Setting a cycle computer

I have been using the cycle computer setting provided in the instruction manual chart for my size
tyres for a while now. Recently whilst checking a couple of routes on a map and computer route
finder I have found that the cycle computer is giving slightly longer distances than actual.

Using the information from Sheldon Brown's article on cycle computers I have come up with a range of
values. My cycle computer requires circumferance in
mm. The tyres are semi-slick with a lightly pimpled centre tread and outer knobs and measure 26x1.95
(53-559), they're pumped to approx. 55psi.

The problem that I'm having is that I am coming up with quite a range of values. I'm not worried
about spot on accuracy but would like to get things as close as possible.

The computer manual's suggested setting is 2089 Using Sheldon Brown's ISO formula gives 2088 Hand
measuring the circ. with a tape measure gives 2070 Hand measuring the diamater and multiplying by
3.14 gives 2070

All these values are pretty close, however when doing a roll out test (marking a point on the ground
and then marking a second one wheel revolution away using the valve as a reference point) I
regularly get a value of 2020 to 2030. The value is pretty similar whether I simply roll the bike or
sit on it at the time.

I appreciate that any of the hand measured approaches have scope for human error and the fact that
the old fabric tape measure might have stretched a bit and that the bike must travel in a completely
straight line for the roll out test but still the difference seems slightly odd.

So my question is, which setting to use?

Thanks for any thoughts

Paul.

 

In article <[email protected]>, [email protected] says...

...

> All these values are pretty close, however when doing a roll out test (marking a point on the
> ground and then marking a second one wheel revolution away using the valve as a reference point) I
> regularly get a value of 2020 to 2030. The value is pretty similar whether I simply roll the bike
> or sit on it at the time.
>
> I appreciate that any of the hand measured approaches have scope for human error and the fact that
> the old fabric tape measure might have stretched a bit and that the bike must travel in a
> completely straight line for the roll out test but still the difference seems slightly odd.
>
> So my question is, which setting to use?

The rollout test, with your weight on it, would be the most accurate,
IMO. That's what I used, and came up only about 2% off. If you really trust your maps' distances,
you could just adjust your entered circumferance by the appropriate percentage to get you to
the mapped distance.

--
David Kerber An optimist says "Good morning, Lord." While a pessimist says "Good Lord,
it's morning".

Remove the ns_ from the address before e-mailing.

 

This may help. It's what I've used a few times.

http://www.sheldonbrown.com/cyclecomputers/index.html

Cheers Paul J Pharr

"archer" <[email protected]_hotmail.com> wrote in message news:[email protected]...
> In article <[email protected]>, [email protected] says...
>
> ...
>
> > All these values are pretty close, however when doing a roll out test (marking a point on the
> > ground and then marking a second one wheel revolution away using the valve as a reference point)
> > I regularly get a value of 2020 to 2030. The value is pretty similar whether I simply roll
the
> > bike or sit on it at the time.
> >
> > I appreciate that any of the hand measured approaches have scope for
human
> > error and the fact that the old fabric tape measure might have stretched
a
> > bit and that the bike must travel in a completely straight line for the
roll
> > out test but still the difference seems slightly odd.
> >
> > So my question is, which setting to use?
>
> The rollout test, with your weight on it, would be the most accurate,
> IMO. That's what I used, and came up only about 2% off. If you really trust your maps' distances,
> you could just adjust your entered circumferance by the appropriate percentage to get you to
> the mapped distance.
>
>
> --
> David Kerber An optimist says "Good morning, Lord." While a pessimist says "Good Lord, it's
> morning".
>
> Remove the ns_ from the address before e-mailing.

 

Paul wrote:
> I have been using the cycle computer setting provided in the instruction manual chart for my size
> tyres for a while now. Recently whilst checking a couple of routes on a map and computer route
> finder I have found that the cycle computer is giving slightly longer distances than actual.

I find the same thing but surely, that's because no-one rides perfectly straight. All the little
curves (that are too small to show up on the map), plus wobbles and dodges and micro-detours add up.

