The best way to demonstrate it (without maths) is to say to people: _"Imagine pushing on the pivot point itself and you will see that the wrench arm won't be able to rotate, causing no click even with the highest pressure you can apply. However, this applied pressure WILL still turn the bolt."_ Basically, the main pivot point simply cannot rotate if you push directly onto it, but the bolt will still rotate. The only way you can rotate the pivot point is to have that force applied at a length away from it, turning the applied force into rotation (torque).
I love when I'm puzzled about something after one explanation, only to find a different explanation that makes everything instantly click into place. Pun intended. Thank you.
Agree...when I was at about 3:15 I imagined applying force exactly half way between the clicker and the pivot point...and saw that I would be applying half my force without going "through" the clicker. But now I agree, your explanation is the best.
He still doesn't get it. If the torque wrench will click at 50 ft. lbs with 5 pounds of force at 10 inches, then the force required to get it to click choked up at 5 inches will be 10 pounds (twice as much)! Clearly someone is trying to pretend that they are smarter than they actually are.
@@JohnAnderson-sm8jl this video maker and those that believe hand position matters just can't see beyoud the principal of the lever but can't get it in context with a torque wrench. They all need to watch this short video it's factual. When the tool clicks no matter where from or how much force was imparted one stops pushing. The only error is oporator error but to get that one has to be a compleat ideot, check it out th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=YbAvHDfQ3x6HyaOX
@@JohnAnderson-sm8jl He is a professor of engineering with a PhD. I don’t think he’s pretending to be smart. As you correctly point out, the torque delivered by a given force depends on the distance between where the force is applied and the torque is measured. In the same way, the torque being produced by that force is different depending on where you measure it. If you put a second ratchet half-way along the beam, you’d measure different torque there than at the end from the same force applied to the handle, just as you said. Two different torques experienced, depending on the distance it is from where the force is applied. The pivot point is that second measuring distance.
Great to see someone get some constructive feedback and actually do some critical thinking of their own beliefs. Well done for putting this together and proofing your initial thoughts, and those of many others, incorrect. You’ve got me as a subscriber.
He still doesn't get it. If the torque wrench will click at 50 ft. lbs with 5 pounds of force at 10 inches, then the force required to get it to click choked up at 5 inches will be 10 pounds (twice as much)! Clearly someone is trying to pretend that they are smarter than they actually are.
Thanks for an excellent and informative video. I had to watch it 3 times to wrap my head around what's going on here. As it did with you, the physics seemed painfully straight forward, and where you hold the handle shouldn't matter. But it does. My way of explaining it in layman's terms would be that if you're choking up on the handle (as compared to the prescribed point at which you're supposed to hold it), you're choking up more (percentage-wise) on the click pivot point than you are on the fastener pivot point. Similarly, if you apply the force further from the pivot points, the percentage with which you're doing so is different for each pivot point. Finally, this can be more concretely demonstrated by applying the force directly at the click pivot point. When doing so, you'll still be applying torque to the fastener, but the wrench will never click because there's no torque being applied to the click mechanism.
@@TheBikeSauce why? Are you really that brain dead. The only time it won't put force on and click the wrench is if you pull dead centre of the nut. But anywhere else and mathematicaly it will always put torqu on the nut nut at fantastic power. But when the tool clicks it will still only tighten your nut to what ever the tool is set to. You all really do need to watch this short video and a longer one on my channel. th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=YbAvHDfQ3x6HyaOX
Extremely clear explanation of a complex topic. And nice drawings to accompany! You made a potentially dry topic fun and engaging...the mark of a great teacher!
Excellent explanation of the physics behind the length-dependent click-type torque wrench! I had thus far never considered the dependence of the applied torque on my grip position when using my torque wrenches. From now on I will be more mindful of it. The engineer in me is also puzzled by the nonlinearity of the torque discrepancy that you calculated by changing d1 from 5 inches to 2 inches and then to 10 inches. My intuition would've been that the error in torque would've been linear with the error in moment arm length. So I worked out the equation for the relative error sensitivity, S, of torque with respect to grip position (ie the fractional error in applied torque per unit distance error in hand placement): S=-(d2/d1)/(d1+d2) S is directly proportional to d2/d1 and inversely proportional the sum of d1 and d2, so the longer the overall length of a torque wrench, the less sensitive it will be, and the longer the handle bar is relative to the head unit, also the less sensitive. The negative sign tells us that a closer hand position to the handle pivot pin will always result in more torque applied at the drive when the click occurs, and a farther hand position always results in less torque being applied, which is somewhat counterintuitive, and quite the opposite behavior of a normal wrench. The relative error sensitivity is independent of Fs, r, or the torque setting selected, as both the absolute error in torque and the torque setting itself scale identically linearly with each of these parameters.
It's super simple, but it's not very practical, since it's useful just for small deviations. It's just the first derivative in Tailor series, but the function is nowhere near close to linear. Mc = (1+d2/d1)*r*Fs Mc'(d1) = - r*Fs*(d2/d1^2) And here you get: S= - (d2/d1)*1/(d1+d2) And there are Slip or Cam-Over Torque Wrenches, which do not have this problem. They use a radial ball clutch. And not only this problem is eliminated, but also when they slip, over torque is not possible.
No one pulls on a torqu wrench anywher but the handle. But if one does it make no difference at all watch this short factual video and learn even if you can't comprehend. th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=YbAvHDfQ3x6HyaOX
To me you are merging two mechanical systems of wrench: the lever ratio of d2/r with tension spring which is defining the click (an internal transfer function of wrench to initiate a click) and d1+d2 torque lever. When applying torque exceeds specified value, equilibrium between friction force (setted up by spring tension of transfer function) and (d1 + d2) F breaks - the transfer function makes click. As mentioned one of viewer one year ago, before click the wrench is the one solid pivot lever with size d1+d2. So, to calibrate the wrench we need correct weight at correct d1+d2 distance. As I see, you are trying to describe all absolute forces that are originating in wrench.
it seems very counterintuitive but I now think you are right. If you apply a force very close to the pivot point you can understand it will snap at a very or extremely high force which would obviously mean an extreme torque to the bolt. If you apply right on the pivot point it will never snap and you will still torque the bolt until you break it for sure unless something else breaks
I couldn't quite get your explanation to click (pun intended), so I ran through the math myself. The key for me was to look at the free body diagram for the handle (ignoring the head of the wrench). The handle sees two torques about point Q, one is applied by the person's hand (F*d1), and the other is applied by the clutch mechanism (Fs*r). Balancing these torques allows us to compute the force required to click the wrench: F = Fs*r/d1. Now, we look at the torque applied to the fastener (Tf), treating the wrench as a solid beam: Tf = F*(d1+d2). We already know the force required to click the wrench, so we can substitute: Tf = Fs*r/d1*(d1+d2). Simplifying we get: Tf = Fs*r*(1+d2/d1). Since Fs, r, and d2 are all constants determined by how the wrench is set up, increasing d1 decreases the torque applied to the fastener before the wrench clicks. What's nice about this explanation, is we can directly calculate the percent error based on where the force is applied. If you have a wrench with d2 = 2" where you're supposed to hold it at 16", if you instead choke up to 6" then you'll apply 18% more torque than indicated.
Ahhh, YES... An M.E. weighs in on the question, and with a beautifully simple derivation takes this counterintuitive, very interesting analysis, and boils it down to very practical, quantifiable terms. Now THIS is something I can apply! If I'm in an awkward configuration that for some reason requires me to choke up on my click-type torque wrench, I understand intuitively that I will have to apply more force (F). I don't really care how much. Using only d2 and the new d1 (both externally measurable) I can calculate my resulting over-torque and back down the setting appropriately. Thank you, sir, and nicely done.
I knew the rudiments of your explanation prior. But, I couldn't get my math to explain why things worked like they do. I was very excited to learn what I didn't know about click type torque wrenches. Thanks so much for opening my eyes!
Great video but I think the point about hanging weights accuracy testing being flawed is, well, flawed. You can absolutely test the accuracy that way, as long as your calibration hanging point is in the middle of the handle. The wrench should click when you hang the weight off that point and should stop clicking once you move the weight as little as 1/4" inwards. What may actually be flawed about the tests you did is that you need to measure the weights before using them (exercise equipment weights are rarely very precise); and you also need to factor in the weight of the wrench itself with this method since it also applies some torque at the pivot point when used in this orientation. Something like half the weight times length of the wrench should provide a good first approximation.
Great explanation. It amazes me how many guys currently have TH-cam vids “testing” the accuracy of torque wrenches and their hands are all over the wrenches and in many cases far closer to the tool heads than the tool handles.
