What did I miss re Ti-frame makers?

[John Forrest Tomlinson]

On Tue, 1 May 2007 09:24:30 -0700, "Bill Sornson" <[email protected]>
wrote:

> if your intent was to further
>harm him (beyond JFT's character assassination)

I'm trying to confine my comments about Hickey and waterboarding, and
about you in general, to threads that are obviously off-topic for this
group, and in the interests of not cluttering up this group would
appreciate your doing the same.

If you want to rag on me, that is your business, but could you try to
similarly confine where you do it?

Other readers -- I am in Sorni's killfile so perhaps one of you could
quote tis verbatim so he sees it. Thanks in advance.

--
JT
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Visit http://www.jt10000.com
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[Michael Press]

In article <[email protected]>,
Gary Young <[email protected]> wrote:

> On Mon, 30 Apr 2007 23:44:57 -0400, David L. Johnson wrote:
>
> > Michael Press wrote:
> >
> >> Stiffness could be overrated as an advantage in a fork.
> >
> > I suggest otherwise.
> >
> >> A steel fork that tapers extemely thin at the drop outs
> >> with all the fork offset in the curve in the thin
> >> sections can soften a lot of road shock.
> >
> > Only to the extent that tubular steel makes a good spring, which is not
> > at all. Tubes do not do well at all in flex. Don't get deceived by the
> > fact that traditional forks curve at the end, it is not to flex. It is
> > to provide the proper trail. Hetchins seems to not believe this, but
> > whether or not their frames look like the bending is there to flex, it
> > better not be, since the tubes would quickly fail if they flexed
> > significantly.
> >
>
> One thing I don't quite understand is why it's important that a disk-brake
> fork have a large cross section along its entire length. If forces
> experienced at the fork ends result in movement at the crown, why would
> the cross section of the fork end matter?
>
> The only explanation I can think of is that the disk brake exerts a force
> that moves the fork to the limit of the crown's range of bending and the
> "excess" force results in bending of the fork end. Is that the case?

The disk brake loads the fork blade in the middle. A fork blade
not designed for a disc brake is optimized to be light, hence has
less material. Braking forces are large and a disk brake mounted
on a typical thin fork blade could buckle the blade.

--
Michael Press

 

[_]

On Tue, 01 May 2007 13:49:57 -0400, John Forrest Tomlinson wrote:

> On Tue, 1 May 2007 09:24:30 -0700, "Bill Sornson" <[email protected]>
> wrote:
>
>> if your intent was to further
>>harm him (beyond JFT's character assassination)
>
> I'm trying to confine my comments about Hickey and waterboarding, and
> about you in general, to threads that are obviously off-topic for this
> group, and in the interests of not cluttering up this group would
> appreciate your doing the same.
>
> If you want to rag on me, that is your business, but could you try to
> similarly confine where you do it?
>
> Other readers -- I am in Sorni's killfile so perhaps one of you could
> quote tis verbatim so he sees it. Thanks in advance.

In addition, it should be pointed out that it's hardly character
assassination to repeat another poster's words; the deed was done by the
poster himself.

Sorni has had trouble figuring out how usenet quoting works, though...

 

[Gary Young]

On Tue, 01 May 2007 10:52:55 -0700, Michael Press wrote:

> In article <[email protected]>,
> Gary Young <[email protected]> wrote:
>
>> On Mon, 30 Apr 2007 23:44:57 -0400, David L. Johnson wrote:
>>
>> > Michael Press wrote:
>> >
>> >> Stiffness could be overrated as an advantage in a fork.
>> >
>> > I suggest otherwise.
>> >
>> >> A steel fork that tapers extemely thin at the drop outs
>> >> with all the fork offset in the curve in the thin
>> >> sections can soften a lot of road shock.
>> >
>> > Only to the extent that tubular steel makes a good spring, which is not
>> > at all. Tubes do not do well at all in flex. Don't get deceived by the
>> > fact that traditional forks curve at the end, it is not to flex. It is
>> > to provide the proper trail. Hetchins seems to not believe this, but
>> > whether or not their frames look like the bending is there to flex, it
>> > better not be, since the tubes would quickly fail if they flexed
>> > significantly.
>> >
>>
>> One thing I don't quite understand is why it's important that a disk-brake
>> fork have a large cross section along its entire length. If forces
>> experienced at the fork ends result in movement at the crown, why would
>> the cross section of the fork end matter?
>>
>> The only explanation I can think of is that the disk brake exerts a force
>> that moves the fork to the limit of the crown's range of bending and the
>> "excess" force results in bending of the fork end. Is that the case?
>
> The disk brake loads the fork blade in the middle. A fork blade
> not designed for a disc brake is optimized to be light, hence has
> less material. Braking forces are large and a disk brake mounted
> on a typical thin fork blade could buckle the blade.
>

The disk-brake situation is easy enough to explain if one believes, as you
do, that there is significant bending at the fork end. What I'm hoping for
is an explanation from someone who believes that bending occurs primarily
at the crown. I'm inclined to agree with the bending-at-the-crown
explanation for a variety of reasons, but it does seem that the need for
large cross-section forks when using disk brakes poses a challenge to that
explanation, at least at first glance.

 

[Matt O'Toole]

On Tue, 01 May 2007 05:48:03 -0400, John Forrest Tomlinson wrote:

> On Mon, 30 Apr 2007 21:28:52 -0700, jim beam
> <[email protected]> wrote:

>> offset is easily achieved with straight
>> fork blades - ask a colnago rider.

> No, at least traditionally tubes were bent for offset. It's much
> easier and flexible to have a single fork crown design and vary rake
> by how much one bends the fork blades, than to have to have slightly
> different crowns to create different rakes.

It's also easier to see if a fork has been damaged in a crash, because the
straight legs will be bent backward and no longer lined up with the head
tube. With a fork offset at the crown (Colnago), there's no reference
point.

Matt O.

 

[[email protected]]

Jay Taylor writes:

>>> if your intent was to further harm him (beyond JFT's character
>>> assassination)

>> I'm trying to confine my comments about Hickey and waterboarding,
>> and about you in general, to threads that are obviously off-topic
>> for this group, and in the interests of not cluttering up this
>> group would appreciate your doing the same.

>> If you want to rag on me, that is your business, but could you try
>> to similarly confine where you do it?

>> Other readers -- I am in Sorni's killfile so perhaps one of you
>> could quote tis verbatim so he sees it. Thanks in advance.

> In addition, it should be pointed out that it's hardly character
> assassination to repeat another poster's words; the deed was done by
> the poster himself.

> Sorni has had trouble figuring out how usenet quoting works,
> though...

Just the same, many contributors cloud their responses by leaving a
difficult to follow list of "so-and-so wrote..." in layers of layers.
The response is to the previous poster, as in this case Jay Taylor.
All those other attributions are not germane to the current response.

....a mild example is the posting to which this one responds.

Jobst Brandt

 

[Bill Sornson]

[email protected] wrote:
> Jay Taylor writes: {snipped}

> Just the same, many contributors cloud their responses by leaving a
> difficult to follow list of "so-and-so wrote..." in layers of layers.
> The response is to the previous poster, as in this case Jay Taylor.
> All those other attributions are not germane to the current response.

"Jay Taylor"? You mean ol' "_"?!? LOL

> ...a mild example is the posting to which this one responds.

Completely ignoring its point, of course. Namely, that character
assassination wasn't enough for the cowardly anonymous OP.