~PB

 

> The rollout test, with your weight on it, would be the most accurate,
> IMO. That's what I used, and came up only about 2% off. If you really trust your maps' distances,
> you could just adjust your entered circumferance by the appropriate percentage to get you to
> the mapped distance.

The rollout test is the most accurate. You must add your usual riding weight to the bike. For best
accuracy, use a smooth surface, go as straight as possible, hold the bike vertical and measure ten
or so revolutions.

--
Ted Bennett Portland OR

 

"Paul" <[email protected]> wrote in message
news:[email protected]...
> I have been using the cycle computer setting provided in the instruction manual chart for my size
> tyres for a while now. Recently whilst checking a couple of routes on a map and computer route
> finder I have found that the cycle computer is giving slightly longer distances than actual.
>
> Using the information from Sheldon Brown's article on cycle computers I
have
> come up with a range of values. My cycle computer requires circumferance
in
> mm. The tyres are semi-slick with a lightly pimpled centre tread and outer knobs and measure
> 26x1.95 (53-559), they're pumped to approx. 55psi.
>
> The problem that I'm having is that I am coming up with quite a range of values. I'm not worried
> about spot on accuracy but would like to get
things
> as close as possible.
>
> The computer manual's suggested setting is 2089 Using Sheldon Brown's ISO formula gives 2088 Hand
> measuring the circ. with a tape measure gives 2070 Hand measuring the diamater and multiplying by
> 3.14 gives 2070
>
> All these values are pretty close, however when doing a roll out test (marking a point on the
> ground and then marking a second one wheel revolution away using the valve as a reference point) I
> regularly get a value of 2020 to 2030. The value is pretty similar whether I simply roll
the
> bike or sit on it at the time.
>
> I appreciate that any of the hand measured approaches have scope for human error and the fact that
> the old fabric tape measure might have stretched a bit and that the bike must travel in a
> completely straight line for the
roll
> out test but still the difference seems slightly odd.
>
> So my question is, which setting to use?

Try the most accurate method which is to ride through a spot of wet paint and measure from spot to
spot, preferably over more than one and average the distance. Also, get a steel tape. That gives you
an actual distance travelled which is exactly what you need to know, without any intermediate
error-prone steps. Oh, and have your usual gear on the bike at your usual inflation pressure to
avoid other errors.

--
Andrew Muzi http://www.yellowjersey.org Open every day since 1 April 1971

 

On Tue, 22 Apr 2003 17:40:08 +0100, "Paul" <[email protected]> wrote:

>So my question is, which setting to use?
>
>Thanks for any thoughts

I know many people swear by the "rolling measurement" method, but for years I've just measured the
diameter and multiplied by pi. When I lived in New Jersey we regularly rode between "measured
mile" signs posted along the roads. I always measured 1.01 miles. I figured this was close enough
and rationalized that the extra hundredth was the inevitable wandering as we try to ride a
straight line.

jeverett3<AT>earthlink<DOT>net http://home.earthlink.net/~jeverett3

 

"Paul" <[email protected]> wrote in message
news:<[email protected]>...
> I have been using the cycle computer setting provided in the instruction manual chart for my size
> tyres for a while now. Recently whilst checking a couple of routes on a map and computer route
> finder I have found that the cycle computer is giving slightly longer distances than actual.
>
> Using the information from Sheldon Brown's article on cycle computers I have come up with a range
> of values. My cycle computer requires circumferance in
> mm. The tyres are semi-slick with a lightly pimpled centre tread and outer knobs and measure
> 26x1.95 (53-559), they're pumped to approx. 55psi.
>
> The problem that I'm having is that I am coming up with quite a range of values. I'm not worried
> about spot on accuracy but would like to get things as close as possible.
>
> The computer manual's suggested setting is 2089 Using Sheldon Brown's ISO formula gives 2088 Hand
> measuring the circ. with a tape measure gives 2070 Hand measuring the diamater and multiplying by
> 3.14 gives 2070
>
> All these values are pretty close, however when doing a roll out test (marking a point on the
> ground and then marking a second one wheel revolution away using the valve as a reference point) I
> regularly get a value of 2020 to 2030. The value is pretty similar whether I simply roll the bike
> or sit on it at the time.
>
> I appreciate that any of the hand measured approaches have scope for human error and the fact that
> the old fabric tape measure might have stretched a bit and that the bike must travel in a
> completely straight line for the roll out test but still the difference seems slightly odd.
>
> So my question is, which setting to use?
>
> Thanks for any thoughts
>
> Paul.

