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There are two popular tools I use to apply torque to a fastener (bolt, screw, etc.): an impact driver and a drill.

The drill is a motor hooked up to some gears and eventually a bit that fits over the fastener. If I want to apply 40 lb/ft of torque, I feel as though I have to brace myself for that amount of torque, like by using both hands and my body.

The impact driver is a similar motor to the drill, but there is a spring-loaded mechanism that applies the same amount of energy, but in short bursts rather than continuously. I can easily apply 40 lb/ft of torque with my wrist barely moving; using two hands or bracing myself doesn't really make a difference.

Why is this the case? Why is there no equivalent force on my wrist when using the impact driver?

This may be similar to using a hammer to drive a fastener into the ground: if I generate force by swinging very fast with a hammer, why isn't there an equivalent force that lifts me off the ground?

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    $\begingroup$ Friction! It all about overcoming friction and achieving sliding. $\endgroup$ – John Alexiou Apr 12 '18 at 19:30
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the reason you can sustain torque bursts easier with the rotary impact tool than steady torque loading from the nonimpact tool with the same average power rating is the rotary inertia of the impact tool reflects most of the impact shock into the tool bit before it can get transmitted out of the tool and into your hands (i.e., it is inertially clamped). Demolition tools like pneumatic and electric jackhammers operate on the same principle.

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  • $\begingroup$ thank you for the info! so the tool bit--even in a big tool like a jackhammer--can handle enough of the reflect force (eg, via the hammer mechanism rotating the opposite direction and losing energy due to friction) to make it barely felt by the user? $\endgroup$ – tau Apr 12 '18 at 18:55
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    $\begingroup$ @tau, yes, that's the advantage of impact tools in general. $\endgroup$ – niels nielsen Apr 12 '18 at 21:30
  • $\begingroup$ sorry to ask again, but im having trouble finding what intertially clamped means. also, does the same apply to the hammer example? you dont fly off the ground because your body is flexible and the reflected force is spread out in a less noticeable way than it is for the fastener, which can usually only get driven inward. $\endgroup$ – tau Apr 12 '18 at 22:08
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    $\begingroup$ @tau, inertial clamping is an engineering term for a system that is mass-controlled (sorry, another jargon term!). Here is an example: if the mass of an anvil is a certain percent greater than that of the hammer that strikes it, the hammer will bounce off the anvil without moving it- and the necessary percentage difference can be calculated. I am not sure what to say about your hammer example- let me think about it... $\endgroup$ – niels nielsen Apr 12 '18 at 22:30
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Sorry to revive this old topic, but all the answers were slightly imprecise and I think I can add more. The best way to understand this phenomenon is with angular impulse. Angular impulse is defined as torque integrated over time:

$$ L = \int{\tau dt} $$

In the case of the drill, the torque is applied continually, and so you feel the reactive torque equal to the torque applied to the fastener. For the impact tool, the force is applied over a very short duration, and so the total impulse is very small.

In order for the impact wrench to not start rotating, you still need to apply some torque continually to counter the angular impulse, such that the net angular impulse over time is zero. The impact wrench actually applies the torque during a very small fraction of the time. There must be enough rotational inertia in the tool, such that each individual impulse does not significantly accelerate the tool. In that case, you can counter the angular impulse (so the net angular momentum stays zero) by applying a continuous but much smaller torque in the opposite direction. The tool will vibrate with each impact, but on average will remain stationary.

For example, let's guess that an impact wrench is loosening a fastener by applying 100 ft-lb of torque with impacts such that torque is actually applied 5% of the time. Let's say the tool does ten impacts per second, and each one lasts 5 ms. So each impact has 0.5 ft-lb-sec of angular impulse (100 ft-lb * 0.005 sec), and you get ten of these every second. So, on average, the impulse applied is 5 ft-lb-sec per second. So you can comfortably apply only 5 ft-lb with your hands to hold the tool steady.

In practice, the tool works by converting the momentum of a hammer into torque. So the angular impulse is actually applied at a constant rate, and the counter-torque you need to apply to hold the tool steady also stays constant. The impulse from each impact is applied to the fastener over a shorter and shorter duration as the torque gets higher. This is why you need to be very careful tightening fasteners with these tools -- you get zero physical feedback of how much torque is actually being applied.

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  • $\begingroup$ thank you. so another hypothetical situation: lets say you have some sort of axle that can rotate with 1,000,000 ft/lbs of torque but only for 1 microsecond. if you strapped the axle in place via a zip tie that can only support 1.1 pounds, 1 ft away from the point of rotation (assuming some sort of weightless, unbendable bar was fixed to the axle, like a hand on a clock), then that zip tie would not break? $\endgroup$ – tau Apr 3 '19 at 16:04
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    $\begingroup$ There would need to be enough rotational inertia and give in the plastic to allow the inertia to oscillate $\endgroup$ – Jonathan Awerbuch Apr 4 '19 at 17:05
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    $\begingroup$ Edited to mention this. I'm not sure if that made it more confusing but at least it's more complete. Thanks for your question, which was obviously not really a question :) $\endgroup$ – Jonathan Awerbuch Apr 4 '19 at 17:09
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To get something to turn you need to apply a minimum amount of torque to overcome friction. The impulse of a collision (eg swinging a hammer at the lever) enables you to apply a high torque for a short time, when a constant push with the maximum stationary force you can provide is less than the required minimum torque. The same principle of expending additional energy to increase the maximum applied force is used in the pile driver and jackhammer.

After a few blows the friction force has reduced sufficiently to enable you to continue applying a constant force. This is more efficient than swinging a hammer.

The mass of the hammer is concentrated at the head. The impact then occurs close to the centre of percussion of the hammer. Minimum reactive shock is felt at the pivot (your hand) when impact occurs at the centre of percussion.

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