As others have already said, this was brilliant. Well done! It's been over 40 years since I took Statics, and I was an EE so I promptly forgot all I learned in the required ME courses. Kudos to your engineering skills!
Don’t feel bad, it’s been 30 years since I graduated as an ME and I’ve forgotten most of both the ME and EE content!! Excellent video - trying to understand how I can still apply the “weight” plate test accurately. Using his example, Am I to assume that if I apply a 10lb weight at a distance of 5” from pivot point and set the wrench to 60in-lb it will pop off if calibrated? Or a 20lb weight with wrench set to 120in-lb? Is that a decent way to calibrate?
Awesome. Your self-depreciation is hilarious and the explanation is excellent. I was torquing up a headset only last week with my hand choked up over the head and I idly wondered if that made a difference to the result. Now I know!
To greatly exaggerate, click style torque wrench would never click if force is apllied directly over pin (or closer to the head) no matter the Force applied (pin that is just bellow the head of the wrench), so logic is that handle is where force should be applied because distance (from pin) obviously matters. I always used click torque wrenches as they should be used, but now i clearly understand why.
Yes indeed, this is probably the simplest way to quickly see that moving the point of applied force must change the resultant torque at which the click will occur. This video delivers the most clearly presented explanation of why that I've seen. Just a pity that so many folk on the internet, like our friend JohnAnderson-sm8jl, still don't get it!
Dude, fantastic work. When Russ at Path Less Pedaled said that grip position mattered, I was sure he was wrong. So, I came over here to see what you said. Thanks for explaining so well. Great illustrations and sample calcs.
Hi Bike Sauce, I think you have this wrong. The second pivot and clutch is a mechanism to apply a consistent and calibrated load to the socket. The load transfer is not changed by the hand position. It is limited by the transfer of load through the clutch to the rigid arm connected to the socket. You are correct that the testing technique you used is flawed (measuring load on the lever arm to make it click is simply measuring the input force) but you need to look at realised torque at the socket. To do this you need to set up a strain gauge on the rotating object - nut or bolt - and see if the realised torque differs with the same wrench settings and different lever lengths - hand positions. I have not tested this but my analysis of load transfer leads me to believe that hand position does not alter the load transferred to the rigid arm through the clutch.
Great video, thanks! Didn't necessarily follow all the math, but understood the basic concepts. I'll have to go back and read my torque wrench manuals.
Nowhere will a manual tell you outright that hand position matters other than in one of my videos they say it make only a difference to the oporator error. Watch this short video. th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=YbAvHDfQ3x6HyaOX
I didn’t understand why hand position mattered until recently because I didn’t understand how clicky torque wrenches worked. And that is not straightforward at all. And manufacturers absolutely do not help with this. If we read the instructions (lol), they say do it this way but they virtually never explain why. And many many people assume that there is just some mechanism built into the ratchet head that is somehow calibrated to click with proportional to torque. The big problem I have is with people who, once shown the mechanics of the wrench, the statics equations, etc., still just say “bah! torque is torque! I’m right and you’re dumb.” But alas, the internet is FULL of such people. Anyway, great video. It’s nice to learn new things.
Thankyou for that, it really helps with the physics understanding and it also means that hand position is actually more important than I thought it was as even using the end of the torque wrench will result in skewed settings.
Excellent explanation. I have always wondered why my torque wrench stays at the back of the toolbox and that I am more confident of my hand force on the wrench. The handle of the click type torque wrench should be articulated, pivoting at a point exactly at the "middle of the handle". As long as the articulated handle is not bumping against the stops of its free range, the torque value should be consistent. Or maybe the torque wrench should not have a handle at all but a ring at its end with which you can use a finger or a chain to pull on it. That is old stuff. Digital seems to be the answer now.
its more often about repeatability than exact torque ,, plastic parts need an even clamping force to help prevent leaks , rubber gaskets wont mind say an extra 5% torque , as long as the fasteners all have the same to prevent distortion.
I simply loved your explanation and the way you approached the problem. What is intuitive during usage is made perfectly clear with math now! Thank you for taking the pains to make this video!
Thanks for this video. I still don't understand the maths, but I do now understand why grip position makes a difference. Also explaining the inner workings of the torque wrench helped a lot. Thanks again.
Nice demostration … there is also another consequence to consider in my opinion. When you use this kind of torque wrenches you should not apply additional torque with your hand. If the hand rotates the applied torque is not accurate. Ideally you should only apply force in a single point in the middle of the handle. I hope this is clear.
Hi. It is true. Torque is torque when you have only ONE lever to move. Here though you have two bodies interacting. The inner shaft and the outer tube. OK plus the bolt which only lends its reactions to the system. Also Fs will always be Fs meaning that the limiting force at which the click happens is always the same since it is dependent only on the spring’s selected stiffness by the dial and we don’t change that. So what else could be changing ? During the click a collapse mechanism is occurring. The two beams change their relative angle. Note that Fs doesn’t exist as an outer force for the system. It exists only for the inner shaft or the outer tube when separated and viewed as stand alone. Therefore the easiest way to prove your point would be to first show the equilibrium for the whole system just prior to the click when the two beams act as one and calculate all the outer reactions. Then show the equilibrium for the inner shaft ONLY using the previous reactions plus the force Fs as mentioned and considering ΣM=0 at the pivot point of the two beams. Actually when the click happens the pivot point is where the true rotation takes place even if the inner part doesn't really move - your hand’s force F moves respectively to it along with the outer tube. STEP 1 The System. Apart from the torque Mbolt {which from now on we consider it as a moment reaction applied to the shaft} a significant vertical reaction from the bolt -F should also be shown {which is opposite to the F you apply by your hand to the OUTER tool and satisfies the equation ΣFy = 0 for the WHOLE initial system of the two bodies that still act as one. A force which correctly is not shown in your diagram since your pole coincides with the bolt and therefore -F doesn't produce any torque there} STEP 2. The shaft. This vertical reaction now produces a real moment -F*d2 for the inner shaft respectively to the pivot point and it should be noted that it is a variable moment that changes every time you place your hand in a different location only because the vertical reaction changes. Again it is important to realize that for the inner shaft this variability occurs ONLY due to the difference in the strength with which you push the whole tool and not due to the distance from the bolt because the inner shaft is isolated and perceives no moving loads {since all the forces acting upon it have their points of application fixed either at its 2 ends or at the pivot point and therefore perceives only the fluctuations in its reactions}. Of course the distance comes into consideration for the whole system only and goes along with the general consensus that when you want to achieve similar magnitudes of torque and you decrease the distance you have to increase the force. This increment is what is transfered to the shaft and plays a big role as we will see - big enough to alter the torque output. For completeness we should add here that there is also one more vertical force applied to the inner shaft at the pivot point by the outer tube and it is equal to F-Fs. It is derived from the equilibrium ΣFy=0 {of the inner shaft this time} since there are 3 vertical forces at its 3 points -F and Fs at the ends and F-Fs at the pivot point. This one though doesn't produce any moments respectively to the pivot point and we won't analyse it any further. Now the net effect is that this aforementioned moment -F*d2 acts as a relieving moment for the shaft and is actually subtracted from the applied moment Mbolt in the inner shaft ‘s equilibrium equation. ΣM=0 => Mbolt-F*d2-Fs*r=0 => Fs*r=Mbolt-F*d2. Hence for the click to happen Mbolt should now be larger to counteract that larger [-F*d2] and yeld the same Fs that releases the spring. To put it simply the constant moment Fs*r is neutralized partly by the vertical reaction's moment and partly by the true moment applied on the bolt as torque! To get a recurrent reliable Mbolt you should therefore combine it with the same vertical force that was present during calibration. It is part of the tool ‘s intrinsic formula. This a formula can also be viewd as Mbolt=F*d2+Fs*r. Both addends on the right side should appear as they were during calibration for the torque result to be reproduced and yeld the Mbolt that the factory had initially gauged. To summarize it all Fs is not the only parameter of the system. The force with which you push the wrench is equally important. So when you hold the tool EXACTLY at the grip of the tool that vertical reaction is the same one that produced the torque Mbolt together with Fs*r during the factory callibration and this is the only way to always yeld the same result. You exert the exact same forces of the calibration and they produce the exact same torque on the bolt. As you approach your hand towards the bolt the vertical reaction on the bolt gets larger. One addent gets larger. That new larger F*d2 is not what the factory had taken into consideration when calibrating Mbolt and since it is again added to Fs * r which is always the same that means that the click happens at a larger Mbolt !
