before anything else.

Two weightlifters, one lifts the weight up and over his head and back down in 3 seconds, second only lifts it to his knees in 3 seconds. Are you telling me that both used the same force, same impulse/strength ? If so, then how did the weightlifter have more impulse/stremgth to be able to lift the weight overhead against gravity and air resistance ? Someone said that here you work out the average in two ways, maybe this is where we are losing each other.

Wayne

 

[Dale]

Two weightlifters, one lifts the weight

I think that the phrase "the weight" means that both weightlifters are lifting the same weight.

up and over his head and back down

I think this means that they both start at rest and stop at rest.

in 3 seconds, second only lifts it to his knees in 3 seconds.

So the time is the same.

Are you telling me that both used the same force, same impulse/strength ?

Let's use what we have learned so far and see if we can work this out.

First, we need to check our three assumptions:
1) Newton's second law: F_{net}=ma
Since we are on Physics Forums, let's go ahead and allow this assumption since it is pretty central to physics, even though it wasn't explicitly mentioned in the problem.

2) Free weights have only two forces acting on them, gravity and lift from the person: F_{net}=F_{grav}+F_{lift}
In the problem description there was no mention of any machine that would cause friction or any other force on the weight, so this one is a valid assumption also.

3) The lift starts and stops at rest: v_i=v_f=0
You explicitly mentioned this in the problem description.

So all three assumptions are valid. We know that whenever the assumptions are valid we can calculate the weightlifter's impulse by:
I_{lift} = -mg(t_f-t_i)

So for lifter 1 we have: I_{lift1} = -mg(t_f-t_i) = -mg (3 s)

And since the mass is the same for both and obviously gravity is the same for both, then for lifter 2 we have: I_{lift2} = -mg(t_f-t_i) = -mg (3 s)

Therefore I_{lift1} = -mg (3 s) = I_{lift2}

If so, then how did the weightlifter have more impulse/stremgth to be able to lift the weight overhead against gravity and air resistance ?

He used a higher peak force, and a lower minimum force, so that the average force was the same.

 

[douglis]

Two weightlifters, one lifts the weight up and over his head and back down in 3 seconds, second only lifts it to his knees in 3 seconds. Are you telling me that both used the same force, same impulse/strength ?

Wayne

Yes...we're telling you exactly that.Both used the same average force and impulse.

 

[waynexk8]

Wayne, there are very few important details:

1) how much is the weight
2) how long does it take
3) what is the change in velocity (in our previous discussions I had always assumed this to be 0, but with your discussion about throwing weights it is no longer certain)

1,
The weight is 80% of a person’s IRM, {Repetition Maximum} we can call that 80 pounds.

2,
.6 of a second to move 20 inch.

3,
This is a tricky one, an in lifting weight you usually grasp the barbell with your hands, and that griping the bar with your hands stops the weight from moving out of your hands if you press to hard and the weight takes on a faster motion than you press it, as of inertia, mind you this is very hard to do with this weight, just about imposable, it could happen you could press that weight with a constant accelerating force for the whole ROM {Range Of Motion} but you cannot as of the biomechanical advantages and disadvantages thought-out the ROM. But what I tried on a Smith machine was pressing the bar as hard/fast as I could with our hands, and the bar did not leave my hands at all, as the only reason you grip the bar is for stability, and to make the bar part of you, and you just need to hold the bar with most lifts, or they would fall out of your hands. So basically the lift is thrown, but your hands are griping the bar, you’re your muscles take any upward motion, inertia there is.

Nothing else matters. With that information you can determine the impulse, without that information you cannot. If those three factors are kept constant then none of the other details about the force matter, the impulse is the same.

Are you taking of impulse on this Earth, with taking gravity, air resistance, the mass/weight of the object, the acceleration components, the action reaction principle, or are you talking of in space ? One more, for my deceleration, are you taking “anything” off my acceleration impulse ? As when I asked D. about the propulsive study, for some unknown reason, he seemed to be taking force from my acceleration force with my deceleration force.

No way to tell from your description. Weight and change in velocity are missing.

The weight is 80 pounds, and the 1RM the most the person can produce is 100 pounds {this is very important} the change in velocity will be moving the weight at an acceleration, as the whole lift is 20 inch that takes .6 of a second, as no force is produced when the weight is not moving, that before the lift and that the transition, we are not concerned with this.

The weight is 200 kg in both cases, the duration is 1 s in both cases, and the change in velocity (0 m/s) is the same in both cases, therefore the impulse is the same in both cases (~2000 Ns). What you are neglecting is the negative impulse from the sudden deceleration which is -2000 Ns for one and -6000 Ns for the other.

Now this is interesting, but where you lose me and where I need help. Not sure what you mean by the change in velocity is 0m/s how can it be 0 ? And why are you taking N’s from both lifts as of the deceleration, I think that cannot be done, as the impulse you use on the acceleration or the constant fierce, is and has been used, you cannot take anything from this phase, on paper you may be able to do this, but in practice you lift the weight, that takes force and impulse, and when its taken force and impulse, you cannot have that back, or take anything away from it, as it has been used and used up, also do not you still use force and impulse when you decelerate, thus I do not understand why you take it away ? Could you please explain, and also just add up the both forces without taking anything from them, I say and these numbers mean nothing, that the fast could be a 10000, and the slow 6000, thus if you take 6000 and 2000 from these both would have the same impulse 4000. But I am sure this is where we are going wrong, as you cannot take anything of the force and impulse I have used.

Why and how can you take anything away ?

Yes. And the average force is ALWAYS equal to the weight if the barbell starts at rest and stops at rest.

In a rep the weight starts at rest at a low point, is lifted upwards to a high point where it is again at rest (at least momentarily), and is then brought back down to the same low point once again at rest. Correct? I have never discussed the time before nor after a rep, but this is my understanding of what a rep is.

I would prefer to return to discussing weightlifting and reps like normal, but we can continue talking about throwing weights if you prefer.

Yes, not only can you, physics demands it. Irrevocably and invariably. If those three important factors are held constant, then everything that happens in the middle must unavoidably average out. No amount of EMG readings or anecdotes about fatigue can change this simple physical fact.

Not sure about that, as that’s why Kinology was added as a branch of physics.

Yes, exactly. That quantity, the force applied for some duration, is the impulse.

Yes, which is why impulse includes both.

Will have to come back to these later, have been very busy.

