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sophiecentaur
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=41 if you follow the bodmas convention. Look it up. Anything X zero is zero.
To make it clear, you should use brackets.
To make it clear, you should use brackets.
Zula110100100 said:In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for
Zula110100100 said:bodmas? I learned please excuse my dear aunt sally (Parenthesis, exponents, multiplication, division, addition, subtraction)
Zula110100100 said:The process that causes it to register more is the "force-applied" impulse being shorter time, more force(higher reading), but this will still be equal to the impulse of gravity if you average in the time that no force was applied. Just before it is moving up again,
Zula110100100 said:it IS at a still start again. Any extra force applied to "get it going back up" from downward motion can be split between stopping it and starting again. The "stopping it part" is part of the downward rep.
douglis said:I see the problem is a little deeper than I thought...
sophiecentaur said:=41 if you follow the bodmas convention. Look it up. Anything X zero is zero.
To make it clear, you should use brackets.
Static hold half way up. Average muscle activation = 92.1.
Slow. Average muscle activation = 159.1.
Fast. Average muscle activation = 191.7.
Also, is not power in mechanics, the combination of forces and movement ? I thought power was the product of a force on an object and the object's velocity ? As we all will agree that the fast uses more power, {even I can work that out with some simple equations} so if power is the of a force on an object and the object's velocity, does that not mean more force more power ?
Zula110100100 said:Yeah...I would say 41..and read it 40+(40*0)+1 = 40+0+1 = 41
Zula110100100 said:I want you to do that expiriment because I think it will show that activity does not equal force, I think if you hold 20lbs for 1 second, it will be some amount that is less than double activity to hold 40lbs for 1 seconds.
Zula110100100 said:In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for, rather than try to figure out how that answer is wrong, perhaps you should look for how that answer relates to what you're looking for.
Zula110100100 said:What are you trying to find out about PS resistance training overall?
Not sure how you get 41 ?
Zula110100100 said:If the weight and time were the same, and they started and ended at v=0, then the force-applied impulses are the same in each.
Zula110100100 said:This shows that Average muscle activation does not directly correlate to fore-applied impulse. Do the same thing with different amounts of weight, each held for the same time, in the same position.
Zula110100100 said:Let’s say we have a force of 1N, we apply that force at two velocities, one is 12m in 6s = 2m/s the other is 2m in 6 seconds = .33m/s, the power is 2(1) = 2watts for the fast one, and .33(1) .33watts for the slow one. The same force, different velocities.
Zula110100100 said:Because it is a well accepted convention that you do the multiplication before you do addition
so 40 + 40 x 0 + 1 you multiply 40 by 0 and have 40+0+1=41
waynexk8 said:Why can't anyone asnswerr this please ?
I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.
Wayne
40 + 40 x 0 + 1 = ?
Yet again you fail to answer.
waynexk8 said:Why can't anyone asnswerr this please ?
1,
We both have a 100 pounds of maximum strength {or a machine is lifting} that we can bench press. We use 80% 80 pounds, I move the weight 1000mm in .5 of a second, I will be accelerating this weight for 60% for the ROM, {range of motion} of for the concentric rep, you will be moving the same weight in .5 of a second for only 166mm.
Question,
If I have moved the weight more distance in the same time frame, does that not that mean that I have used more force ? {force and strength are the same I would have thought, but let’s just go for force now} As even if you just count my acceleration its 600mm, that far moré distance than the slow.
2,
I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.
Question, basically the same as the above.
As a force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1Newton, if you move it futher in the same time from you have to use more force, N’s.
Wayne
sophiecentaur said:You seem to want answers about this without having to make any effort to know the basics of Science or even primary School Arithmetic. We just can't help you on your terms. You have had many answers (the same answer put in different ways, actually) in a attempt to get the message over to you. Yet you refuse to give an inch (or even 3.54cm) so I am unsubscribing from this thread. I suggest you return the the pub / club / locker room and have your conversation with people who have no use for real Science but who just want an hour's worth of idle, anecdotal chit chat.
SumDumGoi said:Wayne,
As asked above, could you please clarify what your argument is? Are you saying that the maximum possible forced produced by a muscle will occur with faster vs. slower contractions?
SumDumGoi said:Also, in regards to your physicist, what is his background in physiology? From a mathematical standpoint some of the things you have said are true.
