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Larry
You are correct. Using your example, at 12" the applied force required is 64lbs. The three inch extension will experience 80 FT-LBS of torque. Your math example assumes a torque wrench that is calibrated at 12 inches. The problem arrises when Tim (380mxc) has a torque wrench that calibrated at 22.57 inches. Torque = force X distance For a 12" torque wrench with a 3" extension, setting 64 FT-LBS results in 80 FT-LBS at the bolt. 64LBS x 1.25feet = 80 FT-LBS For the 22.57" torque wrench with a 3" extension, setting 70.56 FT-LBS results in 80 FT-LBS at the bolt. A force of 37.23LBS at 22.57 inches (1.88feet) = a torque of 70.56 FT-LBS Adding a 3" extension to the equation. 37.23LBS x 25.57 inches (2.13feet) = 80 FT-LBS Tim can't just multiply torque required by .8 and plug that into his torque wrench. His torque wrench manual values line up with the extension formula. Wrench torque = Torque Spec * (Wrench length / (Wrench length + extension length)) |
Mike,
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--Paul |
Now what was for sale?:eek:
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In the video the wrench appears to be made of 2 thinner pieces based on the appearance. is it? Or is it one piece with a line around it? Thanks Marty
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This is supposed to be a prop wrench, how 'bout a prop service bulletin for reference?
http://www.mccauley.textron.com/sb227B.pdf, see the last page. |
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dial reading = torque wrench length (ft) x desired torque (ft lbs) / [torque wrench length + extension length (ft)] Clearly this falls into "the 0.8 multiplier is not a constant" camp. Im back to not knowing what to think. Could this SB really be wrong? erich |
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As I said before, show me the math and I'll be a believer, but I still haven't seen it. All ft pound wrenches are not 1 ft long which is what this "theory" is based on. If every torque wrench was 1 ft long then Allen would be right, end of story. |
Allan,
Looks like the Retail Price of your wrench has gone up... include a 12" 3/8" drive Snap On high quality torque wrench in the wrench kit and then nobody will have the need to do the math. :D |
Take 3 torque wrenchs, one 1 ft long one 2 ft long and ine 3 ft long. All are calibrated for 100 ft lbs torque.
So even though they are different lengths, they all put exactly the same 100 ft lbs of torque at the head. So all three torque wrenches put exactly 100 ft lbs of torque at the head and then you add the exact same 3 " extension to them. Wouldnt the 3 " extension have the same effect on the 100 ft lbs of torque at the head on all of them? Just saying...:D |
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EXACTLY!!!!!!!!:D You got it perfect, Thanks, Allan |
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There are many publications out there that have printed it wrong I guess. It sure would be easy to prove.
After reading the last 10 pages, the only way I would use this product would be with the wrench at 90 degrees from the adapter..... easy to do with this application. At least it's not a primer or F/P issue......................:D |
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You got it wrong, sorry It's already been explained correctly a few times. The nut sees a torque of applied force * distance from handle to nut. The wrench gauge sees a torque of applied force * distance from handle to drive end of wrench. So correction is the ratio of those lengths. Put the crowsfoot at 90 degrees; the lengths are the same so the correction is zero. |
she's a witch!
Many of these posts are pretty humorous, I wonder if I use a large red font if it will have more impact...
It is interesting to see various responses and who makes them; I must be the dumbest guy in the forum; perhaps PI really is 3. "Who are you, who are so wise in the ways of science?" http://youtu.be/zrzMhU_4m-g ![]() |
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I bet there are a ton of AN bolts under and over torqued out there... :eek:
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Larry, this is great, this is the closest I've seen to someone actually laying out a mathematical argument for a constant conversion factor.
Now my question is, if this is a good argument, why not apply to a torque wrench that reads inch-pounds. Just change foot to inch: Quote:
OK so far? --Paul |
Allan,
New Service Bullet SB 2012-2-14 For everyone purchasing this wrench, do your own math calculations for torque value. :D |
So tell me, wise ones...
A click wrench uses a spring and pawl, which releases when the desired torque setting is hit. This compressed spring is calibrated to match ft/lbs., which is based on one pound of force, at one foot from the pivot point. The one foot is only a measurement of force. The wrench does not have to be a foot long. How is this spring going to know, if you're grasping the wrench at the exact same spot on the shaft. Does it care? No it doesn't! The force you apply at the handle depending on length, varies. But the spring just doesn't care. It's going to release at the setting it's set to...........which is based on one pound at one foot, for ft/lb settings. edit: And I'm going to add a 100' extension, pulled by block and tackle, connected to a cement truck.................and the spring still won't care, or notice the difference!!!! It will just merely click, when it reaches the set point.. L.Adamson |
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The torque at the end of the extension, however, doesn't simply scale as a function of torque and distance, it's a function of *force* and distance, the force being the force you've applied and the distance being the distance from the point of application of the force to the end of the extension. It does not matter that the torque in/on the system somewhere in between (say, at the wrench head) is less, the torque at the extension end is a function of total distance (wrench plus extension) times force applied. |
The torque on the nut is not being applied at the drive end of the torque wrench. It is being applied by the force on the handle.