Wayne

 

[waynexk8]

The weight is 200 kg in both cases, the duration is 1 s in both cases, and the change in velocity (0 m/s) is the same in both cases, therefore the impulse is the same in both cases (~2000 Ns). What you are neglecting is the negative impulse from the sudden deceleration which is -2000 Ns for one and -6000 Ns for the other.

Yes. And the average force is ALWAYS equal to the weight if the barbell starts at rest and stops at rest.

If I ran a 100m in 10s, I might be decelerating for some, but if you added my forward forces up, you would not or could not take anything away, as to move a 100m in 10s, takes force on acceleration and deceleration, and it’s the same in lifting a weight up, when it’s on the deceleration, the muscles are still using force, thus which the above taking force off, cannot be done

I do not really understand that sorry. So if the weight was 80 pounds, you are saying if we average all the highs and lows from several fast reps, and one slow rep that takes the same time frame, we will get 80 ? But if we divided 80 by a 1000 we get 12.5, but if we than divided 160 by 2000 we still get 12.5, so if the slow was 80 by a 1000, and the fast was 160 by 2000, both would have the same average, but the fast would have the higher impulse. As I said before, average has no meaning in this debate.

In a rep the weight starts at rest at a low point, is lifted upwards to a high point where it is again at rest (at least momentarily), and is then brought back down to the same low point once again at rest. Correct? I have never discussed the time before nor after a rep, but this is my understanding of what a rep is.

Well that’s right and wrong, as to be honest, the times when the weight is at rest, does nothing for the force being used, thus I do not understand why we need to count this in or mention it ? All we need to add up is the time that the weight is moving.

I would prefer to return to discussing weightlifting and reps like normal, but we can continue talking about throwing weights if you prefer.

As I said before, the rep moved as fast as you can in the fast reps, is a throw, but it’s just your grip around the weight holds it there, or do you see the rep and a throw rep different, if so please say.

Yes, not only can you, physics demands it. Irrevocably and invariably. If those three important factors are held constant, then everything that happens in the middle must unavoidably average out. No amount of EMG readings or anecdotes about fatigue can change this simple physical fact.

Hmm, but I ask this again, are you just talking about the forces from the muscles, or the forces from the muscles and on the muscles, sorry if you missed this. What I mean by the forces on the muscles, to which there are NONE with the slow rep, are the reaction forces when the weight is traveling down at a controlled decent, as you know far better than me, is that when the weight is being lowered under control, if the weight was 100 pounds, and you instantly pushed a scales under the eight when it was being lowered at 20 inch every .5 of a second, that the scales would not register the exact weight of the weight which is 100 pounds, it would register higher, say 110 pounds, and for the whole of the negative, the acceleration components of the weight being lowered under control, the weight or forces are going up and up from a 100 pounds to 105 pounds, 110 pounds and so on, and this extra forces is going on the muscles, than at the bottom transition from negative to positive you have the very high peak forces on the muscles. But as I said, the slow rep does not have any of those forces on the muscles; the slow rep just has forces from the muscles.

If you ever try negative only training, where you just lower the weight, you also fail far far far faster if you lower the weight faster, but NOT too fast as you will then be talking less force than the weight. The trick is on a normal positive negative rep, is on the negative it’s too light, as the way the muscles evolved you can lower say 40% more than you can lift, so as you see the negative is to light, so as I said, the trick is to lower the weight quite fast for a third of the distance, and then when the weight is heavier than normal as of its moving down faster, then slighter slow it, and at the transition give as much upward force as you can, then the peak force will be very high, as high as say 140% To which these forces are then from and on the muscles, you do NOT get these forces on the muscles with the slow rep.

I hold 80 pounds for 3 seconds, what you’re saying, is that if I move this weight up and down for 3 seconds as fast as I can, the high will be far higher 80, but also the lowers will be far lower than 80, so add these up and they are both equal. But I don’t see it like that or working like that, take what I said before, I lift 80 pounds over my head and it takes me 3 seconds, you cannot lift 80 pounds off the ground, but you try for 3 seconds, I don’t see on that, how you think you have produced the same impulse/strength as me in this time frame, as you could not have, as you did not have enough impulse/strength to overcome gravity and lift the weight, so you say produced 79 pounds of force for 3 seconds, but I must have produced more, to accelerate the weight up over my head. As you know you do use more force and impulse over time if you accelerate something rather than just lift it at a constant speed.

Isaac Newton's second law tells us that force cause’s acceleration, and we can write that law as an equation;
F = m a.
In other words, we can define force as the rate of change of momentum: the more quickly something's momentum changes, the more force is acting on it.
Rearranging the equation one more time, we get;
F t = m v
And as you showed me, the change in momentum/movement is equal to the force acting on something multiplied by the time for which it acts, and it's called an impulse. The impulse is simply F × t.

Yes, exactly. That quantity, the force applied for some duration, is the impulse.

Yes, which is why impulse includes both.

Thank you for the impulse again; I don’t understand why people did not tell me that before.

Wayne

 

[waynexk8]

No amount of EMG readings or anecdotes about fatigue can change this simple physical fact.

The EMG does not read, add up or even know or acknowledge fatigue, it only reads muscle activity, that’s muscles force, or better still muscle impulse, and it reads this over a time frame be it high or low and averages them out.

It would be nice to get an expert on EMG on this, I will E-mail someone.

Wayne

 

[Dale]

This is a tricky one

OK, then from now on you need to specify the initial and final velocities for every single scenario that you propose. If we cannot rely on it being 0 then you must tell us what it is in order to analyze a given scenario.

Are you taking of impulse on this Earth, with taking gravity, air resistance, the mass/weight of the object, the acceleration components, the action reaction principle, or are you talking of in space ?

Either way. If g = 9.8 m/s² then you are on earth, if g = 0 then you are in space. You could analyze a weightlifter on the moon, mars, or Jupiter by substituting the appropriate local value of g.

no force is produced when the weight is not moving

This is completely and obviously false. Put a weight on a scale. It is not moving, but the scale measures a force. Or hold a weight stationary at arm's length and tell me that there is no force.

Not sure what you mean by the change in velocity is 0m/s how can it be 0 ?

The change in velocity is the final velocity minus the initial velocity. If the final velocity is 0 and the initial velocity is 0 then the change in velocity is 0-0=0.

And why are you taking N’s from both lifts as of the deceleration, I think that cannot be done, as the impulse you use on the acceleration or the constant fierce, is and has been used, you cannot take anything from this phase, on paper you may be able to do this, but in practice you lift the weight, that takes force and impulse, and when its taken force and impulse, you cannot have that back, or take anything away from it, as it has been used and used up, also do not you still use force and impulse when you decelerate, thus I do not understand why you take it away ?