SumDumGoi said:However, when you are presented with what happens from a physiological standpoint your muscles would not be able to produce the increase in force required for the math to work. In other words there are physiological reasons why you cannot move a heavy weight very quickly and why when you move a light weight quickly muscular force is reduced. You need to apply the math within the constraints of muscle physiology. The person you are talking with needs to know both physics and physiology to understand the problem.
Your above examples violate what the muscle is capable of doing in the physiological world so they are not worth commenting on.
douglis said:Wayne...you got your answer.What's clear is that you don't understand the answer or you don't like it.
In both cases the applied force is equal with gravity's impulse over 6 seconds...so it's the same.
waynexk8 said:Add in the acceleration conponants please, How can the applied force be equal with gravity's impulse over 6 seconds all the time ? Add in time and distance.
Or try and tell me how you move a weight 166mm and I move a weight 1000mm in the same time frame, and I accelerate for about 600mm of that distance. Your trying to tell me that you use the same force moving a weight far less disstynce in the same time frame
Let’s see you try and prove this, with basic numbers
40 + 40 x 0 + 1 = ?
Is not 40 + 40 = 80 ? Now we x 80 by 0, and whatever we x 0 by it will always be 0 right ? So we are now at 0 right ? So add 1 to 0 = 1 right ?
Wayne
douglis said:I really can't figure out what Wayne can't understand.
If we compare 6 reps with .5/.5 and 1 rep with 3/3 the only difference is that the 6 reps have higher fluctuations of force.But in both cases the same average force(the weight) is applied for 6 seconds.
What's so hard in the above to understand?Why does he keep posting the same irrelevant examples?Do we have to use superior maths for something obvious even to a 10 years old child?
douglis said:Irrespective of any of the details along the way (acceleration, deceleration, fraction of time spent accelerating, distance covered, etc.), the average force is always equal to the weight of the object, so long as you start and finish at rest.
douglis said:Instead of trying to prove nonsense learn to do the mathematic acts at the right order or find a kid to show you.
waynexk8 said:How do you work out my EMG states you are wrong ?
And how do you work out my clay example proves you wrong ?
Could you do me a favour please, could you tell me again how you work out this average force. Let’s say I am moving a weight at 1m/s and move it for 1 second, I then move a weight 100m/s and move it for 1 second, how do you work out the averages ? AND WHAT YOUR SAYING IS THAT YOU USE THE SAME TOTAL OR OVERALL FORCE TO MOVE THE SAME WEIGHT 1M IN 1 SECOND, AND TO MOVE IT 100M IN ONE SECOND ?
Wayne
douglis said:Your EMG is not suitable for lifting that involves SSC...I explained why.
douglis said:Your irrelevant clay example just proves that higher peak force is produced with fast lifting.Nothing more.
douglis said:I must have said this more than a hundred of times.
I work out the average from the net impulse delivered.Either you move the weight 1 or 100m the change in momentum is zero so the net impulse delivered is zero too.Concequently the average net force is zero and the applied force equal with the weight.
End of story.
douglis said:It's obvious that you're not able to understand the answer or you don't want to.Either way continue alone.I'm fed up with that nonsense.
The force at work on a Formula 1 car as it starts a race! If the F1 car has a Mass of 600kg and an Acceleration of 20m/s/s then we can work out the Force pushing the car by multiplying the Mass by the Acceleration like this 600 x 20 = 12000N
Wayne says, If the F1 car has a Mass of 600kg and an Acceleration of 40m/s/s then we can work out the Force pushing the car by multiplying the Mass by the Acceleration like this 600 x 40 = 24000N. But your saying that’s wrong ?
Zula110100100 said:@Wayne - If you do the experiment I asked about using a static hold with different weights you would either, A) See why the EMG results don't work in this situation, or B) Prove to Douglis that they do work in this situation.
Zula110100100 said:@My hypothesis is that when you double the force required to keep it still(by doubling the mass) you will see less than a double result from the EMG. This may or may not be true, I don't have a machine to test it but I feel it would bring us much closer to agreement on this issue.
F = ma , Force = mass * acceleration, The mass is constant to it really comes down to the acceleration to decide the force.
Zula110100100 said:@Do you agree that in order to have a velocity that begins and ends on 0, it must both accelerate and decelerate?
Zula110100100 said:@So let's say we accelerates it with a1 acceleration for t1seconds, it's velocity, v1 is then given by
v1 = a1t1
Now, it must decelerate back to 0, so gravity accelerates is a2 for t2, so this time we are starting at v1 velocity and going to 0, well, that is really the same amount as starting at 0 and going to v1, so the equation
v1 = a2t2 must also be true.