If you applied a pure torque to the drive adaptor in the prop wrench (e.g. by using a pneumatic nut driver) you actually wouldn't apply any torque at the nut end of the prop wrench. (Imagine putting a screwdriver handle in the drive hole; it wouldn't matter how long the wrench was, you would just generate a linear force at the other end, not a torque) I think this debate is a bit like the aircraft on a conveyor belt saga. |
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Your correct about force and distance in relation to absolute torque. But it has nothing to do with the spring setting. The spring only cares about the correct torque applied to the nut. Not the potential torque that you can add to it, by the force of leverage. |
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This would be a really nice place for someone with the proper equipment to set up a demonstration. Easier to do here than with "the downwind turn."
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The formulas that all of you are getting stuck on are only for simple levers and pivots.
Torque wrenches when you include the indicating mechanizm are complex machines not simple levers. The key is the calibrated indicating method of the torque being applied. If you take a typical high quality aircraft torque wrench and set it to click at 100ft pounds and pull it on a nut in stock configuration, it should click when there is 100ft pounds on the nut. How much you have to pull is based on how long the wrench handle is. Remember the clicker is calibrated to a certain torque on the nut driver shaft. Now take the same wrench and add a 3 ft extension to the handle. Pull the handle till the clicker clicks.....You still get 100ft pounds at the nut as long as you stop pulling when the clicker clicks. The amount you have to pull is reduced because of the longer handle. The calibrated clicker still clicks at 100ft lbs at the nut driver shaft. If you take all the indicating mechanizms away and use a spring scale on the end of a simple lever, all of these textbook formulas work just fine with including the lever length. |
Well said Brian...
"If you take all the indicating mechanizms away and use a spring scale on the end of a simple lever, all of these textbook formulas work just fine with including the lever length."
I will add that if the attachment was welded on to the torque wrench then the over-all length would matter but it is not, the torque is all measured and applied to the attachmnet at a point 3 inches from the nut and this gives the additional leverage that must be compensated for..... |
Not to ruin the fun, but I think it's way past time for the moderators to move this out of the classifieds section. I asked a question last week (on page 2) as to whether this works on a MT prop and there was so much talk about how to torque a prop that no one answered my question that could actually turn into a sale....
We are on page 11 now, and nothing in the last 8 pages has anything to do with the classifieds. Let's move this to the proper list. And I'm still hoping someone can answer my question on whether this works on an MT 3 bladed RV-10 prop? The web site mentions Hartzell.... Thanks -Mike Kraus RV-10 Flying |
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http://www.hartzellprop.com/public_dl.php?id=4 Again, the .8 factor is ONLY for a 1 ft torque wrench, for any other length you need to do the math for your wrench. |
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How many torque wrenches, that you would be using on a prop.........are just one foot long? I doubt there are any! That one foot figure has to do with 12" meaning a foot pound. That's the distance that torque values are measured from. The drawing isn't even close to being scale. L.Adamson P.S.--- the extensions, crows foot, or whatever, are required, because you can't just place a torque wrench directly over the prop nuts. There is a flange in the way. As Hartzell says, if the extension is at 90 degrees, just use the normal reading. |
Since so many here, are ignoring the mechanical aspects of a torque wrench, as Brantel mentioned.....
We'll just all go out, and take an ordinary wrench & stick 100 lbs of dumbbell weights at the 12" mark, or 50 lbs. at the 24" mark. If the 3" extension is used, we'll put only 80 lbs. at the 12" mark. Note: It doesn't matter how long your wrench actually is. It's those 12", 24" or any calculated mark in between that matters. Figure in a bit of weight, for the shaft, too. If using an actual torque wrench, then ignore the above... L.Adamson |
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It seems that this whole discussion is about whether the 0.8 multiplier with a 3" extension is universal for any torque wrench. I propose it is not.