Impulse is a vector quantity, meaning that it has a magnitude and a direction. The sum of an impulse of 1 N*s up and 1 N*s down is 0.

Could you please explain, and also just add up the both forces without taking anything from them, I say and these numbers mean nothing, that the fast could be a 10000, and the slow 6000, thus if you take 6000 and 2000 from these both would have the same impulse 4000. But I am sure this is where we are going wrong, as you cannot take anything of the force and impulse I have used.

Why and how can you take anything away ?

Because you stopped the weight. If you had thrown the weight then the impulse would have been greater on the weight that was moving faster. I am sorry that you don't like the definition of impulse, but it is what it is.

 

[Dale]

Well that’s right and wrong, as to be honest, the times when the weight is at rest, does nothing for the force being used, thus I do not understand why we need to count this in or mention it ?

We are not counting it, we are just saying that the condition of being at rest defines the start and the end of the rep. While the rep is going on the weight is moving. When the weight is not moving the rep is not going on.

If you like then you can choose alternative end conditions if you wish, but then you must specify the exact velocity at the beginning and the exact velocity at the end.

All we need to add up is the time that the weight is moving.

Which means that the initial and final velocity is 0, because otherwise the weight is still moving when you stop adding it up. I.e. the rep starts at the very last instant that the velocity is 0 and ends at the very first instant that the velocity becomes 0 again.

As I said before, the rep moved as fast as you can in the fast reps, is a throw, but it’s just your grip around the weight holds it there, or do you see the rep and a throw rep different, if so please say.

Yes. According to my view the weight starts and stops at rest in a rep and it starts at rest but ends with some non-zero velocity (as it leaves your hand) in a throw.

What I mean by the forces on the muscles, to which there are NONE with the slow rep

No, this is incorrect. By Newton's 3rd law the forces on the muscles are equal and opposite to the forces from the muscles. That is the "every action has an equal and opposite reaction" law. I don't know where you got the incorrect idea that those forces were ever different under any conditions.

I hold 80 pounds for 3 seconds, what you’re saying, is that if I move this weight up and down for 3 seconds as fast as I can, the high will be far higher 80, but also the lowers will be far lower than 80, so add these up and they are both equal.

Yes.

But I don’t see it like that or working like that, take what I said before, I lift 80 pounds over my head and it takes me 3 seconds, you cannot lift 80 pounds off the ground, but you try for 3 seconds

Check the three assumptions!

You really need to stop proposing irrelevant scenarios and assuming that my analysis of this scenario applies to unrelated scenarios. If the weight is on the ground then the second assumption is explicitly violated and you are discussing a completely different scenario. If you want to analyze a different scenario then start a new thread. Frankly having to repeat this so often is becoming quite irritating.

 

[waynexk8]

Hi waynexk8, you have made this comment enough times that I think that we need to address it head-on. Here is the proof that your claim is incorrect. Please walk through the proof and ask questions. This is something that you need to understand, so let's take the time to do it correctly.

Here is the mathematical proof, it relies only on 3 assumptions:
1) Newton's second law: F_{net}=ma
2) Free weights have only two forces acting on them, gravity and lift from the person: F_{net}=F_{grav}+F_{lift}
3) The lift starts and stops at rest: v_i=v_f=0

I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} ma \; dt = m \int_{t_i}^{t_f} \frac{dv}{dt} \; dt = m (v_f-v_i) = 0
I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} F_{lift} \; dt + \int_{t_i}^{t_f} F_{grav} \; dt = I_{lift} + \int_{t_i}^{t_f} mg \; dt = I_{lift} + mg(t_f-t_i) = 0
I_{lift} = -mg(t_f-t_i)

So the impulse of the lift depends only on the weight (mg) and on the time (t_f-t_i), not on the speed of the lift.

I do not understand the physics equations.

Bur let’s look at some averages, a Man runs the 100m 5 times and they go like this, 10s, 10s, 10s, 10, 30, next Man 14, 14, 14, 14, 14, even thou the averages are the same, in two days time it would be mathematically fact that the first Man would win 80% of all races.

However we have several problems, if you showed on paper that the Bumble bee could not fly, and I showed you it flying, I think you and every one would say there is something wrong with the equations.

The most simple and most compelling proof and evidence to me, that you are using more impulse per unit of time it the fact that you fail faster with a faster rep speed, note the following to which I have said before. Fast 110, 100, 100, 90, 00, slow 80, 80, 80, 80, 80. Now please look at the first four numbers, fast 110, 100, 100, 90, slow 80, 80, 80, 80, now here is what I am trying to say and explain, but no one seems to listen or understand for the first four fifths I use in the fast 110f, {force} 100f, 100f, 90f = 400f, on the other hand the slow, 80f, 80f, 80f, 80f = 320f, this means to this point in time my force is higher, and my impulse is higher, as I only have a temporary amount of force in my muscles until I rest them, when using a higher amount of force and impulse force in this first four fifths, I MUST/DO fatigue the force system, or the limited force I have in my muscles faster than if I just used 80f, 80f, 80f, 80f = 320f as 80 is ONLY 80% of my strength, but a 100f is 100% of my strength/impulse, and if you constantly try and use 100% of your strength/impulse you “are” going to use more strength/impulse up faster, you can try and average this all out on paper, but with a system like the muscles, or an engine, or anything else for that matter, if you try to run anything using its full/maximum force and impulse, it I using force and impulse faster per unit of time, so the muscles will have to fail faster, and the machines engine will burn out faster, next time you use your car, rev it up on full for an hour, and your engine will burn out, but rev it at far less, and you could sit there all day and your engine will be fine. {ho and the last one fifth where I am on the deceleration but still using force and impulse, will be for .1 of a second, will give “no” time at all for these systems to make up the force, impulse or energy}

The other problem we have, is that if you claim that both use the same force and impulse per same time frame, and the 80f makes up, balances out the 100f is that if you were right, why does the fast use 6 times more energy, as its going 6 times as fast in the same time frame, if both were using the same force and impulse per unit of time, then the energy for the same time frame should be the same, but it’s not, the fast uses 6 times more energy. Force and impulse used per unit of time, has a direct usage of the amount of energy used, {see a nutritionist book or kinology} meaning if my impulse used was a 100, then the energy would be a 100, if the impulse was 200 then the energy used would be 200, but in the fast slow examples, the energies are always linear in agreement to the speed you move the weight, and what happens when you move anything faster, like in acceleration the weight, you have to use more force and impulse, if as some here do not agree with this, then please state why the fast uses more energy in a linear motion as of the speed/acceleration its going, I have said why, if I am wrong, this is physics and there will be another reason, that’s if you are right.

