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Torque Wrench Tool Calculation

  1. Sep 26, 2010 #1

    I have a problem with a calculation regarding a mechanical tool, so I hope that this forum section is the proper place to post this message. If not, please let me know where would be best.

    Some background.... In automobile mechanics, construction, etc., a torque wrench is a socket tool used to accurately apply proper torque when tightening fasteners. Occasionally there may be a need to apply more torque than a certain torque wrench can deliver. In this case, there's a simple adapter which can be added to increase the tool's torque capability -- an extender bar.

    An extender bar is a section of pipe or tubing of any length desired, but they all have a male drive at one end to accept a socket, and a female fitting at the other end to allow the torque wrench to be attached. The combined tool looks like this:

    |- Extender (length = E) -|- Torque Wrench (length = W) -|

    O-----------------------O---------------------------======= handle

    The calculation for the resulting torque at the socket is well known and simply multiples the overall length (E + W above) by the force which is applied at the center of the torque wrench handle.

    For example, suppose the torque wrench is 2 feet long (W) and the extender is also 2 feet long (E). Assume the torque wrench setting is adjusted to be 100 ft-lbs of torque. Since the torque wrench is 2 feet long (W), 50 pounds of force must be applied at the center of the torque wrench handle to produce the torque wrench setting of 100 ft-lbs. So when this 50 pound force is multiplied by the overall length of 4 feet (E + W), this results in 200 ft-lb.s of torque at the socket on the end of the extender bar.

    The above is easy to see and makes perfect sense, but the problem is there's another way to look at this calculation which also SEEMS to make sense, yet it's WRONG. It's hard to see why it's wrong.... though it is! Here's this (wrong) alternative calculation:

    Since the definition of torque is just force times distance, it sounds possible to simplify the calculation and just use the basic definition instead. So forget about the actual physical length of the torque wrench (W). If the torque wrench is set to 100 ft-lbs, that seems identical to saying that we effectively have 100 pounds of force at an effective length of 1 foot. If that's so, then adding the extender bar length of 2 feet would then yield an overall length of 3 feet. And 3 feet times the 100 pounds of force gives 300 ft-lbs of torque at the socket. Sounds good, but that's incorrect. As we know, the right answer is 200 ft-lbs.

    Bottom line: Where's the error in the "simplified" way of looking at this? I understand the correct way to do the calculation, but where's the fallacy in the alternate view? Can someone point out what's wrong? Help. Thanks!
  2. jcsd
  3. Sep 26, 2010 #2


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    You are putting 100 ft-lbs of force on a 2 ft extender bar. Shouldn't it be only 2 feet in your calculation instead of 3? The 100 ft-lbs doesn't mean that you simply have 1 ft in length already to start with. For example, take your above calculation. 4 ft of length = 50 lbs of force that you must apply. If you remove the first 2 ft, then it becomes 100 lbs, which matches exactly to what the torque wrench is applying.
  4. Sep 27, 2010 #3
    What type of torque wrench are you using, the type that goes "cracks" at the desired torque or one with an indicator?
  5. Sep 27, 2010 #4
    You are using two totally different scenarios, it's no wonder you are getting different results. The first you are using 50lbf multiplied by the distance, in the second you are using 100lbf multiplied by the distance.

    50 lb @ 2 feet = 100 ft-lb.f
    100 lb @1 foot = 100 ft-lb.f

    Then in the scanarios you with the extender bar you had.

    50lb @ 4 feet = 200 ft-lb.f (1st scenario)
    100lb @ 3 feet = 300 ft-lb.f

    In reality if you set your torque wrench to 'click' at 100 ft-lbf. Then you could apply:

    100 lb @ 1 foot
    50 lb @ 2 feet
    33.3 @ 3 feet
    25 @ 4 feet.

    The desired torque would be reached.
  6. Sep 27, 2010 #5
    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 second method which troubles me, not the first.

  7. Sep 27, 2010 #6
    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 mean "lb-ft" of "ft-lb", in other words torque).

    When you said this....

    " In reality if you set your torque wrench to 'click' at 100 ft-lbf. Then you could apply:

    100 lb @ 1 foot
    50 lb @ 2 feet
    33.3 @ 3 feet
    25 @ 4 feet. "

    It sounds like you think I am trying in my example to always get the same 100 ft-lbs of torque at the socket, but that is not so. The purpose of the extender bar is to be able to apply MORE torque than the basic torque wrench can do on its own, and to be able to do this accurately with that same wrench.