BS (nope)

 

[Michael Press]

In article <[email protected]>,
Gary Young <[email protected]> wrote:

> On Tue, 01 May 2007 10:52:55 -0700, Michael Press wrote:
>
> > In article <[email protected]>,
> > Gary Young <[email protected]> wrote:
> >
> >> On Mon, 30 Apr 2007 23:44:57 -0400, David L. Johnson wrote:
> >>
> >> > Michael Press wrote:
> >> >
> >> >> Stiffness could be overrated as an advantage in a fork.
> >> >
> >> > I suggest otherwise.
> >> >
> >> >> A steel fork that tapers extemely thin at the drop outs
> >> >> with all the fork offset in the curve in the thin
> >> >> sections can soften a lot of road shock.
> >> >
> >> > Only to the extent that tubular steel makes a good spring, which is not
> >> > at all. Tubes do not do well at all in flex. Don't get deceived by the
> >> > fact that traditional forks curve at the end, it is not to flex. It is
> >> > to provide the proper trail. Hetchins seems to not believe this, but
> >> > whether or not their frames look like the bending is there to flex, it
> >> > better not be, since the tubes would quickly fail if they flexed
> >> > significantly.
> >> >
> >>
> >> One thing I don't quite understand is why it's important that a disk-brake
> >> fork have a large cross section along its entire length. If forces
> >> experienced at the fork ends result in movement at the crown, why would
> >> the cross section of the fork end matter?
> >>
> >> The only explanation I can think of is that the disk brake exerts a force
> >> that moves the fork to the limit of the crown's range of bending and the
> >> "excess" force results in bending of the fork end. Is that the case?
> >
> > The disk brake loads the fork blade in the middle. A fork blade
> > not designed for a disc brake is optimized to be light, hence has
> > less material. Braking forces are large and a disk brake mounted
> > on a typical thin fork blade could buckle the blade.
> >
>
> The disk-brake situation is easy enough to explain if one believes, as you
> do, that there is significant bending at the fork end.

It is not what I believe. I _know_ that a disc brake caliper
mounted at mid-span in a typical road fork blade will overload the
designed resistance to bending moment of the fork blade at mid-span

> What I'm hoping for
> is an explanation from someone who believes that bending occurs primarily
> at the crown. I'm inclined to agree with the bending-at-the-crown
> explanation for a variety of reasons, but it does seem that the need for
> large cross-section forks when using disk brakes poses a challenge to that
> explanation, at least at first glance.

--
Michael Press

 

[Gary Young]

On Tue, 01 May 2007 16:18:39 -0700, Michael Press wrote:

> In article <[email protected]>,
> Gary Young <[email protected]> wrote:
>
>> On Tue, 01 May 2007 10:52:55 -0700, Michael Press wrote:
>>
>> > In article <[email protected]>,
>> > Gary Young <[email protected]> wrote:
>> >
>> >> On Mon, 30 Apr 2007 23:44:57 -0400, David L. Johnson wrote:
>> >>
>> >> > Michael Press wrote:
>> >> >
>> >> >> Stiffness could be overrated as an advantage in a fork.
>> >> >
>> >> > I suggest otherwise.
>> >> >
>> >> >> A steel fork that tapers extemely thin at the drop outs
>> >> >> with all the fork offset in the curve in the thin
>> >> >> sections can soften a lot of road shock.
>> >> >
>> >> > Only to the extent that tubular steel makes a good spring, which is not
>> >> > at all. Tubes do not do well at all in flex. Don't get deceived by the
>> >> > fact that traditional forks curve at the end, it is not to flex. It is
>> >> > to provide the proper trail. Hetchins seems to not believe this, but
>> >> > whether or not their frames look like the bending is there to flex, it
>> >> > better not be, since the tubes would quickly fail if they flexed
>> >> > significantly.
>> >> >
>> >>
>> >> One thing I don't quite understand is why it's important that a disk-brake
>> >> fork have a large cross section along its entire length. If forces
>> >> experienced at the fork ends result in movement at the crown, why would
>> >> the cross section of the fork end matter?
>> >>
>> >> The only explanation I can think of is that the disk brake exerts a force
>> >> that moves the fork to the limit of the crown's range of bending and the
>> >> "excess" force results in bending of the fork end. Is that the case?
>> >
>> > The disk brake loads the fork blade in the middle. A fork blade
>> > not designed for a disc brake is optimized to be light, hence has
>> > less material. Braking forces are large and a disk brake mounted
>> > on a typical thin fork blade could buckle the blade.
>> >
>>
>> The disk-brake situation is easy enough to explain if one believes, as you
>> do, that there is significant bending at the fork end.
>
> It is not what I believe. I _know_ that a disc brake caliper
> mounted at mid-span in a typical road fork blade will overload the
> designed resistance to bending moment of the fork blade at mid-span