The method that I use is to measure the circumference of the wheel, set the computer, then go for a
ride around the local 10 mile TT course. Check the mileage after the 10 miles and then adjust the
circumference value on the computer by +/- the percentage that the actual muileage is out by. i.e.
if the computer read 11 miles I would take 10% off of the circumference measurement. Regards, David

 

This last method sounds like the most convoluted and inaccurate technique of all. You can't know which lane the designers of the track used in order to come up with the distance around the track.

Have someone help you with the rollout method by putting a mark on the ground where the valve stem begins in the 6 o'clock position. Sit on the seat, apply weight to the bike as best you can, lean the bike, then use one foot to scoot along until the valve stem has done a 360 and ends up again at 6 o'clock. Put another mark on the ground, measure the distance between the two marks, and convert to metric. One inch equals 25.4 millimeters.

 

dennisg <[email protected]> wrote in message news:<[email protected]>...
> This last method sounds like the most convoluted and inaccurate technique of all. You can't know
> which lane the designers of the track used in order to come up with the distance around the track.

Most of the TT courses near to where I lived in the UK were on single carriageway roads and
sanctioned by the RTTC for competition use, so would be fairly accurately measured to within a few
yards for the *normal* path of the cyclist around the course. Start and finish points are well known
amongst the TT comunity. I have found it possible to get accuracy to within 1% using this method, It
does sometimes take 2 or 3 adjustments to get spot on, but this can all be done as part of normal
riding/training. Regards, David.

 

How accurate do you want it?

Use the size stamped on the tire=95% (if you're lucky)

Measure the diameter/circumference=about the same (maybe a little better)

Roll it on the ground=96% (97% best)

Roll it on the ground for several turns while sitting on the bike=97-98%

Actually ride the bike over a measured distance of known accuracy=I've gotten as close as 99.993%.

May you have the wind at your back. And a really low gear for the hills! Chris

Chris'Z Corner "The Website for the Common Bicyclist": http://www.geocities.com/czcorner

 

dennisg wrote: " This last method sounds like the most convoluted and inaccurate technique of all.
You can't know which lane the designers of the track used in order to come up with the distance
around the track. "

Which is why you carefully measure (and doublecheck) the distance of the course yourself.

Of course, you can always make your own. http://www.geocities.com/czcorner/tech18.html

May you have the wind at your back. And a really low gear for the hills! Chris

Chris'Z Corner "The Website for the Common Bicyclist": http://www.geocities.com/czcorner

 

bobv wrote:
> I have always wondered if the "riding weight" on the bike really made any difference. After all
> the weight doesn't change the circumference of the tire, which is really all that counts in
> distance, not the diameter! Think about it.

You think about the distance travelled in one revolution of a small wheel compared to a large wheel.
The wheel (with tyre) effectively becomes smaller when you sit on the bike. The maximum
circumference of the inflated tyre is irrelevant. To take it to the extreme, think about it with a
flat tyre.

~PB

 

You are confusing the diameter of the tire with the circumference. PI doesn't apply to elliptical
objects. A partially flat tire still has the same circumference as a fully inflated one (disclaimer
below), where a totally flat tire is riding on the rim which is whole different ball game not to
mention diameter AND circumference.

Now inflation pressures could possibly change the circumference of the tire due to stretching of the
tread, but adding weight to the bike would have minimal effect on the circumference in mho.

