Quote by douglis
As always you misunderstood what I admited I was wrong.
I was wrong from biology point of view.From physics point of view I was perfectly right and you're saying nonsense.You can't possibly know if you spend more energy if you lift a weight 10 times up and down or you just hold it for 10 seconds.

YES you can, I don’t understand why you don’t know this or could think other, you use more force on the acceleration and on the higher vilocity, will explain this in full tomorrow. Do a search on physics and enegy used uin distance and or disstnce per unit of time.
Then do another sherach, on room calorimetry. the Energy Expenditure and Nutrient Oxidation will and have been established in a room calorimetry. Many Statistics will be measured eg; Heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, to determine the effects of exercise at different intensities. On every occasion when an activity has been practiced at a faster rate, the heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, has been higher, therefore energy expenditure, nutrient oxidation will be higher.
The increase in the rate of energy expenditure/cost will be due the increase in the intensity/force of muscle activity. When a muscle is subjected to a greater acceleration, speed or velocity, EMG reading will go up, when muscle uses more force, more force will equal more acceleration, speed, velocity, equals more energy.
jarednjames wrote;Let's say it takes 10N to lift the weight in 1 second (a = f/m), if you apply 20N it would only take 0.5 seconds and so on. But the force applied per unit time is the same.
The difference comes when you look at energy. The kinetic energy of moving the weight (100lbs) at an average velocity of 1m/s is 22 joules, but the KE of moving the weight at 2m/s is 90 joules. So the higher the average velocity of the weight, the more the energy use increases  which is why you get tired quicker.
The faster you move something, the more energy it takes. As you can see, simply doubling the velocity (which would halve the rep speed) requires over four times more energy. Each time you double speed of the reps, and as such halve the time for the reps, you are increasing the energy requirement in this manner.
That is why you're using more energy.
* All figures are for guidance only.
jarednjames wrote;
To reduce the time to move an object a distance of 1m, you have to increase the force. By increasing the force, the acceleration becomes larger:
a = f/m therefore if you double to force to 2f you get 2a = 2f/m.
Let's say we have an object that is 1kg. To move that 1m in 1s (1m/s) requires a KE of 0.5mv2 = 0.5*1*1 = 0.5 Joules of energy.
Now, to move it 100m in 1s (100m/s) requires a KE of 0.5*1*10000 = 5000 Joules. So in the first case I need a tiny amount of energy, in the second I need a huge amount in comparison.
For your case to move that 1m in 1s (1m/s) requires a KE as above (0.5 Joules).
Now, to move it 1m in 0.5s (2m/s) requires a KE of 0.5*1*4 = 2 Joules. So again you can see how simply halving the time of the repetition requires you to use more energy to complete it. The time applied is considered in the velocity figure.
For you to move the weight 1 rep in 1s requires 0.5 Joules  that is the energy you must provide to do it.
For you to move the weight 1 rep in 0.5s requires 2 Joules  again, that is the energy you must provide to do it.
If you do not provide that energy, you can't complete the rep in the required time.
Quote by douglis
OMG...you're totally delusional!What world wide debate you idiot?

This has been going on for many years on different forums with different people.
Quote by douglis
It's just a couple of us with superhuman patience trying to explain basic physics to you.

If you as you do, claim you are right, then how do you account for the following.
1,
As when using 80% of your 1RM {Repetition Maximum} when you fail to lift the weight again, or you hit momentary muscular failure 50% faster on the faster reps, this must surely mean you have used up your temporary force faster, it can’t mean you have used up your temporary force slower, or you would still have force left, but you don’t, you don’t have any force left, so that can mean one think only, that on the fast you are using more force per unit of time. I use more energy in the same time frame; I move the weight far far far further in the same time frame.
And if you say I run out of force because of I used up the energy fast, I know that, the question is sort of why did I use up more energy doing something faster ??? Did I use up more because I was using more force per unit of time, the same, or less. But the main issue is the force, the energy is “only” the supply, and we are on about the force, the energy supply. as in the fast rep fails at lifting first, that “must” mean, that the person doing the fast, used more muscle force per unit on time, if not why does the fast have no force to pick up the 80 pounds, but the slow does ??? Also, how does the fast move the weight 6 times further in the same time span, how does the fast use more energy in the same time span, if it’s not as you two say using more force up per unit of time.
If not please explain your way of thinking, please for once answer.
2,
I don’t understand, and no one have tried to explain, is that when my peak acceleration forces are say a 100% or 100 pounds of force, how do you think that your 80% or 80 pounds of force can make up the 100 pounds of force, when its only 80 pounds, how can 80 ever be as high as a 100 ??? The only way the 80 pounds of force could make this extra force up, or the shall we say the 100 pounds of tension on the muscles to the 80 pounds of tension on the muscles, is if the 80 pounds of tension was on the muscle far far far longer, please do you agree there, if not, how does the 80 pounds of tension make up the higher 100 pounds of tension on the muscles, as a 80 pounds tension, can never be as high as the 100 tension, as how could 80 pounds feel like 80 pounds on the muscles ??? Its impulse, force with respect to time. A small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important. But in our case the reps, lifts are done for the same time frame.
If not please explain your way of thinking, please for once answer.
Quote by douglis
The graph is not mine it's from the first site I found when I googled "rms amplitude".

Yes but what do you think it means, and why did you show it ???
Quote by douglis
The RMS is not the average.This is not offered for discussion....it's basic maths.You don't have to mail professors!

I never said it was.
Quote by douglis
Check the below numbers:
2, 5, 8, 9, 4
Their average is 0 but their RMS is 6.16.
The RMS is NOT the average.It's the quadratic mean.You probably don't have a clue what that means so you have to trust the link and me.
http://www.analytictech.com/mb313/rootmean.htm

GOD, I “think” I know what you two are doing wrong. No wonder its what or why the EMG experts use it, no time to explain now, as not 100% sure, but you should be able to see why, if you try to understand why am EMG expert would ”have” to use the RMS or quadratic mean. Problem is, why are you not using this ??? As we are on about the total or overall muscle activity, force.
RMS or quadratic mean will used in the situations, where it is the square of the values that matters; an electrical current squared will be proportional to the power. A quadratic mean, RMS should be used for periodic data (i.e., over time, a graph describes a sine curve), or when both
positive and negative data are included, and what you seek is an "average distance from zero" for the various data point. The quadratic mean, a statistical measure of the magnitude of a varying quantity. It is especially useful when varieties are positive and negative.
You times it by two; you're multiplying a number by itself. Five squared is 25, five times five is 25.
The problem is, why are you two not adding in the negative force ???
Quote by douglis
I've tried so many times to explain that the forceenergy relation is not linear and greater energy expenditure doesn't equate greater force.
I'm sure by now that that's way beyond your intelligence.

I know it’s not liner, try and explain on this physics forum then.
Wayne