Skid-stops on a fixed-wheel?

[Anthony Jones]

Rob Morley wrote:
> I notice you snipped the rest of my post - didn't you want to disagree
> with it?

Nope, I agreed with the rest of it. I didn't think it contradicted anything
I said.

Anthony

 

[Rob Morley]

In article <[email protected]>
Anthony Jones <[email protected]> wrote:
> Rob Morley wrote:
> > I notice you snipped the rest of my post - didn't you want to disagree
> > with it?
>
> Nope, I agreed with the rest of it. I didn't think it contradicted anything
> I said.
>
It must be the way I read your posts then.

 

[DavidR]

"Anthony Jones" <[email protected]> wrote
> Anthony Jones wrote:
>> I'm suggesting that the torque required to skid the rear wheel, under
>> normal conditions, is higher than that that can be applied while the
>> pedals are turning. Hence the rate of deceleration is higher.
>
> Hmmm, that's rather ambiguous, I should have phrased it as 'the torque
> required to *hold* the rear wheel in a skid'.

I don't think you understand the meaning of torque.

 

[Anthony Jones]

DavidR wrote:
> I don't think you understand the meaning of torque.

Perhaps you'd like to explain to me?

Anthony

 

[Anthony Jones]

Rob Morley wrote:
> It must be the way I read your posts then.

Errr, yes, it appears I didn't do a very convincing job of explaining my
logic...

Anthony

 

[soup]

Pete Biggs wrote:


> Once the wheel is skidding, pushing harder back on the pedals doesn't slow
> you down any faster. I /think/ that's all Soup meant.

Yup sort of, what I was saying was the if the wheel is locked you can't
apply more torque to it (consider the cranks are straight up and down
any increased force will just try to elongate the crank) and
disregarding ;blown knees, ability to do so etc (theory rather than
practice ?)a "just" turning wheel provides more braking force than a
locked one. This discussion is losing way under semantics and dodgy
explanations (or rather my limitations of explaining things in a purely
text medium) so will leave it there.


--
www.cheesesoup.myby.co.uk
http://www.youtube.com/watch?v=nileh1ZPGq4

 

[soup]

soup wrote:

> Yup sort of, what I was saying was the if the wheel is locked you can't
> apply more torque to it

Well you can but it makes the wheel rotate "backwards" it doesn't
increase the breaking force.

--
www.cheesesoup.myby.co.uk
http://www.youtube.com/watch?v=nileh1ZPGq4

 

[Anthony Jones]

Clive George wrote:
> See my comment about one song to the tune of another - I suspect he knows
> what's going on, same as you and I do, it's just the explanations which
> are going wrong

Considering the disagreement I seemed to have caused, obviously I've said
*something* stupid, but apart from my posts being a bit
stream-of-consciousness I'm honestly struggling to work out what it is. If
someone can muster the will to explain to me I'd be genuinely interested,
but I suspect this has drifted on for long enough...

Anthony

 

[Tony Raven]

Anthony Jones wrote on 25/11/2006 10:41 +0100:
> Clive George wrote:
>> See my comment about one song to the tune of another - I suspect he knows
>> what's going on, same as you and I do, it's just the explanations which
>> are going wrong
>
> Considering the disagreement I seemed to have caused, obviously I've said
> *something* stupid, but apart from my posts being a bit
> stream-of-consciousness I'm honestly struggling to work out what it is. If
> someone can muster the will to explain to me I'd be genuinely interested,
> but I suspect this has drifted on for long enough...
>

Well I understood it and I think anyone who has ridden a fixed and
experienced trying to slow it down by resisting the pedals turning would
know what you meant (and why a skid stop will be faster)


--
Tony

"...has many omissions and contains much that is apocryphal, or at least
wildly inaccurate..."
Douglas Adams; The Hitchhiker's Guide to the Galaxy

 

[Pete Biggs]

Anthony Jones wrote:
> Clive George wrote:
>> See my comment about one song to the tune of another - I suspect he
>> knows what's going on, same as you and I do, it's just the
>> explanations which are going wrong
>
> Considering the disagreement I seemed to have caused, obviously I've
> said *something* stupid, but apart from my posts being a bit
> stream-of-consciousness I'm honestly struggling to work out what it
> is. If someone can muster the will to explain to me I'd be genuinely
> interested, but I suspect this has drifted on for long enough...

There seemed to be a slight misunderstanding between you and Soup at one
point, but the rest of you wrote seems clear to me.

~PB

 

[soup]

Pete Biggs wrote:
> Anthony Jones wrote:
>> Considering the disagreement I seemed to have caused, obviously I've
>> said *something* stupid, but apart from my posts being a bit
>> stream-of-consciousness I'm honestly struggling to work out what it
>> is. If someone can muster the will to explain to me I'd be genuinely
>> interested, but I suspect this has drifted on for long enough...
> There seemed to be a slight misunderstanding between you and Soup at one
> point, but the rest of you wrote seems clear to me.

I was meaning that in theory a slightly rotating wheel has a greater
retarding force than one which is skidding whilst Anthony was saying
that in practise it is very hard to let the wheel just rotate and it was
easier to lock the wheel than to resist it's rotation so not actually
disagreeing just talking about two different things.

--
www.cheesesoup.myby.co.uk
http://www.youtube.com/watch?v=nileh1ZPGq4

 

[DavidR]

"Anthony Jones" <[email protected]> wrote
> DavidR wrote:

>> I don't think you understand the meaning of torque.
>
> Perhaps you'd like to explain to me?
>
You apply a force on the pedals, which applies a force on the rim
proportional to crank & rim diameters and sprocket ratios to perform
whatever accelerate/decelerate manouevre you are doing.

