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I have two impact sockets. Both of them drive fixtures which have 19mm heads. Both are driven by a 1/2" drive impact gun. Here is an image of the two sockets:

enter image description here

The one on the left is an ordinary impact socket. It weighs in about 10oz (maybe?). The one on the right is one specifically designed to take off the hub bolt on Honda engines (will work in many other applications as well). It weighs in at over 3 lbs. If I use the left impact socket on the Honda hub bolt, in most cases (far in excess of 99+% of the time) it will not take the bolt loose. Yet, with the one on the right, takes only moments to loosen the stubborn bolt. This is with the same impact gun, using the same air pressure (IOW: all other things are equal except the socket itself).

My question is: Can someone explain why this works the way it does? I know the mass of the socket is the real reason ... what I'm looking for is the physics reasons why it works.

EDIT: A newer video on YouTube which tries to answer the question at large, yet stumps the physics instructor who gave them feedback. NOTE: I have no affiliation with the video ... just thought it was pertinent.

Pᴀᴜʟsᴛᴇʀ2
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  • If the application of torque was performed statically as if by a manual wrench the mass, shape or size of the socket wouldn't impact performance so this is definitely an effect associated with how the force and energy is transferred by the impact to the socket and screw. I even doubted whether this is thing but evidently there is quite a bit of online forum activity among mechanics although without any explanation. Looked at https://www.youtube.com/watch?v=FixrSlY2H7g for example where it evidently matters. – Squid Jan 07 '17 at 21:28
  • @Squid - They work, no doubt. I'm curious as to why it does ... I guess that's obvious, lol. – Pᴀᴜʟsᴛᴇʀ2 Jan 07 '17 at 21:34
  • Regarding that video the fact that the torque at the screw is reduced by the length of the socket (extension) I think ought to be associated with the fact that the extension twists in response to the drive torque and thus energy is siphened off to elastic potential energy reducing the kinetic energy of the system and lessening the torque eventually experienced at the screw. A thicker more massive socket will twist less but my energy estimates aren't enough to convince that it's necessarily the explanation for the difference in your situation. – Squid Jan 07 '17 at 21:41
  • @Squid - I quit watching the video after about a minute due to the long extension for just that reason. When wrenching on vehicles, you always use the shortest extension you can get away with because of the flex of the extension. There are also things which are called torque sticks which are designed to flex (or give) at a certain point to only allow so much torque be applied to the fastener. They are fairly accurate, I think to within a couple percent. – Pᴀᴜʟsᴛᴇʀ2 Jan 07 '17 at 22:46
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    Point of the extension was clearly to reduce the effective torque to the point that the difference between the impact sockets would be consistently shown as his driver otherwise was able to operate even with the smaller socket. Half the comments are about people pointing this out and the poster sighing and noting that he already explained in the video description. The point here is really if you, having experience with different extensions and sockets think it is reasonable that it is the difference in rigidity that might account for difference in performance of the sockets? – Squid Jan 08 '17 at 02:00
  • @Squid - Great point. Thank you for pointing it out. I rewatched the entire video. Makes sense now. – Pᴀᴜʟsᴛᴇʀ2 Jan 08 '17 at 02:08
  • THis video shows the principle in action as I've stated above. Thought I'd add it for some extra context. – Pᴀᴜʟsᴛᴇʀ2 Jun 12 '21 at 03:17
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    I was going to guess that the impact socket that best matches the mechanical impedance of the target bolt wins here. This is from my experience in vibration control. So the best socket for one application would not be the best for another application. – John Alexiou Nov 26 '22 at 22:46

