sophiecentaur said:
You say you have come to "talk Physics" but you refuse to do just that. You have your own brand of vocabulary, a lot of which has nothing to do with Physics. I'm sorry but several of us have told, on many occasions, you that what you say is not understandable because you use meaningless combinations of a random set of words that you haven't defined.
Is not force a push or pull ? Is not distance inches or other ? Is time, as in seconds or other ? “Please” I do not understand what you don’t understand, or what I am asking wrong, please could you either tell me what you don’t understand, or how I am saying it wrong ?
sophiecentaur said:
That machine of yours is just telling you something about muscle activity and, hence, about the Energy that you’re Muscles are transferring.
This machine is “NOT” telling me the energy I am using, its telling me the muscle “activity” as in Newton’s used, the machine measures this in μV. Do you know what I force plate is, it basically does the same thing. Electromyography (EMG) is a diagnostic procedure to assess the health of muscles and the nerve cells that control them (motor neurons).
Motor neurons transmit electrical signals that cause muscles to contract. An EMG translates these signals into graphs, sounds or numerical values that a specialist interprets.
An EMG uses tiny devices called electrodes to transmit or detect electrical signals. During a needle EMG, a needle electrode inserted directly into a muscle records the electrical activity in that muscle. A nerve conduction study, another part of an EMG, uses surface electrodes — electrodes taped to the skin — to measure the speed and strength of signals traveling between two or more points. EMG results can reveal nerve dysfunction, muscle dysfunction or problems with nerve-to-muscle signal transmission. An electromyography detects the electrical potential generated by muscle cells when these cells are electrically or neurologically activated. The signals can be analyzed to detect medical abnormalities, activation level, recruitment order or to analyze the biomechanics of human or animal movement.
http://www.nlm.nih.gov/medlineplus/ency/article/003929.htm
Please sophiecentaur, we need to clear this up first, the EMG reads up muscle activity, not the calories you’re using.
http://www.youtube.com/watch?v=B8gtp...ofilepage#t=3s
http://www.youtube.com/watch?v=pd0ZA...ofilepage#t=1s
sophiecentaur said:
It tells you nothing about the amount of (and here I am using the term in the correct way) Work Done on the weights you are shifting.
Yes is does, and this is a well known fact, as if someone is thought to have muscle atrophy, or disuse atrophy, is defined as a decrease in the mass of the muscle; it can be a partial or complete wasting away of muscle, these machines are used.
sophiecentaur said:
You want there to be a definite relationship. There is no reason why there should be, in general. It's often quite clear when you're doing more useful work and less work work - for instance, when running uphill with a pack and the EMG would probably agree. But when you are lifting weights, holding them and lowering them, there can be no useful connection. Why do you insist that you want one?
Right now I am maybe understanding you here. As I think of when I lift a weight up and down as strength used, and physics calls this the force used, as force is a push or pull, so that’s why I said force/strength. However are you saying I have to call force/strength work ? But I thought work is the force times the distance through which it acts, and if a constant force pushes on a object that moves in the direction of the force, then the work done by this force, so the force is doing the work, and the force in this case in the strength of my muscles.
sophiecentaur said:
You won't accept that this is just not Physics. Proper Physics equations will not be "wrong" when applied to the right situation.
Very true, but have all the variables been added in, I don’t think so,
please read this, look at this from 5.00 min, it states I am right.
Fast
P = 695
F = 579
V = 192
Slow
P = 649
F = 546
V = 161
Please also read chapter 4, just the first 2 pages, it says like I have been trying to say all along, these forces that I talk about cannot be easily equated with physics.
http://www.findphysio.com/E-books/Biomechanical%20Evaluation%20of%20Movement%20in%20Sport%20and%20Exercise.pdf
sophiecentaur said:
This is just not one of those situations.
I think it is, as how do you explain I fail faster in the faster repetitions, thus I “HAVE” used up my force faster than the slow ?
sophiecentaur said:
There are questions in the above list that you could answer for yourself just by looking at wikipedia and there are some which no one could answer because they don't relate to Physics.
I have tried to learned physics a little, this may not mean much to you, but I can work out the power myself, see below, but force is far harder.
To determine the force we will need to figure out what the weight of the barbell is (W = mg = 91 kg x 9.81 m/s? = 892 kg.m/s? or 892 N). Now, if work is equal to Force x distance then, U = 892 N x 1.85 meter = 1650 Nm.
Power, takes time into consideration. If for example, it took .5 seconds to complete the concentric lift, then the power generated is 1650 J divided 1.7 s = 3300 J/s.
If it took 2 seconds to complete the concentric lift, then the power generated is 1650 J divided 2s = 825 J/s.
power = force {strength} x velocity. The force is greater if the SPEED is increasing. If the speed is increasing then the weight is accelerating.
Again, the force generated by the muscles is given by the following: F=mg + ma. The first term on the right (mg) is the load the gravitational contribution. The second term on the right (ma) is the contribution due to the acceleration.
If the speed is constant then a=0 and F=mg...equal to the load. If the speed is increasing, then a is not zero and F=mg + ma.
This is Newton's 2nd Law. It cannot be refuted. At least not in this Universe. In the equation F=mg + ma the speed is irrelevant in the first term on the right (mg=load). But it is not irrelevant in the second term (ma). If the speed is increasing, then there is a non-zero acceleration and a=v/t. A CONSTANT acceleration results in a force. This force is added to the load (mg). Acceleration is a change in SPEED.
Wayne