No, not at all, I thought your post was quite clear. Thanks for submitting it.
Regarding your "second wrench" comment, I like that way of putting it much less than your earlier elegant post. What has always been obvious is that merely trying to look only at the torque applied by the wrench...
"Is there anything else we can help you with?" No, not really. BTW, the "we" in that sentence makes sounds like you surely must be the forum moderator. If not, maybe the word should have been "I"?
My post #24 was merely admission that we seemed to be done. There was no criticism intended...
I don't really need to hear your unnecessary criticism. When you start a post that way, it makes me less than interested in whatever else you might have to say. However... yes, I see all that (and saw it earlier). I've shown the proper formula as well as explained the proper calculation in...
I like this way of expressing the difficulty with the second point of view I presented. It shows the problem in a nice closed form way. At this point I will just use the proper calculation already shown and forget the totally wrong alternate point of view. Thanks for jumping in with this.
John
To All-
It seems we aren't getting very far.
I keep repeating the basic question using different words, using different analogies. I showed how to do the calculation properly as well as the correct answer, and then asked, please, that no one re-explain what I already knew and have already...
One more time... there is NO 1 foot torque wrench. There is only a 2 foot torque wrench. All that's happening is looking at that same wrench in a different way. Namely, by using the definition of torque to see what going on with the 2 foot long wrench, that moving on to see how that effects...
Of course you can buy extenders designed for limited space applications. These often end in open or closed wrenches, usually not sockets. But that is not the only purpose for this technique.
If you owned a torque wrench rated for a max of 150 ft-lbs, yet you need to pull 250 ft-lbs, you can...
Of course you CAN "take the wrench in isolation". This is what's done in mechanics all the time (pick up any book on engineering mechanics or what others call frames and machines and you'll see this one by one, "isolation" technique throughout). For example, take analyzing an old fashioned...
First let me say that someone said earlier that I had over-complicated this problem. Let me in turn say that you may be doing that a bit yourself (though I certainly don't think you mean to do this). So rather than changing lengths and ratios, can we just stay with ONE set of numbers and one...
Forget about "stiffness" (if I understand your meaning). Assume the torque wrench and the extender bar are made from totally "stiff" materials -- zero deflection, zero bending. Nothing is moving at all after the required torque is reached.
Not sure how to do this here, so let me do it...
After a quick look, I didn't see any errors. However, it looks like you just restated the problem.
You show in the one case the correct answer of 200 ft-lbs at the socket on the end of the extender bar. Then in another case you show 300 ft-lbs at that same point, the right answer for your...
Okay, I now understand your nomenclature - lbf.
Sorry, but I'm not trying to make anything overly complicated.
First I explained what a torque wrench is (solely because I wasn't sure who the audience is here -- were it a forum for auto mechanics, I would have not added that section.)...
I have both kinds, but I don't believe this matters. If we were trying to put a piece of pipe on the HANDLE end of the torque wrench to make pulling the force required easier on the user, then it most certainly WOULD matter. The older style torque wrench with a pointer rod (so-called "beam...
Look more closely at the two calculations I originally gave and I think you'll see that it is NEVER correct to multiply ~torque~ by distance. Torque is defined as ~force~ times distance. So when you quote me as multiplying 50 lbf by distance, that was never done (assuming that by "lbf" you...
First, the value 100 ft-lbs you mention denotes torque, not force. Also, I am not sure which means of doing the calculation you are discussing, the first method or the known to be wrong second method. The first method is correct and definitely gives the right answer in my example. It is the...