The only way the person using the 80f could put out as much force and impulse as the 100f, is maybe when both use the same energy in the same time frame, but that scientifically impossible.

Then you have a very sophisticated bit of computer, the EMG, stating the faster uses more muscular activity in the same time frame, this it because the peak highs, 110f and the highs 100f cannot be made up by the lower 80f, as the 100f will and does take it out of you far more and faster than the 80f that is why you fail roughly 50% with the faster rep.

I lift the weight up thought the force of gravity and down 6 times in 6 seconds, you just lift it up and down once, I used 100% effort, force strength/impulse, or try to do for 100% of the time, you use 80% efforts, force strength/impulse for 100% of the time.

Let's compare a car traveling different distances in the same period of time.

400M in 10 seconds = 40M/S
2000M in 10 seconds = 200M/S

Are some saying that it do not matter that the car traveled 5 times the distance in the second example compared to the first example ? Are some saying that the car produced the same amount of horsepower in both examples ? Are some, I don’t think so, saying that the car had an equal rate of fuel usage in both examples ? Both examples produced horsepower for the same duration; however they certainly did not produce the same amount of horsepower.

Moving 100 pounds a distance of 10M in 10S is not going to require the same amount of force compared to moving 100 pounds a distance of 50M in 10S. The duration of tension/force is the same; the magnitude of tension/force is not.


Will get back to the other posts shortly, thank you for the debate, it’s very interesting, and I am learning a lot.

Wayne

 

[douglis]

, thank you for the debate

It's not a "debate".It's a guy with superhuman patience trying to teach you simple physics facts.

, it’s very interesting

It's interesting only to see how far your obsession could go.
You have all the mathematical proofs that you're WRONG,you admit that you can't understand them,yet you still have the nerve to try to convince us that you're right.

, and I am learning a lot.

No...you're not learning anything.Every day you get worse.I imagine DaleSpam's disappointment when he'll read that you wrote " I MUST/DO fatigue the force system" and all your other nonsensical and irrelevant examples.

 

[sophiecentaur]

I do not understand the physics equations.
Wayne

Just because you don't understand them doesn't mean that they don't apply. If you don't understand them then you cannot argue against them and any number of made-up scenarios that you propose cannot disprove the basic facts. The reason that Physics is described with Maths is that the Maths is the only way to deal with complicated scenarios. Endless 'stories' about different weight-lifting situations do nothing to make any of this clearer. If you refuse to get involved with the Maths then you are effectively excluding yourself from any argument and you just have to accept the 'authority' of the textbooks and people who can use the knowledge.
There is no way round that, I'm afraid.

 

[waynexk8]

I agree. That is why scientists continuously put the paper equations to the most careful and rigorous experiments that they can imagine, and great honors are bestowed on scientists which prove equations wrong and come up with better ones. The particular equations that we are discussing here have passed the most exacting tests imaginable for over 3 centuries of the most brilliant experimentalists. The real world tests completely agree with the equations I posted.

Aaa, but no one here can show me one of these experiments ? Also, I know sorry I have said this before and you may have answered them, and for first is a new one.

1,
The faster produces more heat.

2,
The faster moves the weight 6 times further in the same time frame, if you are correct, why does not your slow rep move the weight the same distance, if the impulse is the same ? Why does the slow not catch up the distance when I am on the deceleration ? Mind you saying that, the slow must have a deceleration also. Here is how I see it, let’s just say you are holding the weight, as you also say this exerts the same impulse as me, even thou I am moving the weight for the same time frame, you hold the weight and you exert 80 pounds of force for 1 second, I also at first before the test starts, exert 80 pounds to hold the weight, so far all is equal, however to accelerate the weight I have to use more and more force, my maximum force is a 100 pounds, so I use a 100 pounds of force for one second, and move the weight say 1m, what you seem to do and say, is that when I am on the acceleration my impulse is say 100, but when I am on the deceleration, I am using less force than the weight, so you take 20 off, leaving a impulse of 80, just like you holding the weight. But as I said before, you cannot take anything from my acceleration, it’s all ready been used, and when I am on the deceleration, I am still using force, so basically you HAVE to add on to my accelerating impulse, do you see my point ? I Acceleration is directly proportional to force because, when the mass is constant, so the more force I use moving the weight, the more acceleration, thus more impulse, as of Newton's second law states, F=ma, so if we find value of acceleration then formula will become, F/m=a that is a=f/m this indicates that it is directly proportional.
3,
I use more energy/calories, if the impulse was the same, we both should use the same energy/calories, as we do not, why ? All know why I think it is.

4,
EMG.

No, the way of doing the equations is correct. If you disagree then please point out EXACTLY where I made my mistake.

Sorry here, maybe I did not explain right, I am not sure where, to put it simple, the equation you are doing, may be leaving out something, some variable that no one has thought of. It’s like when we did the equations for pushing up a weight fast and then stopping, and then too see how far the weight went on its own, we worked out it would travel 3 inches, and this was totally correct, but when we all tried it, it did not move 3 inch, it did not move at all, but still the equations were right, the reason the equation was put down wrong, or why it was not right in this instance, as the equation was worked out for a constant acceleration force, or should I say impulse force, but, and this is the important part, because as I have said before, as of the biomechanical disadvantages and advantages of the muscles over the range of motion, the muscles could not push with a constant acceleration, as at some parts of the lift they were very strong, and other parts very weak, thus the equation was NOT for muscles pushing with an inconstant force. I am not saying this equation is anything like that, but if you fail faster, move the weight more distance, use more energy, and the EMG stats the fast higher in the same time frame, there must be something wrong. I simply say it requires more force to perform more work within the same period of time, It requires greater acceleration to accelerate an object from 0 to 100m/s in 5 seconds than it does to accelerate the same object from 0 to 50M/S in 5 seconds. In order to move the same load over the same distance within half the time from a standing start {zero velocity} you're going to need greater acceleration. What I am trying to say is that assuming both weights start at zero velocity and they both reach their top speed within the same period of time, the faster weight would require more acceleration than the slower weight, I am not talking about maintaining velocity, but about acceleration, which of the following requires greater acceleration ? Moving 100 pounds a distance of 100M in 1 second from a standing start, moving 100 pounds a distance of 100M in 2 seconds from a standing start ?

