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Car Forum / Honda Cars / November 2005Does smaller friction area cause less friction?
From: http://www.boltscience.com/pages/faq.htm > Typically 50% of the torque is used to overcome friction > under the tightening surface. Hence a smaller friction radius > will result in more torque going into the thread of the bolt > and hence being over tightened. Is this relationship linear? If the friction area is halved so is the friction? > > Typically 50% of the torque is used to overcome friction > > under the tightening surface. Hence a smaller friction radius [quoted text clipped - 3 lines] > Is this relationship linear? > If the friction area is halved so is the friction? The coefficient of friction does not change, but the friction force does. > From: "alt.autos.honda group" <noreply@googlegroups.com> > Date: Tue, 22 Nov 2005 01:47:59 +0000 [quoted text clipped - 18 lines] > The coefficient of friction does not change, but the friction > force does. Actually, neither the coefficient of friction nor the friction force does change. See my next message to Michael Pardee. > From: > [quoted text clipped - 7 lines] > Is this relationship linear? > If the friction area is halved so is the friction? If I understand what they are saying, the smaller radius results in less rotational friction, which would be a linear relationship with respect to parasitic torque. That is, if the original torque was 50% on the threads and 50% on the bolt head, cutting the radius in half would cause only 25% of the torque to be used to overcome bolt head friction and would put the remaining 75% of the torque on the threads. Note that this is not the area but the radius that is changed. In theory, friction is independent of area. What they seem to be describing is leverage. Mike >> From: >> [quoted text clipped - 20 lines] > > Mike 'Frictional Force' is calculated by multiplying the coefficient of friction times the normal force. Normal force is the perpendicular force in a system. It can be quite difficult to determine the correct coefficient of friction to utilize when there is plating involved, two different types of materials, etc. This same problem is why some of these highway patrolmen's estimates of a car's speed that was involved in an accident can be total b.s. Many factors such as road film, tire condition, loose gravel and brake conditions should be taken into account, technically, and there's simply no way that this could be calculated accurately at the scene of an accident. Ron M. >>> From: >>> [quoted text clipped - 34 lines] > > Ron M. As karl clarifies, the linear friction is indeed constant, whatever it is. It is the translation of that into angular resistance (torque) that varies proportionally with the radius (diameter). Mike > From: "alt.autos.honda group" <noreply@googlegroups.com> > Date: Tue, 22 Nov 2005 01:47:59 +0000 [quoted text clipped - 34 lines] > > Mike Got it. It is the integral of the travel. That is, friction at the bolt head (washer) is proportional to, friction = D(l)^2/2 - D(s)^2/2 D(s) = small diameter D(l) = large diameter Note that D(l) is the smaller of bolt head diameter or large diameter of the washer. > > From: "alt.autos.honda group" <noreply@googlegroups.com> > > Date: Tue, 22 Nov 2005 01:47:59 +0000 [quoted text clipped - 3 lines] > > http://groups.google.com/group/alt.autos.honda/browse_thread/thread/a22a08205be38867 > > ============================================================================= snip > Got it. It is the integral of the travel. That is, friction at the bolt > head (washer) is proportional to, [quoted text clipped - 6 lines] > Note that D(l) is the smaller of bolt head diameter or large diameter > of the washer. This is wrong, it relates to the area. But relevant is the length of the travel. Friction then is proportional to, friction = D(l) - D(s) I hope I got it right this time. |
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