I should have been more clear. What I meant is that explaining what
happens with a disk-brake is no difficulty if one already believes that an
ordinary fork undergoes significant bending at places other than the crown.

>
>> What I'm hoping for
>> is an explanation from someone who believes that bending occurs primarily
>> at the crown. I'm inclined to agree with the bending-at-the-crown
>> explanation for a variety of reasons, but it does seem that the need for
>> large cross-section forks when using disk brakes poses a challenge to that
>> explanation, at least at first glance.
>

 

[[email protected]]

On May 1, 12:06 pm, Gary Young <[email protected]> wrote:
>
> > The disk brake loads the fork blade in the middle. A fork blade
> > not designed for a disc brake is optimized to be light, hence has
> > less material. Braking forces are large and a disk brake mounted
> > on a typical thin fork blade could buckle the blade.
>
> The disk-brake situation is easy enough to explain if one believes, as you
> do, that there is significant bending at the fork end. What I'm hoping for
> is an explanation from someone who believes that bending occurs primarily
> at the crown. I'm inclined to agree with the bending-at-the-crown
> explanation for a variety of reasons, but it does seem that the need for
> large cross-section forks when using disk brakes poses a challenge to that
> explanation, at least at first glance.

No it doesn't, because the disk brake applies the force
perpendicular to the small cross-section end. You
run a risk of locally buckling the tube.

What are the forces on a fork and their directions?
Coasting along, you have the axle pushing up on the fork end
and the head bearing pushing down. Say the front wheel is
bearing 35 kg weight out of 90 kg bike+rider, there is a 35 kgf
downforce at the head bearing, but most of that is along the
fork, and only 35*cos(73) = 10 kgf perpendicular to the fork.
The force at the dropout is slightly different due to the angle
of the fork bend being slightly less than 73 deg.

Now apply a front rim brake hard at 0.5 g. That is 0.5*90 =
45 kgf in the forward direction, and about 45*sin(73) =43 kgf
is the component perpendicular to the fork. Fortunately
this is being applied at the nice sturdy fork crown.
The bending moment at the crown is small because
the brake-crown distance is small.

But if you have a disk brake, the force is applied directly
at the small end and perpendicular to the blade. It's loaded
in a different direction than the load from a road bump,
and the bending moment at the crown (or anywhere along
the fork) is much larger because the distance between the
brake and the constrained point at the crown is large. Now,
40+ kgf is likely not enough to bend even a small blade
fork most of the time. But it probably reduces the margin
of safety unacceptably, to where a heavy rider or braking
and hitting a pothole or some other unexpected load
could cause problems.

Ben

 

[jim beam]

[email protected] wrote:
> On May 1, 6:02 am, jim beam <[email protected]> wrote:
>> [email protected] wrote:
>>
>>> Gah. Stiffness of something like a fork depends on the
>>> cross-section of the legs, not the bend.
>> to some extent, that's true about the crown. but regarding the blade, a
>> bend /does/ ensure that flex occurs in the region preferred - buckling.
>> do some experiments with coat hanger wire if you don't believe me.
>
> Coat hanger wire has negligible cross-section. Fork
> tubes are different. You don't want a fork tube to
> approach buckling at all.

as evidence of principle, wire works perfectly. and it's cheap. and
readily available. try it.

as for "buckling tube, let me be more specific, i don't mean buckling
like a tube that kinks on failure, i mean buckling like the bending
motion when compressing a long rod. consult euler on the subject. but
something tells me you already knew that.