10 N*s is indeed more impulse than 5 N*s, nobody is disputing that.

Odd, now you are saying I am right, well the above is all I am saying, so how and why do you say I am wrong in the other instances ? As that’s what I am doing in the repping, using more Ns thus more impulse ?

What you don't seem to realize is that after you have applied 10 N*s and I have applied 5 N*s your weight is moving faster than mine. Over the course of the next second, if we both bring our weights to rest, then you must use LESS force than me such that our average force is the same.

But you too have to decelerate and use less force than the weight ? If I moved a weight 20 inch in .6 of a second, why can I not decelerate and bring it to a complete stop very fast, as I actually do in repping {please just do this, but please watch you don’t jerk too much, just press you empty arm up very fast and bring it to a stop, do this a few times, and you will see its more like a throw, it’s not so much you decelerate, it’s more like you constantly accelerate, but there is a thick steel plate 1mm from your full extent. Also, why do you think I need more time to decelerate than you ? Also I still don’t see how you can say that a 100 force like this 100, 100, 100, 100, 00, and 80, 80, 80, 80, 80, HOW can one 80 make up the force of four 100s ? 80 force can never be as high as 100. Take 5 cars driving into a wall, the first four cars dive in 100mm the fifth, does not start, the 5 other cars drive into the wall, the five cars all drive into the wall 80mm, not one car has drove into the wall a 100mm, all are 20mm short of that force, then we still are leaving out the 5 peak force from the transition from negative to positive.

Good to hear. I certainly have never mentioned anything about any time before the weight has started to move or after it had stopped.

K. but you did mention the weight starting at zero and ending at zero, all I talk is the times between these.

The impulse is the same for fast or slow, the energy is higher for fast than for slow, therefore, the energy does not depend only on the impulse.

However you actually have not answered why the fast uses more energy, you have and are are very clever and just say the energy does not depend only on the impulse, but you still can not say why the fast uses more energy. This could be a part of the equation that you and the other are doing wrong, you maybe have not added in that the fast must and does use far far far more energy in the same time frame as the slow, as do not you agree, there must be a physics answer to the fast using more energy, and all know why I state it is, or what else can the fast be doing or using if it’s not more impulse. It does use more power, its transferring more energy faster because its moving the weight faster, and what do you need to accelerate the weight faster, more force/strength/impulse I say, if not, what ?

The impulse is the same for fast or slow, the EMG is higher for fast than for slow, therefore, the EMG does not depend only on the impulse.

It must, it measures muscle activity, and more activity the higher the reading, and the more muscle activity, means more force/strength/impulse used, what else could it mean ?

The impulse is the same for fast or slow, you fail sooner for fast than for slow, therefore the failure time does not depend only on the impulse.
However again you have not answered why the fast hits failure faster, you have and are very clever and just say failure time does not depend only on the impulse, but you still can not say why the fast hits failure faster. This could be a part of the equation.

That would be correct IF the failure depended only on the average strain, therefore the failure does not depend only on the average strain.

I like what you said there, but its far too late to comment, it’s about 2.

Let's say that the weight is 75 lb and you lift for 1 s at 100 lb and I lift for 1 s at 80 lb. At the end of that second your weight is moving 11 ft/s and my weight is moving just 2 ft/s. Now, let's say we want to decelerate the weight in 0.5 s, in order to do that, you must use a force of 25 lb, and I must use a force of 65 lb. Any other force and the weight will not stop, i.e. the rep will not be at an end.
For me: 80 lb * 1 s + 65 lb * 0.5 s = 112.5 lb*s
For you: 100 lb * 1 s + 25 lb * 0.5 s = 112.5 lb*s

Your opinion is wrong, as proven mathematically, and the real world careful controlled experiments confirm the math. The average force is the same for fast and slow. Momentary muscular failure occurs sooner for fast than for slow. Therefore momentary muscular failure is not due only to the average force.

Yes. And your weight is moving faster than mine. Since the rep doesn't end until the weight is stopped again, this is not a complete description of a rep.

waynexk8, I really think that you should look at the mathematical proof I posted and try to understand it. You are just shrugging your shoulders and ignoring it instead of taking the opportunity to learn something. I cannot help if you don't tell me what part of it you do not understand. Do you understand the 3 assumptions?

Wayne

 

[Dale]

I do not understand the physics equations.

Can you be more descriptive. I know that you have used f=ma, so I know that you understand some equations. Which equations do you not understand here?

Bur let’s look at some averages, a Man runs the 100m 5 times and they go like this, 10s, 10s, 10s, 10, 30, next Man 14, 14, 14, 14, 14, even thou the averages are the same, in two days time it would be mathematically fact that the first Man would win 80% of all races.

Irrelevant scenario, start a new thread.

However we have several problems, if you showed on paper that the Bumble bee could not fly, and I showed you it flying, I think you and every one would say there is something wrong with the equations.

Irrelevant scenario, start a new thread.

waynexk8, you are making a bunch of claims that you have never substantiated. I think that it is time for you to justify your claims:

The most simple and most compelling proof and evidence to me, that you are using more impulse per unit of time it the fact that you fail faster with a faster rep speed

I accept the experimental data that you fail faster with a faster rep speed. However, what exactly is the relationship between impulse and failure and how did you obtain that relationship?

if both were using the same force and impulse per unit of time, then the energy for the same time frame should be the same

Please show your derivation of this. How did you arrive at this relationship?

Then you have a very sophisticated bit of computer, the EMG, stating the faster uses more muscular activity in the same time frame

I also accept the experimental data that EMG readings are higher with a faster rep speed. However, what exactly is the relationship between impulse and EMG readings and how did you obtain that relationship?

Let's compare a car traveling different distances in the same period of time.

Irrelevant scenario. Start a new thread.

 

[Dale]

Aaa, but no one here can show me one of these experiments ?

Sure we can, these experiments are performed every day in introductory physics labs across the world. You can probably do them yourself. Here are a few examples:

http://sdsu-physics.org/physics_lab/p182A_labs/indi_labs/Newtons2ndLaw.pdf
http://swift.sonoma.edu/education/Newton/Newton_2/html/Newton2.html
http://galileo.phys.virginia.edu/outreach/8thgradesol/Newtons2Frm.htm
http://webpages.charter.net/rhutchis/Physics/Unit5/Lab_Newtons_2nd.pdf

The first recorded experimental confirmation of Newton's 2nd law was by Galileo in his famous "ramp and ball" experiment:

http://en.wikipedia.org/wiki/Experiment#Galileo_Galilei

 

[Dale]

The faster produces more heat.