Great illustrations! They explain the problem very nicely. Thank you. However, I think your equation on the left side is off a bit. Your summation shouldn’t include the internal moment “-rFs”. Until the clutch slips the entire wrench could be considered as a simple solid bar when summing the moments about pivot point “o”. The correct summation equation is just: (d1 + d2)F -Mc = 0 Nevertheless, when you perform additional free body summations of the individual wrench parts to capture terms “Fs” and “r” and simplify you will eventually get to the key equation which you have correct on the right side that shows that the clutch slippage force "Fs" is indeed dependent on d1: d1F = rFs or Fs = F(d1/r)
You are absolutely right. But see, this is physics, he can not be a bit off, he is just completely wrong. There is a system of two equations for the torque around the two points of rotation - O and Q respectively: (d1+d2)*F = Mc d1*F = r*Fs And eliminating F, (not Fs as you do) we get: Mc = [(d1+d2)/d1]*r*Fs And we get that the relation between Mc and Fs includes d1, which is not a constant, but a variable. The other two, namely d2 and r do not matter, since they are a constant dimensions of the tool. And we also see, that the naming scheme of the lengths is also completely wrong, since d1 and d2 have completely different meaning - a constant and a variable. Also r is like d2 - a constant, they must be on the contrary - similarly named.
I love your approach to academic discussion 😆. Are you equating torques at 2 different locations? Naming conventions are pretty arbitrary, not sure how choice of variable name is ‘wrong.’ Thanks for your input nonetheless. Would love to hear precisely why this is completely wrong though.
@@TheBikeSauce In the model of the click wrench you have two axes of rotation, namely O and Q. Every one of them produces one equation. Basic naming convention suggests using at least different naming for constants and variables. At school they suggest using the beginning of the alphabet for constants, middle of the alphabet for integer indices, and the end of the alphabet for variables. For example a polynomial is written as: Sum (An*x^n) a, b, c, d - constants i, j, k, l, m, n - integer indices x, y, z - variables But you do not need to follow that exactly. Proper way is: d1 and d2 - constants r - variable Or: r1 and r2 - constants d - variable You see, also, two of these lengths are dimensions of a real physical object, while the third one is an abstract length from a physical point to an imaginary one. The real force applied by the hand is distributed over a surface area. Also, another mistake there. In the simplified model all forces come from points on the axis of the wrench. But you draw the applied force on the surface of the wrench handle. You can have a model like that, but than you must take into account the thickness of the wrench handle. And you get half the thickness of the wrench handle as a third constant there. Going outside the handle and / or having variations of that thickness makes this constant a function of our variable, the position of where the force is applied. So you get a more accurate, so to speak, second order model.
@@TheBikeSauce If you want, I will add this. In the second order model, where the thickness of the wrench is taken into account, it actually happens, that it does not matter. Yes, the thickness of the wrench cancels out. Which is another, seemingly, "strange and non intuitive" fact. So, it's not a problem to have a short wrench with a thick handle. The equations go something like this: Sqrt[(d1+d2)^2 + h^2] * F * cos A cos A = (d1+d2)/Sqrt[(d1+d2)^2 + h^2] Sqrt[d1^2 + h^2] * F * cos B cos B = d1/Sqrt[d1^2 + h^2] where h is half the thickness of the wrench where the force is applied. And there we get the same equations as before.
Really appreciate that you’re interested, but have a bit of perspective maybe. You can always model a system more accurately given the time, but that does not make simpler models wrong. This model is adequate to demonstrate the counterintuitive idea that grip position actually matters on this type of wrench. It is an explanation for similarly counterintuitive empirical evidence seen in other videos. In your more complex model, you gain nothing by considering the geometry of the handle. The cos will be negligible wrt unity and the geometry of the handle is inconsequential.
This explains why I broke one of my valve cover fasteners the first time I did my valve cover. I thought I was slick by "choking up" on the torque wrench. Good thing for helicoil.
I like the explaination. However it provoked more questions like; Where exactly is the "middle" of the handle? If your hand is smaller than my hand is it more/less accurate? How do you apply force to a specific area with a hand, that is much wider than a string with weight?
Man, this was an excellent video you made, and a brilliant explanation of why I find all the torque wrenches that I buy, to be out of calibration! It seems I was making the same mistake as all of us... But now, I know!! A million bravos from me too!! Thank you!!
Ok, this is absolutely lovely, and makes my (physics degree earning) brain glow, thank you. It's exactly the right level of abstraction to explain what's going on. Also, I absolutely respect the patience you have for dealing with both ends of the comment spectrum (this is wrong because it's just a beam vs this is wrong because it doesn't take into account blah blah blah).
I have a split beam torque wrench with the knob on the side (not the micrometer style click wrench), and I also have one of those digital torque adapters. I don't know if the physics are the same as you showed with a split beam, but gripping different parts of the lever produce different torque values at the socket (where the digital adapter is) when the wrench clicks. It doesn't seem to be a linear relationship though. The more I choke up the harder it is to generate torque at the socket (of course), but pulling between the handle and the pivot point produces erratic readings from test to test and causes the wrench to click with 35, or 46, or 41 etc. ft lbs at the socket (when it's set to 50). This is not quite what I understood from the video which (I think) said the higher you choke up the MORE torque should be at the socket when the wrench clicks. But again I have a different style torque wrench. The conclusion is the same, of course, and it's something I eventually figured out when playing around trying to see if both my adapter and wrench were accurate (they are both new). Hanging weights off the shaft 12 inches from center socket caused the wrench to click 5 ft lbs. too early. Hanging the weights off the center of the handle got me within a few ft lbs. (borderline acceptable range). But actually pulling on the handle properly while bracing the socket gets me within less than 1% repeatably. Conclusion: how you grip and apply the torque to the handle matters, so do it properly. Since I'm getting both my devices to agree with a high degree of accuracy I'm going to assume the equipment is good enough to rotate my tires with. lol (because honestly what else am I going to do with it, build a space shuttle in my garage?)
For a beam style wrench, the length is a factor so grip position is also a factor since it’s based on beam deflection. But.. I totally agree - we’re not building space shuttles 😆😆😆
This helps makes sense as to why the hand position matters for a click type torque wrench. One issue with the equation, though. If you use the socket as the pivot there is a force on the pin where the inside pivot holds the handle to the inner beam. Also, the socket itself has a torque which is equal to the torque that is applied to the bolt, and that is the quantity to solve for.
Thx, Kelly. The force you mention is accounted for in the additional couple Mc. The torque at the socket is equal to the applied torque on the bolt as it's in static equilibrium.
@@TheBikeSauce, Mc does not account for the torque added directly at the pivot point by the orthogonal force of the handle. I think you have made some catastrophic simplifications and mistakes in your modeling.
There is a system of two equations for the torque around the two points of rotation - O and Q respectively: (d1+d2)*F = Mc d1*F = r*Fs And eliminating F we get: Mc = [(d1+d2)/d1]*r*Fs Or: Mc = [1+d2/d1]*r*Fs
@@jasonbeisiegel5550 Mc is the reaction of the bolt or nut to be tightened. It's equal to the torque that the wrench applies to the nut or bolt, but has opposite direction and is applied to the wrench. It has nothing to do with any pivot points. Also, the pivot point can not be a source of a force, a torque, or anything. He is wrong, but you are not right too.
Excellent explainer / walkthrough of the math and visualizations … as a data analytics and visualization instructor (and cyclist, of course 😁), this really hit the mark
So you CAN use the weight on a string method as long as you place the string with the weight at the center of the handle? This means to measure from the center of the socket to the center of the handle and multiply by the weight to get the expected torque.
Well "torque is still torque", but changing the position of your hand on the wrench changes the way the forces interact with the internal "clutch mechanism" and the click occurs at the wrong time. Someone who took engineering mechanics in engineering school will immediately understand everything you did. There are other types of mechanisms used in torque wrenches, some use internal beams, some use actual electric resistance load cells, I don't know what else is available. When I grew up working in my father's garage we used a "beam type" torque wrench. There is nothing to calibrate on those, the deflection of the beam will be repeatable if you don't damage the thing somehow, deflection being P times L-cubed divided by (3 times E times I), spoken as pee ell cubed over three eee eye. The beam type torque wrench that we used enforced proper use by having a pivot pin through the handle. You had to pull in just the right place to keep the handle from turning with respect to the beam, therefore the force was always applied at the exact L from the fastener that the wrench was manufactured for. The only trick was getting in a position where you could accurately read the scale while pulling hard on the handle. Sometimes it helped to get a second person involved.
Excellent point! I still have my Craftsman beam torque wrench from the late 60s, when as a teenager I did my first serious engine work. A kindly machine shop tech took the trouble to point out exactly how to use the tool and emphasized putting smooth pressure only on the handle pivot pin for accuracy. I’ve kept that in mind ever since, even with all the newer torque wrench types.