>
>>> We all know that
>>> both curved and straight blade forks have offset. There
>>> is fork flexibility and a lot of the bend occurs at the crown.
>>> If you have two forks, and one has stiffer legs, but they
>>> have the same crown, the stiff-leg one is not necessarily
>>> going to be stiffer overall.
>> if it's the same crown and steerer tube, there's /no way/ a fork with
>> stiffer blades can be anything other than stiffer overall. absolutely
>> no way.
>
> Suppose that you have a system with two parts. One's
> really stiff and one's much less so. The stiff part can
> transmit all the bending moment to the weak link without
> taking any bending up itself. I'm not going to argue that
> it would be measurably less stiff and I think that it's
> unlikely to be significant in bike forks. However, as a
> general design rule, it's a bad idea to let an incredibly
> strong piece transmit stress to a joint that might be
> more vulnerable.

what part of "if it's the same crown and steerer tube, there's /no way/
a fork with stiffer blades can be anything other than stiffer overall."
is confusing you? why all this straw grasping for previously
undisclosed "weak links"? what's next - sawing a blade halfway through
and then claiming it's not stiff? c'mon dude, get a grip.

 

[Ben C]

On 2007-05-02, [email protected] <[email protected]> wrote:
> On May 1, 12:06 pm, Gary Young <[email protected]> wrote:
>>
>> > The disk brake loads the fork blade in the middle. A fork blade
>> > not designed for a disc brake is optimized to be light, hence has
>> > less material. Braking forces are large and a disk brake mounted
>> > on a typical thin fork blade could buckle the blade.
>>
>> The disk-brake situation is easy enough to explain if one believes, as you
>> do, that there is significant bending at the fork end. What I'm hoping for
>> is an explanation from someone who believes that bending occurs primarily
>> at the crown. I'm inclined to agree with the bending-at-the-crown
>> explanation for a variety of reasons, but it does seem that the need for
>> large cross-section forks when using disk brakes poses a challenge to that
>> explanation, at least at first glance.
>
> No it doesn't, because the disk brake applies the force
> perpendicular to the small cross-section end. You
> run a risk of locally buckling the tube.
>
> What are the forces on a fork and their directions?
> Coasting along, you have the axle pushing up on the fork end
> and the head bearing pushing down. Say the front wheel is
> bearing 35 kg weight out of 90 kg bike+rider, there is a 35 kgf
> downforce at the head bearing, but most of that is along the
> fork, and only 35*cos(73) = 10 kgf perpendicular to the fork.
> The force at the dropout is slightly different due to the angle
> of the fork bend being slightly less than 73 deg.
>
> Now apply a front rim brake hard at 0.5 g. That is 0.5*90 =
> 45 kgf in the forward direction, and about 45*sin(73) =43 kgf
> is the component perpendicular to the fork. Fortunately
> this is being applied at the nice sturdy fork crown.
> The bending moment at the crown is small because
> the brake-crown distance is small.
>
> But if you have a disk brake, the force is applied directly
> at the small end and perpendicular to the blade. It's loaded
> in a different direction than the load from a road bump,
> and the bending moment at the crown (or anywhere along
> the fork) is much larger because the distance between the
> brake and the constrained point at the crown is large.

And even larger still because the force itself is about 8x larger-- 4x
because the disk radius is about 1/4 that of the rim, and another 2x
because the disk is only on one side.

> Now, 40+ kgf is likely not enough to bend even a small blade fork most
> of the time. But it probably reduces the margin of safety
> unacceptably, to where a heavy rider or braking and hitting a pothole
> or some other unexpected load could cause problems.