Torque is only internal to the mechanism. For example, applying force to
the left hand crank puts a torque on the axle transfering the drive across
the frame.

 

[Anthony Jones]

DavidR wrote:
> You apply a force on the pedals, which applies a force on the rim
> proportional to crank & rim diameters and sprocket ratios to perform
> whatever accelerate/decelerate manouevre you are doing.
>
> Torque is only internal to the mechanism. For example, applying force to
> the left hand crank puts a torque on the axle transfering the drive across
> the frame.

I'm afraid I still don't get how that contradicts what I was saying.

I felt that a description in terms of the torque resulting from the force on
the pedals was clearer than referring solely to 'force' when there are two
pedals and the forces are in unspecified directions...

Anthony

 

[Bronzie]

Duncan Smith wrote:
> Then I read about skid stops

You mean like this?
http://www.youtube.com/watch?v=en8ESzTmSAM&NR

 

[DavidR]

"Anthony Jones" <[email protected]> wrote in message
news:[email protected]...
> DavidR wrote:
>> You apply a force on the pedals, which applies a force on the rim
>> proportional to crank & rim diameters and sprocket ratios to perform
>> whatever accelerate/decelerate manouevre you are doing.
>>
>> Torque is only internal to the mechanism. For example, applying force to
>> the left hand crank puts a torque on the axle transfering the drive
>> across
>> the frame.

> I'm afraid I still don't get how that contradicts what I was saying.

I wasn't really paying attention to what you were saying in detail. I am
not interested in skid stops on a fixer, either for aesthetic or technical
reasons. It's only that the curious use of the word 'torque' jumped out.

> I felt that a description in terms of the torque resulting from the force
> on
> the pedals

....about what axis?

> was clearer than referring solely to 'force' when there are two
> pedals and the forces are in unspecified directions...

You have a crank describing a diameter of about 13 inches. The sprockets
give you an imaginary wheel of about 60 inches diameter. It's also an
imaginary direct drive - think of a penny-farthing. Force on the pedal
gives 13/60 of that force at the rim. That's all that matters in your
experiment.

But...real wheels are not 60 inches. It's most likely 27 inches. Let's
assume you need 54 pound force (for sake of argument), to lock the back
wheel. That's equivalent to the wheel being driven by 48 foot-pound force.
Now try a Brompton which has 16 inch diameter wheels; to lock you will
need exactly the same 54lb force at the rim. That would be developed from a
torque of 20lbf-ft.

Don't you now see that torque is not meaningful?

 

[Clive George]

"Bronzie" <[email protected]> wrote in message
news:[email protected]...
>
> Duncan Smith wrote:
>> Then I read about skid stops
>
> You mean like this?
> http://www.youtube.com/watch?v=en8ESzTmSAM&NR

aha - thank you!

(I'd been wondering what the competition ones looked like - had a vague
idea, but the video helps somewhat).

cheers,
clive

 

[Anthony Jones]

DavidR wrote:
>> I felt that a description in terms of the torque resulting from the force
>> on
>> the pedals
>
> ...about what axis?

The axis of the bottom bracket spindle.

> You have a crank describing a diameter of about 13 inches. The sprockets
> give you an imaginary wheel of about 60 inches diameter. It's also an
> imaginary direct drive - think of a penny-farthing. Force on the pedal
> gives 13/60 of that force at the rim. That's all that matters in your
> experiment.

You're making an assumption about the direction of the force. That was what
I was trying to avoid.

> But...real wheels are not 60 inches. It's most likely 27 inches. Let's
> assume you need 54 pound force (for sake of argument), to lock the back
> wheel. That's equivalent to the wheel being driven by 48 foot-pound force.
> Now try a Brompton which has 16 inch diameter wheels; to lock you will
> need exactly the same 54lb force at the rim. That would be developed from
> a torque of 20lbf-ft.

I could say exactly the same about force on the pedal. Obviously the wheel
size and gear ratio must be constant.

> Don't you now see that torque is not meaningful?

If I'd used 'force', you could have used exactly the same arguments as
above, substituting 'direction' for 'axis'.

If my mistake was not clearly stating the axis about which the torque was
acting and/or that wheel size and gear ratio were contant, then fair
enough, I see your point.

Anthony

 

[DavidR]

"Anthony Jones" <[email protected]> wrote
> DavidR wrote:

>>> I felt that a description in terms of the torque resulting from the
>>> force
>>> on
>>> the pedals
>>
>> ...about what axis?
>
> The axis of the bottom bracket spindle.

>> You have a crank describing a diameter of about 13 inches. The
>> sprockets
>> give you an imaginary wheel of about 60 inches diameter. It's also an
>> imaginary direct drive - think of a penny-farthing. Force on the pedal
>> gives 13/60 of that force at the rim. That's all that matters in your
>> experiment.
>
> You're making an assumption about the direction of the force. That was
> what
> I was trying to avoid.

The above only describes a ratio...

If I am not mistaken, you are trying to slow a fixed wheel bike by applying
enough force on the rim to cause the wheel to skid. That doesn't imply
direction?


>> Don't you now see that torque is not meaningful?
>
> If I'd used 'force', you could have used exactly the same arguments as
> above, substituting 'direction' for 'axis'.

If you had concentrated on force I wouldn't have responded.

 

[Anthony Jones]

DavidR wrote:
> The above only describes a ratio...
>
> If I am not mistaken, you are trying to slow a fixed wheel bike by
> applying enough force on the rim to cause the wheel to skid. That doesn't
> imply direction?

I was meaning force at the pedal. If the force on the pedal is, for example,
directly towards the BB spindle, the force at the rear wheel is zero.

Anthony