    Let's say you have a torque wrench which has a maximum rating of 150 ft-lbs. Then if you try to develop more than 150 ft-lbs at a socket attached directly to the end of the wrench, you will almost certainly break the tool. However suppose you have some large fastener which requires way more than 150 ft-lbs of torque, say maybe 250 ft-lbs. In that case, you remove the socket from the torque wrench, snap the extender bar onto where the socket had been attached, move the socket to the end of the extender bar, and now you are all set to pull the 250 ft-lbs with accuracy and without coming close to overtaxing the basic torque wrench (of course all this assumes you have an extender bar that is long enough).

    This extender bar idea is very old and very well known. Look around on the web and you'll find them along with online calculators to do the simple math involved. To find them, you may have to play with the search string. I call the thing an "extender bar", but others call it an "extender rod", "torque extender", "torque extension", etc. If you use the phrase "torque multiplier", you will find an entirely different beast. Usually "torque multiplier" refers to a very expensive add-on for torque wrenches which can cost from $300 to $1500 and up. These are expensive because they use planetary gears to do the job. Very elegant, but way beyond the needs for most folks (unless they work in heavy construction or on farms). The little bar I refer to is hyper-simple by comparison.

    Most of the above is really only partially relevant to my problem. The extender bar works. The first way I explained to calculate the resulting final torque is correct. What I am asking here is solely this: The second calculation I gave is very attractive due to its simplicity, it sounds right, but it is definitely wrong -- why??? If anyone can help, thanks in advance.

  8. Sep 27, 2010 #7
    That's tl;dr. Fact is you are trying to overcomplicate something very very simple.

    T = Fd

    This is the relation, no matter what words you use. I don't know what you've done in your own mind in the second example but it makes perfect sense.

    100 lbf @ 3 foot from the fulcrum = 300 ft-lbf
  9. Sep 27, 2010 #8
    lbf is pounds-force. It just differentates from lb mass.

    When you said this....

    " In reality if you set your torque wrench to 'click' at 100 ft-lbf. Then you could apply:

    100 lb @ 1 foot
    50 lb @ 2 feet
    33.3 @ 3 feet
    25 @ 4 feet. "
    If you have a torque wrench, you specify the cut off torque. You can't apply any more torque over and above what is set. It's ratched to a certain torque. So lets eliminate the complexity of a torque wrench as it's pointless for this discussion.

    You just have a breaker bar 1 foot long. If you apply 100 lbf at the end, you have 100 ft-lbf torque atthe fulcrum. If you have an extender bar of 3 feet (edit, the extender is 2 feet making the total length 3 feet). You are now applying 100lbf at 3 feet so you have 300 ft-lbf at the fulcrum.

    The fact is, its hard to answer this bue to the way you've phrased your question. You've made it overly complicated, and confused yourself. That's why it's not working for you.

    EDIT: I'm starting to get the feeling that we need to start again with this from the basics, as I think you are describing something a bit different to what I was thinking. (The solution methods are still the same though).
    Last edited: Sep 27, 2010
  10. Sep 27, 2010 #9
    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 torque wrench") had a handle with a center pin. The user was required to wobble the handle as they pulled the torque to keep the force centered on that pin. Obviously this would be impossible to do with a big piece of pipe over that handle.

    However, never am I doing any of the above. The torque extender bar I refer to (regardless of whether this is the exact proper name for it) snaps onto the drive end of the torque wrench. And the socket in turn is moved to the end of this extender bar where there is another square drive (1/2" drive, 3/4" drive, whatever).

  11. Sep 27, 2010 #10
    OK I've decided I was thinking of the wrong thing. Ignore what I said above.

    I'm just thinking of a way to describe the difference you are seeing. This is to do with the 'effective stiffness' of the beam. Having the pivot point (ie where the torque wrench ratchets about) closer to the point where the force is applied allows for torque to be transferred to the fulcrum (where the socket is).

    Lets say you then had a torque wrench only 6 inches long. You could apply 200 lbf before the ratchet 'clicks'. If you added an extender bar, you are then transferring this 200 lbf to the fulcrum point.

    So an extender of 2 feet would mean 2.5 feet from force to fulcrum. This would mean a max of 500 ft-lbs transferred.

    The torque transferred is a ratio from the ratched point to the fulcrum point. In the 2 foot wrench 2 foot exter case the fulcrum is 1/2. Meaning you can transfer


    100/.5 = 100 ft-lbs.

    The second case is a wrench 1 foot long. So the ratio is 1/3

    100/.33=300 ft-lbs.
    Last edited: Sep 27, 2010
  12. Sep 27, 2010 #11
    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.)

    Next I explained what I am trying to do, why, and what an extender bar is. If you object to that style of presentation, sorry, but I don't see any reason to rewrite it.