Think of the tread (outer circumference) of the tire being a tape of a constant length and even if
you now change the shape of it, it still contacts the ground for the same distance for each
revolution.

Bob

On Fri, 25 Apr 2003 18:16:43 +0100, "Pete Biggs" <pLime{remove_fruit}@biggs.tc> wrote:

>bobv wrote:
>> I have always wondered if the "riding weight" on the bike really made any difference. After all
>> the weight doesn't change the circumference of the tire, which is really all that counts in
>> distance, not the diameter! Think about it.
>
>You think about the distance travelled in one revolution of a small wheel compared to a large
>wheel. The wheel (with tyre) effectively becomes smaller when you sit on the bike. The maximum
>circumference of the inflated tyre is irrelevant. To take it to the extreme, think about it with a
>flat tyre.
>
>~PB

 

bobv wrote:
> You are confusing the diameter of the tire with the circumference.

The effective circumference is calculated from the radius from axle to ground of the loaded
wheel and tyre.

> PI doesn't apply to elliptical objects.

The loaded tyre isn't an ellipse. It's a circle with a flat spot. You don't need to measure the
circumference of the full stationary circle. The relevant circle is formed from the point nearest to
the axle that contacts the ground. So with a totally flat tyre, that's virtually the rim; top of
tyre for a solid one, and somewhere in between for a normal loaded inflated pnuematic tyre.

The rest of the tyre is irrelevant because it will bulge out of the way sideways as it contacts the
ground as the wheel turns.

> A partially flat tire still has the same circumference as a fully inflated one.

The effective circumference is the only thing that is relevant. As far as the calculation is
concerned, the wheel is still a perfect circle, just a smaller one.

> (disclaimer below), where a totally flat tire is riding on the rim

Exactly.

> which is whole different ball game not to mention diameter AND circumference.

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?

> Now inflation pressures could possibly change the circumference of the tire due to stretching of
> the tread, but adding weight to the bike would have minimal effect on the circumference in mho.
>
> Think of the tread (outer circumference) of the tire being a tape of a constant length and even if
> you now change the shape of it, it still contacts the ground for the same distance for each
> revolution.

You don't ride on the whole length of the "tape" at once so that distance is not being travelled
once per revolution. You ride on the constant point where the outside edge of the tyre ends up when
its at the bottom of the wheel. The radius of the rest of the tyre is irrelevant because it will
squash in when at bottom. It's excess and doesn't exist as far as the distance travelled is
concerned.

You ride on a circle that is smaller than the unloaded inflated tyre. Forget the bigger circle
exists. Thats out in mid air doing nothing!

~PB

 

It's you who is confused, Bob.

More weight results in a shorter radius, a shorter diameter and a shorter circumference.

bobv <[email protected]> wrote:

> You are confusing the diameter of the tire with the circumference. PI doesn't apply to elliptical
> objects. A partially flat tire still has the same circumference as a fully inflated one
> (disclaimer below), where a totally flat tire is riding on the rim which is whole different ball
> game not to mention diameter AND circumference.
>
> Now inflation pressures could possibly change the circumference of the tire due to stretching of
> the tread, but adding weight to the bike would have minimal effect on the circumference in mho.
>
> Think of the tread (outer circumference) of the tire being a tape of a constant length and even if
> you now change the shape of it, it still contacts the ground for the same distance for each
> revolution.
>
> Bob
>
> On Fri, 25 Apr 2003 18:16:43 +0100, "Pete Biggs" <pLime{remove_fruit}@biggs.tc> wrote:
>
> >bobv wrote:
> >> I have always wondered if the "riding weight" on the bike really made any difference. After all
> >> the weight doesn't change the circumference of the tire, which is really all that counts in
> >> distance, not the diameter! Think about it.
> >
> >You think about the distance travelled in one revolution of a small wheel compared to a large
> >wheel. The wheel (with tyre) effectively becomes smaller when you sit on the bike. The maximum
> >circumference of the inflated tyre is irrelevant. To take it to the extreme, think about it with
> >a flat tyre.
> >
> >~PB
> >
>

--
Ted Bennett Portland OR

 

On Sat, 26 Apr 2003 00:08:09 +0100, "Pete Biggs" <pLime{remove_fruit}@biggs.tc> wrote:

>bobv wrote:
>> You are confusing the diameter of the tire with the circumference.
>
>The effective circumference is calculated from the radius from axle to ground of the loaded wheel
>and tyre.

