Which wheel travels furthest?

On 9 July, 09:54, TrailRat wrote:
The front one or the back one?

Surely the front one as the back one follows on behind. Ahhh, but the
back starts in the same location as it finishes, the exact same
distance behind the front one.

Anyway, the point of this madness.

I got a little bored and decided a little experiment. Before I set out
I made sure both wheels had the valves pointing straight up. The
wheels, tyres and the inner tubes are identical back and front (apart
from that big block of cogs on the back one, duh!). Both are disc
braked. I also made sure the pressure was the same.

Then I set off on a gentle 10 mile ride. All road and cycle paths.

Any loose paths will make for more rear wheel rotation due to slippage
under drive.

Nothing to strenuous. I made sure to use access ramps and not just
bounce off the kerbs. I kept my speed below 15mph.

Anyway on my return, I inspected the valves expecting to find them
still aligned and was surprised to find that front one was about 4" in
front of the rear one.

and how many rotations? Are you sure rear wheel is behind the
front? About 40" is the usual amount. Typographical error?


So, I turn the scientific analysis to my peers and ask that ultimate
scientific question, huh??

Although the rear wheel usually takes a slightly shorter route around
bends so resulting in fewer revolutions, it is also subject to the
driving force which amounts to a little slippage, resulting in more
revolutions than the front. I don't know whether there is a typical
value on this.

On 09/07/2010 19:58, thirty-six wrote:

Although the rear wheel usually takes a slightly shorter route around
bends so resulting in fewer revolutions, it is also subject to the
driving force which amounts to a little slippage, resulting in more
revolutions than the front. I don't know whether there is a typical
value on this.

I'd guess about the square root of feckall .. if the rear spun that much
then you'd be skidding all over. Maybe on sheet ice ...


--
Paul - xxx

'96/'97 Landrover Discovery 300 Tdi
Dyna Tech Cro-Mo comp

On 9 July, 20:16, Paul - xxx wrote:
On 09/07/2010 19:58, thirty-six wrote:

Although the rear wheel usually takes a slightly shorter route around
bends so resulting in fewer revolutions, it is also subject to the
driving force which amounts to a little slippage, resulting in more
revolutions than the front. * I don't know whether there is a typical
value on this.

I'd guess about the square root of feckall .. if the rear spun that much
then you'd be skidding all over. *Maybe on sheet ice ...


Drive-slip occurs at the centre of the wheel tread only, not like
wheelspin which is slip across the whole of the tread.

On 10/07/2010 04:48, thirty-six wrote:
On 9 July, 20:16, Paul - wrote:
On 09/07/2010 19:58, thirty-six wrote:

Although the rear wheel usually takes a slightly shorter route around
bends so resulting in fewer revolutions, it is also subject to the
driving force which amounts to a little slippage, resulting in more
revolutions than the front. I don't know whether there is a typical
value on this.

I'd guess about the square root of feckall .. if the rear spun that much
then you'd be skidding all over. Maybe on sheet ice ...


Drive-slip occurs at the centre of the wheel tread only, not like
wheelspin which is slip across the whole of the tread.

? How does that work then?

How can the centre of the tread slip without the rest of the tread?


--
Paul - xxx

'96/'97 Landrover Discovery 300 Tdi
Dyna Tech Cro-Mo comp

On 10 July, 06:29, Paul - xxx wrote:
On 10/07/2010 04:48, thirty-six wrote:

On 9 July, 20:16, Paul - *wrote:
On 09/07/2010 19:58, thirty-six wrote:

Although the rear wheel usually takes a slightly shorter route around
bends so resulting in fewer revolutions, it is also subject to the
driving force which amounts to a little slippage, resulting in more
revolutions than the front. * I don't know whether there is a typical
value on this.

I'd guess about the square root of feckall .. if the rear spun that much
then you'd be skidding all over. *Maybe on sheet ice ...

Drive-slip occurs at the centre of the wheel tread only, not like
wheelspin which is slip across the whole of the tread.

? *How does that work then?

How can the centre of the tread slip without the rest of the tread?

It is longer in the centre with a round section tyre. The wheel is
under drive and so any slip will be the centre of tread going
backwards. The sides of the tread still grip the road fully and are
not subjected to the same amount of wear. That the centre of the rear
tyre of a bicycle slips is quite obvious when you examine the wear
pattern.

On Sat, 10 Jul 2010, Phil W Lee wrote:
Paul - xxx considered Sat, 10 Jul 2010
06:29:52 +0100 the perfect time to write:

How can the centre of the tread slip without the rest of the tread?

How can a tyre with a larger diameter in the middle than at the sides
NOT slip, when both the middle and the edge of the tyre are spinning
at the same rate, and the centre of the tread is therefore traveling
faster than the sides?

Are you assuming the tyre leaves a groove in the road surface?

I think all points on a line across the contact patch are at the same
radius (for the case of a wheel perpendicular to the road surface, but
that seems to be the simplified case under discussion).

