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Thank you. @meccanicamg for the answer.the response of 75% of cutting-resistant area comes from basic formulas, where it says that applied a cutting force t on a circular section of diameter d and having area a, the cutting tangential voltage is calculated as:
\( \tau=\frac{4}{3}•\frac{t}{a}{a}=\frac{1}{0,75}•\frac{t}{t}{a}{a}{a}{a}{a}{a}{a}{a}{a}{a}{a}{a}{a}{a}{a}{a}}{a}}{a}{a}{a}}}{a}{a}{a}}{a}{a}{a}{a}{a}}{a}}{a}{a}{a}}{a}{a}{a}{a}{a}{a}{a}}}}{a}}{a}{a}}{a}{a}}{a}}{a}}}}}}}}}}}}}}}{a}{a}}{a}}}}{a}}{a}{a}{a}{a}{a}{a}{a}{ \)
the reverse of 4/3 is 3/4 i.e. 0,75 i.e. 75%.
Forgive me, but I can't figure out the situation.
I had imagined a sort of punctuation verification where 0.6 I imagine is a approximation of 1/sqrt(3) and the thickness of the sheet was the intake part of threads (pass x number rpm)
I thought the section was the cylindrical one given by a strong diameter for the height of the intake threads. and that once developed it was rectangular.I can't see the circular section you're referring to... ?