I don't think so, as the same area of the tire touches the ground for the full rotation in both
cases. More thoughts below however.
>
>> PI doesn't apply to elliptical objects.
>
>The loaded tyre isn't an ellipse. It's a circle with a flat spot.

Good point, but PI still doesn't apply to a circle with a flat spot

>You don't need to measure the circumference of the full stationary circle. The relevant circle is
>formed from the point nearest to the axle that contacts the ground. So with a totally flat tyre,
>that's virtually the rim; top of tyre for a solid one, and somewhere in between for a normal loaded
>inflated pnuematic tyre.

A totally flat tire is a whole different thing. The tire is not even relevant in this case only the
rim of the wheel.

So let us say that we have tire that is inflated only to the point where it is just above the rim
when it contacts the ground. Then what you say makes sense, but we still have the same outer edge of
the tire (which is a constant length for our argument) contacting the ground for a full revolution.
Unless we have major slippage here I guess I have a problem with this.
>
>The rest of the tyre is irrelevant because it will bulge out of the way sideways as it contacts the
>ground as the wheel turns.
>
>> A partially flat tire still has the same circumference as a fully inflated one.
>
>The effective circumference is the only thing that is relevant. As far as the calculation is
>concerned, the wheel is still a perfect circle, just a smaller one.

I don't think this is true unless there is slippage of the tread on the ground which has to be
negligible.
>
>> (disclaimer below), where a totally flat tire is riding on the rim
>
>Exactly.
>
>> which is whole different ball game not to mention diameter AND circumference.
>
>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.
>
>> Now inflation pressures could possibly change the circumference of the tire due to stretching of
>> the tread, but adding weight to the bike would have minimal effect on the circumference in mho.
>>
>> Think of the tread (outer circumference) of the tire being a tape of a constant length and even
>> if you now change the shape of it, it still contacts the ground for the same distance for each
>> revolution.
>
>You don't ride on the whole length of the "tape" at once so that distance is not being travelled
>once per revolution. You ride on the constant point where the outside edge of the tyre ends up when
>its at the bottom of the wheel. The radius of the rest of the tyre is irrelevant because it will
>squash in when at bottom. It's excess and doesn't exist as far as the distance travelled is
>concerned.

As I mentioned above, unless we have slippage of the "tape" then I still have a problem with this.

Think of a hoop ( not a perfect analogy, but not all that bad), while pressing on top of the hoop
you rotate it on the ground. It will still take exactly the same linear distance to rotate once.
>
>You ride on a circle that is smaller than the unloaded inflated tyre. Forget the bigger circle
>exists. Thats out in mid air doing nothing!

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.

Bob

 

On Fri, 25 Apr 2003 16:24:07 -0700, Ted Bennett <[email protected]> wrote:

>It's you who is confused, Bob.
>
>More weight results in a shorter radius, a shorter diameter and a shorter circumference.
>
>
Great technical argument Pete