That is, the middle of the contact patch does not have a larger radius
than the edge of the contact patch. The radius at front and back of
the contact patch varies, but it varies along the patch, not across,
and the centre therefore does not slip relative to the edges.

regards, Ian SMith
--
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|o o|
|/ \|

On 10 July, 22:15, Ian Smith wrote:

I think all points on a line across the contact patch are at the same
radius (for the case of a wheel perpendicular to the road surface, but
that seems to be the simplified case under discussion).

Then you are a fool. The contact patch is flat, it has no radius.


That is, the middle of the contact patch does not have a larger radius
than the edge of the contact patch. *The radius at front and back of
the contact patch varies, but it varies along the patch, not across,
and the centre therefore does not slip relative to the edges.

regards, * Ian SMith
--
* |\ /| * * *no .sig
* |o o|
* |/ \|

On Sun, 11 Jul 2010 04:48:20 +0100, Phil W Lee wrote:
Ian Smith considered Sat, 10 Jul 2010 21:15:31
+0000 (UTC) the perfect time to write:

On Sat, 10 Jul 2010, Phil W Lee wrote:
Paul - xxx considered Sat, 10 Jul 2010
06:29:52 +0100 the perfect time to write:

How can the centre of the tread slip without the rest of the tread?

How can a tyre with a larger diameter in the middle than at the sides
NOT slip, when both the middle and the edge of the tyre are spinning
at the same rate, and the centre of the tread is therefore traveling
faster than the sides?

Are you assuming the tyre leaves a groove in the road surface?

I think all points on a line across the contact patch are at the same
radius (for the case of a wheel perpendicular to the road surface, but
that seems to be the simplified case under discussion).

I hadn't made the assumption that it was the simplified case - on an
average or typical ride the tyre will act as a taper in both
directions, and the tread has to move slightly against the road for
this to happen. This does happen on both tyres, of course, but with
one significant difference:

No, that's not the same simplified case under discussion.

The case I am talking about is that the wheel is perpendicular to the
road surface and there is not scrubbing going on. You are not
addressing scrubbing either - you are talking about the same
simplified case that I am. If the tyre is inclined, then there is a
differing radius, but that's not the case you're talking about
because you say "larger diameter in the middle than at the sides" and
in the inclined wheel case, the diameter is larger at one side than at
the middle.

With drive taking place through the rear, and braking largely through
the front, tyre drift (one part of the contact patch slipping but with
no overall loss of grip) has a tendency to be biased in opposite
directions at the two wheels, and the rear will move very slightly
faster than the front for the same distance covered.

That's all irrelevant to the particular point, being different radii
across the width of the contact patch. I don't think there are.

Were you assuming the tyre indents the road?

If not, how do you conclude that the tyre in contact with the road has
a different radius at the middle of the contact patch than at the
edges?

regards, Ian SMith
--
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|o o|
|/ \|

On 07/11/2010 12:25 AM, thirty-six wrote:
On 10 July, 22:15, Ian wrote:

I think all points on a line across the contact patch are at the same
radius (for the case of a wheel perpendicular to the road surface, but
that seems to be the simplified case under discussion).

Then you are a fool. The contact patch is flat, it has no radius.

The same distance from the hub kind of radius you dolt!

--
www.slowbicyclemovement.org - enjoy the ride

On 11 July, 04:48, Phil W Lee wrote:
Ian Smith considered Sat, 10 Jul 2010 21:15:31
+0000 (UTC) the perfect time to write:



On Sat, 10 Jul 2010, Phil W Lee wrote:
*Paul - xxx considered Sat, 10 Jul 2010
*06:29:52 +0100 the perfect time to write:

How can the centre of the tread slip without the rest of the tread?

*How can a tyre with a larger diameter in the middle than at the sides
*NOT slip, when both the middle and the edge of the tyre are spinning
*at the same rate, and the centre of the tread is therefore traveling
*faster than the sides?

Are you assuming the tyre leaves a groove in the road surface?

I think all points on a line across the contact patch are at the same
radius (for the case of a wheel perpendicular to the road surface, but
that seems to be the simplified case under discussion).

I hadn't made the assumption that it was the simplified case - on an
average or typical ride the tyre will act as a taper in both
directions, and the tread has to move slightly against the road for
this to happen. *This does happen on both tyres, of course, but with
one significant difference:
With drive taking place through the rear, and braking largely through
the front, tyre drift (one part of the contact patch slipping but with
no overall loss of grip) has a tendency to be biased in opposite
directions at the two wheels, and the rear will move very slightly
faster than the front for the same distance covered.

This will be the opposite to the effect of the front tyre covering a
greater distance due to steering, so the overall difference (and
therefore which wheel turns further) will depend on route, tyre
pressure, tyre load, weight distribution, riding style, rider
strength, grip level between the road and tyre, and probably some
other things I've missed.
It's very likely the case that they rarely, if ever, rotate at
precisely the same speed for very long.
If they were to do so over the length of a few miles ride it probably
would be only due to an amazing coincidence.

Like going downhill on the down slope side of a hill, like, if you
know what I mean!




That is, the middle of the contact patch does not have a larger radius
than the edge of the contact patch. *The radius at front and back of
the contact patch varies, but it varies along the patch, not across,
and the centre therefore does not slip relative to the edges.

regards, * Ian SMith