# How much torque does it take to turn a doorknob? [closed]

How much torque does it take to turn a doorknob? I'm not looking for an exact answer, just a ballpark for someone who doesn't have a sense of everyday amounts of torque.

Here's a very ordinary picture of a very ordinary doorknob, the type I'm imagining for this question:

-

## closed as too localized by Brandon Enright, Waffle's Crazy Peanut, dmckee♦Jun 5 '13 at 14:41

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

Is this a round or a handle doorknob? –  ja72 Jun 3 '13 at 19:41
@ja72, I'm not sure what distinction you're trying to draw, so I put up a picture of the most normal-looking doorknob I could find. Hopefully it answers what you're asking. –  Joe Jun 3 '13 at 20:48
with a round doorknob you rely on friction, vs. the handle doorknob you rely on a force at a distance. –  ja72 Jun 4 '13 at 12:29
In principle this kind of question is good for learning to ballpark figures, but the problem is devloping into argument and repetition. –  dmckee Jun 5 '13 at 14:42
@dmckee, How could this question be changed to make it a good ballpark question? It seems to me that having a ballpark question for people who have no sense of everyday torques would be quite useful. –  Joe Jun 5 '13 at 16:27

Hanging my iPhone with case (0.16 kg), on a lever arm 12 cm from the center of the door knob just starts turning it. Therefore,

$$m\cdot g\cdot L = 0.19 \mathrm{\,Nm}.$$

-

How about 5 inch-pounds max (0.56 Nm). It would take a weight of 1 pound hung at a distance of 5 inches to turn some doorknobs. Of course it really varies with the strength of the return spring, the friction and the amount of travel the latch needs.

-
5 inch-pounds = 0.56 Nm –  Philip Gibbs - inactive Jun 3 '13 at 21:51
That sounds high. I'm guessing less for most ordinary door knobs. Imagine the outer diameter of the knob is 1.5 to 2 inches from the center, do you really think you exert more than 1 pound turning force at that radius? –  Olin Lathrop Jun 3 '13 at 22:10
Please, at least on this site, use units from the metric system, or at least state values in both systems. Looking at this figure might make you realize how isolated the USA still is in this respect, especially in the context of the world wide web. –  Rody Oldenhuis Jun 4 '13 at 7:36
@OlinLathrop, I said this is max. It is high but to make sure you can open a door you better be able to supply this torque. –  ja72 Jun 4 '13 at 12:28
@Rody: No, I'm going to use whatever I am comfortable with and convert as necessary to whatever I think is relevant at the time. Personally, I have a reasonable sense of what a pound is, but no such sense for a Newton without doing the conversion. The OP didn't specify units, so inch-pounds, foot-pounds, Newton-meters, or slug-fathom-furlongs per fortnight-squared are all valid. –  Olin Lathrop Jun 4 '13 at 14:15

There is so many different variables in play to really give any answer, how old is the door noob, was there a bit of paint in the door nob assembly from the last time it was painted, has the components worn out or have started to oxidize, just to many different variables that I want to be able to give you an answer.

-
This is why I'm asking for a ballpark number. Is it 0.1 foot-pounds? 1 foot-pound? 10 foot-pounds? 100 foot-pounds? –  Joe Jun 4 '13 at 1:17
OK, 1ft-pound, that be maybe a pefect door noob out of the packet. –  Sheathey Jun 4 '13 at 2:01