Let's use the Hartzell formula: (actual torque required) X (torque wrench length) / (torque wrench length) + (length of adapter) = Torque wrench reading to achieve required actual torque Example 1: Actual torque required- 100 ft-lbs Torque wrench length- 12" (1.0') Length of adapter- 3" (0.25') Formula: (100 ft-lbs x 1 ft) / (1 ft + 0.25 ft) = 100 ft-lbs / 1.25 ft = 80 lbs torque wrench reading Example 2: Actual torque required- 100 ft-lbs Torque wrench length- 18" (1.5') Length of adapter- 3" (0.25') Formula: (100 ft-lbs x 1.5 ft) / (1.5 ft + 0.25 ft) = 150 ft-lbs / 1.75 ft = 85.71 lbs torque wrench reading So, using the math that has been shown before seems to indicate that a 0.80 factor is only valid when the torque wrench is 1' long... There is a relatively easy way to confirm or debunk the 0.8 universal multiplier theory. Three beam type torque wrenches (of at least two different lengths) and some adapters are required. Take one torque wrench and clamp the shaft in a vise such that a torque applied to the end can give a reading (the captured wrench). Using an adapter attach the second torque wrench (the free wrench) to the captured wrench. Press the free wrench to 100 ft-lbs. The captured wrench should also read 100 ft-lbs. Do the same test with the third wrench, of a different length. Again, both readings should be the same, as both wrenches, regardless of their length, are calibrated to read the same torque at the head. Now, do the same test with the 3" extension. If at 100 ft-lbs on the captured wrench, both free wrenches read the same, then the 0.8 multiplier is proven. If at 100 ft-lbs on the captured wrench, both free wrenches read different, then the 0.8 multiplier is busted. Hey, maybe we could get Kari from MythBusters involved!! :) |
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My 2 cents
Wow, I just picked up on this thread and have not been involved in the flurry of opinions up to this point. Seeing the chaos I hesitated to chime in but after considering that this could have safety of flight implications for some relying on the information they get here I decided to weigh in. We all bring different strengths to the discussion on this forum. There is a lot I don?t know but I do have a master's degree in mechanical engineering, so whatever that is worth, here are the key points I would offer:
1) A torque wrench is designed to accurately measure and set torque at the head of the torque wrench. To that extent, it does not matter if the handle is 12 inches or 24 inches; whether you apply the force at the end of the handle or in the middle of the handle. The wrench "clicks" when the torque applied at the head reaches the set value. 2) If an extension (or crows foot) is attached the torque applied at the wrench head will be different than the torque applied at the end of the extension unless the extension is turned 90 degrees to the axis of the wrench handle such that the moment arm for the applied force is the same for the extension as it is for the wrench head. 3) If you use an extension on a torque wrench rotate it 90 degrees to the wrench axis and you will be fine. Set the wrench to the same torque value you would without the extension and the same torque will be applied at the bolt head on the extension. 4) If you must use an extension straight in line with the handle of a torque wrench you should be very careful. The torque applied to the bolt will be greater than the value you set on the torque wrench and the correction factor depends on both the length of the extension and the moment arm of the force application. It is not just the length of the torque wrench handle that counts, it is also dependent on where you apply the force on the handle. The correction formula is C = L / (E + L) If the extension length E is 3 inches and the distance from the wrench head to the point on the handle where you apply the force (L) is 12 inches then the correction factor is 0.8. But realize it is not always simply 0.8 for a 3 inch extension. Notice I said ?the point where you apply the force?. Using an extension this way makes the true torque applied to the bolt dependent on your ?technique?. Not so when a torque wrench is used without any kind of extension. I happen to work in aerospace where we build spacecraft and other aerospace equipment and we follow these guidelines in torquing fasteners. As a rule we avoid using extensions or crows feet whenever possible. If we have use an extension we turn the extension 90 degrees to the torque wrench handle. If we can?t do that, we are very careful to calculate the correction factor accurately based on the specific wrench and the force application technique. I hope this helps.:) |
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Start looking through catalogs, and look for torque wrenches in the 100 ft/lb. category...............and see how many you'll find at just 12" long. I've looked all morning, and found none. That .80 factor applies to 12" (1') as it relates to ft/lb torque. Since the measurement of torque that we're using to set the prop nuts is in ft/lbs, then it all makes sense. |
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No amount of math and physics will "fix" this -- some just don't get it and it's more important to believe than to learn.
Either way, for those "wrenchers" -- RTFM for your torque wrench and follow instructions in it.. Once you realize you were wrong all along, you have an option of creating your own torque wrench that works the way you describe it.. you WILL make a lot of money selling those. Heck, I'll buy one.. cause all the torque wrenches I own work the same way Hartzell, McCauley, AC 43.13, and every other publication written says... |
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