    Next I tried to explain the way the calculation is commonly done. I thought of actually showing the formula, but I figured just explaining it is words might be more straight forward (again, uncertain of the audience). If you like formulas better than words, here:

    Torque at extension = T at wrench x ( 1 + E / W )

    Then I gave an alternate way of looking at this which uses the basic definition of torque to simplify (wrongly) what is going on at the torque wrench end of this combination tool. Again, sorry, but if you don't get what I meant, let me know and I will try to explain it again.

    If you haven't done so already, maybe reading the many responses to others might help clarify what I am asking. By the way, I hope you realize that I did not mean for my original question to be controversial, nor am I trying to "trick" or "trap" anyone or be cute and ask some kind of quiz question. I gave two slightly different ways of looking at a math/physics problem, one right, one wrong, and wish to understand why the second one is incorrect.

  13. Sep 27, 2010 #12
    See above :P

    Very quickly done before I leave work, so there may be errors.

    Attached Files:

    Last edited: Sep 27, 2010
  14. Sep 27, 2010 #13
    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 calculation, but definitely not what happens with the actual tools. The question remains -- why?
  15. Sep 27, 2010 #14
    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 manually.... Quote: "Lets say you then had a torque wrench only 6 inches long. You could apply 200 lbf before the ratchet 'clicks'. If you added an extender bar, you are then transferring this 200 lbf to the fulcrum point."

    If the wrench is set to 100 ft-lbs and its length is only 6", then, yes, 200 pounds of force would be needed at the handle to make the tool "click.

    Quote: "So an extender of 2 feet would mean 2.5 feet from force to fulcrum. This would mean a max of 500 ft-lbs transferred."

    Right. 200 pounds of force times 2.5 feet = 500 ft-lbs of torque. But so what?

    Quote: "The second case is a wrench 1 foot long. So the ratio is 1/3

    100/.33=300 ft-lbs"

    Right again. But I hope it's clear that in the second case, there isn't a second wrench. It is a the exact same wrench and the exact same problem, but looked at in a slightly different way.

    One more time.... the second way I described to do this calculation sounds okay, but unfortunately it gives the answer of 300 ft-lbs, which we know is wrong. 200 ft-lbs is the right answer. So WHY is the second way wrong? Or stated better, where is the error in either logic or physics in how that second calculation is done? This is solely what I am trying to understand.

  16. Sep 27, 2010 #15
    In the second case to get 100 ft-lbf at the ratchet, you MUST be applying the force half way along the wrench (1 foot) meaning that the distance from force to socket is now 3 feet and not 4. Making it a different case. You said this yourself.

    The reason for this is:

    You still have a point in the middle that will 'give' at a set torque. I'm already ignoring material stffness. So for example, if you set the torque wrench to 50 ft-lbf, the arrangement has half the stiffness of the wrench set to 100 ft-lbf. It's basically a joint between two beams, the stiffness of that joint detemines how much torque you can put through the system.

    The second one isn't wrong. You are just misunderstanding whats going on. Lets take the cases to their extremes. In each case the length from for the force applied to the socket is 10 feet.


    X= length of torque wrench
    Z= length of extender
    L= total length = 10.

    The true wrench is set to 'click' at 100 ft-lbf.

    1st case is X = 10 Z = 0.
    We know that the maximum torque at the socket is now 100 ft-lbf. As the socket is at the ratchet point.
    So the maximum force you can apply is 10 lbf at the end of the wrench. from
    T = Fd.

    The other extreme is Z = 10 X = 0
    So this means you grab the torque wrench right at the ratchet mechanism (so in reality you have the handle sticking out of your hand in thin air). You now cannot apply a torque to the rathet, meaning it will never click. As T=Fd. You are applying F at d=0, so T=0
    This means that you can apply any force you want to the system without any risk of the ratchet moving, so the max torque applied to the socket is infinite.

    You've basically made a joint between two solid beams, and set the point at which it will break. The closer that joint is the the force the more load you can apply. You could also model those two beams with a joint as a single beam with a corresponding stiffness.
    Last edited: Sep 27, 2010
  17. Sep 27, 2010 #16
    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 simple example? If we don't, it's hard to keep up with what you (or anyone else) means by "first case" or "second example" --my first case or your first case?

    The above said, let's have ONLY one torque wrench length -- 2 feet. And have only one extender bar -- also 2 feet long. And we are only and always setting the wrench to just one setting -- 100 ft-lbs of torque. All very straight forward. By now I think we all agree that when you attach the bar to the wrench thereby getting an overall length of 4 feet, the resulting torque at the socket is 200 ft-lbs (50 pounds of force at the handle times 4 feet).

