Setting a cycle computer

bobv wrote:

/snip

Quoted message said:

Quoted message said:

Why is it a different ball game? Assuming tyre stays on rim, you are still using the whole of the
tyre's circumference. What about when it's almost flat but not quite so rider is suspended just a
couple of milimeters above the rim? Where is the threshold?

The threshold is where the tire sinks below the rim, and the rim becomes the new circumference.

But the tyre still contacts the road and its full length is still used. Flat tyres don't usually
wrinkle and buckle and fill up the rim cavity, they just get squashed out sideways. This also
happens to a lesser extent with an inflated one. The priciple is the same.

/snip

Quoted message said:

I now have two distinct pictures of this. One is yours with the smaller circle, and I still have
the one of the tread circumference being a constant and touching the ground for the full rotation
with no slippage. How do we explain the second picture.

I explain it as being irrelevant. The distance travelled in one wheel revoloution does not have to
match the total tread circumference. (Please explain exactly why not if you disagree).

The circle formed by the actual tread circumference is not being ridden (so the static circumference
can not be used to calculate distance travelled). Instead, you ride at the point where it the tyre
is pushed to. With a flat tyre, that's the rim - so the effective circumference is taken from that.
With an inflated tyre, the point is a distance away from the rim determined by tyre profile, air
pressure and weight.

The circumference doesn't "touch the ground". Only one point of the tread at any one time contacts
the ground, and it moves in nearer the rim as it contacts because the tyre is not rigid. All of the
length of the tread gets used but the full circle of its circumference does not because the tread is
continually moving, continually being pushed in towards the rim - effectively forming a smaller
circle. That circle doesn't actually exist at any one moment but that is what you ride in one wheel
revolution.

The length of the tread doesn't matter, it's where the tread goes that does.

~PB

bobv said:

The threshold is where the tire sinks below the rim, and the rim becomes the new circumference.

Not all of the tyre sinks below the rim. The rim sidewalls ride on the flat tyre with no slippage*.
This is not a special case, it's just the extreme end of the scale. Pump the tyre a bit, and the gap
between rim and ground increases. This distance depends on air pressure and how much the tyre is
bulging out sideways due to weight, so it's not "all or nothing".

* with many average modern road tyres - which are quite tight-fitting. It's true that other
tyres will become so baggy that they won't behave anything like an inflated tyre. Think of the
former type.

~PB

Pete

I need to give this more thought and possibly do an experiment. The other Peter's experiment is
probably a good one, and I might try that. I don't really see any significant difference between a
car's radial belted tire and that of a highly pressurized bike tire.

I now need to go and get myself and the bike ready for a 85 mile ride up the Columbia Gorge
tomorrow. I will pump the tires up to their maximum and hope that the computer is accurate🙂

Bob

Pete Biggs said:

bobv wrote:

/snip

Quoted message said:

Quoted message said:

Why is it a different ball game? Assuming tyre stays on rim, you are still using the whole of
the tyre's circumference. What about when it's almost flat but not quite so rider is suspended
just a couple of milimeters above the rim? Where is the threshold?

The threshold is where the tire sinks below the rim, and the rim becomes the new circumference.

But the tyre still contacts the road and its full length is still used. Flat tyres don't usually
wrinkle and buckle and fill up the rim cavity, they just get squashed out sideways. This also
happens to a lesser extent with an inflated one. The priciple is the same.

/snip

Quoted message said:

I now have two distinct pictures of this. One is yours with the smaller circle, and I still have
the one of the tread circumference being a constant and touching the ground for the full rotation
with no slippage. How do we explain the second picture.

I explain it as being irrelevant. The distance travelled in one wheel revoloution does not have to
match the total tread circumference. (Please explain exactly why not if you disagree).

The circle formed by the actual tread circumference is not being ridden (so the static
circumference can not be used to calculate distance travelled). Instead, you ride at the point
where it the tyre is pushed to. With a flat tyre, that's the rim - so the effective circumference
is taken from that. With an inflated tyre, the point is a distance away from the rim determined by
tyre profile, air pressure and weight.

The circumference doesn't "touch the ground". Only one point of the tread at any one time contacts
the ground, and it moves in nearer the rim as it contacts because the tyre is not rigid. All of the
length of the tread gets used but the full circle of its circumference does not because the tread
is continually moving, continually being pushed in towards the rim - effectively forming a smaller
circle. That circle doesn't actually exist at any one moment but that is what you ride in one wheel
revolution.

The length of the tread doesn't matter, it's where the tread goes that does.

~PB

Seriously, the way to set a cycle computer is to input the rolling loaded circumference of the
inflated tyre measured with rider on the bike by rolling it out beside a tape. This makes for a
speedo/odo much more accurate than most cars'. Mark Lee