Agreed. I have never claimed that there is a relationship between heat and impulse, so the relationship is your idea. What do you believe is the relationship and how did you come to that?

The faster moves the weight 6 times further in the same time frame, if you are correct, why does not your slow rep move the weight the same distance, if the impulse is the same ?

Why should it? Again, this is your claim, not mine.

Here is how I see it, let’s just say you are holding the weight, as you also say this exerts the same impulse as me, even thou I am moving the weight for the same time frame, you hold the weight and you exert 80 pounds of force for 1 second, I also at first before the test starts, exert 80 pounds to hold the weight, so far all is equal, however to accelerate the weight I have to use more and more force, my maximum force is a 100 pounds, so I use a 100 pounds of force for one second, and move the weight say 1m, what you seem to do and say, is that when I am on the acceleration my impulse is say 100, but when I am on the deceleration, I am using less force than the weight, so you take 20 off, leaving a impulse of 80, just like you holding the weight. But as I said before, you cannot take anything from my acceleration, it’s all ready been used, and when I am on the deceleration, I am still using force, so basically you HAVE to add on to my accelerating impulse, do you see my point ?

I am not taking anything away. In the first second we each exert 80 lb*s. In the second second you exert 100 lb*s for a total of 180 lb*s and I exert 80 lb*s for a total of 160 lb*s. Then (assuming that in the 3rd second you bring your weight to rest and I keep mine at rest) in the third second you exert 40 lb*s for a total of 240 lb*s and I exert 80 lb*s for a total of 240 lb*s. Impulse is never taken away, it simply accumulates. The 180 lb*s that you exerted in the first two seconds is simply added to the impulse that you exerted in the third second, you merely add less that third second.

Also, don't forget that impulse is a vector quantity. So a negative impulse doesn't mean impulse is taken away, just that it is added in the opposite direction.

I Acceleration is directly proportional to force because, when the mass is constant, so the more force I use moving the weight, the more acceleration, thus more impulse, as of Newton's second law states, F=ma, so if we find value of acceleration then formula will become, F/m=a that is a=f/m this indicates that it is directly proportional.

Interesting that you choose to quote Newton's second law at me when you seem to disbelieve it above. If you accept Newton's second law then my claims regarding impulse follow. If not, then it is rather disingenuous to use it here.

Do you accept Newton's second law?

I use more energy/calories, if the impulse was the same, we both should use the same energy/calories, as we do not, why ?

Why should we use the same calories for the same impulse? These are different quantitties. I have never claimed that there is a relationship, that is your claim. Can you justify that claim in any way?

Odd, now you are saying I am right, well the above is all I am saying, so how and why do you say I am wrong in the other instances ?

10 N*s is obviously greater than 5 N*s and 10 N*s is obviously equal to 10 N*s. The problem is that you cannot yet correctly calculate the number of N*s over a rep, so you think that they are unequal when they are equal.

As that’s what I am doing in the repping, using more Ns thus more impulse ?

Nope. At least, not unless your reps consist of throwing the weights across the gym.

But you too have to decelerate and use less force than the weight ? If I moved a weight 20 inch in .6 of a second, why can I not decelerate and bring it to a complete stop very fast

You certainly can. And the faster you bring it to a stop the more your upwards force is reduced and the smaller impulse it contributes.

Take 5 cars driving into a wall

Irrelevant scenario. Start a new thread.

However you actually have not answered why the fast uses more energy, you have and are are very clever and just say the energy does not depend only on the impulse, but you still can not say why the fast uses more energy.

A human becomes less efficient at faster speeds. I believe that I have stated this clearly and explicitly multiple times in the previous thread.

as do not you agree, there must be a physics answer to the fast using more energy,

I do not agree. I believe that there must be a biology answer, but I don't believe that there is a physics answer (except insofar as biologists use physics in their answer).

It does use more power, its transferring more energy faster because its moving the weight faster, and what do you need to accelerate the weight faster, more force/strength/impulse I say, if not, what ?

Again, this is not a claim that I have ever made. It is only you who is making the claim that more energy implies more impulse. It is a nonsense claim since energy is a scalar and impulse is a vector, but it is your nonsense claim, not mine. If you want to make a nonsense claim then you must be the one who justifies it, not me.

It must, it measures muscle activity, and more activity the higher the reading, and the more muscle activity, means more force/strength/impulse used, what else could it mean ?

Can you derive the relationship between muscle activity and impulse?

However again you have not answered why the fast hits failure faster, you have and are very clever and just say failure time does not depend only on the impulse, but you still can not say why the fast hits failure faster.

This is true, I have not ever said why the fast fails faster. I do not know why. I do, however, know that it is not due to greater impulse, since the impulse is not greater.

I would guess that the answer is more likely related to energy than impulse, but again the question of energy is one of biological efficiency, not physics. You can physically build a machine that does reps as fast as you like without using any significant amount of energy.

 

[waynexk8]

We are still not sure how much both reps deccelerat and for how long ?

Just thought of something else in the equation. With this, we are on the fast reps, moving the weight up 20 inch and down 20 inch x 6 times in 6 seconds. The slow is up 20 inch and down 20 inch x 1 time in 6 seconds.

Also as I said earlier about the forces “on” the muscles not “from” them. Think I wrote that down wrong, it should be when lowering the weight, the force will be lower than the weight, until the transition from negative to positive, then the forces “on” the muscles are higher than the weight of the weight. Unless you say the that forces I lose in the lowering, too which are then made up by the high forces of the transition, are equalled out by the constant forces which are the same as the weight with the slow reps ?

Thus basically as same as the fast and slow concentric. However I still stick to what I said at first, that is the higher forces and impulses and the higher peak forces and impulses of the fast, say 100 and 140 are higher than the constant forces of the slow 80, thus, and this is the main crux of this debate, what I am saying is these peak and higher forces of the fast, are higher and will put more tension on the muscles or more strain on an engine, as you are using the muscles at a 100% capacity, and using the engine at a 100% capacity, thus the using of any system be it biological or machine, will and does put more tension/strain on the muscles/engine, because 100% is a higher force/impulse, and the lower force/impulse can and does not put as much tension/strain on the muscles/engine, proven as you fail 50% faster with the faster reps.