I got as far as 8.05 mins into your well presented video. But then i woke up. You go on about two torques but infact right up until the clutch as you call it snaps, clicks, the whole thing is basicaly a lever, after the click one stops pushing on the tool. How far up the handel this force comes from does not matter once the tool clicks you stop otherwise you will force the detent against the outer wall of tube and can still carry on to over torque.
@@iliinsky This is something our friend pete has been unable to get his head around. No-one (in more than one place on TH-cam) has been able to get him to realize that the fact that force applied at the handle pivot point can never produce a click regardless of the amount of torque being delivered, points out that it does indeed matter how far up the handle the force comes from. Still, one can only hope - he may read and have it "click" for him that he has said "once the tool clicks", and in that case the tool will never click.
Brilliantly explained! Could you repeat the test with the proper method on the budget torque wrench, it would be interesting to see how it really performs.
The very first to make perfect sense.👴👌 I've seen many vid's stating and showing that it will make a difference where the force is applied, all empirical...👴🤷♂🧐🤔🤨 This one however is the first one clearly explaining how and why.👴😊🤗👌👌👌 Thank you for that.👴🤗👌👌👌
The click type torque wrench is not very accurate because operator cannot respond quickly enough to the click and will usually over torque the bolt. Through experimentation I found that digital torque wrenches provide the most accuracy because the operator can anticipate when he/she is at the correct torque.
Great video! Now I know how a torque wrench works :) The exact distance is only important during verification; it doesn't matter for the application itself. You just press with force x until it clicks. Or am I wrong about that?
You are correct, and TheBikeSauce is not. After calibration done application distance will define how much force - more, equal, or less - user should apply to ‘reach the click’
Love this video and totally makes sense. My only question would be a practical one for a user of a click-type torque wrench. Mine would not have an 8” range for practically where I could grip it. If I narrowed it down to plus or minus 2 inches (which might still be wider than I could practically hold it), what’s the error and does it matter?
Great explanation! But what are the points where you measure the "middle of the handle"? The S1)socket end or S2)the sort of invisible pivot [start of d1] to E1) absolute end of wrench or E2) end of main handle, before the adjuster bit? I ask because my recent experience includes learning the hard way that most (my) non-Park priced torque wrenches don't work on reverse threaded things like drive-side bottom brackets (which is not entirely applicable to this video but I'm whining) and crank arm bolts (due to user error apparently) :-)
Really interesting to consider then how a crow foot attachement would change the values... It is an extension of "d2", but the d in that situation changes as the crows foot rotates probably? How would that math play out? Thanks for the video!!
Wow, this just made maths interesting! Brilliant video. With this in mind does this alter any of your previous conclusions about the other wrenches in the pervious video?
No. It does not matter too much where you place your hand. The slip mechanism is always calibrated to a fixed rotational force at the pivot point (based on the spring pressure). But it does matter if you are measuring the force applied at a certain point to get the corresponding torque at the end. In such case, you can use the formula normally used for torque wrench extension to back calculate the proper weight at your chosen point of force applied, and to account for the distance between the slip point and the socket point.
That makes sense - so I can still calibrate using a fixed weight at a fixed location using the equations prescribed in video. Just as easy to use middle of handle. So, for this example, setting a 10lb weight at 5” from pivot and assuming 1” distance from pivot to socket attachment center, I would set wrench to 60in-lb and if calibrated, it should “click”?
@@TheBikeSauce I think there is a problem with your maths, though. Turning force is always applied to the inner arm (and therefore the socket) at the clutch mechanism. So the equation for the torque at the socket should be the force experienced at the clutch mechanism (Fs) multiplied by the length of the inner torque arm (d2+r). So the torque wrench will always click at Fs*(d2+r).
Head hurts with the math ,, it would be nice to see it with polarised Perspex , then one could see the forces as the math really hurts my tiny brain, I’m not entirely sure that my suggestion would work …..,due to this tiny brain again, Great explainer all the same ❤
😂if your head hurts now wait till you have seen these two short videos. If this guy is a professor god help science. th-cam.com/video/tTbjLAm7XnQ/w-d-xo.htmlsi=885sZ8Fwt9Iw9fHV th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=vdS4hI08-O6Tjshk
Me too. This is the most understandable explanation I've become aware of. I knew John was correct at Auto Expert (and he measured to demonstrate), but the explanation of the reason behind it wasn't easy to visualize.
Great video and a very good explanation! Something crossed my mind though; The adjustment on some torque wrenches effectively changes the lever length of the wrench from the handle. It's not the case with Park Tools torque wrenches, but for many others, the mechanism is such that the handle moves closer to the socket when increasing the torque setting. I haven't thought about it much yet, but I'd like to hear your ideas about this. Do you think it gets compensated somehow? Assuming the holding point of the wrench is always in the middle of the handle or some other fixed point on the handle. Or is it an error induced on higher torque settings on a given wrench when it's been calibrated at a lower setting?
@Alireza9900 This is from a long time ago, but if you do come back to the topic, the handle does indeed move a modest distance as torque settings are changed. This is part of internal variations that contribute to the tool's tolerance specification, often +/- 4%. At the position of the handle center, the rate of change in torque per distance alteration is comparatively small, at least an order of magnitude less than the 4%. On my torque wrench, the movement over the torque adjustment range is total 16mm, I did the variation calculation once, I think it was under 0.1%. On a tool that can vary over an 8% range and still be within spec, it's not significant. Calibration is done at 20%, 60% and 100% of the tool rating, and the calibration procedure includes a test of the tool's parameters with the Force Loading Point on the calibrator being moved 10mm either side of the correct Loading Length.
Theoretically, the manufacturer/designer of the wrench is supposed to design the dial markings (where the user sets the specified torque) with that effect in mind. The only way to make sure it's to hang different weights off the center of the handle, note the dial position when it starts to click and do the math.
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The best way to demonstrate it (without maths) is to say to people: _"Imagine pushing on the pivot point itself and you will see that the wrench arm won't be able to rotate, causing no click even with the highest pressure you can apply. However, this applied pressure WILL still turn the bolt."_ Basically, the main pivot point simply cannot rotate if you push directly onto it, but the bolt will still rotate. The only way you can rotate the pivot point is to have that force applied at a length away from it, turning the applied force into rotation (torque).
Love that! Thanks
this is a brilliant explanation! so simple. thank you
I love when I'm puzzled about something after one explanation, only to find a different explanation that makes everything instantly click into place. Pun intended.
Thank you.
That's the way I confirmed it in my head. Applying force at the pivot would result in infinite torque. You posted it so I don't have to
Agree...when I was at about 3:15 I imagined applying force exactly half way between the clicker and the pivot point...and saw that I would be applying half my force without going "through" the clicker. But now I agree, your explanation is the best.
This was not only brilliant in substance, it was explained brilliantly. A million bravos.
Thanks for checking it out, Stan
He still doesn't get it. If the torque wrench will click at 50 ft. lbs with 5 pounds of force at 10 inches, then the force required to get it to click choked up at 5 inches will be 10 pounds (twice as much)! Clearly someone is trying to pretend that they are smarter than they actually are.
@@JohnAnderson-sm8jl this video maker and those that believe hand position matters just can't see beyoud the principal of the lever but can't get it in context with a torque wrench. They all need to watch this short video it's factual. When the tool clicks no matter where from or how much force was imparted one stops pushing. The only error is oporator error but to get that one has to be a compleat ideot, check it out th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=YbAvHDfQ3x6HyaOX
🤦🏻♂️
@@JohnAnderson-sm8jl He is a professor of engineering with a PhD. I don’t think he’s pretending to be smart.
As you correctly point out, the torque delivered by a given force depends on the distance between where the force is applied and the torque is measured. In the same way, the torque being produced by that force is different depending on where you measure it. If you put a second ratchet half-way along the beam, you’d measure different torque there than at the end from the same force applied to the handle, just as you said. Two different torques experienced, depending on the distance it is from where the force is applied. The pivot point is that second measuring distance.
Great to see someone get some constructive feedback and actually do some critical thinking of their own beliefs. Well done for putting this together and proofing your initial thoughts, and those of many others, incorrect.
You’ve got me as a subscriber.
Everyone makes mistakes. Key is resolution.
Agree completely, and subscribed as well.
Ditto, subscribed!
He still doesn't get it. If the torque wrench will click at 50 ft. lbs with 5 pounds of force at 10 inches, then the force required to get it to click choked up at 5 inches will be 10 pounds (twice as much)! Clearly someone is trying to pretend that they are smarter than they actually are.
