Do you know the difference between an ideal, pefectly rigid body as used in theories vs. a body which is described as "rigid" by a layman, but is actually an elastic solid? You know, like a steel column? You do know that steel behaves elastically up to a point before plastic deformation also occurs, correct?Do you know the difference between a rigid body and an elastic body?
I don't know about Bazant's approach. I thought you were trying to assail my formula for impact. Since you are not, I will bow out as it has gone beyond my expertise but if you feel comfortable, carry on.
Both Bazant and, almost certainly, your formula incorporated assumptions of rigid masses impacting steel rods or columns. Before you "bow out", you should at least follow up on your word to tell us what assumptions were used in deriving the formula you quoted.
Was the mass that impacted the rod an ideal, rigid body or wasn't it?
Was the base that the rod rested upon an ideal, rigid body or wasn't it?
The difference between us being that I actually studied from teh book that I took that formula from. I actually studied the very chapter and formula in class. Meanwhile, you did take your formula from a book which I doubt you took the class in. The danger being you are misapplying a formula. My point was about impact.
No I didn't take a class in elastic theory. Since you have taken a class, you should be able to answer my question, which you have already said you would do, regarding whether or not the impacting mass was an ideal, rigid body or not. Surely, your class experience informed you as to what that is, no? Surely you are capable of answering this simple question.
Huh? I don't think you you understand what I've written, at all. Perhaps another JREF'er with a technical background can explain it to you, better than me. You know, somebody like Mackay or Newton's Bit.I am not arguing whether the equation is sound, but whether it is applicable. By YOUR equation, if a body is moving at a constant velocity of 100000000 m/s or whether it is moving at 1 m/s makes no difference in the force transferred.
No. See above.Does that sound right to you?
Really? Would Mackay and Newton's Bit agree with this? Frankly, the formula is rather simple. What is there not to understand?No, it is clear you are applying the formula in a way not intended.
I've already done so. I also asked you specific questions to help you clarify your thinking. All for nought.I was waiting for your big reveal on why the impact formula was wrong.
I'm not aware that we have established this, although the scale is quite up to the task of showing me that I failed to follow my diet.Glad we have established that your scale is insufficient to the task.
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