It’s like dropping a 100 pounds on a bridge that has a braking strain of a 100 pounds, if you had two brides the same, dropped 100 pounds on one, the bridge would break, then drop another 100 pounds on the other bridge and that bridge would also break, it’s the same repping very fast, the muscles will tear.

Now you have the same bridges, drop 80 pounds on the bridge and nothing is broke, drop another 80 pounds on the bridge and again nothing is broke, THIS ADDS UP TO MORE THAN THE 100 POUNDS DROPPING, IT ADDS UP TO 160 POUNDS ON PAPER, but still this 160 pounds as NOT done the same damage to the bridge as one drop from the 100 pounds, THIS IS THE SAME WITH THE FASTER REPS, ONE FASTER REP WILL TEAR THE MUSCLES, several slow reps of even higher force/impulse on paper, will and does not put nearly as much TENSION ON THE MUSCLES AS THE FAST REPS, ADD IN SEVERAL FASTER REPS, and you have more impulse faster, it’s like a pneumatic demolition hammer, you can break concrete very easy the higher the beats/bumps per minutes there are, and the higher the Newton’s are each beat/bump.

The other problem with the slow rep is, that at first you have the 3 second positive, with a constant impulse on the muscles, then as the muscles are 40% stronger in the negative, you have 3 seconds to let the muscles sort of recover, with the fast reps there is not time to recover, it’s a constant high force, high impulse, then a very high peak force, very high peak impulse, then and very fast lower force for the negative, the muscles are under a constant very high tension, not like the slow, where the tension is just median all the time.

In other words, the impact force and impulse of the 100, will and does do more damage than the 80, and for the 80 to make up or balance this damage out, it will have to be produced for a far longer time frame. Impulse force with respect to time. When a force is applied to a rigid body it changes the momentum/movement of that body. A small force applied for a long time can produce the same momentum/movement 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 the slow reps do not have a long time, they have the same time as the fast, That’s why the EMG stats higher, you use more energy, and you fail faster with the fast. I do not understand why some think that a 140 and 100 can be make up by 80 ? In basically the same time frame.

Wayne

 

[sophiecentaur]

Oh Wayne.
Are you still here?
When are you going to say something different?
You are running poor old Dale-spam ragged. If you have half the stamina when you're lifting your weights you must be world class.

 

[waynexk8]

Dalepam, will get back as soon as I can. Would like to see the proof on say YouTube, that proves you right in the practical as well, not sure what you meant by that.

Oh Wayne.
Are you still here?
When are you going to say something different?
You are running poor old Dale-spam ragged. If you have half the stamina when you're lifting your weights you must be world class.

I have not come here to mock, but for a serous scientific debate; I know you are a very intelligent person, in your field of intelligence. It's NOT that I have to say anything diffrent, it's you who have first to prove the below in your favor.

So if you want to prove yourself right, you need to say in a scientific way, why I use more energy/calories if as you think both impulses are the same. Then why a very bit of sophisticated computer, to which can add up more and faster than many humans put together states you wrong. Then you need to say why I fail faster in the faster reps.

Wayne

 

[sophiecentaur]

Dalepam, will get back as soon as I can. Would like to see the proof on say YouTube, that proves you right in the practical as well, not sure what you meant by that.



I have not come here to mock, but for a serous scientific debate; I know you are a very intelligent person, in your field of intelligence. It's NOT that I have to say anything diffrent, it's you who have first to prove the below in your favor.

So if you want to prove yourself right, you need to say in a scientific way, why I use more energy/calories if as you think both impulses are the same. Then why a very bit of sophisticated computer, to which can add up more and faster than many humans put together states you wrong. Then you need to say why I fail faster in the faster reps.

Wayne

Not really. I /we are saying that the problem is too basic on the one hand yet too complex on the other. We have been over all this time after time but the definitions of work, force, energy etc etc are all well established. The problem with your request (demand??!) is that the system has just not been characterised properly. Your gizmo tells you about muscle activity and it could be very 'accurate' but muscles aren't springs and motors. They are far too complex and work to their own agenda. The very fact that you get tired just stand holding something tells you that there is no simple answer.
The reason that you fail faster at faster reps is the way that your muscles respire the food. It's nothing to do with the simple mechanics of the situation so you just can't expect an answer in those terms. However many times you come back with this question, the answer will always be the same from PF.
It is possible that a biochemist could give you some answers to some of this but I guess those answers could be above both of our heads. I can guarantee that any full answer to this would be a biological one, so complicated that you are unlikely even to recognise it as the right one.

 

[Dale]

Dalepam, will get back as soon as I can. Would like to see the proof on say YouTube, that proves you right in the practical as well, not sure what you meant by that.

No. I will not do that for you under any rational conditions. YouTube is not a valid scientific resource. If you will not accept the scientific evidence that is in every introductory physics textbook and repeated in introductory physics labs countless times worldwide every semester then you have no business posting questions on Physics Forums.

I need a clear yes/no answer: do you accept Newton's 2nd law as valid here?

Please answer this before responding to any additional posts.

 

[douglis]

So if you want to prove yourself right, you need to say in a scientific way, why I use more energy/calories if as you think both impulses are the same. Then why a very bit of sophisticated computer, to which can add up more and faster than many humans put together states you wrong. Then you need to say why I fail faster in the faster reps.

Wayne

This is total madness!

For God sake...you have the scientific proof and you admitted that you can't understand it!
It's YOUR claims that faster failure and greater energy usage equates greater impulse so it's YOUR job to prove it somehow.
You're obviously unable to do that(that would violate many known laws anyway) so the least thing you should do is to accept the 'authority' and stop bothering us with nonsensical scenarios.

 

[waynexk8]

Let look at the MMMTs, {Momentary Maximum Muscle tensions} the impact/impulse forces of the weight + velocity + distance + time from the transition from negative to positive, these are the highest force that will come from the muscles and will be on the muscles.

The weight will be lowered under control at 2m/s, and just before the transition, the velocity will be at 1.8m/s, we will say the contact/transition {as we are doing equations for a mass dropping and coming into contact with the floor, some little things in my rep transition will have to be change, but here goes my go now} will be for 0.20s and the mass of the weight = 35kg.

What would happen on the every action there is an opposite reaction force/impulse ? 6 balls weighing 35kg are lowered under control at 2m/s {repping the weight up and down, let’s say 1m in .5 of a second this time} into clay, this represents the tension displayed by the muscles, contact with the clay {peak time of tension on the muscles} for 0.20. Then you lower under control, 35kg at .3m/s, do not all agree that the fast = a far far far deeper hole made into the clay than the slow ? This will also be true for the every action there is an opposite reaction force/impulse, when lifting the weights at different velocities, the faster you move the weight up the opposite reaction force/impulse. So lift 80kg up at 2m/s for 6 times, then lift 80kg up at .3m/s for 1 time, contact with the clay {peak time of tension on the muscles} for 0.40, and the opposite reaction force/impulse, “will” be far far far larger for the rep speed of 2m/s.