@@JohnAnderson-sm8jl except it isn’t as the pivot for the torque measurement is not the centre of the torque applied to the bolt
Thanks for an excellent and informative video. I had to watch it 3 times to wrap my head around what's going on here. As it did with you, the physics seemed painfully straight forward, and where you hold the handle shouldn't matter. But it does. My way of explaining it in layman's terms would be that if you're choking up on the handle (as compared to the prescribed point at which you're supposed to hold it), you're choking up more (percentage-wise) on the click pivot point than you are on the fastener pivot point. Similarly, if you apply the force further from the pivot points, the percentage with which you're doing so is different for each pivot point. Finally, this can be more concretely demonstrated by applying the force directly at the click pivot point. When doing so, you'll still be applying torque to the fastener, but the wrench will never click because there's no torque being applied to the click mechanism.
I love that endpoint case of applying force at the pivot point. Still turning the wrench, but it’ll never click! Thx for that
@@TheBikeSauce why? Are you really that brain dead. The only time it won't put force on and click the wrench is if you pull dead centre of the nut. But anywhere else and mathematicaly it will always put torqu on the nut nut at fantastic power. But when the tool clicks it will still only tighten your nut to what ever the tool is set to. You all really do need to watch this short video and a longer one on my channel. th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=YbAvHDfQ3x6HyaOX
Extremely clear explanation of a complex topic. And nice drawings to accompany! You made a potentially dry topic fun and engaging...the mark of a great teacher!
Excellent explanation of the physics behind the length-dependent click-type torque wrench!
I had thus far never considered the dependence of the applied torque on my grip position when using my torque wrenches. From now on I will be more mindful of it.
The engineer in me is also puzzled by the nonlinearity of the torque discrepancy that you calculated by changing d1 from 5 inches to 2 inches and then to 10 inches. My intuition would've been that the error in torque would've been linear with the error in moment arm length.
So I worked out the equation for the relative error sensitivity, S, of torque with respect to grip position (ie the fractional error in applied torque per unit distance error in hand placement):
S=-(d2/d1)/(d1+d2)
S is directly proportional to d2/d1 and inversely proportional the sum of d1 and d2, so the longer the overall length of a torque wrench, the less sensitive it will be, and the longer the handle bar is relative to the head unit, also the less sensitive. The negative sign tells us that a closer hand position to the handle pivot pin will always result in more torque applied at the drive when the click occurs, and a farther hand position always results in less torque being applied, which is somewhat counterintuitive, and quite the opposite behavior of a normal wrench. The relative error sensitivity is independent of Fs, r, or the torque setting selected, as both the absolute error in torque and the torque setting itself scale identically linearly with each of these parameters.
This is fantastic.
It's super simple, but it's not very practical, since it's useful just for small deviations. It's just the first derivative in Tailor series, but the function is nowhere near close to linear.
Mc = (1+d2/d1)*r*Fs
Mc'(d1) = - r*Fs*(d2/d1^2)
And here you get:
S= - (d2/d1)*1/(d1+d2)
And there are Slip or Cam-Over Torque Wrenches, which do not have this problem. They use a radial ball clutch. And not only this problem is eliminated, but also when they slip, over torque is not possible.
No one pulls on a torqu wrench anywher but the handle. But if one does it make no difference at all watch this short factual video and learn even if you can't comprehend. th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=YbAvHDfQ3x6HyaOX
To me you are merging two mechanical systems of wrench: the lever ratio of d2/r with tension spring which is defining the click (an internal transfer function of wrench to initiate a click) and d1+d2 torque lever.
When applying torque exceeds specified value, equilibrium between friction force (setted up by spring tension of transfer function) and (d1 + d2) F breaks - the transfer function makes click.
As mentioned one of viewer one year ago, before click the wrench is the one solid pivot lever with size d1+d2.
So, to calibrate the wrench we need correct weight at correct d1+d2 distance.
As I see, you are trying to describe all absolute forces that are originating in wrench.
it seems very counterintuitive but I now think you are right. If you apply a force very close to the pivot point you can understand it will snap at a very or extremely high force which would obviously mean an extreme torque to the bolt. If you apply right on the pivot point it will never snap and you will still torque the bolt until you break it for sure unless something else breaks
I couldn't quite get your explanation to click (pun intended), so I ran through the math myself. The key for me was to look at the free body diagram for the handle (ignoring the head of the wrench). The handle sees two torques about point Q, one is applied by the person's hand (F*d1), and the other is applied by the clutch mechanism (Fs*r). Balancing these torques allows us to compute the force required to click the wrench: F = Fs*r/d1. Now, we look at the torque applied to the fastener (Tf), treating the wrench as a solid beam: Tf = F*(d1+d2). We already know the force required to click the wrench, so we can substitute: Tf = Fs*r/d1*(d1+d2). Simplifying we get: Tf = Fs*r*(1+d2/d1). Since Fs, r, and d2 are all constants determined by how the wrench is set up, increasing d1 decreases the torque applied to the fastener before the wrench clicks.
What's nice about this explanation, is we can directly calculate the percent error based on where the force is applied. If you have a wrench with d2 = 2" where you're supposed to hold it at 16", if you instead choke up to 6" then you'll apply 18% more torque than indicated.
Hell yea. Nice work
Ahhh, YES...
An M.E. weighs in on the question, and with a beautifully simple derivation takes this counterintuitive, very interesting analysis, and boils it down to very practical, quantifiable terms.
Now THIS is something I can apply!
If I'm in an awkward configuration that for some reason requires me to choke up on my click-type torque wrench, I understand intuitively that I will have to apply more force (F). I don't really care how much. Using only d2 and the new d1 (both externally measurable) I can calculate my resulting over-torque and back down the setting appropriately.
Thank you, sir, and nicely done.
I knew the rudiments of your explanation prior. But, I couldn't get my math to explain why things worked like they do. I was very excited to learn what I didn't know about click type torque wrenches. Thanks so much for opening my eyes!
Great video but I think the point about hanging weights accuracy testing being flawed is, well, flawed. You can absolutely test the accuracy that way, as long as your calibration hanging point is in the middle of the handle. The wrench should click when you hang the weight off that point and should stop clicking once you move the weight as little as 1/4" inwards.
What may actually be flawed about the tests you did is that you need to measure the weights before using them (exercise equipment weights are rarely very precise); and you also need to factor in the weight of the wrench itself with this method since it also applies some torque at the pivot point when used in this orientation. Something like half the weight times length of the wrench should provide a good first approximation.
Great explanation. It amazes me how many guys currently have TH-cam vids “testing” the accuracy of torque wrenches and their hands are all over the wrenches and in many cases far closer to the tool heads than the tool handles.
As others have already said, this was brilliant. Well done! It's been over 40 years since I took Statics, and I was an EE so I promptly forgot all I learned in the required ME courses. Kudos to your engineering skills!
Nice! Thankfully, I don’t believe Statics has changed much.
@@TheBikeSauce Because sum(rF) = 0!!!!!! in Statics :) - Seriously Great Video (from a ChE who also forgot all his Statics decades ago) Bravo!
Don’t feel bad, it’s been 30 years since I graduated as an ME and I’ve forgotten most of both the ME and EE content!! Excellent video - trying to understand how I can still apply the “weight” plate test accurately.
Using his example, Am I to assume that if I apply a 10lb weight at a distance of 5” from pivot point and set the wrench to 60in-lb it will pop off if calibrated? Or a 20lb weight with wrench set to 120in-lb? Is that a decent way to calibrate?
Never too nerdy. Great video.
This is fantastic content. It’s fun to take it in as a super bike nerd.
Ha thx. Few will appreciate it, but I think people should know
A video I didn’t even know I needed to watch until I watched it, good one!
The best kind of feedback. Thanks!
Please don’t stop making these type of videos, I learned so much from this then at high school . Shit I feel smart !
Awesome. Your self-depreciation is hilarious and the explanation is excellent.
I was torquing up a headset only last week with my hand choked up over the head and I idly wondered if that made a difference to the result.
Now I know!
Excellent as always, love the demonstration - great visualization!
Thanks, Paul!
To greatly exaggerate, click style torque wrench would never click if force is apllied directly over pin (or closer to the head) no matter the Force applied (pin that is just bellow the head of the wrench), so logic is that handle is where force should be applied because distance (from pin) obviously matters. I always used click torque wrenches as they should be used, but now i clearly understand why.
Yes indeed, this is probably the simplest way to quickly see that moving the point of applied force must change the resultant torque at which the click will occur. This video delivers the most clearly presented explanation of why that I've seen. Just a pity that so many folk on the internet, like our friend JohnAnderson-sm8jl, still don't get it!
Dude, fantastic work. When Russ at Path Less Pedaled said that grip position mattered, I was sure he was wrong. So, I came over here to see what you said. Thanks for explaining so well. Great illustrations and sample calcs.
Ha nice. Thanks for following his shout out to my video. I was surprised at the result as well.
