Does smaller friction area cause less friction?
11-21-2005 | 06:12 AM
#1
Guest
Posts: n/a
Does 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?
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?
11-21-2005 | 08:26 AM
#2
Guest
Posts: n/a
Re: Does smaller friction area cause less friction?
> > 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?
The coefficient of friction does not change, but the friction force does.
> > 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?
The coefficient of friction does not change, but the friction force does.
11-21-2005 | 06:05 PM
#3
Guest
Posts: n/a
Re: Does smaller friction area cause less friction?
"karl" <ottokarl@cognisurf.com> wrote in message
news:1132571575.180298.25790@g49g2000cwa.googlegro ups.com...
> 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?
>
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
news:1132571575.180298.25790@g49g2000cwa.googlegro ups.com...
> 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?
>
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
11-21-2005 | 09:44 PM
#4
Guest
Posts: n/a
Re: Does smaller friction area cause less friction?
"Michael Pardee" <michaeltnull@cybertrails.com> wrote in message
news:C_SdnTBQEpspzx_enZ2dnUVZ_sqdnZ2d@sedona.net.. .
> "karl" <ottokarl@cognisurf.com> wrote in message
> news:1132571575.180298.25790@g49g2000cwa.googlegro ups.com...
>> 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?
>>
> 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
'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.
news:C_SdnTBQEpspzx_enZ2dnUVZ_sqdnZ2d@sedona.net.. .
> "karl" <ottokarl@cognisurf.com> wrote in message
> news:1132571575.180298.25790@g49g2000cwa.googlegro ups.com...
>> 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?
>>
> 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
'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.
11-21-2005 | 10:42 PM
#5
Guest
Posts: n/a
Re: Does smaller friction area cause less friction?
> From: "alt.autos.honda group" <noreply@googlegroups.com>
> Date: Tue, 22 Nov 2005 01:47:59 +0000
>
> ================================================== ===========================
> TOPIC: Does smaller friction area cause less friction?
> http://groups.google.com/group/alt.a...2a08205be38867
> ================================================== ===========================
>
> == 2 of 3 ==
> Date: Mon, Nov 21 2005 1:26 pm
> From: "Steve Mackie"
>
> > > 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?
>
>
> 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.
11-21-2005 | 10:43 PM
#6
Guest
Posts: n/a
Re: Does smaller friction area cause less friction?
> From: "alt.autos.honda group" <noreply@googlegroups.com>
> Date: Tue, 22 Nov 2005 01:47:59 +0000
>
> ================================================== ===========================
> TOPIC: Does smaller friction area cause less friction?
> http://groups.google.com/group/alt.a...2a08205be38867
> ================================================== ===========================
>
> == 3 of 3 ==
> Date: Mon, Nov 21 2005 4:05 pm
> From: "Michael Pardee"
>
> "karl" <ottokarl@cognisurf.com> wrote in message
> news:1132571575.180298.25790@g49g2000cwa.googlegro ups.com...
> > 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?
>
>
> 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
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.
11-21-2005 | 11:11 PM
#7
Guest
Posts: n/a
Re: Does smaller friction area cause less friction?
"Ron M." <nobody@nowhere.net> wrote in message
news:11o51gbhnd2nld1@corp.supernews.com...
> "Michael Pardee" <michaeltnull@cybertrails.com> wrote in message
> news:C_SdnTBQEpspzx_enZ2dnUVZ_sqdnZ2d@sedona.net.. .
>> "karl" <ottokarl@cognisurf.com> wrote in message
>> news:1132571575.180298.25790@g49g2000cwa.googlegro ups.com...
>>> 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?
>>>
>> 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
>
> '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.
>
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
news:11o51gbhnd2nld1@corp.supernews.com...
> "Michael Pardee" <michaeltnull@cybertrails.com> wrote in message
> news:C_SdnTBQEpspzx_enZ2dnUVZ_sqdnZ2d@sedona.net.. .
>> "karl" <ottokarl@cognisurf.com> wrote in message
>> news:1132571575.180298.25790@g49g2000cwa.googlegro ups.com...
>>> 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?
>>>
>> 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
>
> '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.
>
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
11-22-2005 | 01:28 AM
#8
Guest
Posts: n/a
Re: Does smaller friction area cause less friction?
karl wrote:
> > From: "alt.autos.honda group" <noreply@googlegroups.com>
> > Date: Tue, 22 Nov 2005 01:47:59 +0000
> >
> > ================================================== ===========================
> > TOPIC: Does smaller friction area cause less friction?
> > http://groups.google.com/group/alt.a...2a08205be38867
> > ================================================== ===========================
snip
> 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.
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.
Thread
Thread Starter
Forum
Replies
Last Post
17years
honda / acura
0
08-09-2008 02:02 AM
Sharon Peters
Motorcycle Section
3
12-04-2007 06:05 PM
BE
Hyundai Mailing List
0
09-04-2003 10:19 AM
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)