If it gave a different velocity it would not cancel the other velocity right to 0, it would still be in motion.
With that being the case, how can a1t1 not equal a2t2
The clay situation doesn't work for total force because like you said, it is a measure of peak force, it will only measure what the most force applied for an instant was, not in the same time frame.
Zula110100100 said:that is the difference, when asking for who applies most in the same time frame you are requiring that we average in the time that less or no force is applied, had the question been, "Which person generates a larger force" the answer is the fast one, hands down, it must have been a larger acceleration to still have time to decelerate and make it the same distance faster.
But that was not the question. Which is why I say you should re-evaluate what you are trying to learn or show. There may be another factor to look at.
Zula110100100 said:This is a completely different situation, if you include a requirement that they both came to rest again from the force of friction and air-resistance, the again, the overall force would be 0, this example does not look at that, it looks only at the force to accelerate it, not decelerate it, if the question was about two cars accelerating an unknown amount, but both coming to rest after an equal time of an equal force acting on them the answer would be that the same force was applied.
The important this is doing that experiment to see what the correlation between force and EMG activity is.
However why is this zero movement at both ends of the transitions so important in this debate ? As they will be so small compeered to the rest of the reps, they will be like ? 1 part in 10,000.
Hmm, not sure on this, as I did not say the clay would just measure peak force; I think it would measure all the forces, why don’t you or anyone think it will measure all the forces ? As a scales would if you stood on them, and if you were lifting overhead and put the clay under your hands and feet, would not this measure all the forces ? As soon as you lift the weight, the clay would squash a little, then if you accelerated it slow or fast, or very slow, the clay would then squash more. Dammed it getting late.
waynexk8 said:I did not read why ?
So you’re now saying that all the sport science people all over the World are wrong when they use EMG ? D. you know very well the EMG shows the muscle activity, and it makes no difference if I have more SSC than you, as that’s part of the lift, YOU CANT TAKE IT OUT ? God, you don’t like something because its right, so you say it’s wrong, pathetic.
Wayne
Ha...just saw that...no wonder why in the study you post they use isometric contractions.The RMS is accurate here since with isos the force doesn't fluctuate.
In dynamic contractions where the stretch-shortening cycle is used the raw EMG must be normalized by using maximal voluntary isometric contractions (MVIC) just like they did with the two push ups studies and the Elliot et al study(the one that found 52% deceleration).
Here's explained better:
"EMG signal amplitude normalization technique in stretch-shortening cycle movements
Normalization using maximal voluntary isometric contractions (MVIC), irrespective of rms processing (total, mean or peak), demonstrated greater CV above the raw data for both muscle actions. "
http://www.sciencedirect.com/science/article/pii/105064119390013M
Zula110100100 said:Equation 3 from that source points out that a linear impulse is equal to change in linear acceleration
If we take an force of 6.2N over 10.9 seconds, that is an impulse of 67.58N*s
This means we must see a change in momentum of 67.58kg*m/s The weight presumably started at a velocity of 0, so this means it is left with a velocity of .4m/s, thus, it has not yet stopped moving, if you stop applying force, it will take gravity another .04s to stop it, and during that time there is a negative propulsive force, an applied force less than the force required to hold the object up.
So in my opinion the data is not good, because they did not wait for it to stop to take their readings.
Zula110100100 said:If you take 1672.2N(the applied force required to get a propulsive force of 6.2N) * 10.9+ 0N*.04s = 1672.2 and then divide that by the total time 10.94 = 1666.0859N avg which is really close to the 1666N that were required to hold it up.
Zula110100100 said:For the faster rep, we have an avg propulsive force of 45.3N * 2.8s = 126.8N*s, so we know it is left with a velocity of .75m/s which would take gravity .076s to decelerate, so again, add the 45.3 to 1666 = 1711.3*2.8s = 4791.64N*s from the force applied
+0n*.076s =4791.64/(2.8+.076) = 1666.078 once again really close to the force required to hold it still in the first place.
Zula110100100 said:So overall, sure, it is more force to get the same weight going faster, anyone will agree there,
Zula110100100 said:the problem is that unless you apply a force to pull the weight down toward you before it stop moving up, then comparing the force for the whole cycle should be the same.
Zula110100100 said:in other words, if you apply more than required to hold it up, and then never less than required to hold it up, it will keep moving up, since that is not the case there is more going on than listed in that expiriment