Hi Bike Sauce, I think you have this wrong. The second pivot and clutch is a mechanism to apply a consistent and calibrated load to the socket. The load transfer is not changed by the hand position. It is limited by the transfer of load through the clutch to the rigid arm connected to the socket. You are correct that the testing technique you used is flawed (measuring load on the lever arm to make it click is simply measuring the input force) but you need to look at realised torque at the socket. To do this you need to set up a strain gauge on the rotating object - nut or bolt - and see if the realised torque differs with the same wrench settings and different lever lengths - hand positions. I have not tested this but my analysis of load transfer leads me to believe that hand position does not alter the load transferred to the rigid arm through the clutch.
The linked video shows this exact phenomenon using a strain gauge torque sensor. This is just the math to explain it. I know.. really counterintuitive
Very interesting analysis! Also, really liked your “click” annotation.
Great video, thanks! Didn't necessarily follow all the math, but understood the basic concepts. I'll have to go back and read my torque wrench manuals.
Nowhere will a manual tell you outright that hand position matters other than in one of my videos they say it make only a difference to the oporator error. Watch this short video. th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=YbAvHDfQ3x6HyaOX
I didn’t understand why hand position mattered until recently because I didn’t understand how clicky torque wrenches worked. And that is not straightforward at all. And manufacturers absolutely do not help with this. If we read the instructions (lol), they say do it this way but they virtually never explain why. And many many people assume that there is just some mechanism built into the ratchet head that is somehow calibrated to click with proportional to torque. The big problem I have is with people who, once shown the mechanics of the wrench, the statics equations, etc., still just say “bah! torque is torque! I’m right and you’re dumb.” But alas, the internet is FULL of such people. Anyway, great video. It’s nice to learn new things.
Bike tool + math/physics. +1. Looking forward to Bike Sauce / 3B1B collaboration.
Thankyou for that, it really helps with the physics understanding and it also means that hand position is actually more important than I thought it was as even using the end of the torque wrench will result in skewed settings.
Thanks for making this video. I was often choking up on the handle to improve stability of the wrench on the bolt - I'll stop doing this now!
Excellent explanation. I have always wondered why my torque wrench stays at the back of the toolbox and that I am more confident of my hand force on the wrench.
The handle of the click type torque wrench should be articulated, pivoting at a point exactly at the "middle of the handle". As long as the articulated handle is not bumping against the stops of its free range, the torque value should be consistent. Or maybe the torque wrench should not have a handle at all but a ring at its end with which you can use a finger or a chain to pull on it. That is old stuff. Digital seems to be the answer now.
You're intuitively grasping something the TH-cam personality overlooked.
its more often about repeatability than exact torque ,, plastic parts need an even clamping force to help prevent leaks , rubber gaskets wont mind say an extra 5% torque , as long as the fasteners all have the same to prevent distortion.
I simply loved your explanation and the way you approached the problem. What is intuitive during usage is made perfectly clear with math now! Thank you for taking the pains to make this video!
Thanks for this video. I still don't understand the maths, but I do now understand why grip position makes a difference. Also explaining the inner workings of the torque wrench helped a lot. Thanks again.
Nice demostration … there is also another consequence to consider in my opinion. When you use this kind of torque wrenches you should not apply additional torque with your hand. If the hand rotates the applied torque is not accurate. Ideally you should only apply force in a single point in the middle of the handle. I hope this is clear.
Hi. It is true. Torque is torque when you have only ONE lever to move. Here though you have two bodies interacting. The inner shaft and the outer tube. OK plus the bolt which only lends its reactions to the system. Also Fs will always be Fs meaning that the limiting force at which the click happens is always the same since it is dependent only on the spring’s selected stiffness by the dial and we don’t change that. So what else could be changing ? During the click a collapse mechanism is occurring. The two beams change their relative angle. Note that Fs doesn’t exist as an outer force for the system. It exists only for the inner shaft or the outer tube when separated and viewed as stand alone. Therefore the easiest way to prove your point would be to first show the equilibrium for the whole system just prior to the click when the two beams act as one and calculate all the outer reactions. Then show the equilibrium for the inner shaft ONLY using the previous reactions plus the force Fs as mentioned and considering ΣM=0 at the pivot point of the two beams. Actually when the click happens the pivot point is where the true rotation takes place even if the inner part doesn't really move - your hand’s force F moves respectively to it along with the outer tube. STEP 1 The System. Apart from the torque Mbolt {which from now on we consider it as a moment reaction applied to the shaft} a significant vertical reaction from the bolt -F should also be shown {which is opposite to the F you apply by your hand to the OUTER tool and satisfies the equation ΣFy = 0 for the WHOLE initial system of the two bodies that still act as one. A force which correctly is not shown in your diagram since your pole coincides with the bolt and therefore -F doesn't produce any torque there} STEP 2. The shaft. This vertical reaction now produces a real moment -F*d2 for the inner shaft respectively to the pivot point and it should be noted that it is a variable moment that changes every time you place your hand in a different location only because the vertical reaction changes. Again it is important to realize that for the inner shaft this variability occurs ONLY due to the difference in the strength with which you push the whole tool and not due to the distance from the bolt because the inner shaft is isolated and perceives no moving loads {since all the forces acting upon it have their points of application fixed either at its 2 ends or at the pivot point and therefore perceives only the fluctuations in its reactions}. Of course the distance comes into consideration for the whole system only and goes along with the general consensus that when you want to achieve similar magnitudes of torque and you decrease the distance you have to increase the force. This increment is what is transfered to the shaft and plays a big role as we will see - big enough to alter the torque output. For completeness we should add here that there is also one more vertical force applied to the inner shaft at the pivot point by the outer tube and it is equal to F-Fs. It is derived from the equilibrium ΣFy=0 {of the inner shaft this time} since there are 3 vertical forces at its 3 points -F and Fs at the ends and F-Fs at the pivot point. This one though doesn't produce any moments respectively to the pivot point and we won't analyse it any further. Now the net effect is that this aforementioned moment -F*d2 acts as a relieving moment for the shaft and is actually subtracted from the applied moment Mbolt in the inner shaft ‘s equilibrium equation. ΣM=0 => Mbolt-F*d2-Fs*r=0 => Fs*r=Mbolt-F*d2. Hence for the click to happen Mbolt should now be larger to counteract that larger [-F*d2] and yeld the same Fs that releases the spring. To put it simply the constant moment Fs*r is neutralized partly by the vertical reaction's moment and partly by the true moment applied on the bolt as torque! To get a recurrent reliable Mbolt you should therefore combine it with the same vertical force that was present during calibration. It is part of the tool ‘s intrinsic formula. This a formula can also be viewd as Mbolt=F*d2+Fs*r. Both addends on the right side should appear as they were during calibration for the torque result to be reproduced and yeld the Mbolt that the factory had initially gauged. To summarize it all Fs is not the only parameter of the system. The force with which you push the wrench is equally important. So when you hold the tool EXACTLY at the grip of the tool that vertical reaction is the same one that produced the torque Mbolt together with Fs*r during the factory callibration and this is the only way to always yeld the same result. You exert the exact same forces of the calibration and they produce the exact same torque on the bolt. As you approach your hand towards the bolt the vertical reaction on the bolt gets larger. One addent gets larger. That new larger F*d2 is not what the factory had taken into consideration when calibrating Mbolt and since it is again added to Fs * r which is always the same that means that the click happens at a larger Mbolt !
Great illustrations! They explain the problem very nicely. Thank you.
However, I think your equation on the left side is off a bit. Your summation shouldn’t include the internal moment “-rFs”. Until the clutch slips the entire wrench could be considered as a simple solid bar when summing the moments about pivot point “o”. The correct summation equation is just:
(d1 + d2)F -Mc = 0
Nevertheless, when you perform additional free body summations of the individual wrench parts to capture terms “Fs” and “r” and simplify you will eventually get to the key equation which you have correct on the right side that shows that the clutch slippage force "Fs" is indeed dependent on d1:
d1F = rFs or Fs = F(d1/r)
You are absolutely right. But see, this is physics, he can not be a bit off, he is just completely wrong. There is a system of two equations for the torque around the two points of rotation - O and Q respectively:
(d1+d2)*F = Mc
d1*F = r*Fs
And eliminating F, (not Fs as you do) we get:
Mc = [(d1+d2)/d1]*r*Fs
And we get that the relation between Mc and Fs includes d1, which is not a constant, but a variable.
The other two, namely d2 and r do not matter, since they are a constant dimensions of the tool.
And we also see, that the naming scheme of the lengths is also completely wrong, since d1 and d2 have completely different meaning - a constant and a variable.
Also r is like d2 - a constant, they must be on the contrary - similarly named.
