Maybe you can teach your daughter about hyperbole too.Quote
nitewing76
hercek...as I constantly have to tell my 11yr old daughter, "It is ok to be wrong, but it is NOT ok to assume you are correct." And there are no "hogshead" units in that document.Quote
hercek
Hell, that documents must be written by heretics of physics. Not only the authors use "hogshead" units, but they cannot even get it right, even with them. Ok, so they define modulus as lb/in² but (based on the formula (in note 7) for usage of their modulus values) the unit is actually only lb. Or at least I hope this is what they intended (one cannot be completely sure when they have contradictions even in such a simple document).
Well, if you wanted to indicate that I should use only precise terms in a somewhat technical discussion then I can agree with you and you scored a point.
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nitewing76
Though it should read, "((lb/in2)/in)" or (psi/in). Remember...when there is no value in front of a unit of measure, it is understood to have a value of one. Therefore if his GT2 belts were 2" wide, then the value would have been 36,000 psi/2in.This is a unique case where the units of measure do not cancel because the author must show the derivative value of E to avoid confusion.
- In the expression lb/in2/in there is no cancelling (even if you would want to cancel anything) because the only way you can simplify that is by writing it as lb/in3.
- Imperial users like to specify Young's modulus in psi (force per square inch of material cross-section area). Since the Gates document deals only with belts of specified profiles and variable width then it makes sense to use unit lbf/in (force per inch width of a belt). That is because the belt profile height is fixed. Or if we consider only one specific belt then the unit can be lbf only (foce for given belt). But hardly it makes sense to define it as force per cubic inch ... which is what you proposed.
- Notice you used unit lb when talking about Young's modulus above. In the Young's modulus, the lb actually stands for force (not for mass). You should have used lbf if you want me to even consider first half of your later message (where you complain about that) as anything else than an attempt to confuse me. Finally attention to detail is what you preach, right?
OK, attention to detail! You indicated I used somewhere equation BSL = TM / ((BSL) x (TL). I have two problems with that:Quote
nitewing76
The author gives an equation of BE = ((BSL) x (TL)) / TM. Your equation above is BSL = TM / ((BSL) x (TL). Why did you use the reciprocal of the author's equation? ATTENTION TO DETAIL!!Quote
hercek
Ok, so for GT2 belt it would be 18000 lb for 1" wide belt. Based on note 4, that is 18000/25.4*6*0.82 ≅ 3486.6 lb (or about 15509 N) for the common 6mm belt repraps often use. That means the elongation for 1m long belt and 57N force change is 1 * 57 / 15509 ≅ 0.0037 m = 3.7 mm.
- First problem with that claim is that you do not have parentheses balanced there so I'm not sure what you mean.
- The second problem is that I think I did not use it. Can you cite me precisely where I used it. Leave out the text around, cite only my symbolic equations or my computations with numbers.