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.
Ahh, now I get this part a little more. So you’re saying if the ROM was 1m, what part do you think they took the readings ? But again, the slow was done for 10.9 seconds, what difference could this make ? How much more force would that add to the slow ? As remember the lift did take 10.9 seconds. So I now don’t get you again.
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.
YES, but that’s just about RIGHT. As basically if a lift took 10 seconds, all you are doing is moving it very slow, so all you’re doing is basically holing it up. This is my basic debate, as to move the weight much faster, you have to use more force, and more, and then a very fast deceleration.
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.
Have to read that over, getting late, and don’t get what each numbers are for yet.
Zula110100100 said:
So overall, sure, it is more force to get the same weight going faster, anyone will agree there,
Yes.
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.
But I am using the force of the opposite muscles to bring it to a stop and reverse the weight, but up until the last Miily second, I use as much force as I can from the muscles.
If I used a light weight and accelerated it fast, and was not holding it, it would move out of my hands, I would have to wait Milly second for it to return, but that’s BASICALLY what happens in my reps, but as I am holding onto the weight it does not go out of my hands. But I am using a weight that is too heavy to move out of my hands anyway, thus I have to push will full force right and UNTIL THE VERY END of the lift. Will video some very heaver fast lifts on a Smith machines with my hands open, that a machine that like this.
What even if I accelerate for say 3 times the distance as the other, and accelerate longer ? Sorry don’t get that, let me read and think it over, my fault for starting so late again.
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
That sounds interesting, need to read this all over.
ANOTHER THING TWO THINGS WANT TO CLEAR UP, we did go over this before, but could we clear this up for once and for all, deceleration. These numbers are just for this, so don’t take them as 100% fact. We break the fast concentric up into 15 parts of force, and if D. is right on deceleration it goes like this, as in a person using 80% {80 pounds} and the rep taking 1.5 seconds, the study says they decelerate at 60% of the ROM.
60% is on part 9.
100 100 100 100 95 90 85 80 75 70 65 60 30 10 zero on transition and reverse for the eccentric.
What about this. As your accelerating, and then accelerating slower, that your still going faster and faster.
100 100 100 100 100 100 100 100 90 80 70 50 40 20 zero on transition and reverse for the eccentric.
As does deceleration in EVERY case, mean you are then using less force than the weight, in this case 79 pounds, or can it mean you are not accelerating so fast, thus decelerating, so using less and less force as in 2.
Slow would be more like this.
80 80 80 80 80 80 80 80 80 80 80 70 60 40 zero
No need for an explanation, just say 2 can happen or it can’t. However to be honest I always would have thought 2 would have been deceleration, but I never thought about it or read about it before.
So next, the above on the deceleration at 60% was taken from a study that used 80% but the concentric rep took 1.5 seconds, what would happen to the deceleration if the rep took 3 seconds, and what would happen if it took .5 of a second, opinions and facts please. AND WITH THE SLOW REP, HOW DO WE NOT KNOW THIS REP IS DECELERATING FOR 40% OF THE REP AS WELL ?
http://books.google.co.uk/books?id=...the final 52% of the range of motion.&f=false
Thx for all the help and time
Wayne