I love your approach to academic discussion 😆. Are you equating torques at 2 different locations? Naming conventions are pretty arbitrary, not sure how choice of variable name is ‘wrong.’ Thanks for your input nonetheless. Would love to hear precisely why this is completely wrong though.
@@TheBikeSauce In the model of the click wrench you have two axes of rotation, namely O and Q. Every one of them produces one equation.
Basic naming convention suggests using at least different naming for constants and variables.
At school they suggest using the beginning of the alphabet for constants, middle of the alphabet for integer indices, and the end of the alphabet for variables. For example a polynomial is written as:
Sum (An*x^n)
a, b, c, d - constants
i, j, k, l, m, n - integer indices
x, y, z - variables
But you do not need to follow that exactly. Proper way is:
d1 and d2 - constants
r - variable
Or:
r1 and r2 - constants
d - variable
You see, also, two of these lengths are dimensions of a real physical object, while the third one is an abstract length from a physical point to an imaginary one. The real force applied by the hand is distributed over a surface area.
Also, another mistake there. In the simplified model all forces come from points on the axis of the wrench. But you draw the applied force on the surface of the wrench handle. You can have a model like that, but than you must take into account the thickness of the wrench handle. And you get half the thickness of the wrench handle as a third constant there. Going outside the handle and / or having variations of that thickness makes this constant a function of our variable, the position of where the force is applied. So you get a more accurate, so to speak, second order model.
@@TheBikeSauce If you want, I will add this. In the second order model, where the thickness of the wrench is taken into account, it actually happens, that it does not matter. Yes, the thickness of the wrench cancels out. Which is another, seemingly, "strange and non intuitive" fact. So, it's not a problem to have a short wrench with a thick handle.
The equations go something like this:
Sqrt[(d1+d2)^2 + h^2] * F * cos A
cos A = (d1+d2)/Sqrt[(d1+d2)^2 + h^2]
Sqrt[d1^2 + h^2] * F * cos B
cos B = d1/Sqrt[d1^2 + h^2]
where h is half the thickness of the wrench where the force is applied.
And there we get the same equations as before.
Really appreciate that you’re interested, but have a bit of perspective maybe. You can always model a system more accurately given the time, but that does not make simpler models wrong. This model is adequate to demonstrate the counterintuitive idea that grip position actually matters on this type of wrench. It is an explanation for similarly counterintuitive empirical evidence seen in other videos.
In your more complex model, you gain nothing by considering the geometry of the handle. The cos will be negligible wrt unity and the geometry of the handle is inconsequential.
This explains why I broke one of my valve cover fasteners the first time I did my valve cover. I thought I was slick by "choking up" on the torque wrench. Good thing for helicoil.
well done diagrams. thanks for the video to help me with my random wonder of how torque wrenches work
Thank you for explaining tourque wrench use 🙂
Excellent explanation of a super nerdy stuff. Awesome video
Not nerdy enough in my opinion. But excellent explanation, very well done!
Great video in terms of describing the previous misconceptions, retesting, and results! Also TesseracT 👌🏻
As a tutor of college mathematics and physics, I love practical applications--that aren't necessarily intuitive. Thanks.
I like the explaination. However it provoked more questions like; Where exactly is the "middle" of the handle? If your hand is smaller than my hand is it more/less accurate? How do you apply force to a specific area with a hand, that is much wider than a string with weight?
Great video! I had no idea Thank you! Very thorough explanation.
Very helpful video and pleasingly nerdy as well. Thanks!
The best kind of nerdy !
Absolutely amazing video!
I fully appreciated every second of it, on the edge of my seat!
PS - And I just now became your newest loyal subscriber!!
Man, this was an excellent video you made, and a brilliant explanation of why I find all the torque wrenches that I buy, to be out of calibration! It seems I was making the same mistake as all of us... But now, I know!! A million bravos from me too!! Thank you!!
Completely counterintuitive but you did a good job of explaining the real physics. Thank you.
Ok, this is absolutely lovely, and makes my (physics degree earning) brain glow, thank you. It's exactly the right level of abstraction to explain what's going on.
Also, I absolutely respect the patience you have for dealing with both ends of the comment spectrum (this is wrong because it's just a beam vs this is wrong because it doesn't take into account blah blah blah).
I have a split beam torque wrench with the knob on the side (not the micrometer style click wrench), and I also have one of those digital torque adapters. I don't know if the physics are the same as you showed with a split beam, but gripping different parts of the lever produce different torque values at the socket (where the digital adapter is) when the wrench clicks. It doesn't seem to be a linear relationship though. The more I choke up the harder it is to generate torque at the socket (of course), but pulling between the handle and the pivot point produces erratic readings from test to test and causes the wrench to click with 35, or 46, or 41 etc. ft lbs at the socket (when it's set to 50). This is not quite what I understood from the video which (I think) said the higher you choke up the MORE torque should be at the socket when the wrench clicks. But again I have a different style torque wrench.
The conclusion is the same, of course, and it's something I eventually figured out when playing around trying to see if both my adapter and wrench were accurate (they are both new). Hanging weights off the shaft 12 inches from center socket caused the wrench to click 5 ft lbs. too early. Hanging the weights off the center of the handle got me within a few ft lbs. (borderline acceptable range). But actually pulling on the handle properly while bracing the socket gets me within less than 1% repeatably. Conclusion: how you grip and apply the torque to the handle matters, so do it properly. Since I'm getting both my devices to agree with a high degree of accuracy I'm going to assume the equipment is good enough to rotate my tires with. lol (because honestly what else am I going to do with it, build a space shuttle in my garage?)
For a beam style wrench, the length is a factor so grip position is also a factor since it’s based on beam deflection. But.. I totally agree - we’re not building space shuttles 😆😆😆
Just wow. Love the honesty, transparency, and research. Cheers
Thx, Henry!
Thank you for the explanation I really appreciate it.
Can you do another video explaining the physics of the ball bearings ???
where were you when i needed a college tutor for this class... i would have aced the course!
Brings back engineering school memories!
Hopefully good ones
Seriously amazing video man
very well explained, this made so much sense. Thankyou very much.
Great presentation, well done.
This helps makes sense as to why the hand position matters for a click type torque wrench. One issue with the equation, though. If you use the socket as the pivot there is a force on the pin where the inside pivot holds the handle to the inner beam. Also, the socket itself has a torque which is equal to the torque that is applied to the bolt, and that is the quantity to solve for.
Thx, Kelly. The force you mention is accounted for in the additional couple Mc. The torque at the socket is equal to the applied torque on the bolt as it's in static equilibrium.
@@TheBikeSauce, Mc does not account for the torque added directly at the pivot point by the orthogonal force of the handle. I think you have made some catastrophic simplifications and mistakes in your modeling.
There is a system of two equations for the torque around the two points of rotation - O and Q respectively:
(d1+d2)*F = Mc
d1*F = r*Fs
And eliminating F we get:
Mc = [(d1+d2)/d1]*r*Fs
Or:
Mc = [1+d2/d1]*r*Fs
@@jasonbeisiegel5550 Mc is the reaction of the bolt or nut to be tightened. It's equal to the torque that the wrench applies to the nut or bolt, but has opposite direction and is applied to the wrench. It has nothing to do with any pivot points. Also, the pivot point can not be a source of a force, a torque, or anything. He is wrong, but you are not right too.
👍
very cool! Very well explained.
Not everyday one learn something new; thanks!
A most excellent video! From one engineer to another: Thank you!
*
Excellent explainer / walkthrough of the math and visualizations … as a data analytics and visualization instructor (and cyclist, of course 😁), this really hit the mark
Right on! You appear to represent the precise target audience for this video
That was a fun video. Thanks! Curious on your view of crank arm lengths and climbing. That might be a great video as well (?).
So you CAN use the weight on a string method as long as you place the string with the weight at the center of the handle? This means to measure from the center of the socket to the center of the handle and multiply by the weight to get the expected torque.
Well "torque is still torque", but changing the position of your hand on the wrench changes the way the forces interact with the internal "clutch mechanism" and the click occurs at the wrong time. Someone who took engineering mechanics in engineering school will immediately understand everything you did.
There are other types of mechanisms used in torque wrenches, some use internal beams, some use actual electric resistance load cells, I don't know what else is available. When I grew up working in my father's garage we used a "beam type" torque wrench. There is nothing to calibrate on those, the deflection of the beam will be repeatable if you don't damage the thing somehow, deflection being P times L-cubed divided by (3 times E times I), spoken as pee ell cubed over three eee eye. The beam type torque wrench that we used enforced proper use by having a pivot pin through the handle. You had to pull in just the right place to keep the handle from turning with respect to the beam, therefore the force was always applied at the exact L from the fastener that the wrench was manufactured for. The only trick was getting in a position where you could accurately read the scale while pulling hard on the handle. Sometimes it helped to get a second person involved.