Bob

>bobv <[email protected]> wrote:
>
>> You are confusing the diameter of the tire with the circumference. PI doesn't apply to elliptical
>> objects. A partially flat tire still has the same circumference as a fully inflated one
>> (disclaimer below), where a totally flat tire is riding on the rim which is whole different ball
>> game not to mention diameter AND circumference.
>>
>> Now inflation pressures could possibly change the circumference of the tire due to stretching of
>> the tread, but adding weight to the bike would have minimal effect on the circumference in mho.
>>
>> Think of the tread (outer circumference) of the tire being a tape of a constant length and even
>> if you now change the shape of it, it still contacts the ground for the same distance for each
>> revolution.
>>
>> Bob
>>
>> On Fri, 25 Apr 2003 18:16:43 +0100, "Pete Biggs" <pLime{remove_fruit}@biggs.tc> wrote:
>>
>> >bobv wrote:
>> >> I have always wondered if the "riding weight" on the bike really made any difference. After
>> >> all the weight doesn't change the circumference of the tire, which is really all that counts
>> >> in distance, not the diameter! Think about it.
>> >
>> >You think about the distance travelled in one revolution of a small wheel compared to a large
>> >wheel. The wheel (with tyre) effectively becomes smaller when you sit on the bike. The maximum
>> >circumference of the inflated tyre is irrelevant. To take it to the extreme, think about it with
>> >a flat tyre.
>> >
>> >~PB
>>

 

Sorry Pete, meant Ted here

On Sat, 26 Apr 2003 05:08:35 GMT, bobv <[email protected]> wrote:

>On Fri, 25 Apr 2003 16:24:07 -0700, Ted Bennett <[email protected]> wrote:
>
>>It's you who is confused, Bob.
>>
>>More weight results in a shorter radius, a shorter diameter and a shorter circumference.
>>
>>
>Great technical argument Pete
>
>Bob
>
>>bobv <[email protected]> wrote:
>>
>>> You are confusing the diameter of the tire with the circumference. PI doesn't apply to
>>> elliptical objects. A partially flat tire still has the same circumference as a fully inflated
>>> one (disclaimer below), where a totally flat tire is riding on the rim which is whole different
>>> ball game not to mention diameter AND circumference.
>>>
>>> Now inflation pressures could possibly change the circumference of the tire due to stretching of
>>> the tread, but adding weight to the bike would have minimal effect on the circumference in mho.
>>>
>>> Think of the tread (outer circumference) of the tire being a tape of a constant length and even
>>> if you now change the shape of it, it still contacts the ground for the same distance for each
>>> revolution.
>>>
>>> Bob
>>>
>>> On Fri, 25 Apr 2003 18:16:43 +0100, "Pete Biggs" <pLime{remove_fruit}@biggs.tc> wrote:
>>>
>>> >bobv wrote:
>>> >> I have always wondered if the "riding weight" on the bike really made any difference. After
>>> >> all the weight doesn't change the circumference of the tire, which is really all that counts
>>> >> in distance, not the diameter! Think about it.
>>> >
>>> >You think about the distance travelled in one revolution of a small wheel compared to a large
>>> >wheel. The wheel (with tyre) effectively becomes smaller when you sit on the bike. The maximum
>>> >circumference of the inflated tyre is irrelevant. To take it to the extreme, think about it
>>> >with a flat tyre.
>>> >
>>> >~PB
>>> >
>

 

bobv wrote:
> On Sat, 26 Apr 2003 00:08:09 +0100, "Pete Biggs" <pLime{remove_fruit}@biggs.tc> wrote:
>
>
>>bobv wrote:
>>
>>>You are confusing the diameter of the tire with the circumference.
>>
>>The effective circumference is calculated from the radius from axle to ground of the loaded wheel
>>and tyre.
>
>
> I don't think so, as the same area of the tire touches the ground for the full rotation in both
> cases. More thoughts below however.

I've done the experiment with a car tire (belted radial) where there is a more substantial 'flat
spot' on the bottom of the tire. I measured three things: 1) radius of the tire where it was
unloaded, 2) distance from the axle to the ground, and 3) circumference by rolling the car for one
tire revolution. I don't have the exact numbers anymore, but the circumference came out to just
barely under 2*pi*(unloaded radius) and was much greater than 2*pi*(distance from axle to ground).
This was what I expected on the basis that the tread can't squirm much, especially with an
essentially fixed-length steel belt directly underneath it - and it confirms your hypothesis above.
I haven't tried it with a bias-ply construction which might allow for more squirming of the tread
and consequent reduced circumference.