Fast,
Impulse will = mass x change in velocity/time

{1.8m/s will represent the change in direction}

I = 35kg {2m/s – {-1.8m/s} / 0.20 s

I = 1N.

X this by 2 for the positive and negative = 2N, then x it by 6 for 6 reps = 12N.

Impact force will be roughly 70N.

Slow,
Impulse will = mass x change in velocity/time

{.2m/s will represent the change in direction}

I = 35kg {.3m/s – {-.2m/s} / 0.40 s

I = .25N.
X this by 2 for the positive and negative = .50N.

Impact force will be roughly 10N

So the fast rep in 6 seconds has an impulse of 2300% more than the slow.

Suppose I have done several things wrong there, {however drop 6 balls into a piece of clay like I said 6 times at 2m/s and then drop 1 ball at .3m/s and you will see what I mean about the peak forces from the transition from negative to positive will/are like. then you also have the higher impulse on the muscles for the fast 6 reps up} but at least I am having a go, like a lot have asked me.

Wayne

 

[waynexk8]

I have stated so many times why I think your examples are irrelevant that I'm sure it would be a waste of time to do it once again.But what the hell...

I showed you studies that found that the greater fluctuations of the same average muscle force are more energy demanding.

Where ? Which one was that. HOWEVER that is NOT an answer, the question was “WHY” is more energy/calories used in the faster rep. Please answer my questions directly, and not with answer that have nothing to do with an actual answer to my question.

For example,if the applied force varies from 0 to 10N in a second requires more energy than an applied force that varies from 4 to 6N in a second.Despite the fact that impulse and the average force is identical.

Yes but why, yet again you give no answer, you are basically saying the same question as me, your just say the fast uses more energy in the same time frame, I know that, look in a nutritionist or kinology book and it will tell you that, and why. What I ask is why the faster uses more energy.

That's the ONLY reason that fast reps lead to failure faster.They're more energy demanding(due to their greater fluctuations of force) even though their average force per second is identical with slow reps.

If it was because of greater fluctuations of force, than as these forces are you say average out over the same time frame, then fluctuations of force, must have nothing to do with it. And yet again, you did not actually say why you think greater/more or/and different amounts of fluctuations of force need to use more energy ?

Your example proves that you didn't even read or you didn't understand(most probably) the assumptions that DaleSpam is talking about.

Let me rephrase, you hit a Snooker ball {move a weight as in repping it up}with a force of 10N, the cue {arm/hand/muscles} comes in contact with the ball {weight} for .1 of a second, it moves 1m {arm moves the weight up 1M} hits another Snooker ball, {the weight goes full ROM, and that jolt goes on the muscles as tension} the original Snooker ball stops, and the second Snooker ball goes 1m. You hit a Snooker ball with a force of 5N, the cue comes in contact with the ball for .1 of a second, it moves 1m hits another Snooker ball, the original Snooker ball stops, and the second Snooker ball goes 500mm. This will be the same as your muscles moving a weight, as the reaction, reaction, will then be the muscles moving the weight, and the reaction, reaction from the force and impulse will be the tension on the muscles.

All you did above is again get out of answering the question, how can you say that a force of 5N is equal to 10N.

We're talking about cases where the load starts and stops at rest hence there's zero change in momentum.

By momentum do you mean movement ? If so there is a change, it’s about 1m

In your example,in the first case the one ball hits the other with double speed than the second case(I assume that the 10N and 5N are average values) and in .1sec the balls stop in both cases.
So the first case has double change in momentum than the second.Nothing like our weight lifting example where the change in momentum is always zero.

Not sure why you say there is no momentum/movement ? Let me show you on the fast, it starts still, then has momentum/movement for 1m then it changes direction and has momentum/movement for 1m, this happens 6 times in 6 seconds, the slow just goes up once and down once in the same time frame.

Please try and think other than you must use more impulse as you fail faster, or something else, it can only be this I think ? If as you think it is more fluctuations of force, then why would that be ? It is because the impulses do not even out at the end ?

Wayne

 

[Dale]

Please answer my question of post 37.

 

[douglis]

If it was because of greater fluctuations of force, than as these forces are you say average out over the same time frame, then fluctuations of force, must have nothing to do with it. And yet again, you did not actually say why you think greater/more or/and different amounts of fluctuations of force need to use more energy ?

The force and the impulse "average out"...not the energy usage.Are you able to understand the difference?
The fact is that fluctuations of the SAME force are more energy demanding.Why?I don't know and I don't care.The answer is irrelevant because is purely biological.
I could search at the discussion page of the studies to see the opinion of the biologists but I won't waste a minute of my life for something irrelevant.

By momentum do you mean movement ? If so there is a change, it’s about 1m

Wayne

Oh God...No...I don't have patience to deal with this!

 

[Dale]

I need a clear yes/no answer: do you accept Newton's 2nd law as valid here?

Hi waynexk8, since your name is crossed out, I assume that you answered this question in the negative and therefore got banned. I am sorry to see that, but if you do not accept basic physics principles like Newton's 2nd law then there is nothing that can or should be done to answer you in terms that you will accept. In fact, with that outlook, I have no idea why you are even bothering to ask your questions on Physics Forums. Any answer you receive here will obviously be based on experimentally validated physical laws like Newton's 2nd.

 

[Dale]

As one final follow-up, let me simply list the take-home messages.

1) Impulse over any number of reps depends only on the weight and the total time, not the speed of the rep.
2) Reps start and stop at rest, the above does not apply to scenarios where the weight does not start and stop at rest.
3) Reps involve only the lifting force and gravity, the above does not apply to scenarios involving another force on the weight.
4) The force from the muscles is always equal and opposite to the force on the muscles.
5) Energy consumption has no unique relationship to impulse.
6) More energy is consumed in the faster rep because the human is less efficient at higher speeds.
7) 6 does not contradict 1 because of 5
8) Fatigue has no unique relationship to impulse.
9) Fatigue occurs sooner in the faster rep for biological reasons (unknown to me).
10) 9 does not contradict 1 because of 8

I think that is a pretty complete characterization of the things I had hoped you would learn.