Excellent point! I still have my Craftsman beam torque wrench from the late 60s, when as a teenager I did my first serious engine work. A kindly machine shop tech took the trouble to point out exactly how to use the tool and emphasized putting smooth pressure only on the handle pivot pin for accuracy. I’ve kept that in mind ever since, even with all the newer torque wrench types.
This is great! Does this mean the cheaper tool was actually less accurate?
Nice!
Thx, Russ!
I got as far as 8.05 mins into your well presented video. But then i woke up. You go on about two torques but infact right up until the clutch as you call it snaps, clicks, the whole thing is basicaly a lever, after the click one stops pushing on the tool. How far up the handel this force comes from does not matter once the tool clicks you stop otherwise you will force the detent against the outer wall of tube and can still carry on to over torque.
Now imagine what happens if you're pushing the handle at the pivot point. Or on the far side of the pivot point. It's not just a beam.
@@iliinsky This is something our friend pete has been unable to get his head around. No-one (in more than one place on TH-cam) has been able to get him to realize that the fact that force applied at the handle pivot point can never produce a click regardless of the amount of torque being delivered, points out that it does indeed matter how far up the handle the force comes from. Still, one can only hope - he may read and have it "click" for him that he has said "once the tool clicks", and in that case the tool will never click.
Brilliantly explained! Could you repeat the test with the proper method on the budget torque wrench, it would be interesting to see how it really performs.
Good idea. I guess the budget wrench results wouldn’t change though because the weight was more or less placed at the handle center
Excellent! Thanks for the explanation. It clicked to me this time.
😆
The very first to make perfect sense.👴👌
I've seen many vid's stating and showing that it will make a difference where the force is applied, all empirical...👴🤷♂🧐🤔🤨
This one however is the first one clearly explaining how and why.👴😊🤗👌👌👌
Thank you for that.👴🤗👌👌👌
should have more views, great video
Ha thx. Small channel, just posted it.
The click type torque wrench is not very accurate because operator cannot respond quickly enough to the click and will usually over torque the bolt. Through experimentation I found that digital torque wrenches provide the most accuracy because the operator can anticipate when he/she is at the correct torque.
Thanks Bro for the enlightenment. This will put the rocket scientist at work in their place.😅✌
Every single part of that went so far over my head! but I thought the video was really cool.
excellent video! now I get it! thanks.
So, 12 months on, I know, but how about a beam style torque wrench? They would be less hand position dependent, right?
That was great!!! Thank you!
My head hurts, thank you 👍🏻
Ha there’s an apology at the end..
I have done my own test’s and it doesn’t matter where I hold my hand it gives the same results
You measured the applied torque with a digital strain gauge at the socket?
Great video! Now I know how a torque wrench works :) The exact distance is only important during verification; it doesn't matter for the application itself. You just press with force x until it clicks. Or am I wrong about that?
It matters during application too
Well, after giving it a second thought, it somehow makes sense - the tool needs to be calibrated to a specific reference point. 🤯
You are correct, and TheBikeSauce is not. After calibration done application distance will define how much force - more, equal, or less - user should apply to ‘reach the click’
@@pavand7410the math doesn’t lie
Just felt a strong urge to change my click style torque wrenches to digital ones, I assume that they use strain gauges around the actual socket point.
Love this video and totally makes sense. My only question would be a practical one for a user of a click-type torque wrench. Mine would not have an 8” range for practically where I could grip it. If I narrowed it down to plus or minus 2 inches (which might still be wider than I could practically hold it), what’s the error and does it matter?
Great explanation! But what are the points where you measure the "middle of the handle"?
The S1)socket end or S2)the sort of invisible pivot [start of d1]
to
E1) absolute end of wrench or E2) end of main handle, before the adjuster bit?
I ask because my recent experience includes learning the hard way that most (my) non-Park priced torque wrenches don't work on reverse threaded things like drive-side bottom brackets (which is not entirely applicable to this video but I'm whining) and crank arm bolts (due to user error apparently) :-)
Excellent video
Lol... Nice to find that I still feel at home amongst "The Nerdiest Of Nerds"!
Hello secret friends 😁
Really interesting to consider then how a crow foot attachement would change the values... It is an extension of "d2", but the d in that situation changes as the crows foot rotates probably? How would that math play out? Thanks for the video!!
Well done !!
This is making me think I should just get the beam type torque wrench.
Haha, there are other types of non length dependent ones too
Fantastic video.
Does your revised opinion on the accuracy of the "nice" Park torque wrench mean the less fancy (no plastic) budget model is now inaccurate?
Wow, this just made maths interesting! Brilliant video.
With this in mind does this alter any of your previous conclusions about the other wrenches in the pervious video?
Yes I did just pass Statics--- dynamics also :) But I'm an EE...so it doesn't matter :)
Haha, apparently not the only EE in the comments.
Thank you, it was very helpful :-)
This was absolutely entertaining
Ha, only to a select few. Thanks
No. It does not matter too much where you place your hand. The slip mechanism is always calibrated to a fixed rotational force at the pivot point (based on the spring pressure). But it does matter if you are measuring the force applied at a certain point to get the corresponding torque at the end. In such case, you can use the formula normally used for torque wrench extension to back calculate the proper weight at your chosen point of force applied, and to account for the distance between the slip point and the socket point.
Jet Li, I agree and well said. Thanks for explaining what I intuitively knew but could not put into words.
It seems intuitive, but the math (and experimental evidence) is counterintuitive
That makes sense - so I can still calibrate using a fixed weight at a fixed location using the equations prescribed in video. Just as easy to use middle of handle. So, for this example, setting a 10lb weight at 5” from pivot and assuming 1” distance from pivot to socket attachment center, I would set wrench to 60in-lb and if calibrated, it should “click”?
@@TheBikeSauce I think there is a problem with your maths, though. Turning force is always applied to the inner arm (and therefore the socket) at the clutch mechanism. So the equation for the torque at the socket should be the force experienced at the clutch mechanism (Fs) multiplied by the length of the inner torque arm (d2+r). So the torque wrench will always click at Fs*(d2+r).
Head hurts with the math ,, it would be nice to see it with polarised Perspex , then one could see the forces as the math really hurts my tiny brain, I’m not entirely sure that my suggestion would work …..,due to this tiny brain again,
Great explainer all the same ❤
😂if your head hurts now wait till you have seen these two short videos. If this guy is a professor god help science. th-cam.com/video/tTbjLAm7XnQ/w-d-xo.htmlsi=885sZ8Fwt9Iw9fHV th-cam.com/video/mk95F0hHS3U/w-d-xo.htmlsi=vdS4hI08-O6Tjshk
A certain someone now needs to go and re torque all his bolts 😬
I'm here because of auto expert.
I still didn't understand why, hoping this explains it.
Me too. This is the most understandable explanation I've become aware of. I knew John was correct at Auto Expert (and he measured to demonstrate), but the explanation of the reason behind it wasn't easy to visualize.
Wow.. I actually understood everything you said... Crap... I AM A NERD.... =) .. Great insightful explanation!!!
Great video and a very good explanation! Something crossed my mind though; The adjustment on some torque wrenches effectively changes the lever length of the wrench from the handle. It's not the case with Park Tools torque wrenches, but for many others, the mechanism is such that the handle moves closer to the socket when increasing the torque setting. I haven't thought about it much yet, but I'd like to hear your ideas about this. Do you think it gets compensated somehow? Assuming the holding point of the wrench is always in the middle of the handle or some other fixed point on the handle. Or is it an error induced on higher torque settings on a given wrench when it's been calibrated at a lower setting?
Interesting point.
@Alireza9900 This is from a long time ago, but if you do come back to the topic, the handle does indeed move a modest distance as torque settings are changed. This is part of internal variations that contribute to the tool's tolerance specification, often +/- 4%. At the position of the handle center, the rate of change in torque per distance alteration is comparatively small, at least an order of magnitude less than the 4%. On my torque wrench, the movement over the torque adjustment range is total 16mm, I did the variation calculation once, I think it was under 0.1%. On a tool that can vary over an 8% range and still be within spec, it's not significant. Calibration is done at 20%, 60% and 100% of the tool rating, and the calibration procedure includes a test of the tool's parameters with the Force Loading Point on the calibrator being moved 10mm either side of the correct Loading Length.
Theoretically, the manufacturer/designer of the wrench is supposed to design the dial markings (where the user sets the specified torque) with that effect in mind. The only way to make sure it's to hang different weights off the center of the handle, note the dial position when it starts to click and do the math.