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440SIXPACK Can I inform this prof that he is full of it & that all of his higher education was a waste

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Another thing that concerns me is an extension between the ratchet and the socket. The length and diameter of this type of extension can make a difference. It will twist like the torsion bars on our mopars. How much of my ft-lbs applied force are absorbed by the twisting motion of this type of extension? The absorbed amount is not applied to the fastener.Tav

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This statement is ALSO not true, unless you are talking about an impact wrench, in which case a "rubbery" extension will indeed absorb torque. If you were to take something--like a small diameter torsion bar, and two identical torque wrenches, clamp one wrench at one end, and apply X amount of force with the other torgue wrench, you will find that REGARDLESS of stretch or twist in the torsion bar, the two wrenches will read THE SAME torque on the gauges.

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So then, as far as the extension twisting and absorbing force, it will only affect things if...

1. The torque level you're trying to reach is less than what the extension will absorb. Example : if the twisting motion of your extension will absorb 20 inch pounds, then you couldn't use it to tighten something to 15 inch pounds. Once you get to 20 inch pounds and the extension stops twisting, then you'll have 20 inch pounds applied to the fastener. You could only use that extension to torque amounts higher than 20 inch pounds. In all practicality the amount of force absorbed by a twisting extension is probably negligible.

2. The torque level your trying to reach exceeds the breaking point of your extension. Example: Using a 1/4 inch drive extension to torque down your main caps will twist it right in two!

Tav

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Roger,
Thanks for posting the picture. Sometimes my explanations aren't as clear to everyone else as they are o me!

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a 3" C to C extension onto a 12" C to C torque wrench would increase your leverage by 25% so reduce your reading by that amount.

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Jack I am wondering if I am correct or not in my original(above quoted) post? At this point in time I do not know but I am going to stay with this until I find out.

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I can't take it anymore. A picture is work 1K words.

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Not exactly. A crows foot at 90 degrees like in the picture requires no torque wrench setting change. Any angle besides exactly 90 degrees requires the torque wrench setting to be calculated. Basically anything that adds leverage AFTER the head of the torque wrench requires the setting to be recalculated.

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exactly!

let's take the torque wrench out of this for a minute.

lever A is 1 foot long.

put 100 lbs on it, and you get 100 ft lbs of torque.

if you add another foot to that lever, it does not matter if the new length is between the pivot point and the original lever, or simply on the end of the old lever, you now have 200 ft lbs.

the issue comes in with the torque wrench and how it "snaps" at the set torque.

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This is what I was saying. My examples used a ratchet NOT a torque wrench.

Tav

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the issue comes in with the torque wrench and how it "snaps" at the set torque.

if you have a 1 foot torque wrench and set it to 100 ft lbs, you will have to put 100 ft lbs on the end to make it click.

if you add a 1 foot extension to the end of the torque wrench that's set at 100, the wrench will click 50 lbs of force, because 50 x 2 = 100

if you add a 1 foot extension to the pivot point, with say, the extended crows foot, and the torque wrench is set to 100, you will have to put 100 lbs of force on the end of the torque wrench to get it to click, however, this is actually putting 200 ft lbs of force to the pivot point.

so, to compensate, you have to re-calculate based on the length of your extension. and in this case, with a 1 foot extension, you will have to set the torque wrench to 50, and then put 50 lbs on the end of the wrench to get 100 ft lbs at the pivot point, because again, 50 x 2 = 100

but, the picture shows that clearly.

in order to tell your professor friend that he was wrong, make sure you understand what he meant, because he might have told you that "it makes no difference, the torque is the same" meaning that the torque at the pivot point is the same, and NOT that the calibrated wrench will work properly.

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with this also!

By the way, Excellent pictures 6PakBee & 446acuda!!

Tav

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An extension down the centerline of the socket drive affects NOTHING. There is no "absorbsion of energy", because the extension is hooked from one end to the wrench, and to the other end to the socket. ALL THE TORQUE must "go" somewhere.

STILL ANOTHER EXAMPLE:

If you were to take a rubber band, like in the old wind up airplanes, and put a lever and scale at the tail, then wind up the rubber, then put an IDENTICAL lever and scale at the prop end, THE TWO SCALES WOULD READ EXACTLY THE SAME, assuming they are both accurate to each other.

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I go through this topic at least once a month with one of our maintenance mechanics who always claims "you always lose torque through an extension". One even said he'd show me. Got out three 6" extensions old enough for Columbus to have used, puts them together with an electric impact and starts rattling away on a bolt. Whole lot of shaking going on but no action. Then he puts the impact on the socket and triumphantly shows me that he can spin the bolt right out. I went to the crib, got a new socket and a new longer extension (can't remember exact length) and asked him to try that. Worked like a charm. People seem to forget that an impact wrench only travels a fixed amount of rotation with every 'blow'. If the extensions/sockets/impacts are sloppy, all that travel is consumed before the fastener ever gets rapped.

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Hmmm, I see. So even if the extension twists it's still applying that amount of twist to the fastener. Any loss is only momentary.
Thank you for explaining my concern!

Tav