Phantomjay wrote;More force goes into the muscles only during the accelerating phase.[/i]

True.
Phantomjay wrote;For the slow case, you might have 110 pounds at the beginning during the accelerating phase...for the fast case, it might be 150 pounds during the accelerating phase, so more squooshh...and clay doesn't rebound.

Yes right, more squash, or should I say more compression on the clay for the fast. No, the clay does not rebound; it does not whatever you do to it, not sure why you say this.
Phantomjay wrote;The averrage force is still 100 pounds for both cases. But yes, at some point during the fast case, your muscles are receiving (and giving) more force during the accelerating phase, because it requires more force to accelearte to the higher speed in the same distance. Now I bet your really confused.

Yes this what it seems to say on paper, “however” I have contended from the beginning of this debate on other forums, “how” can you know, that as we all agree the fast repetition is using more and a LOT more force on the first say 60% {remember the fast does 6 repetitions in the same time frame as the slow moving the weight 6 times more the distance, using far more energy} than the slow, then when the fast is on its deceleration, it’s using less force than the slow, {and here is my main point} “how” do you know/think or work out, that the slow force makes up or balances out the forces here ??? as the slow force is “never” as high as the higher force of the fast acceleration forces, in my opinion the whole forces are not linear and cannot make up or be balanced out, as of my EMG reading, the more work/distance moved and more energy used.
It’s like a impulse; 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. So I say the small force would need far far far more longer time to make up or balance out the large force made by the faster accelerations of the faster repetitions.
That’s why you “always” fail far far far faster doing faster repetitions to slow, as the high acceleration forces, put far far far more tension on the muscles than the slow in the long run.
Quote by PhanthomJay
Your body is not telling you to decelerate it; the laws of physics demand it if you are to slow it.

Yes very true.
Quote by PhanthomJay
Your body tells you that your muscles are getting strained with too much force, and it will tell you that even if you were to hold the weight still and motionless for a long period of time, without speed , acceleration, or deceleration,

Again true.
Quote by PhanthomJay
only a force of 100 pounds on your muscles.Sorry, i misspoke, the average force (for what it's worth) is always just 100 pounds.

I do not think average means much.
The average net force is equal to the weight, regardless to duration and time ??? So if I move a weight for an hour, for a 1000m and then for 1 sec, 1mm. And if the net force is the same, the moving of the weight for an 1 hour will put a very lot of force out from and onto the muscles, even thou the average is the same.
Quote by PhanthomJay
Even though for most of the time the applied force is 150 pounds while accelerating, there is very little force that you exert when decelerating (the weight does the workfor you),

I contend this, the weight does “no” work for me, I am moving the weight, nothing else is, what else could be moving the weight, I do not understand what you mean or think ??? If I stopped at anytime in the lift, the weight “would” stop moving, it would “not” travel or move on its own, how could it ??? Someone worked out below, that if you lift 200 pounds 15 inch at 1m/s that the weight itself would
{but it does not, please read on as too why} move 20% or 3 inch if you immediately stopped the pushing force at the top. However this does “not” happen, because of the biomechanical advantages and disadvantages thought the ROM, {Range Of Motion} so this means that there in “not” very little force, but a “very” lot of force being used thought the ROM. Even if a machine was pushing it, the weight cannot move on its own ???
OK, let's see if I can reproduce this. Let's assume we're doing a bench press with 200 pounds, and our 1RM is 250 pounds (80% 1RM). Furthermore, the ROM is 15 inches.
The first thing we must do is convert pounds into the English unit of mass  the "slug". That is  200 pounds/32 ft/sec^2=6.25 slugs.
Now, from F(net)=ma, we have 250200=6.25 x a, or a=8ft/sec^2. This is the acceleration of the bar during the concentric.
Now, let's assume that we will push with our maximal force (250 pounds) up to the 12" point. We next need to find the velocity of the bar at this point from the equation v^2=2ad. Plugging in "a" from above and "d"=1ft, v=4 ft/sec. Note that this is not the top of the lift, but 3" from the top.
Next we want to find the time it takes to get to the 12" point, from the equation d=1/2at^2. Plugging in our values of "a" and "d", t=.7 seconds (again not the top).
Now, we assume that we stop pushing the bar at the 12" point, and let gravity slow it down, so that it comes to rest at the top. From v^2=2ad, we plug in our value of "v", but here we use a=32 ft/sec^2 (acceleration of gravity). From this we find that d=3", so that the bar comes to a perfect halt right at the top of the ROM  the 15" point.
Finally, we want to find out how long it takes gravity to stop the bar, from d=1/2at^2. Plugging in d=3" and a=32 ft/sec^2, t is found to be about .1 seconds.
Therefore, the total time for the concentric in this case would be .7 seconds (for the acceleration phase, or "onloading") plus .1 seconds (for the decceleration or "offloading" phase), for a total of .8 seconds. This is a bit faster than 1/1, but the best I could do this late at night. If your ROM were a bit longer then 15", then the speed would be closer to 1/1.
So, we see that in this case the offloading relative to the ROM is 3" out of a total of 15", or 20%. The offloading relative to the time is .1 sec out of .8 sec, or about 12%. Note that this is worst case...that is the first rep. As the set progresses, the offloading will reduce with each rep, due to fatigue, and towards the end of the set will be negligible. Therefore, you could state that the "average" offloading of the entire set relative to the ROM would be about 10%.
Note that this also assumes we can push with maximal force all the way to the 12" point. If our strength curve is such that our force output diminishes towards the top, then the offloading will be less than given above.
Quote by PhanthomJay
so the average force averages out, if you know what I mean (in eight equal intervals on the upstroke, force you apply might be 125, 150,150,150,150, 25, 25, 25; average = 800/8 = 100 pounds). You can be moving at constant speed slowly (say 1m/s) or at constant speed quickly (say 3m/s),and the force you exert is still the same 100 pounds for both cases.

As I said above, I can’t see how the weight would move on its own ??? it would be more like 125, 150, 150, 150, 150, 125, 125, 80, 20, zero movement for the transition.
“If” the forces averaged out, “why/how” is the EMG readings higher, why/how is power higher why/how do you use more energy, why/how does the fast move the weight 6 times further, why/how do you fail on the fast far faster ???
Play from 5.00
http://www.youtube.com/watch?v=6clCe76uDQ
Note that the differences.
Slow,
Power 649
Force 546
Velocity 161.
Fast,
Power 829
Force 579
Velocity 192.
Imagine if the person had done 6 repetitions fast and one slow, the force on the fast would/could be far far far higher, as on the video, the speeds were basically quite close.
Quote by PhanthomJay
Your muscles are tiring more easily in the high powered fast case, because internal energy is being lost from the complex muscular activity..

Yes, that’s because I have to be using more force ??? If not how/why else would I need to use more energy ???
Quote by PhanthomJay
if your muscles were robotic, it wouldn't matter..or a machine getting powered electrically...but humans are not machines, they need chemical/biological/food energy to operate.

The machine too would need more energy to use more force, its impossible otherwise to use more force without more energy, is it not ???
Quote by PhanthomJay
Don't confuse the two.You are not correct.

So why is more energy used when force goes up then, as its “not” a coincident that the “exact” moment you use more force you use more energy, is it ???
Quote by PhanthomJay
However, the power required is greater when you are moving faster with the same applied force.

You would have to use more force to move faster. Where Fnet is the total external force.
Wayne