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Jan13-12, 07:13 PM
P: 399
Quote Quote by douglis View Post
Terminal velocity is the velocity at the end of ROM which is zero in the case of lifting(not throwing)
Please explain how my lifting it different to throwing ???

Yes, but please let us know your actual meaning of terminal velocity here, also why you include it, if what you think it is, is zero. As in weightlifting and physics its different.

Lifting and throwing a weight up in say the bench press, in my form of explosive lifting with 80% is basically the same. As I try to get the weight from a to b as fast as possible, that the same as if I was trying to throw the weight on the bench press.

Quote Quote by douglis View Post
The change in momentum is zero so the average net force is zero...hence the forces "make up".Nothing more to talk about...I don't know what you can't understand.
The change in concentric momentum/movement is from zero velocity, to acceleration, {80% moved 20 inch in .5 of a second} to a deceleration, then zero for the transition, then reverse down for the eccentric.

We are not looking for the net force the way you are saying it, and never have been, why do you mention this ??? You are saying if both forces are equal they cancel each other out the net force is zero. The force we are talking about are the forces from the muscle for holding the weight, for moving the weight, for accelerating the weight and for decelerating the weight, then add in the opposite reaction force, as in the more force you produce the more the clay would get squashed, or the more tension on the muscles. Like my EMG reading proved.

Quote Quote by douglis View Post
If you're going to doubt it without any argument you'll just be ignored.
Doubt what, I dont understand what you think I will not doubt ??? What would be nice if you understood my far higher reading in a practical World experiment, my EMG reading ???

Quote Quote by douglis View Post
I also don't understand why you're repeating that irrelevant clay example.It just shows the higher peak force that causes the deformation of clay...nothing more.
You are quite wrong there, the deformation of clay in both lifts, will show all forces, or if you would like to state how/why you think that the force of you moving very slowly will not be shown in the clay ??? As the law states that for every reaction there is an opposite reaction. Hold the weight with the clay, then move the weight very slow up and down with the clay, you will see the clay is deformed more when you move the weight. It has to show all, please explain why you think it will not ???

Quote Quote by douglis View Post
The length of the deceleration phase is defined by the laws of physics...not by you.
I am, and we are all physics. I can, as I did earlier, choose the length of the acceleration and deceleration. Simple test, punch with as much force and velocity upwards as you can, and slow and stop as close to arms length as you can, now with as much force and velocity upwards as you can, but try and stop half way up, you have just changed both acceleration and deceleration.

The strength of the person will work out the acceleration and deceleration. As I said, I lift explosively and its basically the same as if I was throwing explosively, I would try to do both as fast as possible.

Quote Quote by douglis View Post
Sorry it "is" 1 sec lifting.Check page 459 figure 6a....although it's of little impotance.
Sorry, on the graph it clearly sates 1.5, also on your page, where are you looking ???

Its of the greatest impotence, why do you think not ??? If I move the weight 1m in .5/.5 of a second 6 times to your 1m in 3/3 seconds, I have created 6 large force accelerations and peak forces. I will tell you why more, but no time now.

Why do you think your lower forces can make up or balance out my higher force ??? You still have not explained this, how can say 80 force make up or balance out a 100 force, if both forces are used for close to the same time frame ???

Did you try the weight on a barbell with string, to show very little deceleration ???