    Okay, now do NOT change anything. The example remains the same -- same mechanical drawing, same lengths, same force applied, etc. However.... ask what is actually happening at the end of the torque wrench (not the overall combined tool, just the torque wrench). Well if the wrench "clicks", we assume it is telling us that 100 ft-lbs of torque has been developed at that point. Knowing that, why can't we just say that no matter what the actual length of the wrench may be, we can MODEL it as being equivalent to a 100 pound force at a distance of 1 foot?

    I don't care if you own a 12 foot long torque wrench, if you set it to 100 ft-lbs, it creates the same equivalent force at the same distance I just mentioned.

    So if 100 pounds at 1 foot is what we have no matter the actual wrench length, then why not just add the extender bar length to that? And if we do, we get a 3 feet total, times the 100 pound force, and that comes to 300 ft-lbs -- which is DEAD WRONG.

    I get that, believe me I do. And I know you and others get it too. However it's a logical tautology to say only that looking at the calculation that way it is wrong because the right way is right. This second way of disregarding the actual wrench length seems to be a fine approach -- just to say you always have the EQUIVALENT of 100 pounds at 1 foot is so simple. And it is obviously perfectly accurate. But the minute you add the extender bar, that 's when that approach fails miserably.

    The correct calculation was stated in the very first post (and since then I have given the simple formula as well). Both show the correct number to be 200 ft-lbs. But once again, I'd sure love to hear someone explain the error in the second way of looking at this. Maybe you feel you just did that. If so, sorry, I didn't see anything that explained the error. Maybe I overlooked it?

  18. Sep 27, 2010 #17
    You are just modelling it totally wrong. There's little more I can say than that.

    The main point here is you CAN'T say 'lets take the wrench in isolation', develop an alternative method of describing that. Then use that in a different application. Becuase that's simply not what is happening. Your torque value may be the same, but that's not whats critical in this case, it's the force and distance.

    There is no error, what you are saying is precisely what you would expect to see. You are inveting the error out of a misunderstanding of how the system should work.

    It's your assumption that because the 50 lb at 4 feet gives 200 ft-lb, the other should have the samevalue is where you are going wrong.

    This kind of thing always happens when people post an assumption that leads to an incorrect conclusion. Then ask why the conclusion is wrong. The fact is, in your second case you WOULD expect to see a torque of 300 ft-lb. It's dead right. Seriously draw the picture, look at the figures and try to think why it would work.

    Even better just do it, if you have a torque wrench and an extender bar in the configuration you state. Torque something up holding the handle half way along with 100 lb until the ratchet clicks. Then try to undo it with the same torque wrench set up but pulling at the end. Unless you cheat in some way you'll never be able to get the bolt undone.

    This thread is stanting to need moderation, the posts are just becoming wordfests.

    If you'd like to just post the question, no expinations or assumptions. We can work through it from the beginning so I can show you where you are going wrong with your thinking.

    Unless anyone else wants to attempt an explanation.
    Last edited: Sep 27, 2010
  19. Sep 27, 2010 #18
    When you add an extension you have to adjust the torque setting on the torque wrench. If the extension increases the overall length (L+E) in the diagram you must reduce the reading set on torque wrench. Conversely if extension decreases the overall length (extension turned 180 L-E) you have to add a higher torque value on the wrench. Applies only if extension is inline.

    Setting the extension to 90 degrees – you can use the same torque setting as called for. Any angle other than 90 or 180 degrees requires using the cos of the angle like second link.

    The purpose of the extension is not for adding torque, it should be used where clearance is a problem.

  20. Sep 27, 2010 #19
    If you look I did the very same proof (although not quite as neat as I was in a rush) as in the engineers edge link.

    This is the reason for the phenomena you stated, it's not a mistake (or restating the problem) it's acutally what happens.

    As Nucleus pointed out normally you'd make an adjustment to the mechanism to prevent over torquing. Although you could leave the mechanism as is and increase the torque output. The output is defined by the ratios of the lengths. If you alter the ratio, then you alter drive torque accordingly.
  21. Sep 27, 2010 #20


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    Hows this:

    If you put 100 foot pounds of force on a 2 foot extension, you WOULD get 200, which is just the way it should be. Now, since you are using a 1 foot torque wrench, you must calculate the extra 1 foot in there. 200/3ft = 66 2/3. You would be applying 66 2/3 pounds of force at the handle of your torque wrench which would put 200 foot pounds of force at the socket on the end of the extension.
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