3 Answers3

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It's the greater rigidity of the larger socket rather than its greater mass which makes it more effective.   The physics of the situation is that the change in angular velocity ($\Delta \omega$) of the socket and bolt (which is what we want) is equal to the impulse delivered by the tool, which is torque ($\tau$) times the amount of time it is appled ($\Delta t$), divided by the combined moment of inertia ($I$) of the socket plus the bolt. $$\Delta\omega=\frac{\tau\Delta t}{I}$$ The greater mass of the big socket makes $I$ bigger, which makes $\Delta \omega$ smaller.   That doesn't help.   and we can't make $\tau \Delta t$ bigger, because that's a constant determined by the tool.   But $\Delta \omega$ is zero anyway until we get it to start turning, so what we need is the most torque we can get.   What we can do is to make the $\tau$ part, the force part, of that constant impulse bigger if we can make the $\Delta t$ part smaller.   Changing either the force or the time part of an impulse to make the other one bigger or smaller is an engineering principle that gets used a lot, for instance, in air bags.   If the bulky socket is more rigid, then in slow motion we might see that it is considerably less springy than the small one. So as the flywheel in the tool comes to a stop in a shorter time trying to turn it, it delivers a bigger jolt of force to get the bolt moving.

D. Ennis
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  • Are you saying that turning energy/torque from the drive is absorbed by the twist of the lighter socket, in effect? –  Jan 08 '17 at 13:32
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    Not so much energy, but yes, because of the twisting and annular stretching of the small socket, the tool's flywheel and other turning masses take longer to stop. The result of that is the impulse delivers less force for more time. We want more force for less time. The twisting and stretching of the socket is quite elastic, so the most of the energy temporarily stored in that is not lost, but is returned to the tool assembly. Some is lost to friction in that action, but the energy story is not the focus in torque-hammering the bolt head, force is the focus. – D. Ennis Jan 08 '17 at 14:02
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    The energy story is that electrical energy or the energy of air pressure is transferred to kinetic energy in the spinning flywheel associated masses. Some is also transferred to the large coiled spring which is responsible for controlling engagement of the flywheel with the shaft. When the flywheel is thrown in and engaged, its energy plus the spring's stored energy are very briefly transferred by great force to compressive elastic energy in the flywheel, shaft and socket. Some is lost to particle friction, but most is returned to the assembly. – D. Ennis Jan 08 '17 at 14:21
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    For my fellow gearheads out there, here is a truly great youtube video with a slow motion cutaway view of an actual impact driver: https://www.youtube.com/watch?v=f0gSJa3L_7c – D. Ennis Jan 08 '17 at 14:26
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I don't think the explanation is to do with the rigidity of the bigger socket. The difference is that to work efficiently, the rotational inertia of the socket has to match the inertia of what is driving it.

What happens at each "stroke" of the impact is basically the same as a collision between the parts of the gun which would rotate if there was nothing to stop them, and the socket. If you have a large gun and a small socket, most of the energy is wasted trying to turn the gun, not the socket. The best you can achieve is when the two are correctly matched, and (theoretically) all the energy from the gun motor gets into the socket while the "recoil" of the gun is just enough to stop it turning at all. If the socket is too heavy for the gun, again you waste energy when the gun recoils and tries to reverse its direction of rotation.

Of course there is no guarantee your large socket is "correctly" matched to the gun you are using, but apparently it is a better match than the small one.

As an analogy of this, think about hitting a ball with a hammer to start it rolling. If the hammer is much heavier than the ball, most of the energy you put into the hammer remains in the hammer, which continues to swing after you hit the ball. If the hammer is much too light, it will "bounce off" the ball after the impact. The most efficient situation is somewhere in between those two extremes.

alephzero
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    This line of thinking is initially appealing, because it's like impedance matching for maximum power transfer in electric circuits. But until the bolt turns, don't we have to consider the socket, the bolt, and the engine to be a single mass that the tool is attached to? Also, isn't true that there can be no no energy transfer (aka no work done), until the bolt and socket turn, and we have force exerted through a distance? So, getting the bolt to turn is about force and impulse, not energy transfer. – D. Ennis Jan 08 '17 at 13:32
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Look at the width of the casing on the big socket, compared to the smaller one. The torque is, in effect higher for the right hand side socket, especially at relatively high speeds. While you are not applying the force in the same way as you would by using a long lever on a wheelbrace, once you start up the drill, you are getting some of the lever effect.

It's a guess for sure, but between that and the extra mass overcoming the intertia/initial stickiness of the bolt, I think that's the reason.

If not, I will be delighted if someone comes up with a better (or the correct) one.